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List of integers without any arithmetic progression of n terms Let's consider a positive integer $n$ and the list of the $n^2$ integers from $1$ to $n^2$. What is the minimum number $f(n)$ of integers to be cancelled in this list so that it is impossible to form any arithmetic progression of $n$ terms with the remaining integers?

Extending my comment to an answer for small $n$: a brute force program gives $$ \begin{align} f(1)&=1 &&\{1\} \\ f(2)&=3 &&\{1,2,3\} \\ f(3)&=4 &&\{3,4,5,7\} \\ f(4)&=6 &&\{3,4,5,6,10,13\} \\ f(5)&=7 &&\{3, 7, 9, 10, 11, 16, 21\} \\ f(6)&=9 &&\{5, 8, 12, 14, 15, 16, 21, 26, 31\} \\ f(7)&=11 &&\{3, 9, 11, 12, 13, 14, 15, 22, 29, 36, 43\} \\ f(8)&=13 &&\{7, 10, 16, 18, 19, 20, 21, 22, 29, 36, 43, 50, 57\} \end{align} $$

From a (not positive definite) Gram matrix to a (Kac-Moody) Cartan matrix Suppose I am given a symmetric matrix $G_{ij}$ with $G_{ii} = 2$. Can I always find an invertible integer matrix $S$ such that $(S^T G S)_{ii}=2$ and $(S^T G S)_{ij} \leq 0$ for $i \neq j$? Is there a practical algorithm to do so? If you'd like a particular challenge, I'd like to know the answer for $$G = \begin{pmatrix} 2 & -4 & 3 \\ -4 & 2 & -2 \\ 3 & -2 & 2 \\ \end{pmatrix}.$$

Wondering how well this would work out in $4$ by $4,$ given that David says the individual steps in the "algorithm" preserve something important: $$ G = \left( \begin{array}{cccc} 2 & a & b & c \\ a & 2 & d & e \\ b & d & 2 & f \\ c & e & f& 2 \end{array} \right) $$ $$ \det G = a^2 f^2 + b^2 e^2 + c^2 d^2 - 2 (a b e f + a c d f + b c d e) + 4 (a b d + a c e + b c f + d e f) -4 ( a^2 + b^2 + c^2 + d^2 + e^2 + f^2 ) + 16 $$ where the degree four terms can be regarded as the indefinite ternary form $\langle 1,1,1,-2,-2,-2 \rangle$ in $x=af, y=be,z=cd.$ $$ S = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & f \\ 0 & 0 & 0& -1 \end{array} \right) $$ $$ S^t G S = \left( \begin{array}{cccc} 2 & a & b & fb-c \\ a & 2 & d & fd-e \\ b & d & 2 & f \\ fb-c & fd-e & f& 2 \end{array} \right) $$ So, this basic operation (an involution) flips two positions at once.. $$ T = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & f& 1 \end{array} \right) $$ $$ T^t G T = \left( \begin{array}{cccc} 2 & a & fc-b & c \\ a & 2 & fe-d & e \\ fc-b & fe-d & 2 & f \\ c & e & f& 2 \end{array} \right) $$ hmmmm....... Just checking, $$ N_1 = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0& -1 \end{array} \right) $$ $$ N_1 G N_1 = \left( \begin{array}{cccc} 2 & a & b & -c \\ a & 2 & d & -e \\ b & d & 2 & -f \\ -c & -e & -f& 2 \end{array} \right) $$ $$ N_2 = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0& -1 \end{array} \right) $$ $$ N_2 G N_2 = \left( \begin{array}{cccc} 2 & a & -b & -c \\ a & 2 & -d & -e \\ -b & -d & 2 & f \\ -c & -e & f& 2 \end{array} \right) $$ Right, that is enough to know on these because $-I$ changes nothing. Really ought to see what a permutation, a single transposition, does $$ T_2 = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) $$ $$ T_2 G T_2 = \left( \begin{array}{cccc} 2 & a & c & b \\ a & 2 & e & d \\ c & e & 2 & f \\ b & d & f& 2 \end{array} \right) $$ Swaps two pairs

Finding matrices $A$ such that the entries of $A^n$ have specified signs What techniques are there for ensuring nonnegativity of various entries of matrix powers? Specific Question: Consider a matrix $A\in SL_2(\mathbb R)$. Let $(A^n)_{i,j}$ denote the $(i,j)$ entry of the matrix power $A^n$. Under what conditions on $A$ does the following hold: $$ \text{sgn}\ (A^n)_{i,j}=(-1)^{j+1} $$

To follow up on Terry's comment (actually, I did the computation before seeing it, but whatever): in the parabolic case, where the matrix has the form $$A = \begin{pmatrix} d & -b \\ -c & a\end{pmatrix} \begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix} \begin{pmatrix} a & b \\ c & d\end{pmatrix},$$ Then $$A^n = \begin{pmatrix}1 + c d n x & d^2 n x\\ - c^2 n x & 1 - c d n x\end{pmatrix}.$$ So, if $c \in \mathbb{R},$ then $x < 0$ (from the bottom left), and so $c < 0$ from the bottom right, and we are fine (that is, if $|c d x| > 1,$ then any $a, b$ works). Since $x$ is real, it is clear from the bottom left that $c$ is also, so this finishes the parabolic case. In the hyperbolic case, $$A = \begin{pmatrix} d & -b \\ -c & a\end{pmatrix} \begin{pmatrix} x & 0 \\ 0 & 1/x\end{pmatrix} \begin{pmatrix} a & b \\ c & d\end{pmatrix},$$ where $x>1.$ and $$A^n = \begin{pmatrix}=b c x^{-n} + a d x^n & b d x^{-n}(-1 + x^{2 n})\\ a c(x^{-n}+ x^n) & ad x^{-n} - b c x^n\end{pmatrix}.$$ For $n \gg 1,$ we have $$A^n \sim x^n \begin{pmatrix} ad & b d\\ -ac & -bc\end{pmatrix},$$ which implies that either $a < 0, b>0, c> 0, d<0,$ or $a > 0, b< 0, c<0, d> 0.$ Neither of which is compatible with the base case $n=1$ satisfying your condition. In the elliptic case, the eigenvalues are either roots of unity, so $A^n = I,$ infinitely often, or not, in which case $A^n$ is very close to $I$ infinitely often, so either way, your condition fails.

Difference between maxima of random variables Given four independent, identically distributed Gaussian random variables with zero mean and unit variance $x_1$, $x_2$, $y_1$, $y_2$, consider \begin{equation} u \equiv \max(x_1+C\, y_1, x_2+C \, y_2) - \max(x_1-C \, y_1, x_2-C \, y_2), \end{equation} where $C$ is a real number. Do you know how to compute the PDF of $u$, or a least its variance?

Using $\max(a,b) = \dfrac{a+b}{2} + \left| \dfrac{a-b}{2}\right|$, write $u = w_1 + |w_2| - |w_3|$ where $$ \eqalign{ w_1 &= C (y_1 + y_2) \cr w_2 &= \dfrac{1}{2} (x_2 - x_1 + C (y_2 - y_1))\cr w_3 &= \dfrac{1}{2} (x_2 - x_1 - C (y_2 - y_1))\cr}$$ are jointly normal with mean $0$ and covariance matrix $$ V = \pmatrix{2 C^2 & 0 & 0 \cr 0 & (1+C^2)/2 & (1-C^2)/2\cr 0 & (1-C^2)/2 & (1+C^2)/2\cr}$$ In particular, $w_1$ is independent of $w_2$ and $w_3$. Thus $\text{Cov}(w_1, |w_2|) = 0$ and $\text{Cov}(w_1, |w_3|) = 0$. The only nontrivial computation is $ \text{Cov}(|w_2|, |w_3|)$. If $|C| < 1$ get $$ \eqalign{ {\mathbb E} [|w_2| |w_3|] &= \dfrac{1-C^2}{\pi} \arctan\left(\frac{2|C|}{C^2-1}\right) + \dfrac{1-C^2}{2} + 2 \frac{|C|}{\pi}\cr {\mathbb E}|w_2| &= {\mathbb E} |w_3| = \sqrt{\dfrac{C^2+1}{\pi}}\cr \text{Cov}(|w_2|,|w_3|) &= \dfrac{1-C^2}{\pi} \arctan\left(\frac{2|C|}{C^2-1}\right) + \dfrac{1-C^2}{2} - \dfrac{(1-|C|)^2}{\pi} } $$ and then $$ \text{Var}(w_1 + |w_2| - |w_3|) = 2\dfrac{C^2-1}{\pi} \arctan\left(\frac{2|C|}{C^2-1}\right) + 4 C^2 - \frac{4|C|}{\pi} $$ EDIT: That formula was only for $|C|<1$ because of branch problems with the arctan. Here's one that should work for all $C \ne 0$: $$ \text{Var}(w_1 + |w_2| - |w_3|) = 2 \dfrac{C^2-1}{\pi} \arctan \left(\frac{1-C^2}{2|C|}\right) + 3 C^2 - \frac{4|C|}{\pi} + 1 $$

The coefficient of a specific monomial of the following polynomial Let the real polynomial $$f_{a,b,c}(x_1,x_2,x_3)=(x_1-x_2)^{2a+1}(x_2-x_3)^{2b+1}(x_3-x_1)^{2c+1},$$ where $a,b,c$ are nonnegative integers. Let $m_{a,b,c}$ be the coefficient of the monomial $x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}$ in the expansion of $f_{a,b,c}(x_1,x_2,x_3)$. It is easy to see $m_{a,b,c}=0$ when two of $a,b,c$ are equal. I want to ask whether $m_{a,b,c}=0$ or not when $a,b,c$ are pairwise unequal.

The coefficient is always zero. I use the superb conventions of Concrete Mathematics: Sums are over all integers unless otherwise indicated, and $\binom{n}{k}$ is $0$ if $k<0$ or $>n \geq 0$. Expanding by the binomial theorem, $$f_{abc} = \sum_{i,j,k} (-1)^{i+j+k} \binom{2a+1}{i} \binom{2b+1}{j} \binom{2c+1}{k} x_1^{2a+1-i+k} x_2^{2b+1-j+i} x_3^{2c+1-k+j}.$$ We want $$\begin{array}{r@{}c@{}lcr@{}c@{}l} a+c+1 &=& 2a+1-i+k &\implies& c-k &=& a-i\\ a+b+1 &=& 2b+1-j+i &\implies& a-i &=& b-j\\ b+c+1 &=& 2c+1-k+j &\implies& b-j &=& c-k \\ \end{array}$$ so $(i,j,k) = (a+r, b+r, c+r)$ for some $r$. We need to evaluate $$\sum_r (-1)^{a+b+c+3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r}.$$ Pair off the $r$ and $1-r$ terms to get $$(-1)^{a+b+c} \sum_{r \geq 0} \left[ (-1)^{3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r} + \right.$$ $$\phantom{(-1)^{a+b+c} \sum_{r \geq 1}} \left. (-1)^{3-3r} \binom{2a+1}{a+1-r} \binom{2b+1}{b+1-r} \binom{2c+1}{c+1-r} \right].$$ The two terms in the square brackets cancel, and the sum is zero.

Eigenvectors of a matrix with entries involving combinatorics In the question Eigenvalues of a matrix with entries involving combinatorics No_way asked about eigenvectors of $n\times n$ matrix $M$ with entries \begin{eqnarray*} M_{ij}=(-1)^{i+j}F(n, l, i, j), \end{eqnarray*} where $F(n,l,i,j)$ is the cardinality of the set \begin{eqnarray*} \{(k_1, \cdots, k_n)\in\mathbb{Z}^{n}|0\leq k_r\leq l-1\text{ for }1\leq r\leq n\text{, }k_1+\cdots+k_n=lj-i\}. \end{eqnarray*} These eigenvalues are known to be $1, l, l^2, \cdots, l^{n-1}$. Let's remove signs and consider the matrix $M$ with $M_{ij}=F(n, l, i, j)$. According to my numerical experiments eigenvectors do not depend on $l$ for $l\ge 2$ and they are polynomials. Q1: Why do eigenvectors not depend on $l$? For $l=2$ we have $M_{ij}=\binom n{2j-i},$ and first examples are (eigenvectors of $M$ are rows of $V$) $$n=2,\qquad M=\left( \begin{array}{cc} 2 & 0 \\ 1 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right);$$ $$n=3,\qquad M=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 3 & 0 \\ 0 & 3 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{ccc} 1 & 1 & 1 \\ -1 & 1 & 3 \\ 0 & 0 & 1 \\ \end{array} \right);$$ $$n=4,\qquad M=\left( \begin{array}{cccc} 4 & 4 & 0 & 0 \\ 1 & 6 & 1 & 0 \\ 0 & 4 & 4 & 0 \\ 0 & 1 & 6 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ -1 & 0 & 1 & 2 \\ 2 & -1 & 2 & 11 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);$$ $$n=5,\qquad M=\left( \begin{array}{ccccc} 5 & 10 & 1 & 0 & 0 \\ 1 & 10 & 5 & 0 & 0 \\ 0 & 5 & 10 & 1 & 0 \\ 0 & 1 & 10 & 5 & 0 \\ 0 & 0 & 5 & 10 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ -3 & -1 & 1 & 3 & 5 \\ 11 & -1 & -1 & 11 & 35 \\ -3 & 1 & -1 & 3 & 25 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right).$$ Denote by $v_m=(v_m(1),\ldots,v_m(n))$ rows of $V$ ($0\le m\le n-1$). They defined up to multiplicative constant and $v_m(k)=\mu_m P_m(k)$ where $P_m(x)$ are some special polynomials of degree $m$. In particular for $m=0,1,2,3,4$ we have $$P_0(x)=1,\quad P_1(x)=2x-n,\quad P_2(x)=3x^2-3nx+\frac{n(3n-1)}{4},$$ $$P_3(x)=4x^3-6nx^2+n(3n-1)x-\frac{n^2(n-1)}{2},$$ $$P_4(x)=5x^4-10nx^3+\frac{5n(3n-1)}{2}x^2-\frac{5n^2(n-1)}{2}x+\frac{n(15n^3-30n^2+5n+2)}{48}.$$ Q2: What is the generating function for these polynomials?

In fact for a fixed $n$, the matrices $M(l, n)$ for $l>0$ commute with each other and thus are simultaneously diagonalisable. For your second question, if $\{p_j(y)\}$ is a sequence of polynomials satisfying \begin{eqnarray} \left(\frac{t}{\sinh t}\right)^y=\sum_{j=0}^\infty p_j(y)t^{2j}. \end{eqnarray} then the $i$-th entry of an eigenvector corresponding to the eigenvalue $l^{n-k-1}$ is \begin{eqnarray} (-1)^{i-1}\sum_{j=0}^{\lfloor\frac{k}{2}\rfloor}\frac{p_j(n)}{(k-2j)!}(n-2i)^{k-2j}. \end{eqnarray} I believe from here we can work out what the generating function of your $P_m(x)$ is.

An extremal problem Let $f:[0,\pi]\to [0,\pi]$ be a diffeomorphism. How to prove that $$P[f]:=\int_0^\pi \sin^2(x) \left(3+2 \frac{\sin^2(f(x))}{\sin^2 x}+(f'(x))^2\right)^2dx $$ attains its minimum for $f(x)\equiv x$?

The following SymPy script from sympy import * x = Symbol('x') f = Function('f')(x) # define Lagrangian L = (sin(x))**2 * (3 + 2*((sin(f))**2/(sin(x))**2) + (Derivative(f,x))**2)**2 # Euler-Lagrange equation print euler_equations(L,f,x) produces the output [Eq(-4*(2*(Derivative(f(x), x)*Derivative(f(x), x, x) + 2*sin(f(x))*cos(f(x))*Derivative(f(x), x)/sin(x)**2 - 2*sin(f(x))**2*cos(x)/sin(x)**3)*sin(x)*Derivative(f(x), x) + (Derivative(f(x), x)**2 + 3 + 2*sin(f(x))**2/sin(x)**2)*sin(x)*Derivative(f(x), x, x) + 2*(Derivative(f(x), x)**2 + 3 + 2*sin(f(x))**2/sin(x)**2)*cos(x)*Derivative(f(x), x))*sin(x) + 8*(Derivative(f(x), x)**2 + 3 + 2*sin(f(x))**2/sin(x)**2)*sin(f(x))*cos(f(x)), 0)] Using function latex to print the output in a more friendly form, we obtain the scary-looking ODE $$\left(- 4 \left(2 \frac{d}{d x} f{\left (x \right )} \frac{d^{2}}{d x^{2}} f{\left (x \right )} + \frac{4 \frac{d}{d x} f{\left (x \right )}}{\sin^{2}{\left (x \right )}} \sin{\left (f{\left (x \right )} \right )} \cos{\left (f{\left (x \right )} \right )} - \frac{4 \sin^{2}{\left (f{\left (x \right )} \right )}}{\sin^{3}{\left (x \right )}} \cos{\left (x \right )}\right) \sin{\left (x \right )} \frac{d}{d x} f{\left (x \right )} - 4 \left(\frac{d}{d x} f{\left (x \right )}^{2} + 3 + \frac{2 \sin^{2}{\left (f{\left (x \right )} \right )}}{\sin^{2}{\left (x \right )}}\right) \sin{\left (x \right )} \frac{d^{2}}{d x^{2}} f{\left (x \right )} - 4 \left(2 \frac{d}{d x} f{\left (x \right )}^{2} + 6 + \frac{4 \sin^{2}{\left (f{\left (x \right )} \right )}}{\sin^{2}{\left (x \right )}}\right) \cos{\left (x \right )} \frac{d}{d x} f{\left (x \right )}\right) \sin{\left (x \right )} + \left(8 \frac{d}{d x} f{\left (x \right )}^{2} + 24 + \frac{16 \sin^{2}{\left (f{\left (x \right )} \right )}}{\sin^{2}{\left (x \right )}}\right) \sin{\left (f{\left (x \right )} \right )} \cos{\left (f{\left (x \right )} \right )} = 0$$ If we make $f (x) := x$, we obtain the following equation $$ - \left(16 x + 40\right) \cos{\left (x \right )} \sin{\left (x \right )} + \left(16 x + 40\right) \sin{\left ( x\right )} \cos{\left (x \right )} = 0$$ which holds for all $x$. Hence, $f (x) = x$ is a solution to the Euler-Lagrange ODE and, thus, it makes the functional stationary. Proving that it minimizes the functional is left as an exercise for the reader.

Is this BBP-type formula for $\ln 31$, $\ln 127$, and other Mersenne numbers also true? In this post, a binary BBP-type formula for Fermat numbers $F_m$ was discussed as (with a small tweak), $$\ln(2^b+1) = \frac{b}{2^{a-1}}\sum_{n=0}^\infty\frac{1}{(2^a)^n}\left(\sum_{j=1}^{a-1}\frac{2^{a-1-j}}{an+j}+\sum_{k=1}^{a/b-1}(-1)^{k+1}\frac{2^{a-1-bk}}{an+bk}\right)\tag1$$ where $a=2^b$ and $b=2^m$. I was then trying to find patterns for $3\cdot2^{m}+1$ and $9\cdot2^{m}+1$, but only have tentative results so far. However, it seems Mersenne numbers are "easier" as, $$\ln(2^b-1) = \frac{b}{2^{a-1}}\sum_{n=0}^\infty\frac{1}{(2^a)^n}\left(\sum_{j=1}^{a-1}\frac{2^{a-1-j}}{an+j}-\sum_{k=1}^{\lfloor a/b-1 \rfloor }\frac{2^{a-1-bk}}{an+bk}\right)\tag2$$ where $a=2^b-2$, $b$ an odd integer, and floor function $\lfloor x\rfloor$. Notice its satisfying similarity to $(1)$. For example, with $b=5$ then, $$\ln 31 = \frac{5}{2^{29}}\sum_{n=0}^\infty\frac{1}{(2^{30})^n}\left(\sum_{j=1}^{29}\frac{2^{29-j}}{30n+j}-\sum_{k=1}^{5}\frac{2^{29-5k}}{30n+5k}\right)$$ Like $(1)$, I found $(2)$ using the integer relations algorithm of Mathematica (and a lot of patience and doodling). Q: But how to rigorously prove $(2)$, and does it in fact hold for all integers $b>1$?

I'll give a proof of the corrected version of your conjectured series expansion and then comment at the end about the limits of this method for finding base 2 BBP formulas for $\log n$. Let's denote $\alpha_b=2^b-2-b\lfloor\frac{2^b-2}{b}\rfloor$, notice that $\alpha_b=0$ for all primes $b$ but it can be nonzero for othervalues. Start by writing $$\log(2^b-1)=b\log 2+\log\left(1-\frac{1}{2^b}\right)$$ $$=b\sum_{k=1}^{\infty} \frac{1}{k2^k}-\sum_{k=1}^{\infty}\frac{1}{k2^{bk}}=\sum_{n=0}^{\infty}\frac{1}{2^{n(2^b-2-\alpha_b)}}\left(\sum_{j=1}^{2^b-2-\alpha_b}\frac{b2^{-j}}{n(2^b-2-\alpha_b)+j}-\sum_{h=1}^{\lfloor\frac{2^b-2}{b}\rfloor}\frac{2^{-bh}}{n\lfloor\frac{2^b-2}{b}\rfloor+h}\right)$$ $$=\sum_{n=0}^{\infty}\frac{1}{2^{n(2^b-2-\alpha_b)}}\left(\sum_{j=1}^{2^b-2-\alpha_b}\frac{b2^{-j}}{n(2^b-2-\alpha_b)+j}-\sum_{h=1}^{\lfloor\frac{2^b-2}{b}\rfloor}\frac{b2^{-bh}}{n(2^b-2-\alpha_b)+bh}\right)$$ $$=\frac{b}{2^{a-1}}\sum_{n=0}^\infty\frac{1}{(2^a)^n}\left(\sum_{j=1}^{a-1}\frac{2^{a-1-j}}{an+j}-\sum_{k=1}^{ a/b-1 }\frac{2^{a-1-bk}}{an+bk}\right)$$ where $a=2^b-2-\alpha_b$ which gives the correct version of your identity (and agrees with it whenever $b$ is prime). As for the general question of which numbers $n$ give $\log n$ with a BBP formula in base $2$, such identities can be manipulated to work with all numbers $n$ such that $$n=\frac{(2^{m_1}-1)(2^{m_2}-1)\cdots(2^{m_k}-1)}{(2^{r_1}-1)(2^{r_2}-1)\cdots(2^{r_s}-1)}.$$ The first prime not expressible in this form is 23 (this can be proven using Zsigmondy's theorem, since $2^{t}-1$ always has a primitive prime factor for large enough $t$).

Inverse of special upper triangular matrix Consider the following $n \times n$ upper triangular matrix with a particularly nice structure: \begin{equation}\mathbf{P} = \begin{pmatrix} 1 & \beta & \alpha+\beta & \dots & (n-3)\alpha + \beta & (n-2)\alpha + \beta\\ 0 & 1 & \beta & \dots & (n-4)\alpha + \beta & (n-3)\alpha + \beta\\ 0 & 0 & 1 & \dots & (n-5)\alpha + \beta & (n-4)\alpha + \beta\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & \beta & \alpha+\beta\\ 0 & 0 & 0 & \dots & 1 & \beta\\ 0 & 0 & 0 & \dots & 0 & 1 \end{pmatrix} \end{equation} i.e. \begin{equation} p_{i,j}=\begin{cases} 0, &i>j,\\ 1, &i=j,\\ (j-i-1)\alpha+\beta, &i<j. \end{cases} \end{equation} Would it be possible to find an explicit expression for the elements of the inverse of $\mathbf{P}$?

Let $A$ be the nilpotent matrix $$\begin{pmatrix}0 & 1 & 1 & \cdots & 1 \\ & 0 & 1 & \cdots & 1 \\ & & \cdots & \cdots & \cdots \\ & & & 0 & 1 \\ & & & & 0\end{pmatrix},$$ then the matrix $P$ is equal to $1 + \beta A + \alpha A^2$. This gives the inverse: \begin{eqnarray*}P^{-1} & = & (1 + \beta A + \alpha A^2)^{-1} \\ & = & (1 - \lambda A)^{-1} (1 - \mu A)^{-1} \\ & = & \sum_{k \geq 0} \frac{\lambda^{k + 1} - \mu^{k + 1}}{\lambda - \mu} A^k,\end{eqnarray*} where $\lambda$ and $\mu$ are the two roots of the equation $x^2 + \beta x + \alpha = 0$. Since we have $A^n = 0$, the sum essentially ranges through $0 \leq k < n$.

A property of 47 with respect to partitions into five parts Is 47 the largest number which has a unique partition into five parts (15, 10, 10, 6, 6), no two of which are relatively prime?

Yes. Suppose $n>47$. If $2\mid n$, we can take $(n-8,2,2,2,2),(n-10,4,2,2,2)$, which are distinct partitions for $n\geq 14$. If $3\mid n$, we can take $(n-12,3,3,3,3),(n-15,6,3,3,3)$, which are distinct partitions for $n\geq 21$. If $n\equiv 1\pmod 6$, we can take $(n-37,15,10,6,6),(n-43,15,12,10,6)$, which are distinct partitions for $n\geq 55$, and $(21,7,7,7,7),(14,14,7,7,7)$ for $n=49$. If $n\equiv 5\pmod 6$, we can take $(n-41,15,10,10,6),(n-47,15,12,10,10)$, which are distinct partitions for $n\geq 59$, and $(20,15,6,6,6),(15,12,10,10,6)$ for $n=53$. (Thanks to Gerhard for pointing out we can finish the argument quickly from what I wrote before)

How to estimate a total variation distance? Let $X_1, \ldots, X_n$ be independent Bernoulli random variables. Then $Pr[X_i=1]=Pr[X_i=0]=1/2$. Let $X = (X_1, \ldots, X_n)$ and $v \in \{0,1\}^n$, $Y=v \cdot X$, $Z=Y-1$. Let \begin{align} \mu_1(x) = Pr[Y=x], \ \mu_2(x)=Pr[Z=x], \ x \in [n]=\{1, \ldots, n\}. \end{align} Are there some method to estimate the total variation distance? \begin{align} d_{TV}(\mu_1, \mu_2) & = \frac{1}{2} \sum_{x \in [n]} | \mu_1(x) - \mu_2(x) | \\ & = \frac{1}{2} \sum_{x \in [n]} | Pr[Y=x] - Pr[Z=x] |. \end{align} Without loss of generality, we many assume that $v=(1,\ldots, 1,0,\ldots,0)$, where the number of $1$'s is $k$ ($0 < k \leq n$). Then \begin{align} d_{TV}(\mu_1, \mu_2) & = \frac{1}{2} \sum_{x \in [n]} | Pr[Y=x] - Pr[Z=x] | \\ & = \frac{1}{2} \sum_{x \in [n]} | Pr[X_1+\cdots+X_k=x] - Pr[X_1+\cdots + X_k=x+1] |. \end{align} Assume that $n$ is sufficient large. Do we have $d_{TV}(\mu_1, \mu_2)< 1-\epsilon$ for some $0 \leq \epsilon \leq 1$? Thank you very much.

For the TV-distance, we have \begin{align} d_{TV}= \frac{1}{2} \sum_{x=0}^k |d_x| \end{align} (with the summation actually beginning at $x=0$), where \begin{multline*} d_x:=P(S_k=x) - P(S_k=x+1)=\frac1{2^k}\,\Big(\binom kx-\binom k{x+1}\Big) \\ =\frac1{2^k}\,\frac{k!}{(x+1)!(k-x)!}\,(2x-(k-1)) \end{multline*} and $S_k:=X_1+\cdots+X_k$, so that $d_x\le0$ for $x=0,\dots,m:=\lfloor (k-1)/2\rfloor$ and $d_x>0$ for $x=m+1,\dots,k$. Note also that $d_x=c_x-c_{x+1}$, where $c_x:=\frac1{2^k}\,\binom kx$. So, \begin{align*} 2d_{TV}&=-\sum_{x=0}^m d_x+\sum_{x=m+1}^k d_x \\ &=-\sum_{x=0}^m c_x+\sum_{x=0}^m c_{x+1}+\sum_{x=m+1}^k c_x-\sum_{x=m+1}^k c_{x+1} \\ &=-\sum_{x=0}^m c_x+\sum_{x=1}^{m+1} c_x+\sum_{x=m+1}^k c_x-\sum_{x=m+2}^{k+1} c_x \\ &=c_{m+1}-c_0+c_{m+1}-c_{k+1} \\ &=c_{m+1}-1+c_{m+1}-0 \\ &=\frac1{2^k}\,\Big(2\binom k{m+1}-1\Big) \sim\sqrt{\frac 8{\pi k}} \end{align*} as $k\to\infty$, by Stirling's formula, so that \begin{equation} d_{TV} \sim\sqrt{\frac 2{\pi k}}. \end{equation}

Solution for $Xa + X^Tb = c$ where $X^TX = I$? There are three known $n\times1$ vectors: $a, b, c$, along with one unknown $n\times n$ matrix: $X$. I am only interested in the $n={2,3}$ cases. $X$ is $2\times 2$ or $3\times 3$ rotation matrix with an unusual domain specific constriant: * *$X^TX = XX^T = I$ *$Xa + X^Tb = c$ Is there a solution for $X$ in terms of $a,b,c$? Based on where the problem came from, I know there isn't always a solution, but I have stumped myself trying to figure out how to solve it when there is one. I have tried working out the $2\times 2$ case element-wise, and arrived at the following, equally(?) difficult problem: $X = \begin{bmatrix}x_{11} & x_{12} \\ -x_{12} & x_{11}\end{bmatrix}$ $\begin{bmatrix}a_1+b_1 & b_2-a_2 \\ a_2+b_2 & a_1-b_1\end{bmatrix}\begin{bmatrix}x_{11}\\x_{12}\end{bmatrix}=c$ $Ax = c$ where $x^Tx=1$

\begin{equation} Xa = \begin{bmatrix} a_1x_{1,1} + a_2x_{1,2}\\ a_1x_{2,1} + a_2x_{2,2}\\ \end{bmatrix} \end{equation} and \begin{equation} X^Tb = \begin{bmatrix} b_1x_{1,1} + b_2x_{2,1}\\ b_1x_{1,2} + b_2x_{1,2}\\ \end{bmatrix} \end{equation} and \begin{equation} Xa+X^Tb = \begin{bmatrix}a_1x_{1,1} + a_2x_{1,2} + b_1x_{1,1} + b_2x_{2,1}\\ a_1x_{2,1} + a_2x_{2,2}+ b_1x_{1,2} + b_2x_{2,2}\\ \end{bmatrix} = \begin{bmatrix}(a_1+b_1)x_{1,1} + a_2x_{1,2} + b_2x_{2,1}\\ (a_2+b_2)x_{2,2} + a_1x_{2,1} + b_1x_{1,2}\\ \end{bmatrix} = \begin{bmatrix}c_1 \\ c_2\\ \end{bmatrix} \end{equation} but $X$ is a rotation matrix gives us that \begin{equation} Xa+X^Tb = \begin{bmatrix}(a_1+b_1)\cos\theta - a_2\sin\theta + b_2\sin \theta \\ (a_2+b_2)\cos\theta + a_1\sin\theta - b_1\sin\theta\\ \end{bmatrix} = \begin{bmatrix}(a_1+b_1)\cos\theta +(b_2- a_2)\sin\theta \\ (a_2+b_2)\cos\theta + (a_1 - b_1)\sin\theta\\ \end{bmatrix} \end{equation} (if you only have $X^TX= I$ then you have to consider the extra case there $\sin \theta \to -\sin \theta $; i.e. rotation composed with reflection. Notice that $X$ is a rotation implies $X^TX= I$!) and therefore that \begin{equation} Xa+X^Tb = \begin{bmatrix}(a_1+b_1)\cos\theta +(b_2- a_2)\sin\theta \\ (a_2+b_2)\cos\theta + (a_1 - b_1)\sin\theta\\ \end{bmatrix} = \begin{bmatrix}c_1 \\ c_2\\ \end{bmatrix} \end{equation} and therefore (by using the triangle inequality) you don't have a solution if for example \begin{equation} \frac{|(a_1+b_1)|}{\sqrt{2}} +\frac{|(b_2- a_2)|}{\sqrt{2}} < |c_1| \end{equation} and like wise for the second condition \begin{equation} \frac{|(a_2+b_2)|}{\sqrt{2}} +\frac{|(a_1- b_1)|}{\sqrt{2}} < |c_2| \end{equation} and you can probably come up with all kinds of other tests for failure, but here is the most general one: An alternative/equivalent way to look at it is that you have an overdetermined system of 3 equations and 2 unknowns of the form \begin{equation} \begin{array} & ax & + & by & =& c_1\\ cx & + & dy & =& c_2 \\ x^2 & + & y^2 & =& 1 \\ \end{array} \end{equation} where $a = a_1+b_1$ , $b=a_2- b_2$, $c = a_2+b_2$, and $d = a_1- b_1$; which is highly unlikely to have solutions. Therefore you have solutions iff the solution to the system of equations \begin{equation} \begin{array} & ax & + & by & =& c_1 \\ cx & + & dy & =& c_2 \\ \end{array} \end{equation} also satisfies the condition $ x^2 + y^2 = 1 $. If you want to work out the $n=3$ case you can do the same exact thing but use the Euler angles; it will be long and tedious but you can probably get some kind of condition on the solutions.

"Laurent phenomenon"? Define the recurrence \begin{align*} n(2n+x-3)u_n(x) &=2(2n+x-2)(4n^2+4nx-8n-3x+3)u_{n-1}(x) \\ &-4(n+x-2)(2n-3)(2n+2x-3)(2n+x-1)u_{n-2}(x) \end{align*} with initial conditions $u_0(x)=0$ and $u_1(x)=x+1$. The subject of "Laurent phenomenon" was motivated by Somos sequences. In the same spirit, I ask: QUESTION. Is it true that each $u_n(x)$ is a polynomial in $x$? In fact, with positive integer coefficients. EXAMPLES. $u_2(x)=5x^2 + 13x + 6$ and $u_3(x)=22x^3 + 114x^2 + 164x + 60$ and \begin{align*} u_4(x)&=93x^4 + 814x^3 + 2367x^2 + 2606x + 840 \\ u_5(x)&=386x^5 + 5140x^4 + 25030x^3 + 54500x^2 + 51024x + 15120. \end{align*}

In fact, $$ u_n(x) = {2}^{n-1}\prod _{k=0}^{n-1}(2x+2k+1) -{2\,n-1\choose n-1}\prod _{k=0}^{n-1}(x+k) , \tag1$$ which is a polynomial with integer coefficients. P.S. the proof rests on a routine verification that (1) satisfies the given recurrence relation and initial conditions.

Power of $2$ dividing a specialized Mittag-Leffler polynomial While studying the so-called Mittag-Leffler Polynomials, denoted $M_n(x)$, I was looking into the sequence $\frac1{n!}M_n(n)$ which takes the following form $$a_n:=\sum_{k=1}^n\binom{n-1}{k-1}\binom{n}k2^k.$$ QUESTION 1. Let $\nu_2(m)$ denote the $2$-adic valuation of $m\in\mathbb{N}$. Is this true? $$\nu_2(a_n)=\begin{cases} \,\,\,\,\,\,1 \qquad \,\,\,\,\,\text{if $n$ is odd} \\ 3\nu_2(n) \qquad \text{if $n$ is even}. \end{cases} $$ QUESTION 2. Is it true that $a_n$ is never divisible by $5$, for $n\geq1$? Postscript. I've recently solved QUESTION 2, so only QUESTION 1 remains.

I will confine myself to Question 1 since you mentioned that you know how to do Question 2. Also the case when $n$ is odd is easy, and let us restrict to $n$ being divisible exactly by $2^r$ with $r\ge 1$, and we need to show that the exact power of $2$ dividing $a_n$ is $3r$. Thus in what follows we may discard any terms that are divisible by a power of $2$ larger than $3r$. For example, we can restrict attention to the terms with $k\le 3r$ in the sum. The case $r=1$ can be easily checked by hand, and we assume below that $r\ge 2$, and $k \le 3r$. Consider the $k$-th term in the sum defining $a_n$: $$ 2^{k} \binom{n}{k} \binom{n-1}{k-1} = 2^k \frac{n}{k} \prod_{j=1}^{k-1} \Big(1 -\frac{n}{j}\Big)^2. $$ Let us now expand the product above, discarding any terms that are divisible by $2^{3r+1}$. Note that $n/j$ is a $2$-adic integer, since the power of $2$ dividing $j$ is at most $\lfloor \log_2(k-1)\rfloor \le \lfloor \log_2 (3r-1)\rfloor \le r$. We first observe that when expanding the product out terms that ``have $n^4$" in them (so using $3$ factors of $n/j$ in the product) can be omitted, and therefore terms that have higher powers of $n$ may also be omitted. Indeed a term using $3$ factors of $n/j$ in the product is divisible by a power of $2$ $$ \ge k + r -v_2(k) + 3r - 3\lfloor \log_2(k-1)\rfloor. $$ A small calculation shows that $k-v_2(k)-3\lfloor \log_2(k-1) \rfloor \ge -1$ for all $k\ge 1$, and therefore the above is $\ge 3r+r-1 \ge 3r+1$, as desired. Now consider terms with $n^3$ in them (so using $2$ terms of the form $n/j$ from the product). We claim that there is exactly one such term that is relevant, and this happens only in the case $k=4$, and the two terms from the product are (both) $n/2$. Indeed the power of $2$ in term using two factors $n/j$ is $$ \ge k + r-v_2(k) + 2 r -2 \lfloor \log_2 (k-1)\rfloor, $$ and we can check that $k-v_2(k) -2 \lfloor \log_2(k-1)\rfloor \ge 1$ for $k\ge 1$, except in the case $k=4$. In the case $k=4$ we can quickly check that the only relevant term with two factors $n/j$ must be $(n/2)(n/2)$. Putting all these observations together, we find that $a_n$ equals (after omitting any terms divisible by $2^{3r+1}$) \begin{align*} &\sum_{k\le 3r} 2^k \frac{n}{k} \Big( 1 - 2 \sum_{j=1}^{k-1} \frac{n}{j}\Big) + 2^4 \frac{n}{4} \frac n2 \frac n2 \\ &= n \sum_{k\le 3r} \frac{2^k}{k} -n^2 \sum_{k\le 3r} \frac{2^k}{k} \sum_{j=1}^{k-1}\Big( \frac 1j + \frac{1}{k-j}\Big) +n^3\\ &= n \sum_{k\le 3r} \frac{2^k}{k} -n^2 \sum_{k\le 3r} \sum_{j=1}^{k-1} \frac{2^k}{j(k-j)} + n^3. \end{align*} Now here it is convenient to extend the sums over $k$ to infinity, noting that the extra terms are divisible by $2^{3r+1}$. (Thus one checks that for $k>3r$ one has $r+k -v_2(k) \ge 3r+1$, and that $2r + k -2 \lfloor \log_2 (k-1)\rfloor \ge 3r+1$.) Now comes the magical bit. In the $2$-adics one has $$ \sum_{k=1}^{\infty} \frac{2^k}{k} =0, $$ and therefore also $$ \sum_{k=1}^{\infty} 2^k \sum_{j=1}^{k-1} \frac{1}{j(k-j)} = \Big( \sum_{k=1}^{\infty} \frac{2^k}{k} \Big)^2 =0. \tag{*} $$ We are then left with $a_n$ being $n^3$ up to multiples of $2^{3r+1}$, which is what we want. For the evaluation of $\sum_{k=1}^{\infty} 2^k/k$, note that if $|x|_2 <1$ then the series $$ \sum_{k=1}^{\infty} \frac{x^k}{k} $$ converges in the $2$-adics, and this clearly "looks like" $-\log(1-x)$. In our case we'd get $-\log (1-2) = \log (-1) =0$ (upon "using" $\log (-1) + \log (-1) = \log 1 = 0$). This last bit can all be made precise; a lovely write up of what is involved can be found in Keith Conrad's notes (see Example 8.10 there).

When is $2^{2n}-2^n+1$ prime? Is it known when $2^{2n}-2^n+1$ is prime? It seems to be only when n is 1,2,4 or 32.

Expanding on the remarks by Pace Nielsen (with an additional result that we need only consider $n>2$ a multiple of $4.$): $x^n-1=\prod_{d \mid n}\phi_d(x)$. The factors are irreducible over the rationals. In particular $$x^6-1=\phi_1(x)\phi_2(x)\phi_3(x)\phi_6(x)=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$$ And $$\phi_6(x^n)=x^{2n}-x^n+1$$ is a factor of $x^{6n}-1.$ It is an irreducible rational polynomial exactly if $n$ is of the form $n=2^i3^j,$ then $$\phi_6(x^n)=\phi_{6n}(x)$$ . But if $p$ is prime to $6$ then $$\phi_6(u^p)=\phi_p(u)\phi_{6p}(u)$$ so for $n=pm$ $$\phi_{6}(x^n)=\phi_p(x^m)\phi_{6p}(x^m).$$ Accordingly, a necessary condition for $b^{2n}-b^n+1$ to be prime is that $n$ is of the form $2^i3^j.$ In the event that $b \bmod 3=-1$ we also need $n$ even lest $b^{2n}-b^n+1 \bmod 3=0.$ Finally, in the given case that $b=2,$ we have that for $n=4k-2>2$ the integer $2^{2n}-2^n+1$ has two factors of roughly equal size. For example $2^{36}-2^{18}+1=68719214593=246241\cdot 279073.$ This is due to the following interesting Aurifeuillean factorization $$\left(x^{4 k -2}+x^{2 k -1}+1+(x^{3 k -1}+x^{k})\right) \left(x^{4 k -2}+x^{2 k -1}+1-(x^{3 k -1}+x^{k})\right)\\ =x^{8 k -4}+(2 x^{6 k -3}-x^{6 k -2})-(x^{2 k}- 2 x^{2 k -1})+(3 x^{4 k -2}-2 x^{4 k -1})+1$$ Putting $x=2,$ the first two pairs become $0$ and the third $2^{4k-2}.$ So the question might be phrased "when does the Mersenne number $2^n-1$ have factors other than those suggested by algebra?" Here restricted to $n=6m.$ That is the subject of the Cunningham Project where much information (in very condensed form) can be found.

Growth of the "cube of square root" function Hello all, this question is a variant (and probably a more difficult one) of a (promptly answered ) question that I asked here, at Is it true that all the "irrational power" functions are almost polynomial ?. For $n\geq 1$, let $f(n)$ denote the "integer part" (largest integer below) $n^{\frac{3}{2}}$, and let $g(n)=f(n+2)-2f(n+1)+f(n)$. Question : Is it true that $g(n)$ is always in $\lbrace -1,0,1\rbrace$ (excepting the initial value $g(1)=2$) ? I checked this up to $n=100000$. It is not too hard to check that if $t$ is large enough compared to $r$ (say $t\geq \frac{3r^2+1}{4}$), $f(t^2+r)$ is exactly $t^3+\frac{3rt}{2}$ (or $t^3+\frac{3rt-1}{2}$ if $t$ and $r$ are both odd ) and similarly $f(t^2-r)$ is exactly $t^3-\frac{3rt}{2}$ (or $t^3-\frac{3rt+1}{2}$ if $t$ and $r$ are both odd ). So we already know that the answer is "yes" for most of the numbers.

It's not too hard to put a bound on the size of second differences (since without the truncation, they are bounded above by a constant times $n^{-1/2}$), but getting the bound down to one seems delicate. It looks like it can be done with mindless brute force, though. I won't write all of the cases, but here is a start. Write $n = t^2 + r$, for integers $t,r$ satisfying $|r| \leq t$. The binomial theorem implies $n^{3/2} = t^3 + (3/2)tr + (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)r^4/t^5 - \dots$. I'll look at the case where $t$ is even. Then * *$f(n) = t^3 + (3/2)tr + \lfloor (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)r^4/t^5 - \dots \rfloor$ *$f(n+1) = t^3 + (3/2)t(r+1) + \lfloor (3/8)(r+1)^2/t - (1/16)(r+1)^3/t^3 + (3/128)(r+1)^4/t^5 - \dots \rfloor$ *$f(n+2) = t^3 + (3/2)t(r+2) + \lfloor (3/8)(r+2)^2/t - (1/16)(r+2)^3/t^3 + (3/128)(r+2)^4/t^5 - \dots \rfloor$. $g(n)$ then has no contributions from the first two terms of each series, since they cancel. Therefore: $g(n) = \lfloor (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)(r+1)^4/t^5 - \dots \rfloor +$ $\qquad + 2\lfloor (3/8)r^2/t + (3/4)r/t - (1/16)(r+1)^3/t^3 + (3/8t) + (3/128)(r+2)^4/t^5 - \dots \rfloor +$ $\qquad + \lfloor (3/8)r^2/t + (3/2)r/t - (1/16)(r+2)^3/t^3 + (3/2t) + (3/128)(r+2)^4/t^5 - \dots \rfloor$. At this point, you can break the analysis into more cases involving the fractional part of $(3/8)r^2/t$ and the size of $r/t$, and then invoke some estimates about the remaining parts of the sum.

Continuous or analytic functions with this property of sinc function This question is motivated by my previous post in SE (math.stackexchange.com). Prove or disprove that $\frac{\sin x}{x}$ is the only nonzero entire (i.e. analytic everywhere), or continuous, function, $f(x)$ on $\mathbb{R}$ such that $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx=\sum_{-\infty}^\infty f(n) =\sum_{-\infty}^\infty f(n)^2 $$

Consider the function $f:\mathbb R\to\mathbb R$ which vanishes outside of $[-1,1]$ and such that for all $x \in [-1,1]$ has $$ f(x) = \frac{5 x^4}{4}+\frac{1}{2} \sqrt{\frac{37}{6}} x^3-\frac{9 x^2}{4}-\frac{1}{2} \sqrt{\frac{37}{6}} x+1. $$ Then $$\int_\mathbb Rf(x)\,\mathrm dx=\int_\mathbb Rf(x)^2\,\mathrm dx=\sum_{n\in\mathbb Z}f(n)=\sum_{n\in\mathbb Z}f(n)^2=1.$$ Moreover, the functions with the same support which on $[-1,1]$ coincide with the polynomials $$ \begin{array}{l} -\frac{1}{32} (x-1)^2 (x+1) \left(x \left(\left(11+\sqrt{3245}\right) x+\sqrt{3245}-29\right)-32\right); \\\\ \frac{(x-1)^3 (x+1) \left(x \left(7 \left(\sqrt{2657109}-923\right) x+5 \sqrt{2657109}-8431\right)-3392\right)} {3392}; \\\\ -\frac{1}{64} (x-1)^4 (x+1) \left(\left(3 \sqrt{3855}-175\right) x^2+2 \left(\sqrt{3855}-98\right) x-64\right); \\\\ \frac{(x-1)^5 (x+1) \left(3 \left(\sqrt{3040433}-2550\right) x^2+2 \left(\sqrt{3040433}-4571\right) x-3008\right)}{3008}; \\\\ -\frac{(x-1)^6 (x+1) \left(x \left(25 \left(3 \sqrt{622687}-6365\right) x+51 \sqrt{622687}-202885\right)-67328\right )}{67328}; \end{array} $$ have the same properties and are $C^1$, $C^2$, $C^3$, $\dots$, respectively, and it seems one can go on indefinitely. There are no other polynomials satisfying the conditions and of degree less than or equal to theirs. The unique $C^k$ polynomial of degree $4+k$ satisfying the conditions is the product of $(x-1)^{k+1}(x+1)$ and a degree $2$ factor: one should be able to describe this last factor explicitly somehow...

the following inequality is true，but I can't prove it The inequality is \begin{equation*} \sum_{k=1}^{2d}\left(1-\frac{1}{2d+2-k}\right)\frac{d^k}{k!}>e^d\left(1-\frac{1}{d}\right) \end{equation*} for all integer $d\geq 1$. I use computer to verify it for $d\leq 50$, and find it is true, but I can't prove it. Thanks for your answer.

This is true for large $d$, and probably for all $d$. I'll prove that the sum is $$e^d(1-1/d+1/d^2+O(1/d^{2.5+\epsilon}))$$ and leave the explicit bounds to you. Set $k=d+\ell$. For $|\ell| \leq d^{0.5+\epsilon}$, we have $$1-1/(2d+2-k) = 1-\frac{1}{d} \frac{1}{1-(\ell-2)/d} = 1-\frac{1}{d} - \frac{\ell-2}{d^2} - \frac{(\ell-2)^2}{d^3} + O(d^{-2.5+3 \epsilon})$$ $$=1-\frac{1}{d} - \frac{\ell}{d^2} + \frac{2 d - \ell^2}{d^3} + O(d^{-2.5+\epsilon}).$$ We will show later that the difference between your sum and $$\sum_{k=0}^{\infty} \frac{d^k}{k!} \left( 1-\frac{1}{d} - \frac{\ell}{d^2} + \frac{2 d - \ell^2}{d^3} \right)$$ is very small, where $\ell = k-d$. Assuming that, let's compute the new sum. With the aid of Mathematica, $$\sum_{k=0}^{\infty} \frac{x^k}{k!} \left( 1-\frac{1}{d} - \frac{k-d}{d^2} + \frac{2 d - (k-d)^2}{d^3} \right) = e^x \left( 1-\frac{1}{d}+\frac{x}{d^2}-\frac{x^2}{d^3} + \frac{2}{d^2} - \frac{x}{d^3} \right). $$ Plugging in $x=d$, $$\sum_{k=0}^{\infty} \frac{d^k}{k!} \left( 1-\frac{1}{d} - \frac{k-d}{d^2} + \frac{2 d - (k-d)^2}{d^3} \right) = e^d \left( 1-\frac{1}{d} + \frac{1}{d^2} \right).$$ The error coming from $O(d^{-2.5+\epsilon}) \sum d^k/k!$ is $e^d O(d^{-2.5+\epsilon})$. We now just need to think about the error coming from discarding the terms with $|\ell|>d^{0.5+\epsilon}$. No matter what $\ell$ is, that error is no worse than $d^k/k! \cdot ( O(\ell^2/d^3) + O(1))$. But, for $|\ell| > d^{0.5 + \epsilon}$, we have $d^k/k! \leq e^{-d^{2 \epsilon}}$ as I pointed out on math.SE, so the contribution from these terms is exponentially small. I have not found a slick way to give a proof for all $d$, rather than an asymptotic result.

Upper limit on the central binomial coefficient What is the tightest upper bound we can establish on the central binomial coefficients $ 2n \choose n$ ? I just tried to proceed a bit, like this: $ n! > n^{\frac{n}{2}} $ for all $ n>2 $. Thus, $ \binom{2n}{n} = \frac{ (n+1) \ldots (2n) }{n!} < \frac{\left(\frac{\sum_{k=1}^n (n+k) }{n}\right)^n }{n^{n/2}} = \frac{ \left( \frac{ n^2 + \frac{n(n+1)}{2} }{n} \right) ^n}{n^{n/2}} = \left( \frac{3n+1}{2\sqrt{n}} \right)^n $ But, I was searching for more tighter bounds using elementary mathematics only (not using Stirling's approximation etc.).

Here's a way to motivate and refine the argument that Péter Komjáth attributes to Erdős. Start by computing the ratio between the $n$-th and $(n-1)$-st central binomial coefficients: $$ {2n \choose n} \left/ {2(n-1) \choose n-1} \right. = \frac{(2n)! \phantom. / \phantom. n!^2}{(2n-2)! \phantom. / \phantom. (n-1)^2} = \frac{(2n)(2n-1)}{n^2} = 4 \left( 1 - \frac1{2n} \right). $$ For large $n$, this ratio approaches $4$, suggesting that $2n \choose n$ grows roughly as $4^n$. If the factor $1 - \frac1{2n}$ were $1 - \frac1n = (n-1)/n$, the growth would be exactly proportional to $n^{-1} 4^n$. Since $1 - \frac1{2n}$ is (for large $n$) nearly the square root of $1 - \frac1n$, the actual asymptotic should be proportional to $n^{-1/2} 4^n$. So we introduce the ratio $$ r_n := \left( {2n \choose n} \left/ \frac{4^n}{\sqrt n} \right. \right)^2 = \frac{n}{16^n} {2n \choose n}^2. $$ Then $$ \frac{r_n}{r_{n-1}} = \left( 1 - \frac1{2n} \right)^2 \left/ \left( 1 - \frac1n \right) \right. = \frac{(2n-1)^2}{(2n-2)(2n)} \gt 1. $$ Thus $r_{n-1} < r_n$; and since $r_1 = (2/4)^2 = 1/4$ we have by induction $$ r_1 \lt r_2 \lt r_3 \lt r_4 \lt \cdots \lt r_n = \frac12 \frac{1 \cdot 3}{2 \cdot 2} \frac{3 \cdot 5}{4 \cdot 4} \frac{5 \cdot 7}{6 \cdot 6} \cdots \frac{(2n-3)(2n-1)}{(2n-2)(2n-2)} \frac{2n-1}{2n}. $$ Each $r_{n_0}$ gives a lower bound on $r_n$, and thus on $2n\choose n$, for all $n \geq n_0$. The OP asked for upper bounds, so consider $$ R_n := \frac{2n}{2n-1} r_n = \frac{n}{\left(1-\frac 1{2n}\right)16^n} {2n \choose n}^2. $$ Now $R_{n+1}/R_n = (2n-1)(2n+1) \phantom. / \phantom. (2n)^2 = (4n^2-1) \phantom. / \phantom. (4n^2) \lt 1$, so $$ \frac12 = R_1 \gt R_2 \gt R_3 \gt R_4 \gt \cdots \gt R_{n+1} = \frac12 \frac{1 \cdot 3}{2 \cdot 2} \frac{3 \cdot 5}{4 \cdot 4} \frac{5 \cdot 7}{6 \cdot 6} \cdots \frac{(2n-3)(2n-1)}{(2n-2)(2n-2)}. $$ It follows that $R_n \geq r_{n'}$ for any $n,n'$, so $R_1=1/2$, $R_2=3/8$, $R_3=45/128$, etc. are a series of upper bounds on every $r_n$. Since moreover $r_n / R_n = 1 - \frac1{2n} \rightarrow 1$ as $n \rightarrow \infty$, both $r_n$ and $R_n$ converge to a common limit that is an upper bound on every $r_n$. If we accept Wallis's product (which is classical though not as elementary as everything else in our analysis), then we can evaluate this common limit as $1/\pi$ and thus recover the asymptotically sharp upper bound ${2n \choose n} < 4^n / \sqrt{\pi n}$.

A binomial determinant fomula Is there an existing or elementary proof of the determinant identity $ \det_{1\le i,j\le n}\left( \binom{i}{2j}+ \binom{-i}{2j}\right)=1 $?

I don't have a proof yet, but it's likely that this matrix has a simple LU decomposition which makes the determinant obviously $1$. \begin{multline} \scriptsize\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ \scriptsize\hskip1cm= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 4 & 9 & 16 & 25 & 36 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{multline} The first factor appears to be A054142 as a lower diagonal matrix. The second factor appears to be A156308 as an upper diagonal matrix.

Computing $\prod_p(\frac{p^2-1}{p^2+1})$ without the zeta function? We see that $$\frac{2}{5}=\frac{36}{90}=\frac{6^2}{90}=\frac{\zeta(4)}{\zeta(2)^2}=\prod_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\prod_p \left(\frac{(p^2-1)^2}{(p^2+1)(p^2-1)}\right)=\prod_p\left(\frac{p^2-1}{p^2+1}\right)$$ $$\implies \prod_p \left(\frac{p^2-1}{p^2+1}\right)=\frac{2}{5},$$ But is this the only way to compute this infinite product over primes? It seems like such a simple product, one that could be calculated without the zeta function. Note that $\prod_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\prod_p(\frac{p-1}{p-i})\prod_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\prod_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdots$.

I found a proof of $$5 \sum_{m=1}^{\infty} \frac{1}{m^4} = 2 \left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right)^2$$ by rearranging sums and wrote it up. The argument is just 1.5 pages, the other 4.5 are explanations and context. Here is a summary using divergent sums; see the write up for a correct version. Set $$h(m,n) = \begin{cases} \frac{1}{m^3 (n-m)} & m \neq n,\ m \neq 0, \\ 0 & m=n \ \text{or} \ m=0. \end{cases}$$ Then we should have $$\sum_{(m,n) \in \mathbb{Z}^2} h(m,n) - h(n,2n-m) =0$$ as every value occurs twice with opposite signs. So $\sum_{(m,n) \in \mathbb{Z}^2} g(m,n)=0$ where $$g(m,n) := h(m,n) - h(n,2n-m) = \begin{cases} \frac{m^2+mn+n^2}{m^3 n^3} & m \neq n,\ m, n \neq 0 \\ - \frac{1}{m^4} & n=0,\ m \neq 0 \\ - \frac{1}{n^4} & m=0,\ n \neq 0 \\ 0 & m=n \end{cases}$$ Group together the terms where $(|m|, |n|)$ have a common value; we get $\sum_{(m,n) \in \mathbb{Z}_{\geq 0 }^2} f(m,n) =0$ where $$f(m,n) = \begin{cases} \frac{4}{m^2 n^2} & m \neq n,\ m,n >0, \\ - \frac{2}{m^4} & m>n=0, \\ - \frac{2}{n^4} & n>m=0, \\ - \frac{2}{m^4} & m=n>0, \\ 0 & m=n=0. \end{cases} $$ Writing this out, $4\zeta(2)^2 - 6 \zeta(4) - 4 \zeta(4)=0$, as desired. Has anyone seen this? If this is new, I'm thinking of sending it to the Monthly.

Is this system always solvable in radicals by quartics, octics, $12$-ics, etc? While considering this post, it made me wonder about its generalization in another direction and from the perspective of Galois theory. Question: Is it true that, given four constants ($\alpha,\beta,\gamma,\delta$), then the system, $$\begin{aligned} x_1^a+x_2^a+x_3^a+x_4^a &= \alpha\\ x_1^b+x_2^b+x_3^b+x_4^b &= \beta\\ x_1^c+x_2^c+x_3^c+x_4^c &= \gamma\\ x_1^d+x_2^d+x_3^d+x_4^d &= \delta \end{aligned}\tag1$$ can be solved in radicals $x_i$ for any positive integer power ($a,b,c,d$) and $a<b<c<d$? Example: Let's choose the year Back to the Future came out, $$\begin{aligned} x_1+x_2+x_3+x_4 &= \color{brown}1\\ x_1^2+x_2^2+x_3^2+x_4^2 &= \color{brown}9\\ x_1^4+x_2^4+x_3^4+x_4^4 &= \color{brown}8\\ x_1^8+x_2^8+x_3^8+x_4^8 &= \color{brown}5 \end{aligned}$$ Two solutions are given by the roots of the octic, $$4707 + 9036 x - 7495 x^2 - 5096 x^3 + 3024 x^4 + 1848 x^5 - 896 x^6 - 256 x^7 + 128 x^8 = 0$$ Magma says this is $8T47$, has order $1152=2^7\cdot3^2$ and is the largest solvable order for deg $8$. (In fact, it factors over $\sqrt{1441}$). Using the root $r_i$ numbering system of Mathematica, we find that, $$x_1,\, x_2,\, x_3,\, x_4 = r_1,\, r_2,\, r_5,\, r_6\quad \text{(Solution 1)}\\ \text{or}\quad\quad\\ x_1,\, x_2,\, x_3,\, x_4 = r_3,\, r_4,\, r_7,\, r_8\quad \text{(Solution 2)}$$ Remarks: Using generic ($\alpha,\beta,\gamma,\delta$) * *For exponents $a,b,c,d = 1,2,c,8$ and $c=3,4,5$, one ends up with a $8T47$ octic with order $1152=2^7\cdot3^2$. *For $a,b,c,d = 1,2,6,7$, one gets a $12T294$ with order $82944 = 2^{10} \cdot 3^4$. *For $a,b,c,d = 1,2,c,8$ and $c=6,7$, one now gets a $16T1947$ with order $7962624 = 2^{15}\cdot3^5$. These three are the largest solvable orders for their respective degrees. *Using others yields deg $20,24,$ etc. So, is the system $(1)$ always solvable in radicals?

For $(a,b,c,d)=(1,2,3,20)$ and $(\alpha,\beta,\gamma,\delta)=(1,1,1,0)$ one computes that $x_1$ is a root of the irreducible polynomial $4 x^{20} - 20 x^{19} + 40 x^{18} - 40 x^{17} + 195 x^{16} - 704 x^{15} + 1050 x^{14} - 700 x^{13} + 475 x^{12} - 900 x^{11} + 900 x^{10} - 300 x^{9} + 130 x^{8} - 260 x^{7} + 130 x^{6} + 20 x^{4} - 20 x^{3} + 1$, which according to magma, has a non-solvable Galois group, namely the full wreath product $S_4^5\rtimes S_5$. Without Magma: With the same values of $a,b,c,d,\alpha,\beta,\gamma,\delta$ as above, one computes (e.g. using Sage) that $x_1x_2x_3x_4$ is a root of $h(x)=4 x^{5} - 175 x^{4} + 300 x^{3} - 130 x^{2} + 20 x - 1$. Now $h(x-1)$ is an Eisenstein polynomial with respect to $5$, and modulo $3$ $h(x)=(x + 1) \cdot (x + 2) \cdot (x^{3} + 2 x^{2} + x + 1)$ is a factorization into irreducibles. Thus the Galois group of $h$ contains the alternating group $A_5$.

A question about (unicity of certain cycles in a Cayley graph of a) symmetric group Let $S=\{(1,2),(1,2,3,\ldots,n),(1,2,3,\ldots,n)^{-1}=(1,n\ldots,2)\}$ be a subset of the symmetric group $S_n$. We know that $(1,2,\ldots,n)(1,2)=(2,3,\ldots,n)$, and thus $$[(1,2,\ldots,n)(1,2)]^{n-1}=(1,2,\ldots,n)\overbrace{(1,2)\cdots(1,2,\ldots,n)}^{2n-2}(1,2)=(1).$$ We want to know whether or not there exists another sequence of elements $a_1,a_2,\ldots,a_{2n-4}\in S$ such that $$(12\ldots n)a_{2n-4}a_{2n-5}\cdots a_2a_1(12)=(1),$$ where $a_{i+1}\neq a_i^{-1}$ for $i=0,1,2,\ldots,2n-4$ (putting $a_0=(12)$, $a_{2n-3}=(1,2,\ldots,n)$). Equivalently, I want to ask if, in the cubic Cayley graph $Cay(S_n,S)$, there is a unique cycle of length $2(n-1)$ passing through $(1)$, $(1,2,\ldots,n)$ and $(1,2)$.

The smallest $n$ for which there exist sequences as asked for is $n = 7$: * *$(1,2,3,4,5,6,7) \cdot (1,2) \cdot (1,7,6,5,4,3,2) \cdot (1,2) \cdot (1,2,3,4,5,6,7) \cdot (1,2) \cdot$ $(1,7,6,5,4,3,2) \cdot (1,2) \cdot (1,2,3,4,5,6,7) \cdot (1,2) \cdot (1,7,6,5,4,3,2) \cdot (1,2) = ()$, and *$(1,2,3,4,5,6,7) \cdot (1,2,3,4,5,6,7) \cdot (1,2) \cdot (1,7,6,5,4,3,2) \cdot (1,7,6,5,4,3,2) \cdot$ $(1,2) \cdot (1,2,3,4,5,6,7) \cdot (1,2,3,4,5,6,7) \cdot (1,2) \cdot (1,7,6,5,4,3,2) \cdot$ $(1,7,6,5,4,3,2) \cdot (1,2) = ()$. For $n = 8$ there is no such sequence other than the trivial one mentioned in the question, for $n = 9$ there is one, for $n = 10$ there are $18$, for $n = 11$ there are $5$ and for $n = 12$ there are $104$ such sequences. This has been found with the following GAP function: SearchXueyiSequences := function ( n ) local sequences, search, S; search := function ( sequence, a ) local b; if Length(sequence) = 2*n-3 then if sequence[2*n-3] <> (1,2) and Product(sequence)*(1,2) = () then Add(sequences,Concatenation(sequence,[(1,2)])); fi; return; fi; for b in Difference(S,[a^-1]) do search(Concatenation(sequence,[a]),b); od; end; S := Set(GeneratorsAndInverses(SymmetricGroup(n))); sequences := []; search([],S[2]); sequences := Set(sequences); sequences := sequences{[2..Length(sequences)]}; # exclude trivial sequence return sequences; end;

seeking proofs: infinite series inequalities Question. Numerically, the following is convincing. However, is there a proof? $$\left(\sum_{k\geq1}\frac1{\sqrt{2^k+3^k}}\right)^4 <\pi^2\left(\sum_{k\geq1}\frac1{2^k+3^k}\right)\left(\sum_{k\geq1}\frac{k^2}{2^k+3^k}\right).$$ This comes up in some recent work and the inequality seems needed.

This may serve as a different approach. By Cauchy-Schwarz inequality, $$\left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2\leq \left(\sum_{k\geq 1}\frac 1{k^2}\right)\left(\sum_{k\geq 1}\frac{k^2}{2^k+3^k}\right),$$ which shows that $$\left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2\leq \frac{\pi^2}6\left(\sum_{k\geq 1}\frac{k^2}{2^k+3^k}\right).$$ So it suffices to show that $$\left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2<6\cdot \sum_{k\geq 1}\frac 1 {2^k+3^k}.$$ But \begin{align} \left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2 &<\left(\sum_{k\geq 1}\frac 1{\sqrt{2\cdot 6^{k/2}}}\right)^2 \\ &<6\left(\frac 1{2+3}+\sum_{k\geq 2}\frac 1{2\cdot 3^k}\right) \\ &<6\left(\sum_{k\geq 1}\frac 1{2^k+3^k}\right), \end{align} where in the first inequality, the AM-GM inequality was used.

special values of symmetric functions at powers of $\frac1j$ Let $e_n(x_1,x_2,x_3,\dots)$ denote the $n$-th elementary symmetric function in the infinite variables $x_1,x_2,x_3,\dots$. Let $u$ and $v$ be the roots of $z^2-6z+1=0$. Question. Let $x_j=\frac1{j^8}$. The following seems to be true, but can one prove or disprove? $$e_n(x_1,x_2,x_3,\dots)=\frac{4^{3n+1}\pi^{8n}(u^{2n+1}+v^{2n+1})}{(8n+4)!}.$$ For example, $e_0=1$ and $e_1=\sum_{j\geq1}\frac1{j^8}=\zeta(8)=\frac{\pi^8}{9450}$.

As Gro-Tsen suggests in the comments, we have to expand the infinite product $$f(t)=\prod_j \left(1-\frac{t^8}{j^8}\right)=\prod_{j;\,w^4=1} \left(1-\frac{\omega t^2}{j^2}\right)=\prod_{w^4=1}\frac{\sin\pi\sqrt{w}t}{\pi\sqrt{w}t},$$ we expand product of four sines as an alternating sum of cosines $$\sin a\sin b\sin c\sin d=\frac18\sum (\prod \pm)\cdot\cos(a\pm b\pm c\pm d),$$ and write Taylor series for cosines. Powers of $1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)$ appear. To be more concrete, $c_n:=e_n(x_1,\dots)$ equals $(-1)^n\times [t^{8n}]f(t)$, where $[t^n]g$ denotes a coefficient of $t^n$ in $g$. Thus $$c_n=(-1)^{n+1}i\pi^{8n}\frac1{8(8n+4)!}\sum \pm\left(1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)\right)^{8n+4}.$$ We have $$\left(1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)\right)^4=8i(\pm 3\pm 2\sqrt{2}),$$ that gives your formula after careful simplifications.

A problem of divisibility I came across the following problem. Find two integers, $u_{n}$ and $v_{n}$, such that $$a_{n}=4u_{n}v_{n}+(6n-1)v_{n}+(6n-1)u_{n}+8n^{2}-4n$$ divides $$b_{n}=-(2n-1)u_{n}v_{n}-(2n^{2}-3n)v_{n}-(2n^{2}-3n)u_{n}+8n^{2}.$$ For example, for $n=2$, the choices $u_{2}:=-2$ and $v_{2}:=0$ yield $a_{2}=2$ and $b_{2}=36$.

If $a_n\mid b_n$, then $a_n$ also divides $$Q := (2n-1)a_n + 4b_n = (2n+1)^2 (u_n + v_n+4n).$$ We will blatantly require $a_n = Q$. Notice that $$a_n-Q = 4u_nv_n - 2(2n^2-n+1)u_n - 2(2n^2-n+1)u_n - 8n(2n^2+n+1) = (2u_n-(2n^2-n+1))(2v_n-(2n^2-n+1)) - (n+1)^2(2n+1)^2.$$ Thus, it's enough to take $d\mid (n+1)^2(2n+1)^2$ and set $$u_n = \frac{1}{2}(d + 2n^2-n+1),$$ $$v_n = \frac{1}{2}(\frac{(n+1)^2(2n+1)^2}{d} + 2n^2-n+1).$$ In particular, we can always take $d=(n+1)(2n+1)$ and get $$u_n = v_n = 2n^2+n+1.$$ UPDATE. As pointed out in the comments, the above construction works only for odd $n$. For an even $n$, we have $Q/a_n\equiv 3\pmod{4}$ and thus we can require that $a_n = -Q$. Then $$0 = a_n+Q = 2u_nv_n + 2(2n^2+5n)u_n + 2(2n^2+5n)u_n + 8n^2(2n+3) = (2u_n + (2n^2+5n))(2v_n + (2n^2+5n)) + n^2(2n+1)^2.$$ Again, let $d\mid \frac{n^2(2n+1)^2}4$ and set $$u_n = d - \frac{2n^2+5n}2,$$ $$v_n = \frac{n^2(2n+1)^2}{4d} - \frac{2n^2+5n}2.$$ E.g., taking $d=\frac{n}{2}$, we get $$u_n = -n^2-2n\qquad\text{and}\qquad v_n=n(2n^2+n-2),$$ for which $a_n = -2n^3(2n+1)^2$ divides $b_n=n^4(2n+1)^2$.

Rational approximations of $\sqrt{2}$ in $\mathbb{R} \times \mathbb{Q}_7$ Note: this question was updated (2) after GNiklasch's answer was posted, and taking Gro-Tsen's comment into account. The initial question (1) dealt with $\mathbb{Q}_3$. Original post (1). Let's try to solve the equation $x^2 - 2 = 0$ with $x = \frac{a}{b} \in \mathbb{Q}$. We can't have $x^2 \neq 2$, so the best we can do is minimize $|x^2 - 2|$. Let's try to find an approximation that works over two different completions. Can we have this? $$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 2 |_3 \ll \frac{1}{a} $$ I'm trying to write an $S$-adic approximate solution over two places $S = \{ 3, \infty\}$ and $x = \frac{a}{b} \mapsto (\frac{a}{b}, \frac{a}{b}) \in \mathbb{Q} \times \mathbb{Q} \subset \mathbb{R} \times \mathbb{Q}_3 $. What are the correct exponents? Edit (2). Perhaps i need to find a problem statement that has a solution. gro-tsen suggests I change $p=3$ to $p=7$ so that $\sqrt{2}\in \mathbb{Q}_7$. $$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 2 |_7 \ll \frac{1}{a} $$ possibly I can leave the places the same and change the thing I'm approximating. $\sqrt{7}\in \mathbb{Q}_3$ so perhaps I can find a rational number $\frac{a}{b}\in \mathbb{Q}$ such that $$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 7 |_3 \ll \frac{1}{a} $$ Excuse me while I try to state an instance of weak approximation that's not vacuous.

Here is a full solution for the modified problem, inspired by Gro-Tsen's valuable comment. 1. There are infinitely many rational numbers $a/b\in\mathbb{Q}$ in lowest terms such that $$ \left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{1}{b}\qquad\text{and}\qquad \left|\frac{a^2}{b^2}-2\right|_7\ll\frac{1}{b}.$$ To see this, we work in $\mathbb{Z}[\sqrt{2}]$, the ring of integers of $\mathbb{Q}(\sqrt{2})$. In this ring, the rational prime $(7)$ splits as $(3+\sqrt{2})(3-\sqrt{2})$, while the totally positive units are $(3+2\sqrt{2})^m$. For any positive integer $n$, we can choose the positive integer $m$ so that \begin{align*}a+b\sqrt{2}&=(3+2\sqrt{2})^m(3+\sqrt{2})^n\asymp 7^n,\\ a-b\sqrt{2}&=(3-2\sqrt{2})^m(3-\sqrt{2})^n\asymp 1.\end{align*} This is because $a^2-2b^2=7^n$ holds regardless of $m$. By basic arithmetic in $\mathbb{Z}[\sqrt{2}]$, the integers $a$ and $b$ are relatively prime to each other and to $7$ as well. Moreover, $a$ and $b$ are positive and of size $\asymp 7^n$. It follows that $$\left|\frac{a^2}{b^2}-2\right|_\infty=\frac{7^n}{b^2}\asymp\frac{1}{b} \qquad\text{and}\qquad \left|\frac{a^2}{b^2}-2\right|_7=7^{-n}\asymp\frac{1}{b}.$$ 2. By modifying the above argument, we can achieve that $$ \left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{c}{b}\qquad\text{and}\qquad \left|\frac{a^2}{b^2}-2\right|_7\ll\frac{c^{-1}}{b}$$ for any constant $c>0$. This is essentially best possible, because it is straightforward to see that $$ \left|\frac{a^2}{b^2}-2\right|_\infty\left|\frac{a^2}{b^2}-2\right|_7\geq\frac{1}{|b^2|_\infty|b^2|_7}\geq\frac{1}{b^2}.$$

Given a polynomial constraint equation in $n$ variables, can one conclude that the sum of the variables is non-negative? Currently I'm stuck as follows; at least a positive proof if $n=3$ would be a great nice-to-have! Consider real numbers $x_1,x_2,\dots,x_n$ satisfying $$\prod^n_{k=1}\left(1-x_k^2\right)\:=\:\prod^n_{k=1}\,(2x_k+1)\,.$$ If $\,-\frac12 < x_1,\dots,x_n < 1\,$ holds (then all the preceding factors are positive), does it follow that $$\sum^n_{k=1}\,x_k\,\geqslant\,0\;\;?$$ For fixed $n$ denote this statement by $S(n)$. Three easy observations are * *$S(1)$ is true. *If $\,S(N)\,$ holds true, then $S(n)$ is true for every $n<N,\,$ just set $\,x_{n+1}, x_{n+2},\dots, x_N=0$. *If $\,S(M)\,$ is wrong, then $S(m)$ cannot hold for any $m>M\,$ because if $(x_1, x_2,\dots, x_M)$ does not satisfy $S(M)$, then it can be padded with zeros to yield a counter-example to $S(m)$ for any $m>M$. Next look at $S(2)$ which is true as well: Let $$\left(1-x^2\right)\left(1-y^2\right)\:=\:(2x +1)(2y+1)$$ and assume $x+y<0$. Thus, $\,1-y>1+x\,$ and $\,1-x>1+y$, and one gets the contradiction $$\begin{align}\left(1-x^2\right)\left(1-y^2\right) & = (1-y)(1-x)(1+x)(1+y)\\ & >\:(1+x)^2(1+y)^2\:\geq\: (2x +1)(2y+1)\,. \end{align}$$ My attempt to analogously prove $S(3)$: Let $$\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)\:=\: (2x +1)(2y+1)(2z+1)$$ and assume $x+y+z<0\,$. Then $\,1-x > 1+y+z\,$ and cyclically, hence $$\begin{align}(1-x^2)(1-y^2)(1-y^2) &> (1+x)(1+x+y)(1+y)(1+y+z)(1+z)(1+z+x) \\[1ex] & \stackrel{?}{\geqslant} (2x +1)(2y+1)(2z+1) \end{align}$$ Can this be finalised? My guess: The 'validity gap' lies between $S(3)$ and $S(4)$. This is a Cross-post from math.SE after a decent waiting period (lasting 21 days, resulting in no comments, no downvotes, no answers, and 59 views).

A counterexample: $n=4$, $$(x_1,\dots,x_4)=\left(-\frac{1}{4},-\frac{3}{8},-\frac{3}{8},\frac{\sqrt{1970156929}-2048}{45375}\right). $$ So, by what you noted, your conjecture fails to hold for any $n\ge4$. The validity gap is indeed between $n=3$ and $n=4$. Indeed, for $n=3$ Mathematica confirms your conjecture: I think this can also be easily checked using other computer algebra packages. Alternatively, one can try to do the case $n=3$ manually, for instance, as follows: solve the (quadratic) equation $(1-x^2)(1-y^2)(1-z^2)= (2x +1)(2y+1)(2z+1)$ for (say) $z$, then rewrite the corresponding inequalities as polynomial ones (in $x,y$), and finally use (say) resultants -- as explained and done e.g. in Section 5. Here is a simpler manual, almost ingenuity-free solution for $n=3$: Consider the minimization of $x+y+z$ given \begin{equation*} (1-x^2)(1-y^2)(1-z^2)=(2x+1)(2y+1)(2z+1) \tag{1} \end{equation*} and the non-strict inequality constraints \begin{equation*} -1/2\le x,y,z\le1. \tag{2} \end{equation*} If one of these non-strict inequalities is an equality, then without loss of generality (wlog) $x=1$ and $y=-1/2$, whence $x+y+z\ge1-1/2-1/2=0$. So, it remains to consider possible interior extrema, with all inequalities (2) being strict. Then (1) can be rewritten as \begin{equation*} f(x)+f(y)+f(z)=0, \end{equation*} where $f(u):=\ln\frac{1-u^2}{2u+1}$. Suppose that $(x,y,z)$ is an interior extremum point. Then, by Lagrange multipliers, \begin{equation*} f'(x)=f'(y)=f'(z). \tag{3} \end{equation*} It is easy to see that \begin{equation*} f'''(u)=-\frac{4 \left(4 u^6+12 u^5+42 u^4+37 u^3+6 u^2+3 u+4\right)}{(2 u+1)^3 \left(1-u^2\right)^3}<0 \end{equation*} if $-1/2<u<1$, so that $f'$ is strictly concave on $(-1/2,1)$. Hence, for any real $t$, equation $f'(u)=t$ has at most two real roots $u$. So, by (3), wlog $y=x$. Solving now (1) (with $y=x$) for $z$, we see that $x+y+z$ becomes \begin{equation*} s:=\frac{-p+\sqrt d}{(1-x^2)^2}, \end{equation*} where \begin{equation*} p:=1 + 2 x + 4 x^2 + 4 x^3 - 2 x^5 \end{equation*} and \begin{equation*} d:=p^2+x^2 (1-x^2)^2 (6+16 x+9 x^2-4 x^4)>p^2, \end{equation*} so that $s$ is manifestly positive, as claimed.

Primality test for specific class of $N=8k \cdot 3^n-1$ This question is related to my previous question. Can you prove or disprove the following claim: Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$ Let $N=8k \cdot 3^n-1$ such that $n>2$ , $k>0$ , $8k <3^n$ and $\begin{cases} k \equiv 1 \pmod{5} \text{ with } n \equiv 0,1 \pmod{4} \\ k \equiv 2 \pmod{5} \text{ with } n \equiv 1,2 \pmod{4} \\ k \equiv 3 \pmod{5} \text{ with } n \equiv 0,3 \pmod{4} \\ k \equiv 4 \pmod{5} \text{ with } n \equiv 2,3 \pmod{4} \end{cases}$ Let $S_i=S_{i-1}^3-3S_{i-1}$ with $S_0=P_{18k}(3)$ , then $N$ is prime iff $S_{n-2} \equiv 0 \pmod N$ . You can run this test here. I have verified this claim for $k \in [1,300]$ with $n \in [3,1000]$ .

Assume that $N$ is prime. Then we prove $S_{n-2}\equiv 0\pmod N$, the assumption that $8k<3^n$ is not used. I do not know how to prove it in the opposite direction. We have $P_m(2\cos t)=2\cos mt$, so they are Chebyshev polynomials and satisfy $P_{mn}=P_n\circ P_m$. Note that $x^3-3x=P_3$, thus $S_{i}=P_{18k\cdot 3^i}(3)$, $S_{n-2}=P_{2k\cdot 3^n}(3)=P_{(N+1)/4}(3)=P_{(N+1)/2}(\sqrt{5})$. We prove that $2^{3(N+1)/2}P_{(N+1)/2}(\sqrt{5})$ is divisible by $N$ in the ring $\mathbb{Z}[\sqrt{5}]$. This would imply $$\frac{2^{3(N+1)/2}P_{(N+1)/2}(\sqrt{5})}N\in \mathbb{Z}[\sqrt{5}]\cap \mathbb{Q}=\mathbb{Z},$$ thus $N$ indeed divides $P_{(N+1)/2}(\sqrt{5})$. Further we write congruences modulo $N$ in the ring $\mathbb{Z}[\sqrt{5}]$. Using quadratic reciprocity and the explicit calculations of powers of 3 modulo 4, we see that your additional condition means that 5 is a quadratic non-residue modulo $N$. This means that $5^{(N-1)/2}\equiv -1$, or $5^{N/2}\equiv-\sqrt{5}$. We have $$ 2^{3(N+1)/2}P_{(N+1)/2}(\sqrt{5})=2^{N+1}\left(\left(\sqrt{5}+1\right)^{(N+1)/2}+\left(\sqrt{5}-1\right)^{(N+1)/2}\right)\\ =\left(\sqrt{5}-1\right)^{(N+1)/2}\cdot\left( \left(\sqrt{5}+1\right)^{N+1}+2^{N+1}\right)\\ \equiv\left(\sqrt{5}-1\right)^{(N+1)/2}\cdot\left( (\sqrt{5}+1)(5^{N/2}+1)+2^{N+1}\right)\\\equiv \left(\sqrt{5}-1\right)^{(N+1)/2}\cdot\left( (\sqrt{5}+1)(1-\sqrt{5})+2^{N+1}\right)\equiv 0. $$

A constant bizarrely related to the Fibonacci Numbers For roughly the past month, I have been studying denesting radicals. For example: the expression $\sqrt[3]{\sqrt[3]2-1}$ is a radical expression that contains another radical expression, so this radical is nested. Is there a way to express this with radicals that are not (or not as) nested? Writing it in this way is referred to as denesting, and indeed, there is one such way. Ramanujan mysteriously found that $$\sqrt[3]{\sqrt[3]2-1}=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}.$$ I found that this is also equal to $\sqrt{\sqrt[3]{\frac 43}-\sqrt[3]{\frac 13}}$, which does not denest it, but it does write the expression under a radical of a coprime degree, which fascinates me. I have found an abundance of results, one of which fascinates me the most, and is on the constant, $$1-\sqrt[3]{\frac 12}+\sqrt[3]{\frac 14}.$$ This constant is equal to all of the expressions below. Note that the degree of the radicals are, lo and behold, the Fibonacci numbers! $$\sqrt{\frac 32\bigg(\sqrt[3]2-\frac 1{\sqrt[3]2}\bigg)}$$ $$\sqrt[3]{\frac{3^2}{2^2}\big(\sqrt[3]2-1\big)}$$ $$\sqrt[5]{\frac{3^3}{2^3}\bigg(\frac 3{\sqrt[3]2}-\sqrt[3]2-1\bigg)}$$ $$\sqrt[8]{\frac{3^5}{2^5}\bigg(4-\frac{5}{\sqrt[3]2}\bigg)}$$ $$\sqrt[13]{\frac{3^8}{3^8}\bigg(1+\frac{17}{\sqrt[3]2}-\frac{23}{\sqrt[3]4}\bigg)}$$ and, slightly breaking the pattern, $$\sqrt[21]{\frac{3^{14}}{2^{14}}\big(41-59\sqrt[3]2+21\sqrt[3]4\big)}$$ and presumably, this list goes on forever (but the numbers start becoming pretty big). So... what on earth is going on here? It appears that for some $n$th Fibonacci number $F_n$, this is equal to (for at least most of the time), $$\sqrt[F_n]{\frac{3^{F_{n-1}}}{2^{F_{n-1}}}\big(a+b\sqrt[3]2+c\sqrt[3]4\big)}$$ for some $\{a, b, c\}\subset \mathbb R$, with a minimal polynomial of $4x^3-12x^2+18x-9$. Can anybody explain these wild affairs? Don't know of any other appropriate tags

First of all we get $A=1-\sqrt[3]{\frac{1}{2}}+\sqrt[3]{\frac{1}{4}}=\frac{3\sqrt[3]{2}}{2(\sqrt[3]{2}+1)}$......(1) And from here you can proof by induction... $A^{F_n}=A^{F_{n-1}}.A^{F_{n-2}}$ If $A^{F_{n-1}}=(\frac{3}{2})^{F_{n-2}}(a_{n-1}+b_{n-1}\sqrt[3]{2}+c_{n-1}\sqrt[3]{4})$ and $A^{F_{n-2}}=(\frac{3}{2})^{F_{n-3}}(a_{n-2}+b_{n-2}\sqrt[3]{2}+c_{n-2}\sqrt[3]{4})$ From (1) we get, $A^{F_2}, A^{F_3}$, so we shall get the rest by induction. Where, $F_2=2, F_3=3$. $\frac{3}{2}(\frac{\sqrt[3]{2}}{\sqrt[3]{2}+1})^2=\sqrt[3]{2}-\sqrt[3]{\frac{1}{2}}$ And, $\frac{3}{2}(\frac{\sqrt[3]{2}}{\sqrt[3]{2}+1})^3=\sqrt[3]{2}-1$

Number of d-Calabi-Yau partitions This problem arises from algebraic geometry/representation theory, see https://arxiv.org/pdf/1409.0668.pdf (chapter 2). We call a partition $p=[p_1,...,p_n]$ with $2 \leq p_1 \leq p_2 \leq ... \leq p_n$ d-Calabi-Yau (for $d \geq 1$) when $n-d-1=\sum\limits_{i=1}^{n}{\frac{1}{p_i}}$. For example for $d=1$ there are 4 1-Calabi-Yau partitions: [2,2,2,2],[3,3,3],[2,4,4] and [2,3,6]. For $d=2$ there are 18, see example 2.15. (b) in https://arxiv.org/pdf/1409.0668.pdf . Question 1: Is there a nice formula for the $d$-Calabi-Yau partitions for a fixed $d$? It seems hard to calculate the number for $d \geq 3$ even with a computer since a term $p_i$ can be pretty big. Question 2: For a given $d$, what is the maximal value a term $p_i$ can have? For $d=1$ it is 6 and for $d=2$ it is 42.

Following up on @pbelmans's answer and researching an article mentioned in OEIS A007018, I believe your Question 2 was answered just under 100 years ago by David Curtiss (On Kellogg's Diophantine problem, Amer. Math. Monthly 29 (1922) 380-387).* Curtiss confirms Kellogg's conjecture that the maximum $x_i$ in any $$\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} = 1$$ is given by the sequence $u_1 = 1$ and $u_{k+1} = u_k(u_k+1)$ which begins 1, 2, 6, 42, 1806, 3263442. On p386 he explains, "But the value $u_n$ is actually attained by giving to the $x$'s the values $u_k+1$, so that $u_n$ is the maximum of $f_{n-1}(x)$." (That last expression is something he defined to simplify the proof.) Indeed, \begin{gather*} \frac{1}{1+1} + \frac{1}{2+1} + \frac{1}{6} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1, \\ \frac{1}{1+1} + \frac{1}{2+1} + \frac{1}{6+1} + \frac{1}{42} = \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{42} = 1, \\ \frac{1}{1+1} + \frac{1}{2+1} + \frac{1}{6+1} + \frac{1}{42+1} + \frac{1}{1806} = \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} + \frac{1}{1806} = 1, \dots \end{gather*} * There's also a sci.math.research thread from 1996 where a claim that Curtiss's proof is wrong is retracted, but the proof does seem to be difficult to follow. Gerry Myerson provided a reference to a simpler 1995 proof by Izhboldin and Kurliandchik. I'm including a comment to @katago here so that answers to both questions are in one place. Looking through MathSciNet, the most recent upper bound may be Browning & Elsholtz 2011, something like $$(1.264085...)^{(5/12)^{2^{n−1}}}$$ where apparently the 1.26 is a known constant that arises in these problems. There are also lower bounds in the literature. So the short answer to Q1 is no, at least not yet.

Alternative proofs sought after for a certain identity Here is an identity for which I outlined two different arguments. I'm collecting further alternative proofs, so QUESTION. can you provide another verification for the problem below? Problem. Prove that $$\sum_{k=1}^n\binom{n}k\frac1k=\sum_{k=1}^n\frac{2^k-1}k.$$ Proof 1. (Induction). The case $n=1$ is evident. Assume the identity holds for $n-1$. Then, \begin{align*} \sum_{k=1}^{n+1}\binom{n+1}k\frac1k-\sum_{k=1}^n\binom{n}k\frac1k &=\frac1{n+1}+\sum_{k=1}^n\left[\binom{n+1}k-\binom{n}k\right]\frac1k \\ &=\frac1{n+1}+\sum_{k=1}^n\binom{n}{k-1}\frac1k \\ &=\frac1{n+1}+\frac1{n+1}\sum_{k=1}^n\binom{n+1}k \\ &=\frac1{n+1}\sum_{k=1}^{n+1}\binom{n+1}k=\frac{2^{n+1}-1}{n+1}. \end{align*} It follows, by induction assumption, that $$\sum_{k=1}^{n+1}\binom{n+1}k\frac1k=\sum_{k=1}^n\binom{n}k\frac1k+\frac{2^{n+1}-1}{n+1}=\sum_{k=1}^n\frac{2^k-1}k+\frac{2^{n+1}-1}{n+1} =\sum_{k=1}^{n+1}\frac{2^k-1}k.$$ The proof is complete. Proof 2. (Generating functions) Start with $\sum_{k=1}^n\binom{n}kx^{k-1}=\frac{(x+1)^n-1}x$ and integrate both sides: the left-hand side gives $\sum_{k=1}^n\binom{n}k\frac1k$. For the right-hand side, let $f_n=\int_0^1\frac{(x+1)^n-1}x\,dx$ and denote the generating function $G(q)=\sum_{n\geq0}f_nq^n$ so that \begin{align*} G(q)&=\sum_{n\geq0}\int_0^1\frac{(x+1)^n-1}x\,dx\,q^n =\int_0^1\sum_{n\geq0}\frac{(x+1)^nq^n-q^n}x\,dx \\ &=\int_0^1\frac1x\left[\frac1{1-(x+1)q}-\frac1{1-q}\right]dx=\frac{q}{1-q}\int_0^1\frac{dx}{1-(x+1)q} \\ &=\frac{q}{1-q}\left[\frac{\log(1-(1+x)q)}{-q}\right]_0^1=\frac{\log(1-q)-\log(1-2q)}{1-q} \\ &=\frac1{1-q}\left[-\sum_{m=1}^{\infty}\frac1mq^m+\sum_{m=1}^{\infty}\frac{2^m}mq^m\right]=\frac1{1-q}\sum_{m=1}^{\infty}\frac{2^m-1}m\,q^m \\ &=\sum_{n\geq1}\sum_{k=1}^n\frac{2^k-1}k\,q^n. \end{align*} Extracting coefficients we get $f_n=\sum_{k=1}^n\frac{2^k-1}k$ and hence the argument is complete.

One can also use the binomial transform. (If $A(z)=\sum_{i\geq 0} a_i z^i$ is a (formal) power series, the (formal) power series $B(z):=\frac{1}{1-z} A(\frac{z}{1-z})$ has coefficients $[z^n] B(z)=\sum_{j=0}^n {n \choose j} a_j$). We have $\log(\frac{1}{1-z})=\sum_{k\geq 1} \frac{z^k}{k}$. Thus \begin{align*} \sum_{k=1}^n {n \choose k}\frac{1}{k}&=[z^n] \frac{1}{1-z}\,\log\big(\frac{1}{1-\frac{z}{1-z}}\big)\\ &=[z^n] \frac{1}{1-z}\,\log\big(\frac{1-z}{1-2z}\big)\\ &=[z^n] \frac{1}{1-z}\,\Big(\log\big(\frac{1}{1-2z}\big)-\log\big(\frac{1}{1-z}\big)\Big)\\ &=\sum_{k=1}^n\frac{2^k}{k} -\sum_{k=1}^n \frac{1}{k}\end{align*}

What is the limit of $a (n + 1) / a (n)$? Let $a(n) = f(n,n)$ where $f(m,n) = 1$ if $m < 2 $ or $ n < 2$ and $f(m,n) = f(m-1,n-1) + f(m-1,n-2) + 2 f(m-2,n-1)$ otherwise. What is the limit of $a(n + 1) / a (n)$? $(2.71...)$

Here is a derivation for an explicit formula for $a(n)$. The generating function for $f(m,n)$ is $$F(x,y) := \sum_{m,n\geq 0} f(m,n)x^m y^n = \big(1 + \frac{3x^2y^2}{1-xy(1+2x+y)}\big)\frac{1}{1-x}\frac{1}{1-y}.$$ It follows that \begin{split} a(n) &= 1 + \sum_{i,j=0}^n [x^iy^j]\ \frac{3x^2y^2}{1-xy(1+2x+y)} \\ &= 1 + 3\sum_{i,j=0}^{n-2} [x^iy^j]\ \frac{1}{1-xy(1+2x+y)} \\ &= 1 + 3\sum_{i,j=0}^{n-2} [x^iy^j]\ \sum_{k=0}^{n-2} x^ky^k(1+2x+y)^k \\ &= 1 + 3\sum_{k=0}^{n-2} \sum_{i,j=0}^{n-2-k} [x^iy^j]\ (1+2x+y)^k \\ &= 1 + 3\sum_{k=0}^{n-2} \sum_{i,j=0}^{n-2-k} \binom{k}{i,j,k-i-j} 2^i \\ &= 1 + 3\sum_{i,j=0}^{n-2} \binom{i+j}{i} 2^i \sum_{k=i+j}^{n-2-\max(i,j)} \binom{k}{i+j} \\ &= 1 + 3\sum_{i,j=0}^{n-2} \binom{i+j}{i}\binom{n-1-\max(i,j)}{i+j+1}2^i. \end{split}

Definite integral of the square root of a polynomial ratio I found myself with the following integral $$ \int_{b_1}^{b_2} \sqrt{\frac{(b-b_1)(b_2-b)(b_3-b)}{(b_4-b)}} \ db $$ with $ b_1 < b_2 < b_3 < b_4 $. I know that $$ \int_{b_1}^{b_2} \frac{db}{\sqrt{(b-b_1)(b_2-b)(b_3-b)(b_4-b)}} $$ is equal to $$ \frac{2}{(b_4-b_2)(b_3-b_1)} K(k) $$ where $K(k)$ is the complete elliptic integral of first kind, so I suspect that this integral is somehow reducible to a linear combination of elliptic integrals, but I can't find the right way.

We may as well rescale and translate so $b_1 = 0$ and $b_2 = 1$, leaving just two parameters instead of four. Maple produces a result involving a limit of elliptic integrals: $$ \frac{\underset{t \rightarrow 1-}{\mathrm{lim}}\frac{i \left(b_{3}^{2}+\left(2 b_{4}-2\right) b_{3}-3 b_{4}^{2}+2 b_{4}+1\right) \left(\sqrt{t}-t^{\frac{3}{2}}\right) \sqrt{b_{3}-t}\, \sqrt{b_{4}-t}\, \sqrt{b_{3}-1}\, \Pi \left(\frac{i \sqrt{b_{3}-1}\, \sqrt{t}}{\sqrt{b_{3}}\, \sqrt{1-t}}, \frac{b_{3}}{b_{3}-1}, \frac{\sqrt{b_{4}-1}\, \sqrt{b_{3}}}{\sqrt{b_{4}}\, \sqrt{b_{3}-1}}\right)+\left(i \left(b_{3}-3 b_{4}+1\right) \left(\sqrt{t}-t^{\frac{3}{2}}\right) b_{4} \sqrt{b_{3}-t}\, \sqrt{b_{4}-t}\, \sqrt{b_{3}-1}\, E\left(\frac{i \sqrt{b_{3}-1}\, \sqrt{t}}{\sqrt{b_{3}}\, \sqrt{1-t}}, \frac{\sqrt{b_{4}-1}\, \sqrt{b_{3}}}{\sqrt{b_{4}}\, \sqrt{b_{3}-1}}\right)-i \left(b_{3}-b_{4}-1\right) \left(\sqrt{t}-t^{\frac{3}{2}}\right) \sqrt{b_{3}-t}\, \sqrt{b_{4}-t}\, \sqrt{b_{3}-1}\, F\left(\frac{i \sqrt{b_{3}-1}\, \sqrt{t}}{\sqrt{b_{3}}\, \sqrt{1-t}}, \frac{\sqrt{b_{4}-1}\, \sqrt{b_{3}}}{\sqrt{b_{4}}\, \sqrt{b_{3}-1}}\right)+t \left(-2 \sqrt{b_{4}}\, \sqrt{b_{3}-t}\, \sqrt{b_{4}-t}\, \left(t -1\right) \sqrt{-\left(t -1\right) \left(-b_{4}+t \right) \left(-b_{3}+t \right)}+\left(-b_{3}+t \right) \sqrt{1-t}\, \left(\left(-3 t -b_{3}-1\right) b_{4}^{\frac{3}{2}}+3 b_{4}^{\frac{5}{2}}+\left(t b_{3}+t \right) \sqrt{b_{4}}\right)\right)\right) \left(b_{3}-1\right)}{\sqrt{b_{3}-t}\, \sqrt{t}\, \sqrt{b_{4}-t}\, \left(t -1\right)}}{\sqrt{b_{4}}\, \left(4 b_{3}-4\right)} $$ Let's try a special case: $b_3 = 2$, $b_4 = 3$. Numerical evaluation of the above result and numerical integration, both using 20 decimal digits, agree on the value $0.30296476900449078284$.

Full-rank matrix I have a sparse square matrix and want to see if it is full rank (so that I can apply the implicit function theorem). $$\left[\begin{array}{cccccccccc} 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\ x_{1}^{2} & Nx_{1} & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0\\ 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{1} & 0 & 0\\ x_{2}^{2} & 0 & Nx_{2} & 0 & 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{2} & 0\\ 0 & 0 & 0 & 0 & z_1 & 0 & 0 & -1 & 1 & 0\\ x_{3}^{2} & 0 & 0 & Nx_{3} & 0 & 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{3}\\ 0 & 0 & 0 & 0 & z_2 & z_2 & 0 & 0 & -1 & 1 \end{array}\right]$$ where all variables are strictly positive, and $\sum x_i=1$. Given that it is sparse, one approach that we considered is to do row-reductions and rearranging to reduce it to a block matrix. This is possible and yields: $$\left[\begin{array}{ccc} A & B & 0 \\ 0 & C & D \\ E & 0 & F\end{array}\right]=\left[\begin{array}{ccc|ccc|ccc} N & 0 & 0 & x_{1}-1 & x_{1} & x_{1} & 0 & 0 & 0\\ 0 & N & 0 & x_{2} &x_2-1 & x_{2} & 0 & 0 & 0\\ 0 & 0 & N & x_{3} & x_{3} & x_3-1 & 0 & 0 & 0\\ \hline 0 & 0 & 0 & x_1z_1 & 0 & 0 & -1 & 1 & 0\\ 0 & 0 & 0 & x_1z_2 & x_2z_2 & 0 & 0 & -1 & 1\\ 0 & 0 & 0 & x_1 & x_2 & x_3 & 0 & 0 & 0\\ \hline c & 0 & 0 & 0 & 0 & 0 & -x_{1}^2 & 0 & 0\\ 0 & c & 0 & 0 & 0 & 0 & 0 & -x_{2}^2 & 0\\ 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x_{3}^2 \end{array}\right]$$

OK, let's call the block matrix above $M$. First eliminate $N$ by a substitution $c\mapsto d N$. Then substitute $z_i \mapsto d y_i$ to eliminate $d$. Then you can construct the Schur complement w.r.t. the first and last rows/columns of $M$ to get $\det(M) = -c^2 N x_1^2 x_2^2 x_3^2 \det P$, with $$ P = \begin{pmatrix} x_1 y_1 + x_1^{-1}-x_2^{-1}-{x_1^{-2}} & x_1^{-1}-x_2^{-1}+x_2^{-2} & x_1^{-1}-x_2^{-1} \\ x_1 y_2+x_2^{-1}-x_3^{-1} & x_2 y_2+x_2^{-1}-x_3^{-1}-x_2^{-2} & x_2^{-1}-x_3^{-1}+x_3^{-2} \\ x_1 & x_2 & x_3 \\ \end{pmatrix} $$ It might be easier to discuss this determinant.

An algebraic inequality in three real variables Is it true that $$(v-u)^2+(w-u)^2+(w-v)^2 \\ +\left(\sqrt{\frac{1+u^2}{1+v^2}} +\sqrt{\frac{1+v^2}{1+u^2}}\right) (w-u)(w-v) \\ -\left(\sqrt{\frac{1+u^2}{1+w^2}}+\sqrt{\frac{1+w^2}{1+u^2}}\right) (w-v)(v-u) \\ -\left(\sqrt{\frac{1+w^2}{1+v^2}}+\sqrt{\frac{1+v^2}{1+w^2}}\right) (v-u) (w-u)>0$$ for any real $u,v,w$ such that $u<0<v<w$? Certain numerical evidence suggests it is true. This inequality arose in the previous answer.

Rewrite the inequality in terms of $a,b,c>1$ via $u=(\frac{1}{a}-a)/2$, $v=(b-\frac{1}{b})/2$, and $w=(c-\frac{1}{c})/2$. Note that $\sqrt{1+u^2}=\frac{1+a^2}{2a}$ and so on. Then the conjectured inequality is equivalent to \begin{equation} ((c - b)(a + c)(a + b))^2(a^2b^2c^2-a^2bc+ab^2c+abc^2+ab+ac-bc+1)>0. \end{equation} Note that the second factor equals \begin{equation} a^2bc(bc-1) + abc(b+c-1) + ab + ac +1, \end{equation} which is positive since $b,c>1$.

Generalized Vieta-product It's known that $$S_2={2\over\pi} = {\sqrt{2}\over 2}{\sqrt{2+\sqrt{2}}\over 2}{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\dots$$ The terms in the product approaches 1, the same holds for the following convergent series, with $\phi$ the golden ratio $$S_1 = {\sqrt{1}\over\phi}{\sqrt{1+\sqrt{1}}\over\phi}{\sqrt{1+\sqrt{1+\sqrt{1}}}\over\phi}\dots$$ Let $$S_n = {\sqrt{n}\over c_n}{\sqrt{n+\sqrt{n}}\over c_n}\dots$$ Where $c_n$ is the solution to the equation $x=\sqrt{n+x}$ Is there a simpler formula for $S_n$? What is the asymptotic behavior (Big-O) of $S_n$ as $n->\infty$?

I doubt there exists a closed formula for $n\ne 2$. In the case $n=2$ such formula exists only thanks to the double-angle formula for cosine. Let $n$ be fixed and $c=c_n$. Notice that $n=c^2-c$ and $c\to\infty$ as soon as $n\to\infty$. Denote by $p_k$ the $k$-th multiplier in the product $S_n$. It can be easily seen that $$c\cdot (p_k^2-1) = p_{k-1} - 1$$ Consider the functional equation: $$c\cdot(f(x)^2-1)=f(2cx)-1$$ with $f(0)=1$ and $f'(0)=1$. Its solution can be expressed as a series: $$f(x) = 1 + x + \frac{x^2}{2(2c-1)} + \frac{x^3}{2(2c-1)^2(2c+1)} + \frac{x^4(2c+5)}{8(2c-1)^3(2c+1)(4c^2+2c+1)} + \dots.$$ Then $$p_k = f\left(\frac{x_0}{(2c)^k}\right)$$ where $x_0$ is a solution to $f(x_0)=0$. Now $$S_n = \prod_{k=1}^{\infty} p_k = \exp \sum_{k=1}^{\infty} \ln\left(1 + \Theta\left(\frac{x_0}{(2c)^k}\right) \right) = \exp \sum_{k=1}^{\infty} \Theta\left(\frac{x_0}{(2c)^k}\right) = \exp \Theta\left(\frac{x_0}{2c-1}\right)$$ which tends to $1$ as $c\to\infty$. Therefore, $S_n\to 1$ as $n\to\infty$. Example. For $n=2$, the functional equation admits the analytic solution $f(x)=\cosh(\sqrt{2x})$ for which $x_0=\frac{-\pi^2}{8}$.

Exact Value of a Series It is very easy to show that the series $$\frac{1-1/2}{1\times2} - \frac{1-1/2+1/3}{2\times3} + \frac{1-1/2+1/3-1/4}{3\times4} - ...$$ i.e. $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}[1-\frac{1}{2} + \frac{1}{3} - ...+ \frac {(-1)^{n}}{n+1}]$$ is convergent. Can one find its exact value? Or is it unreasonable to hope for such a thing? Thank you for your answers.

The interior sum is equal to $\int_0^1\frac{1-(-x)^{n+1}}{1+x}dx$ and $$ \sum _{n=1}^{\infty } \frac{(-1)^{n+1} \left(1-(-x)^{n+1}\right)}{n (n+1) (x+1)}= \frac{(x-1) \log (1-x)-x+\log (4)-1}{1+x}. $$ So the answer is eqaul to $$ \int_0^1 \frac{(x-1) \log (1-x)-x+\log (4)-1}{1+x}dx=\frac{\pi ^2}{6}+\log ^2(2)-2. $$

Diophantine equation $2(x - 1/x) = y - 1/y$ Does the Diophantine equation $2(x - \frac{1}{x}) = y - \frac{1}{y}$ have only trivial rational solutions, i.e, $x=\pm1, y = \pm1$?

This question is actually the first example, in a different formulation, of the following problem: Find two integer right-angled triangles with a common base and altitudes in the integer ratio $N:1$, which was considered by the late John Leech (in the 1980s I think). We have to find integer solutions to $B^2+A^2=C^2 \hspace{2cm} B^2+(NA)^2=D^2$ We can express $B=\alpha 2mn , A = \alpha (m^2-n^2)$ and $B=\beta 2pq , NA=\beta (p^2-q^2)$ giving $\frac{NA}{B}=\frac{p^2-q^2}{2pq}=N\frac{A}{B}=N\frac{m^2-n^2}{2mn}$ and defining $X=m/n$ and $Y=p/q$ gives $N(X-1/X )=(Y-1/Y)$ The original form of the problem is also just that of Euler's Concordant Numbers which is related to the elliptic curve $V^2=U(U+1)(U+N^2)$ which has $3$ torsion points of order $2$ when $U=0,-1,-N^2$ and $4$ torsion points of order $4$ when $U=\pm N$, all of which give the trivial answer or lead to undefined quantities. The first curve of rank greater than $0$ occurs for $N=7$ leading to $X=3/2, Y=6$ or $B=12, A=5$.

Eigenvectors of asymmetric graphs Let $G$ be an asymmetric connected graph. Then is it always the case that at least one of the eigenvectors of its adjacency matrix $A$ consists entirely of distinct entries? Thanks!

I think the answer is no. Take the Frucht Graph, the simplest nontrivial asymmetric graph. Its adjacency matrix is \begin{equation*} \left( \begin{array}{cccccccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\\\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\\\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\\\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \end{array} \right) \end{equation*} none of whose eigenvectors seems to have distinct entries.

Approximation of a given function by rational functions Given a function $1/\sqrt{x^2 -k^2}$ where k is a constant with a small imaginary part, how do you go about constructing a rational approximation? I am interested in the L_p (p=2 or $\infty$) norm of the difference being small on the real line. Both the theoretical and the practical implementation is of interest.

Eleven years too late. Using @Robert Israel's approach, the $[n,n]$ Padé approximants $P_n$ $$\sqrt s =\frac{1}{\sqrt{2}}\,\frac {1+\sum_{i=1}^n a_i\,\left(s-\frac{1}{2}\right)^i} {1+\sum_{i=1}^n b_i\,\left(s-\frac{1}{2}\right)^i}+O\left(\left(s-\frac{1}{2}\right)^{2n+1}\right) $$ can easily be built starting from the infinite series $$\sqrt s=\sum_{i=1}^\infty 2^{i-\frac{1}{2}} \binom{\frac{1}{2}}{i}\left(s-\frac{1}{2}\right)^i$$ The table below reports, for a given $n$ the coefficients as well as the norm $$\Phi_n=\int_0^1 \big[\sqrt s-P_n\big]^2\,ds$$ $$\left( \begin{array}{cccc} n & a_i & b_i & \Phi_n \\ 1 & \left\{\frac{3}{2}\right\} & \left\{\frac{1}{2}\right\} & 9.6\times 10^{-4} \\ 2 & \left\{\frac{5}{2},\frac{5}{4}\right\} & \left\{\frac{3}{2},\frac{1}{4}\right\} & 1.35\times 10^{-4} \\ 3 & \left\{\frac{7}{2},\frac{7}{2},\frac{7}{8}\right\} & \left\{\frac{5}{2},\frac{3}{2},\frac{1}{8}\right\} & 3.62\times 10^{-5} \\ 4 & \left\{\frac{9}{2},\frac{27}{4},\frac{15}{4},\frac{9}{16}\right\} & \left\{\frac{7}{2},\frac{15}{4},\frac{5}{4},\frac{1}{16}\right\} & 1.34\times 10^{-5} \\ 5 & \left\{\frac{11}{2},11,\frac{77}{8},\frac{55}{16},\frac{11}{32 }\right\} & \left\{\frac{9}{2},7,\frac{35}{8},\frac{15}{16},\frac{1}{32}\right\} & 6.05\times 10^{-6} \\ \end{array} \right)$$

Computing $\prod_p(\frac{p^2-1}{p^2+1})$ without the zeta function? We see that $$\frac{2}{5}=\frac{36}{90}=\frac{6^2}{90}=\frac{\zeta(4)}{\zeta(2)^2}=\prod_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\prod_p \left(\frac{(p^2-1)^2}{(p^2+1)(p^2-1)}\right)=\prod_p\left(\frac{p^2-1}{p^2+1}\right)$$ $$\implies \prod_p \left(\frac{p^2-1}{p^2+1}\right)=\frac{2}{5},$$ But is this the only way to compute this infinite product over primes? It seems like such a simple product, one that could be calculated without the zeta function. Note that $\prod_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\prod_p(\frac{p-1}{p-i})\prod_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\prod_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdots$.

A proof of this identity not using properties of the Riemann zeta function is listed as an unsolved problem in section B48 of Guy's Unsolved Problems in Number Theory. An amusing observation: this identity implies the infinitude of primes. If the product over $p$ were finite, the final answer would be a rational number with a factor of $3$ in its numerator --- the very first term in the product has a $3$ in the numerator, and $3$ cannot divide $u^2+1$ for any integer $u$, so the $3$ can never cancel.

For $x,y\ge 2$ does $x^4+x^2y^2+y^4$ ever divide $x^4y^4+x^2y^2+1$? For a problem in combinatorics, it comes down to knowing whether there exist integers $x,y\ge 2$ such that $$ x^4+x^2y^2+y^4\mid x^4y^4+x^2y^2+1. $$ Note that $x^6(x^2-y^2)(x^4+x^2y^2+y^4)+(x^2y^2-1)(x^4y^4+x^2y^2+1)=x^{12}-1$ and so we can look at a perhaps simpler problem: Are the only solutions to $$ x^4+x^2y^2+y^4\mid x^{12}-1 $$ (where $x,y\ge 2$) given by $(x,y)=(5,6)$ and $(x,y)=(6,5)$?

It seems to me that you need both $$p=x^2+xy+y^2$$ and $$q=x^2-xy+y^2$$ to divide $$r=x^4y^4+x^2y^2+1$$ Considering all expressions as polynomials in $x$, the remainder when you divide $r$ by $p$ is $$(y^7-y^3)x-y^4+1$$ and the remainder when you divide $r$ by $q$ is $$(y^3-y^7)x-y^4+1$$ if Maple and I are on the same page. These remainders are both zero if and only if $y=\pm1$.

set of centers of sphere inscribed in tetrahedron Having a sphere and three diffrent point $A,B,C$ on this sphere. Find set of all centers of spheres inscribed in a tetrahedron $ABCD$, where $D$ is some point on the given sphere. The problem reduced to 2-dimensions is trivial it's just sum of two arcs of some circle, but in 3-dimensions the set is not so simple. Checking in geogebra it's not sum of some parts of sphere. I don't know what this set looks and how it's described.

Here is a depiction of Robert Bryant's surface defined by the $56$-term polynomial he details:           Note one component is the "triangular tea bag" discernable in my empirical investigation. Here is the polynomial: $$ 2 x^5+x^4 y+x^4 z-7 x^4+2 x^3 y^2+4 x^3 y z-10 x^3 y+2 x^3 z^2-10 x^3 z+8 x^3+2 x^2 y^3-12 x^2 y^2 z-15 x^2 y^2-12 x^2 y z^2+6 x^2 y z+15 x^2 y+2 x^2 z^3-15 x^2 z^2+15 x^2 z-2 x^2+x y^4+4 x y^3 z-10 x y^3-12 x y^2 z^2+6 x y^2 z+15 x y^2+4 x y z^3+6 x y z^2-6 x y z-4 x y+x z^4-10 x z^3+15 x z^2-4 x z-2 x+2 y^5+y^4 z-7 y^4+2 y^3 z^2-10 y^3 z+8 y^3+2 y^2 z^3-15 y^2 z^2+15 y^2 z-2 y^2+y z^4-10 y z^3+15 y z^2-4 y z-2 y+2 z^5-7 z^4+8 z^3-2 z^2-2 z+1$$ The plot above restricts $(x,y,z)$ to lie on or in the sphere.

For which divisors $a$ and $b$ of $n$ does there exist a Latin square of order $n$ that can be partitioned into $a \times b$ subrectangles? There exists a Latin square of order $8$ which can be partitioned into $2 \times 4$ subrectangles: $$ \begin{bmatrix} \color{red} 1 & \color{red} 2 & \color{red} 3 & \color{red} 4 & \color{purple} 5 & \color{purple} 6 & \color{purple} 7 & \color{purple} 8 \\ \color{red} 2 & \color{red} 3 & \color{red} 4 & \color{red} 1 & \color{purple} 6 & \color{purple} 7 & \color{purple} 8 & \color{purple} 5 \\ 7 & 8 & \color{blue} 1 & \color{blue} 2 & \color{blue} 3 & \color{blue} 4 & 5 & 6 \\ 8 & 5 & \color{blue} 2 & \color{blue} 3 & \color{blue} 4 & \color{blue} 1 & 6 & 7 \\ \color{pink} 5 & \color{pink} 6 & \color{pink} 7 & \color{pink} 8 & \color{green} 1 & \color{green} 2 & \color{green} 3 & \color{green} 4 \\ \color{pink} 6 & \color{pink} 7 & \color{pink} 8 & \color{pink} 5 & \color{green} 2 & \color{green} 3 & \color{green} 4 & \color{green} 1 \\ \color{orange} 3 & \color{orange} 4 & \color{brown} 5 & \color{brown} 6 & \color{brown} 7 & \color{brown} 8 & \color{orange} 1 & \color{orange} 2 \\ \color{orange} 4 & \color{orange} 1 & \color{brown} 6 & \color{brown} 7 & \color{brown} 8 & \color{brown} 5 & \color{orange} 2 & \color{orange} 3 \\ \end{bmatrix} $$ If we take the row-symbol parastrophe of this Latin square (i.e., replace each entry $(i,j,l_{ij})$ with $(l_{ij},j,i)$), then the entry colors define a decomposition of $K_{8,8}$ into $2$-regular spanning subgraphs of $K_{4,4}$. I would like to generalize this. Question: For which divisors $a$ and $b$ of $n$ does there exist a Latin square of order $n$ that can be partitioned into $a \times b$ subrectangles? (Note: We assume $a \leq b$. Subrectangles must have $b$ symbols. We don't assume that the boundaries of the subrectangles align.) Observations: * *It's trivially possible when * *$a$ divides $b$ (construct a Latin square with blocks that are $b \times b$ subsquares) *$a=1$ or $b=n$. *The first non-trivial case is when $a=2$, $b=3$, and $n=6$. If my code is correct, then it is impossible (by exhaustive search). (I'm tempted to think this is just because the parameters are too small.) The best possible is $4$ subrectangles, which is straightforward to construct. *My code found a random Latin square which gave an $a=2$, $b=5$, and $n=10$ example: $$ \begin{bmatrix} \color{red} 7 & \color{red} 1 & 9 & 5 & \color{red} {10} & 4 & 8 & \color{red} 2 & 6 & \color{red} 3 \\ \color{red} 3 & \color{red} {10} & 5 & 8 & \color{red} 2 & 6 & 4 & \color{red} 1 & 9 & \color{red} 7 \\ \hline \color{blue} 9 & \color{blue} 6 & 4 & 7 & 1 & \color{blue} 2 & \color{blue} {10} & \color{blue} 3 & 5 & 8 \\ \color{blue} 6 & \color{blue} 3 & 7 & 1 & 5 & \color{blue} 9 & \color{blue} 2 & \color{blue} {10} & 8 & 4 \\ \hline \color{pink} 4 & 5 & 1 & \color{pink} 2 & \color{pink} 8 & 3 & 9 & \color{pink} 7 & \color{pink} {10} & 6 \\ \color{pink} 2 & 9 & 6 & \color{pink} {10} & \color{pink} 4 & 1 & 3 & \color{pink} 8 & \color{pink} 7 & 5 \\ \hline \color{purple} 1 & 2 & \color{purple} 8 & 3 & 7 & \color{purple} {10} & 6 & \color{purple} 5 & 4 & \color{purple} 9 \\ \color{purple} 8 & 4 & \color{purple} {10} & 6 & 3 & \color{purple} 5 & 7 & \color{purple} 9 & 2 & \color{purple} 1 \\ \hline \color{brown} 5 & 8 & \color{brown} 2 & 9 & 6 & 7 & \color{brown} 1 & 4 & \color{brown} 3 & \color{brown} {10} \\ \color{brown} {10} & 7 & \color{brown} 3 & 4 & 9 & 8 & \color{brown} 5 & 6 & \color{brown} 1 & \color{brown} 2 \\ \end{bmatrix} $$

I eventually co-authored a paper which includes this topic. The non-trivial results are: * *There exists a Latin square of order $n$ which decomposes into $2 \times (n/2)$ subrectangles for all even $n \not\in \{2,6\}$. *There exists a Latin square of order n which decomposes into $3 \times (n/3)$ subrectangles if and only if $3$ divides $n$ and $n > 9$. *For sufficiently large n, there exists a Latin square of order $n$ which partitions into $d \times (n/d)$ subrectangles if and only if $d$ divides $n$. Akbari, Marbach, Stones, Wu, Balanced equi-n-squares Electron. J. Combin, 2020. (pdf) The smallest cases we didn't resolve are * *$20 \times 20$ Latin squares into $4 \times 5$ subrectangles; *$28 \times 28$ Latin squares into $4 \times 7$ subrectangles; and *$30 \times 30$ Latin squares into $5 \times 6$ subrectangles.

When $\frac{a^2}{b+c}＋\frac{b^2}{a+c}＋\frac{c^2}{a+b}$ is integer and $a,b,c$ are coprime natural numbers, is there a solution except (183,77,13)? Given $a,b,c\in \Bbb{N}$ such that $\{a,b,c\}$ are coprime natural numbers and $a,b,c>1$. When $$\frac{a^2}{b+c}＋\frac{b^2}{c+a}＋\frac{c^2}{a+b}\in\mathbb Z\,?$$ I know the solution $\{183,77,13\}$. Is there any other solution?

Yes, there is another solution. The next one I found is a bit big, namely $$ a = 15349474555424019, b = 35633837601183731, c = 105699057106239769. $$ This solution also satisfies the property that $$ \frac{a^{2}}{b+c} + \frac{b^{2}}{a+c} + \frac{c^{2}}{a+b} = \frac{31}{21} (a+b+c), $$ which was true of $a = 13$, $b = 77$ and $c = 183$. If you clear denominators in the equation above, you find that both sides are multiples of $a+b+c$, and dividing out gives a plane cubic. One wishes to search for rational points on this plane cubic, which by scaling can be assumed to be relatively prime integers. We also need points where $a, b, c > 0$ and $a+b+c$ is a multiple of $21$. This plane cubic is isomorphic to the elliptic curve $$ E : y^2 + xy + y = x^3 - 8507979x + 9343104706, $$ and $E(\mathbb{Q})$ has rank $3$ and the torsion subgroup is $\mathbb{Z}/6\mathbb{Z}$. Let $E(\mathbb{Q}) = \langle T, P_{1}, P_{2}, P_{3} \rangle$ where $6T = 0$. I searched for points of the form $c_{1}T + c_{2}P_{1} + c_{3}P_{2} + c_{4}P_{3}$ where $0 \leq c_{1} \leq 5$ and $|c_{2}|, |c_{3}|, |c_{4}| \leq 5$. Of these $7986$ points, there are $1464$ where $a$, $b$ and $c$ are all positive, and these yield $11$ solutions (where $a < b < c$), the smallest of which is $(13,77,183)$. The second smallest is the one I gave above. It should be straightforward to prove that one can get infinitely many solutions in this way. EDIT: I searched for solutions that satisfy $$ \frac{a^{2}}{b+c} + \frac{b^{2}}{a+c} + \frac{c^{2}}{a+b} = t(a+b+c) $$ for all rational numbers $t = d/e$ with $d, e \leq 32$. This yields a number of smaller solutions. In particular, for $t = 9/5$ the rank of the curve is $2$ and one finds $$ a = 248, b = 2755, c = 7227. $$ For $t = 21/17$ the rank is $2$ and one finds $$ a = 35947, b = 196987, c = 401897. $$ Interestingly, $t = 31/21$ is the only case where the rank is $\geq 3$.

Proving an homogenous sextic product never a square Can anyone prove that $$2(a^2+b^2)(a^4+a^2b^2+b^4) \ne \square$$ for $a,b$ positive integers?

I am a little confused. First, assume that $a, b$ are relatively prime (dividing through by the gcd divides the lhs by a square). Second, note that $a^2+b^2$ is relatively prime to $2(a^4 + b^2a^2+b^4)$ (since the expression within the last set of parentheses is $(a^2+b^2)^2 - a^2 b^2.$) So, if the thing has any hope of being a square, $a, b$ are two sides of a pythagorean triangle, and $2(a^4 + b^2a^2+b^4)$ is a square. Since, $a, b$ are two sides of a pythagorean triangle, one is even, the other is odd, so the expression within the last paren is odd, but twice an odd number is not a square. UPDATE in the case, $a, b$ are both odd, $2(a^2 + b^2)$ is divisible by $4,$ but the expression in the last paren is $3$ mod $4,$ so again, not a square.

Mean minimum distance for M and N uniformly random points on reals between 0 and 1 Similar to Mean minimum distance for N random points on a one-dimensional line, but instead of only N random points, choose N and M random points and find the mean minimum distance between points of N and M, but NOT M and M or N and N. So, given $x_1,x_2,...,x_N \in [0,1]$ and $y_1,y_2,...,y_M \in [0,1]$ (uniformly random), find the mean distance: $min(abs(x_i - y_j))$ with $i\in[1,N]$ and $j\in[1,M]$. Ideally, the answer would include the probability of the points being a minimum of $d$ distance apart, like in the referenced question.

Let $D_1$ be the answer conditioned on the leftmost point being at 0 and the rightmost point at 1 (colors irrelevant). Let $0 = x_1 \leq \ldots \leq x_{n + m} = 1$ be the ordered coordinates of points. We can see that $s_i = x_{i + 1} - x_i$ are equidistributed subject to $s_1 + \ldots + s_{n + m - 1} = 1$, and coloring is independent of $s_i$. The probability that there are exactly $k$ adjacent color changes in the sequence is $$\frac{{n - 1 \choose \lceil \frac{k - 1}{2} \rceil}{m - 1 \choose \lfloor \frac{k - 1}{2} \rfloor} + {m - 1 \choose \lceil \frac{k - 1}{2} \rceil}{n - 1 \choose \lfloor \frac{k - 1}{2} \rfloor}}{n + m \choose n},$$ since the sequence either starts with a white group, and has $\lceil (k + 1) / 2 \rceil$ white groups and $\lfloor (k + 1) / 2 \rfloor$ black groups, or vice versa. Conditioned on the colouring, the smallest distance is equal to the minimum of $k$ instances of $s_i$ (it is irrelevant which ones due to symmetry). For $0 \leq t \leq 1 / k$, the probability of $\min(s_1, \ldots, s_k) \leq t$ is $1 - (1 - tk)^{n + m - 2}$. Integrating with density, we have $$E\min(s_1, \ldots, s_k) = \int_0^{1 / k} k(n + m - 2)(1 - tk)^{n + m - 3} tdt =$$ $$(n + m - 2) \int_0^1 z^{n + m - 3}\frac{1 - z}{k} dz = \frac{1}{k(n + m - 1)}$$ (note that the answer is correct in the special case $k = n = m = 1$ where some transformations were illegal). Hence $$D_1 = \frac{1}{(n + m - 1){n + m \choose n}}\sum_{k = 1}^{\infty}\frac{{n - 1 \choose \lceil \frac{k - 1}{2} \rceil}{m - 1 \choose \lfloor \frac{k - 1}{2} \rfloor} + {m - 1 \choose \lceil \frac{k - 1}{2} \rceil}{n - 1 \choose \lfloor \frac{k - 1}{2} \rfloor}}{k}$$ Finally, integrating by the span $s$ between extreme points we obtain the complete answer $$D = \int_0^1 (n + m)(n + m - 1)(1 - s)s^{n + m - 2} \cdot sD_1 ds = \frac{n + m - 1}{n + m + 1}D_1 =$$ $$\frac{1}{(n + m + 1){n + m \choose n}}\sum_{k = 1}^{\infty}\frac{{n - 1 \choose \lceil \frac{k - 1}{2} \rceil}{m - 1 \choose \lfloor \frac{k - 1}{2} \rfloor} + {m - 1 \choose \lceil \frac{k - 1}{2} \rceil}{n - 1 \choose \lfloor \frac{k - 1}{2} \rfloor}}{k}$$ When $n = m = 2$, we have $D = \frac{1}{5 \cdot 6}(\frac{2}{1} + \frac{2}{2} + \frac{2}{3}) = \frac{1}{30} \cdot \frac{11}{3} = \frac{11}{90}$, which matches the computations in the comments. Not sure if the sum can be simplified further, still the answer is feasible to compute. The probability of the smallest distance being at least $d$ can be found along the same lines by computing $$\frac{(n + m)(n + m - 1)}{n + m \choose n}\sum_{k = 1}^{\infty}\left({n - 1 \choose \lceil \frac{k - 1}{2} \rceil}{m - 1 \choose \lfloor \frac{k - 1}{2} \rfloor} + {m - 1 \choose \lceil \frac{k - 1}{2} \rceil}{n - 1 \choose \lfloor \frac{k - 1}{2} \rfloor}\right) \times \\ \int_0^1 (1 - s)s^{n + m - 2} \cdot \max(0, 1 - \frac{kd}{s})^{n + m - 2}ds$$

Adventure with infinite series, a curiosity It is easily verifiable that $$\sum_{k\geq0}\binom{2k}k\frac1{2^{3k}}=\sqrt{2}.$$ It is not that difficult to get $$\sum_{k\geq0}\binom{4k}{2k}\frac1{2^{5k}}=\frac{\sqrt{2-\sqrt2}+\sqrt{2+\sqrt2}}2.$$ Question. Is there something similarly "nice" in computing $$\sum_{k\geq0}\binom{8k}{4k}\frac1{2^{10k}}=?$$ Perhaps the same question about $$\sum_{k\geq0}\binom{16k}{8k}\frac1{2^{20k}}=?$$ NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.

Mathematica says, for the first: $$\, _4F_3\left(\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8};\frac{1}{4},\frac{1}{2},\frac{ 3}{4};\frac{1}{4}\right)$$ and $$\, _8F_7\left(\frac{1}{16},\frac{3}{16},\frac{5}{16},\frac{7}{16},\frac{9}{16},\frac{11}{16} ,\frac{13}{16},\frac{15}{16};\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2},\frac{5}{8}, \frac{3}{4},\frac{7}{8};\frac{1}{16}\right) $$ for the second.

Irreducibility of a polynomial when the sum of its coefficients is prime I came up with the following proposition, but don't know how to prove it. I used Maple to see that it holds when $ a + b + c + d <300 $. Let $a,b,c$ and $d$ be non-negative integers such that $d\geq1$ and $a+b+c\geq1$. If $a+b+c+d$ is a prime number other than $2$, the polynomial $ax^3+bx^2+cx+d$ is irreducible over $Z[x]$.

Francesco Polizzi's idea is enough to solve the problem: Lemma. Let $f = ax^3 + bx^2 + cx + d \in \mathbf Z_{\geq 0}[x]$ nonconstant with $d > 0$, such that $f(1)$ is a prime $p > 2$. Then $f$ is irreducible in $\mathbf Z[x]$. Proof. Suppose $f = gh$ for $g,h \in \mathbf Z[x]$; we must show that $g \in \{\pm 1\}$ or $h \in \{\pm 1\}$. First assume $\deg g > 0$ and $\deg h > 0$. Since $\deg f \leq 3$, without loss of generality we may assume $\deg g = 1$ and $m = \deg h \in \{1,2\}$; say \begin{align*} g = ux + v, & & h = \alpha x^2 + \beta x + \gamma , \end{align*} with $u > 0$ and $\alpha > 0$ or $\alpha = 0$ and $\beta > 0$. If $v \leq 0$, then $f$ has a nonnegative real root, which is impossible because $f$ has non-negative coefficients and $f(0) > 0$. Thus, $v > 0$, and therefore $g(1) \geq 2$, forcing $g(1) = p$ and $h(1) = 1$. By symmetry, this rules out the case $\deg h = 1$. Thus, we may assume $\alpha > 0$. The formulas \begin{align*} a = \alpha u, & & b = \beta u + \alpha v, & & c = \gamma u + \beta v, & & d = \gamma v \end{align*} give \begin{align*} \alpha > 0, & & \alpha v \geq -\beta u, & & \gamma u \geq -\beta v, & & \gamma > 0.\label{1}\tag{1} \end{align*} We will show that $h(1) \geq 2$, contradicting the earlier assertion that $h(1) = 1$. This is clear if $\beta \geq 0$, since we already have $\alpha > 0$ and $\gamma > 0$. If $\beta < 0$, then \eqref{1} shows $$\frac{\gamma}{-\beta} \geq \frac{v}{u} \geq \frac{-\beta}{\alpha} > 0,\label{2}\tag{2}$$ hence \begin{align*} h(1) = \alpha + \beta + \gamma &= \left(1 - \left(\frac{-\beta}{\alpha}\right) + \left(\frac{\gamma}{-\beta}\right)\left(\frac{-\beta}{\alpha}\right)\right)\alpha\\ &\geq \left(1 - \left(\frac{-\beta}{\alpha}\right) + \left(\frac{-\beta}{\alpha}\right)^2 \right)\alpha. \end{align*} The function $1-x+x^2$ attains a minimum of $\tfrac{3}{4}$ at $x = \tfrac{1}{2}$, and is bigger than $1$ on $(1,\infty)$. Thus, we only have to consider $x = \tfrac{-\beta}{\alpha} \in (0,1]$. For $x < 1$, we must have $\alpha \geq 2$, so $(1-x+x^2)\alpha \geq \tfrac{3}{2} > 1$. Finally, for $x = 1$, i.e. $\beta = -\alpha$, \eqref{2} shows that $v \geq u$. Since $p = v+u$ is odd, we must have $v > u$, so \eqref{2} gives $\gamma > -\beta$, so $h(1) = \gamma \geq 2$. Thus, we conclude that $\deg g = 0$ or $\deg h = 0$; say $g = n$ for $n \in \mathbf Z_{>0}$ without loss of generality. If $n > 1$, then $n = p$, which is impossible since $f(0) = d \in \{1, \ldots, p-1\}$ is not divisible by $p$. So we conclude that $g = 1$. $\square$ Remark. The argument above relies crucially on the assumption $\deg f \leq 3$. More importantly, for higher degree polynomials the same result is false without a lower bound on $p$ that grows at least linearly in $\deg f$: Example. Let $p$ be an odd prime. Then the polynomial $f = x^{2p-2} + x^{2p-4} + \ldots + x^2 + 1$ has $f(1) = p$, but it factors as $$\Big(x^{p-1}+x^{p-2}+\ldots+x+1\Big)\Big(x^{p-1}-x^{p-2}+\ldots-x+1\Big).$$ Indeed, the above reads $$\zeta_p(x^2) = \zeta_p(x)\zeta_{2p}(x) = \zeta_p(x)\zeta_p(-x),$$ which is true because both sides are monic and have the same roots in $\mathbf C$ (namely the $2p^\text{th}$ roots of unity except $\pm 1$, all with multiplicity $1$).

Analytical solution to a specific differential equation I was wondering whether there is an analytical solution to the ODE \begin{equation} -n\int xy(x)dx + ihy'(x) + (x^2+k)y(x) = 0, \end{equation} where $n=0,1,2,...$, $h \in \mathbb{R}$, and $k=+1,0$ or $-1$. For $n=0$ this can be solved exactly, and the solution is \begin{equation} y(x) = C \exp\left[ih\left(\frac{x^3}{3}+kx\right)\right] \equiv f(x). \end{equation} However, I struggled to find a general solution for different $n$. Differentiating the ODE gives \begin{equation} ihy''(x)+(x^2+k)y'(x)+(2-n)xy(x)=0. \end{equation} I have tried the ansatz \begin{equation} y(x) = f(x)^{(n-2)/2} + f(x)^{(n-2)/2}\int f(x)^{1-n} dx, \end{equation} but it only satisfies the equation when $n=2$. I was able to make some progress by separating the equation into its real and imaginary parts and integrate the coupled equations numerically, but I was hoping for an analytical solution. The other possible solution is the triconfluent Heun function \begin{equation} y(x) = e^{-\frac{x^{3}+3 kx}{3 h}} C_{2} \text { HeunT }\left[0, \frac{-4+n}{h},-\frac{k}{h}, 0,-\frac{1}{h}, x\right] + C_{1} \text { HeunT }\left[0, \frac{-2+n}{h}, \frac{k}{h}, 0, \frac{1}{h}, x\right]. \end{equation} My question is whether there is an analytical solution of a simpler form?

For $k=0$ the solution is a hypergeometric function, $$y(x)=C_1 \, _1F_1\left(\frac{2}{3}-\frac{n}{3};\frac{2}{3};\frac{i x^3}{3 h}\right)-(3h)^{-1/3}(-1)^{5/6} C_2 x \, _1F_1\left(1-\frac{n}{3};\frac{4}{3};\frac{i x^3}{3 h}\right),$$ which at least for some values of $n$ can be reduced to a Bessel function and/or an incomplete gamma function. $$n=1:\qquad y(x)=\frac{\sqrt[6]{-\frac{1}{3}} \sqrt{x} e^{\frac{i x^3}{6 h}} \left(3 \sqrt[3]{2} C_1 \Gamma \left(\frac{5}{6}\right) J_{-\frac{1}{6}}\left(-\frac{x^3}{6 h}\right)-i C_2 \Gamma \left(\frac{1}{6}\right) J_{\frac{1}{6}}\left(-\frac{x^3}{6 h}\right)\right)}{3\ 2^{2/3} \sqrt[6]{h}},$$ $$n=2:\qquad y(x)=\frac{\sqrt[3]{-1} C_2 h^{2/3} \Gamma \left(\frac{4}{3}\right) \left(-\frac{i x^3}{h}\right)^{2/3}}{x^2}-\frac{\sqrt[3]{-1} C_2 h^{2/3} \left(-\frac{i x^3}{h}\right)^{2/3} \Gamma \left(\frac{1}{3},-\frac{i x^3}{3 h}\right)}{3 x^2}+C_1,$$ $$n=3:\qquad y(x)=\frac{C_1 \Gamma \left(\frac{2}{3}\right) \sqrt[3]{-\frac{i x^3}{h}}}{\sqrt[3]{3}}+\frac{C_1 \sqrt[3]{-\frac{i x^3}{h}} \Gamma \left(-\frac{1}{3},-\frac{i x^3}{3 h}\right)}{3 \sqrt[3]{3}}-\frac{(-1)^{5/6} C_2 x}{\sqrt[3]{3} \sqrt[3]{h}}.$$

Do identities exist for the binomial series $\sum_{k=m+1}^{n+1} \binom{k}{m} \binom{n+1}{k-1} $? While examining the product of two upper triangular matrices, I've found that the $(m,n)$'th entry of the resulting matrix amounts to: $$\sum_{k=m+1}^{n+1} \binom{k}{m} \binom{n+1}{k-1} $$ when $n \geq m$ (all other entries are zero). Although I have found some summations of products of binomial coefficients here, identities for the sum-product as described above have so far eluded me. Do you know whether identities for this series -- or perhaps even for generalizations of it -- exist?

$$\sum_{k}{{k\choose m} {n+1\choose k-1}}=\frac{1}{m}\sum_{k}k{k-1\choose m-1}{n+1\choose k-1} = \frac{n+1 \choose m-1}{m}\sum_{k}k{n-m+2 \choose k-m} = \frac{n+1 \choose m-1}{m}(\sum_{k}(k-m){n-m+2 \choose k-m}) + {n+1 \choose m-1}\sum_{k}{n-m+2 \choose k-m} = \frac{{n+1 \choose m-1}(n-m+2)}{m}(\sum_{k}{n-m+1 \choose k-m-1}) + {n+1 \choose m-1}2^{n-m+2} = \frac{{n+1 \choose m-1}2^{n-m+1}(n+m+2)}{m}$$ The identities I have used are ${n \choose k} = \frac{n}{k}{n-1 \choose k-1}, \sum_{k}{n \choose k} = 2^n, {a \choose b}{b \choose c} = {a \choose c}{a-c \choose b-c}$. The summation in the question excludes $k=n+2,m$, so your answer is actually $$\frac{{n+1 \choose m-1}2^{n-m+1}(n+m+2)}{m}-{n+2\choose m}-{n+1\choose m-1} $$

Throwing a fair die until most recent roll is smaller than previous one I roll a fair die with $n>1$ sides until the most recent roll is smaller than the previous one. Let $E_n$ be the expected number of rolls. Do we have $\lim_{n\to\infty} E_n < \infty$? If not, what about $\lim_{n\to\infty} E_n/n$ and $\lim_{n\to\infty} E_n/\log(n)$?

The answer may be expressed more simply, in fact $E_n = \left( \frac{n}{n-1}\right)^n$. Update 1: (The following was independently obtained by Pierre PC, I just found out after I finished typing.) The following is a simple way to see this. Let $E_{n,\ell}$ denote the expected number of rolls if the last roll is $\ell$. We then have the formula \begin{equation*} E_{n,\ell} = \underbrace{1\cdot \frac{1}{n} + \dotsb + 1\cdot \frac{1}{n}}_{\ell-1} + \sum_{k=\ell}^n \left(1+E_{n,k}\right)\cdot \frac{1}{n}. \end{equation*} Solving these backwards (i.e. solve for $E_{n,n}, E_{n,n-1}, \dotsc, E_{n,1}$) gives $E_{n,\ell}=\left( \frac{n}{n-1}\right)^{n-\ell+1}$. It then follows that \begin{equation*} E_n = 1 + \sum_{\ell=1}^n E_{n,\ell}\cdot \frac{1}{n} = \left( \frac{n}{n-1}\right)^n. \end{equation*} Update 2: Here is a derivation starting from mathworker21's formula. We have \begin{align*} E_n & {}= \sum_{k=2}^{\infty} k \sum_{m=1}^n n^{-(k-1)} \binom{m+k-3}{k-2} \frac{m-1}{n}\\ & {}= \sum_{m=1}^n (m-1) \sum_{k=2}^{\infty} k \binom{m+k-3}{k-2} n^{-k}\\ & {}= \sum_{m=1}^n \frac{m-1}{n^2} \sum_{k=0}^{\infty} (k+2) \binom{m+k-1}{k} n^{-k}\\ & {}= \sum_{m=1}^n \frac{m-1}{n^2} \sum_{k=0}^{\infty} (k+2) \binom{-m}{k} \left( -\frac{1}{n} \right)^k.\\ \end{align*} Now massaging the sum and using the binomial series twice gives \begin{align*} \sum_{k=0}^{\infty} (k+2) \binom{-m}{k} x^k & {}= 2 \sum_{k=0}^{\infty} \binom{-m}{k} x^k + \sum_{k=1}^{\infty} k \binom{-m}{k} x^k\\ & {}= 2 (1+x)^{-m} + \sum_{k=1}^{\infty} (-m) \binom{-(m+1)}{k-1} x^k\\ & {}= 2 (1+x)^{-m} - m x (1+x)^{-(m+1)}\\ & {}= \frac{2-(m-2)x}{(1+x)^{m+1}}. \end{align*} Inserting $x=-\frac{1}{n}$ then gives \begin{align*} E_n & {}= \sum_{m=1}^n \frac{m-1}{n^2} \left( \frac{2+\frac{m-2}{n}}{(1-\frac{1}{n})^{m+1}}\right)\\ & {}= \frac{1}{n^2 (n-1)} \sum_{m=1}^n (m-1)(2 n+m-2) \left( \frac{n}{n-1} \right)^m\\ & {}= -\frac{2}{n^2} \left(\sum_{m=1}^n \left( \frac{n}{n-1} \right)^m\right) + \frac{2 n-3}{n^2 (n-1)} \left(\sum_{m=1}^n m \left( \frac{n}{n-1} \right)^m\right) + \frac{1}{n^2 (n-1)} \left(\sum_{m=1}^n m^2 \left( \frac{n}{n-1} \right)^m\right).\\ \end{align*} Finally using \begin{equation*} \sum_{m=1}^n x^m = \frac{x}{1-x} \left( 1-x^n \right), \qquad \sum_{m=1}^n m x^m = \frac{x}{(1-x)^2} \left( n x^{n+1} - (n+1) x^n + 1\right) \qquad\text{ and }\qquad \sum_{m=1}^n m^2 x^m = \frac{x}{(1-x)^3} \left( (1+x) - x^n \left( n^2 (1-x)^2 +2 n (1-x) + x + 1 \right) \right) \end{equation*} one then obtains $E_n=\left(\frac{n}{n-1}\right)^n$ after heavy simplification.

Squares of the form $2^j\cdot 3^k+1$ Squares of the form $2^j\cdot 3^k+1$, for j,k nonnegative. Is it known if there are infinitely many? And if $2^j\cdot 3^k+1=N$ is a square, then it must be necessarly a semi-prime? Do you think that 3-smooth neighbour squares is a good name for these squares?

We may find them all by elementary methods. Assume that $2^j3^k=(n-1)(n+1)$. Since $\gcd(n-1,n+1)\leqslant 2$, we get either $j=0$, $n-1=1$, $n+1=3$, or $j\geqslant 1$, $(n-1)/2$ and $(n+1)/2$ are consecutive 3-smooth numbers. They must be a power of 2 and a power of 3, so we should solve the equations $2^a=3^b\pm 1$. This is not hard. * *$2^a=3^b+1$. Either $b=0,a=1$ or $b\geqslant 1$, 3 divides $2^a-1$, so $a$ is even and $3^b=2^a-1=(2^{a/2}-1)(2^{a/2}+1)$, the multiples are powers of 3 which differ by 2, so $a=2$, $b=1$. *$2^a=3^b-1$. Either $a=1$, $b=1$ or $a\geqslant 2$, 4 divides $3^b-1$, $b$ is even, $2^a=(3^{b/2}-1)(3^{b/2}+1)$, the multiples are powers of 2 which differ by 2, $b=2$, $a=3$.

On Markoff-type diophantine equation Do there exist integers $x,y,z$ such that $$ x^2+y^2-z^2 = xyz -2 \quad ? $$ Why this is interesting? First, this equation arose in an answer to the previous Mathoverflow question What is the smallest unsolved diophantine equation? but was not asked explicitly as a separate question. The context is that, in a well-defined sense for the notion of "smallness", the equation above is the "smallest" open Diophantine equation. Second, this equation is one of the simplest non-trivial representative of the family of equations $ax^2+by^2+cz^2=dxyz+e$, which generalises a well-known Markoff equation $x^2+y^2+z^2=3xyz$. The well-known methods for the former (Vieta jumping) has been extended to the general case if $a,b,c$ are all natural numbers and are divisors of $d$ (see, for example, Fine, Benjamin, et al. "On the Generalized Hurwitz Equation and the Baragar–Umeda Equation." Results in Mathematics 69.1-2 (2016): 69-92). The question seems to be much more challenging when $a,b,c$ have different signs. The simplest case with different signs is $a=b=d=1$ and $c=-1$, which leads to the family of equations $x^2+y^2-z^2=xyz+e$. The equation above is the first non-trivial example from this family.

There is no solution. Fix a solution $(x,y,z)$ with $|x|+|y|+|z|$ minimal. We will show a contradiction. We can't have $xyz=0$ as we would then obtain one of the unsolvable equations $x^2+y^2= -2$, $x^2-z^2=-2$, $y^2-z^2=-2$. If $xyz>0$, then by swapping the signs of two of $x,y,z$ if necessary we can assume $x,y,z>0$, and switching $x$ and $y$ we can assume $x \geq y$. We have the Vieta jump $x \to yz-x$, so if this is minimal we have $x \leq yz/2$. Since $f(x)=x^2+y^2-z^2 - xyz + 2 $ is convex and vanishes at $x$, we must $f(y) \geq 0$ or $f(yz/2) \geq 0$. But $$f(y)= (2-z)y^2 -z^2 + 2$$ so $f(y) \leq 0$ imply $z<2$ and $z=1$ gives the impossible $x^2+y^2-xy=-1$ and $$f(yz/2) = y^2 -z^2 - y^2 z^2/4 +1= (1-z^2/4)y^2 - z^2 +1$$ which again is nonnegative only if $z<2$ which is impossible. If$xyz<0$, then by swapping the signs if necessary we can assume $x,y,z<0$. We have the Vieta jump $z \to -xy-z$, so if this is minimal we have $z \geq -xy/2$. We have $g(z)=z^2 + xyz - x^2-y^2-2$ is convex and vanishes at $z$, we must have $g(0) \geq 0$ or $g(-xy/2) \geq 0$. But $g(0) = -x^2 - y^2 - 2 <0$ and $g(-xy/2) = - x^2 y^2/4 - x^2 - y^2 -2 <0$. So neither case is possible.

Monotonicity of a parametric integral For real $x>0$, let $$f(x):=\frac1{\sqrt x}\,\int_0^\infty\frac{1-\exp\{-x\, (1-\cos t)\}}{t^2}\,dt.$$ How to prove that $f$ is increasing on $(0,\infty)$? Here is the graph $\{(x,f(x))\colon0<x<3\}$: So, it even appears that $f$ is concave. Since $f>0$, the concavity would of course imply that $f$ is increasing. This question arises in a Fourier analysis of a certain probabilistic problem.

Here is another solution: Noting the identity $$\sum_{n=-\infty}^{\infty} \frac{1}{(t+2\pi n)^2} = \frac{1}{2(1-\cos t)}$$ and taking advantage of the fact that the integrand is non-negative, we may apply Tonelli's theorem to get \begin{align*} f(x) &= \frac{1}{2\sqrt{x}} \int_{-\pi}^{\pi} \left( \sum_{n=-\infty}^{\infty} \frac{1}{(t+2\pi n)^2} \right) \bigl( 1 - e^{-x\left(1-\cos t\right)} \bigr) \, \mathrm{d}t \\ &= \frac{1}{4\sqrt{x}} \int_{-\pi}^{\pi} \frac{1 - e^{-x\left(1-\cos t\right)}}{1-\cos t} \, \mathrm{d}t \\ &= \frac{1}{8\sqrt{x}} \int_{-\pi}^{\pi} \frac{1 - e^{-2x\sin^2(t/2)}}{\sin^2(t/2)} \, \mathrm{d}t. \tag{1} \end{align*} Then by applying the integration by parts, we also get $$ f(x) = \frac{\sqrt{x}}{2} \int_{-\pi}^{\pi} \cos^2(t/2) e^{-2x\sin^2(t/2)} \, \mathrm{d}t. \tag{2} $$ Now using $\text{(1)}$, $$ \bigl( \sqrt{x}f(x) \bigr)' = \frac{1}{4} \int_{-\pi}^{\pi} e^{-2x\sin^2(t/2)} \, \mathrm{d}t. \tag{3} $$ Therefore by $\text{(2)}$ and $\text{(3)}$ altogether, \begin{align*} \sqrt{x}f'(x) &= \bigl( \sqrt{x}f(x) \bigr)' - \frac{1}{2\sqrt{x}} f(x) \\ &= \frac{1}{4} \int_{-\pi}^{\pi} e^{-2x\sin^2(t/2)} \, \mathrm{d}t -\frac{1}{4} \int_{-\pi}^{\pi} \cos^2(t/2) e^{-2x\sin^2(t/2)} \, \mathrm{d}t \\ &= \frac{1}{4} \int_{-\pi}^{\pi} \sin^2(t/2) e^{-2x\sin^2(t/2)} \, \mathrm{d}t \\ &> 0. \end{align*} This identity can also be used to show that $f''(x) < 0$, and hence $f$ is concave as expected.

Does $\{\tau(1)\tau(2)+\cdots+\tau(n-1)\tau(n)+\tau(n)\tau(1):\ \tau\in S_n\}$ contain a unique multiple of $n^2$ for each $n\ge6$? Motivated by Question 397575, here I pose a related question. Question. Does the set $$T_n:=\{\tau(1)\tau(2)+\cdots+\tau(n-1)\tau(n)+\tau(n)\tau(1):\ \tau\in S_n\}$$ contain a unique multiple of $n^2$ for each $n\ge6$? I conjecture that the answer is positive. I have verified this for $n=6,\ldots,10$. For $n=6$, we have \begin{align*}&2\times4+4\times1+1\times3+3\times5+5\times6+6\times2 \\=&3\times5+5\times1+1\times2+2\times4+4\times6+6\times3=2\times6^2. \end{align*} For $n=7$, we have $$1\times3+3\times4+4\times5+5\times6+6\times2+2\times7+7\times1=2\times7^2.$$ For $n=8$, we have $$1\times5+5\times3+3\times6+6\times4+4\times7+7\times2+2\times8+8\times1=2\times8^2.$$ For $n=9$, we have $$1\times2+2\times3+3\times5+5\times4+4\times6+6\times8+8\times7+7\times9+9\times1=3\times9^2.$$ For $n=10$, we have \begin{gather*}1\times2+2\times3+3\times6+6\times8+8\times4+4\times9+9\times7+7\times5+5\times10+10\times1 \\=3\times10^2.\end{gather*}

No, for $n = 11$ this fails: 363 = 3 * 11^2 with [7, 2, 8, 5, 3, 4, 6, 9, 10, 1, 11] 484 = 2^2 * 11^2 with [10, 9, 6, 3, 1, 2, 4, 5, 7, 8, 11] Running the code I wrote to check this a little more, there is more than one multiple of $n^2$ in the set you describe for all $11\leq n \leq 50$, see here for two permutations leading to different multiples of $n^2$ for each such $n$.

How many cubes are the sum of three positive cubes? Are there infinitely many integer positive cubes $x^3 = a^3 + b^3 + c^3$ that are equal to the sum of three integer positive cubes? If not, how many of them are there?

There are bivariate coprime polynomial parametrizations: https://sites.google.com/site/tpiezas/010 $$(a^4-2ab^3)^3 + (a^3 b+b^4)^3 + (2a^3 b-b^4)^3 = (a^4+a b^3)^3$$ Added If you drop the positivity constraint then there is another identity for $x=v^4$. $$v^{12}=(v^4)^3=(9u^4)^3+(3uv^3-9u^4)^3+(v(v^3-9u^3))^3$$ I believe, but can't find it at the moment, that for all positive $x$ exist integers $a,b,c$ such that $x^3=a^3+b^3+c^3$.

Splitting the integers from $1$ to $2n$ into two sets with products as close as possible For each positive integer $n$, split the integers $1$ to $2n$ into two sets of $n$ elements each, and such that the products of the elements in each of these sets are as close as possible, say they differ by $a(n)$. It can be checked that $a(1)=1$, $a(2)=2$, $a(3)=6$, $a(4)=18$, and $a(5)=30$. Is this sequence strictly increasing? What about if it is not required that the two sets contain the same number of elements?

Here are optimal solutions for $n \le 10$, and the two sets happen to be equicardinal even if you don't enforce that: \begin{matrix} n & a_n & \text{solution} \\ \hline 1 & 1 & \{2\},\{1\} \\ 2 & 2 & \{2,3\},\{1,4\} \\ 3 & 6 & \{1,5,6\},\{2,3,4\} \\ 4 & 18 & \{1,5,6,7\},\{2,3,4,8\} \\ 5 & 30 & \{2,3,4,8,10\},\{1,5,6,7,9\} \\ 6 & 576 & \{1,4,7,8,9,11\},\{2,3,5,6,10,12\} \\ 7 & 840 & \{2,4,5,6,8,11,14\},\{1,3,7,9,10,12,13\} \\ 8 & 24480 & \{1,5,6,7,8,13,14,15\},\{2,3,4,9,10,11,12,16\} \\ 9 & 93696 & \{2,3,6,8,9,11,12,13,18\},\{1,4,5,7,10,14,15,16,17\} \\ 10 & 800640 & \{2,3,4,8,9,11,12,18,19,20\},\{1,5,6,7,10,13,14,15,16,17\} \\ \end{matrix} I obtained these via integer linear programming by instead minimizing the difference in log products (sum of logs) rather than difference in products.

Writing $1-xyzw$ as a sum of squares Can you write $1 - xyzw$ in the form $p + q (1 - x^{2}-y^{2}-z^{2}-w^{2})$ where $p$ and $q$ are polynomials that are of the form $\sum g_{i}^{2}$ where $g_{i}$ $\in$ $\mathbb{R}[x,y,z,w]$? For instance, in the two variable case, $1 - xy = \frac{1}{2} + \frac{1}{2}(x-y)^{2} + \frac{1}{2}(1-x^{2}- y^{2})$. In this example, $q = \frac{1}{2}$, so we have a very simple expression. I'm looking for an analogous expression in four variables ($p$ and $q$ here can be anything as long as they are sums of squares). Also can we say anything in general for $2n$ variables?

The question would be even more interesting if the factor $1-x^2-y^2-w^2-z^2$ was replaced by $4-x^2-y^2-w^2-z^2$. The reason is that a necessary condition for the existence of such $p,q$ is that whenever $xywz>1$, the last factor is also negative. And it turns out that $$(xywz>1)\Longrightarrow(x^2+y^2+w^2+z^2<4),$$ by AM-GM inequality. Notice that if the answer is positive with this constant $4$, then it is so with your constant $1$. Now the answer: $p$ and $q$ exist for the sharp constant $4$. You may take $q=\frac1{16}(4+x^2+y^2+w^2+z^2)$ and $$p=\frac1{16}(x^2+y^2+w^2+z^2)^2-xywz.$$ To see $p$ is a sum of squares, just notice $$48p=(x^2+y^2-w^2-z^2)^2+\cdots+4(x-y)^2(w+z)^2+\cdots,$$ where the first sum consists of $3$ terms, and the second one consists in $6$ terms.

Full-rank matrix I have a sparse square matrix and want to see if it is full rank (so that I can apply the implicit function theorem). $$\left[\begin{array}{cccccccccc} 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\ x_{1}^{2} & Nx_{1} & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0\\ 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{1} & 0 & 0\\ x_{2}^{2} & 0 & Nx_{2} & 0 & 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{2} & 0\\ 0 & 0 & 0 & 0 & z_1 & 0 & 0 & -1 & 1 & 0\\ x_{3}^{2} & 0 & 0 & Nx_{3} & 0 & 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{3}\\ 0 & 0 & 0 & 0 & z_2 & z_2 & 0 & 0 & -1 & 1 \end{array}\right]$$ where all variables are strictly positive, and $\sum x_i=1$. Given that it is sparse, one approach that we considered is to do row-reductions and rearranging to reduce it to a block matrix. This is possible and yields: $$\left[\begin{array}{ccc} A & B & 0 \\ 0 & C & D \\ E & 0 & F\end{array}\right]=\left[\begin{array}{ccc|ccc|ccc} N & 0 & 0 & x_{1}-1 & x_{1} & x_{1} & 0 & 0 & 0\\ 0 & N & 0 & x_{2} &x_2-1 & x_{2} & 0 & 0 & 0\\ 0 & 0 & N & x_{3} & x_{3} & x_3-1 & 0 & 0 & 0\\ \hline 0 & 0 & 0 & x_1z_1 & 0 & 0 & -1 & 1 & 0\\ 0 & 0 & 0 & x_1z_2 & x_2z_2 & 0 & 0 & -1 & 1\\ 0 & 0 & 0 & x_1 & x_2 & x_3 & 0 & 0 & 0\\ \hline c & 0 & 0 & 0 & 0 & 0 & -x_{1}^2 & 0 & 0\\ 0 & c & 0 & 0 & 0 & 0 & 0 & -x_{2}^2 & 0\\ 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x_{3}^2 \end{array}\right]$$

Thanks very much to Fred Hucht for getting me to think about $c$ and its relationship with $N$. The following is an approach for small $c$: As suggested by the mention of the implicit function theorem, the variables $x_1,x_2,x_3,z_1,z_2,N$ are implicit solutions to a complicated set of equations and $c$ is a parameter. Above it was mentioned that the variables are strictly positive, but furthermore, they do not converge to $0$ as $c \rightarrow 0$ nor do they blow up. This means that for $c$ sufficiently small, the above matrix is "almost" block diagonal and we can use the Gershgorin circle theorem to bound the eigenvalues away from $0$ (or we could argue continuity as $c\rightarrow 0$, or weight rows/columns in such a way that the matrix is diagonally dominant). So, this resolves the "small" $c$ case. We think that the result is generically true for any $c$, but that remains open.

Is it possible to simplify the coefficient matrix for large values of $x$? If I have a system of $8$ linear equations for the eight variables $\{\alpha ,\beta ,\gamma ,\delta ,\eta ,\lambda ,\xi ,\rho \}$ and with the three parameters $\{x,y,z\}$ reals and $x>0$. I want to work on nontrivial solutions of the system, therefore, I need to compute the determinant of the coefficient matrix (given below). My question: Assuming that the determinant of the coefficient matrix is of a polynomial form in $x$ as $f(x)=c_1 x^8+c_2 x^7+c_3 x^6+...+c_0$; I need to evaluate the problem only for large values of $x \to \infty$, in other words, I only need to find the coefficient $c_1$. My question is if it is possible to simplify (manipulate) this system of equations (or equivalently, the coefficient matrix) from the beginning for large values of $x$ so that computing the determinant is simpler. $$ \gamma +\delta +\beta \; e^{\frac{i x}{2}} (x-1)+\delta \; x-\gamma\; x-\alpha \; e^{-\frac{1}{2} (i x)} (x+1)=0, \\ \gamma +\delta +x (\gamma +\lambda )-\eta -\lambda -x (\delta +\eta )=0, $$ $$ \eta (x-1)+\delta (x-1) e^{-i (x+y)}-(x+1) \left(\lambda +\gamma \;e^{i (x-y)}\right)=0, \\\alpha +e^{-\frac{1}{2} i (x+2 y)} \left((x+1) \left(\delta +\beta \; e^{\frac{3 i x}{2}+i y}\right)-\gamma \; e^{2 i x} (x-1)\right)=\alpha \; x, $$ $$ \rho +\xi\; e^{2 i x} (x+1)-\rho \; x+\alpha (x-1) e^{\frac{3 i x}{2}+i y}-\beta (x+1) e^{\frac{1}{2} i (x+2 y)}=0,\\ \xi \; e^{2 i x} (x-1)-\rho (x+1)+\eta (x-1) \left(-e^{i (y+z)}\right)+\lambda (x+1) e^{i (2 x+y+z)}=0, $$ $$ \xi +\rho +\rho \; x+\lambda (x-1) e^{i (x+z)}-\xi \; x-\eta (x+1) e^{-i (x-z)}=0,\\\rho \;x+\alpha e^{\frac{i x}{2}} (x+1)-\xi-\rho -\beta \; e^{-\frac{1}{2} (i x)} (x-1)-\xi \; x=0. $$ Coefficient Matrix $$\scriptsize\left( \begin{array}{cccccccc} -e^{-\frac{1}{2} (i x)} (x+1) & e^{\frac{i x}{2}} (x-1) & 1-x & x+1 & 0 & 0 & 0 & 0 \\ 0 & 0 & x+1 & 1-x & -x-1 & x-1 & 0 & 0 \\ 0 & 0 & (x+1) \left(-e^{i (x-y)}\right) & (x-1) e^{-i (x+y)} & x-1 & -x-1 & 0 & 0 \\ 1-x & e^{i x} (x+1) & (x-1) \left(-e^{\frac{3 i x}{2}-i y}\right) & (x+1) e^{-\frac{1}{2} i (x+2 y)} & 0 & 0 & 0 & 0 \\ e^{\frac{i x}{2}} (x-1) & -e^{-\frac{1}{2} (i x)} (x+1) & 0 & 0 & 0 & 0 & (x+1) e^{i (x-y)} & (1-x) e^{-i (x+y)} \\ 0 & 0 & 0 & 0 & (1-x) e^{-i (x-z)} & (x+1) e^{i (x+z)} & (x-1) e^{i (x-y)} & (x+1) \left(-e^{-i (x+y)}\right) \\ 0 & 0 & 0 & 0 & (x+1) \left(-e^{-i (x-z)}\right) & (x-1) e^{i (x+z)} & 1-x & x+1 \\ e^{\frac{i x}{2}} (x+1) & -e^{-\frac{1}{2} (i x)} (x-1) & 0 & 0 & 0 & 0 & -x-1 & x-1 \\ \end{array} \right)$$ P.S. I have already asked this in MathStackExchange but did not receive any answers. Any comments are highly appreciated.

Let $M:=M(x,y,z)$ be the $8\times8$ matrix in question. Let $m(x):=M(x,0,0)$. We have $$\det m(x)= -256 e^{i x/2} x^2 \cos (2 x) \big((x^2+1)^2 \cos (2 x)-(x^2-1)^2\big).$$ So, $|\det m(x)|$ will be oscillating between the value $0$ (attained when $\cos (2 x)=0$ -- that is, when $x=(2k-1)\pi/4$ for natural $k$) and the value $512 x^2 (1 + x^4)\sim512 x^6$ (attained when $\cos (2 x)=-1$ -- that is, when $x=(2k-1)\pi/2$ for natural $k$). The value $0$ will also be taken by $\det m(x)$ at the occurring almost periodically roots of the equation $\cos (2 x)=r(x):=\dfrac{(x^2-1)^2}{(x^2+1)^2}$, as $r(x)$ is strictly increasing from $0$ to $1$ for $x$ increasing from $1$ to $\infty$, whereas $\cos (2 x)$ is periodically oscillating between $-1$ and $1$. In general, $$\det M(x,y,z)=32 x^2 e^{i (x-4 y+2 z)/2} \\ \times\big(-4 (x^2-1)^2 \sin ^2(x) [\cos (2 y)+\cos (z)] \\ +4 (x^2-1)^2 \cos (2 x)-4 (x^2+1)^2 \cos (4 x)-16 x^2\big).$$ So, by the triangle inequality, $$|\det M(x,y,z)|\le32 x^2 \\ \times\big(4 (x^2-1)^2 \times2 \\ +4 (x^2-1)^2+4 (x^2+1)^2+16 x^2\big)=512 x^2 (1 + x^4).$$ Thus, $|\det M(x,y,z)|\le512 x^2 (1 + x^4)$ for all $x,y,z$, whereas $|\det M(x,0,0)|$ will be oscillating between $0$ and $512 x^2 (1 + x^4)\sim512 x^6$. In particular, it follows that your assumption Assuming that the determinant of the coefficient matrix is of a polynomial form in $x$ as $f(x)=c_1 x^8+c_2 x^7+c_3 x^6+...+c_0$; I need to evaluate the problem only for large values of $x \to \infty$, in other words, I only need to find the coefficient $c_1$. is quite counterfactual. Here are the calculations, done in Mathematica (click on the image below to magnify it):

Vanishing of a product of cyclotomic polynomials in characteristic 2 Let $n$ be a positive integer and $1\leq j\leq n$. Consider the following polynomial: $$p_{n,j}(x)=\frac{\prod\limits_{i=1}^{n+1}\frac{x^{i}+1}{x+1}}{\prod\limits_{i=1}^{j}\frac{x^{i}+1}{x+1}\prod\limits_{i=1}^{n-j+1}\frac{x^{i}+1}{x+1}}\in\mathbb{F}[x]$$ This polynomial can be computed for each $j,n$ given. What I want to do is to understand in which cases $p(1)=0$ and in which cases $p(1)=1$ when working over a field $\mathbb{F}$ of characteristic $2$. I checked some cases by hand and this is what I got: $$p_{2,1}(1)=p_{2,2}(1)=1$$ $$p_{3,1}(1)=p_{3,2}(1)=p_{3,3}(1)=0$$ $$p_{4,1}(1)=p_{4,4}(1)=1~~p_{4,2}(1)=p_{4,3}(1)=0$$ $$p_{5,2}(1)=p_{5,4}(1)=1~~p_{5,1}(1)=p_{5,3}(1)=p_{5,5}=0$$ $$p_{6,1}(1)=p_{6,2}(1)=p_{6,3}(1)=p_{6,4}(1)=p_{6,5}(1)=p_{6,6}(1)=1$$ So apparently there is no easy pattern. It is also easily expressible as a product of cyclotomic polynomials, but I'm not sure if this is of any help. I would like (if possible) to have a way to know if $p_{n,j}(1)=0$ or $p_{n,j}(1)=1$ just in terms of $n$ and $j$. If you are courious, these are the coefficients that has appeared in the computation of the homology groups of a certain family of Artin groups. I want to understand the behaviour of these polynomials over a field of characteristic $2$ in $x=1$ because this is the situation where I can assure that the homology groups are infinite dimensional. Thanks for your help.

Claim: $p_{n,j}(1)$ vanishes modulo $2$ if and only there is a carry when adding $j$ and $n+1-j$ in base-$2$. Proof: Let us write $x^n+1 \equiv x^n-1= \prod_{d \mid n} \Phi_d(x)$ where $\Phi_d$ is the $d$th cyclotomic polynomial. Then, since there are $\lfloor m/d\rfloor$ multiplies of $d$ in $\{1,2,\ldots,m\}$, $$p_{n,j}(x) = \prod_{d=2}^{n} \Phi_d(x)^{\lfloor \frac{n+1}{d} \rfloor- \lfloor \frac{j}{d}\rfloor - \lfloor \frac{n-j+1}{d}\rfloor }.$$ Note that $\lfloor a+b \rfloor \ge \lfloor a\rfloor + \lfloor b \rfloor$ so your expression is necessarily a polynomial (and not a strictly rational function). The value of $\Phi_d(x)$ at $x=1$ is well-understood. One can use the property $x^n-1= \prod_{d \mid n} \Phi_d(x)$ to show $\Phi_d(1)$ is $1$ unless $d$ is a prime power $p^k$ in which case $\Phi_d(1)=p$ (this is classical, see here for proofs). This implies $$p_{n,j}(1) \equiv \prod_{p^k \le n} p^{\lfloor \frac{n+1}{p^k} \rfloor- \lfloor \frac{j}{p^k}\rfloor - \lfloor \frac{n-j+1}{p^k}\rfloor }.$$ Since $2$ is the only even prime, $$p_{n,j}(1) \equiv \prod_{2^k \le n} 2^{\lfloor \frac{n+1}{2^k} \rfloor- \lfloor \frac{j}{2^k}\rfloor - \lfloor \frac{n-j+1}{2^k}\rfloor }.$$ So your expression vanishes modulo $2$ if and only if $$ \sum_{2 \le 2^k \le n} \left\lfloor \frac{n+1}{2^k} \right\rfloor- \left\lfloor \frac{j}{2^k}\right\rfloor- \left\lfloor \frac{n-j+1}{2^k}\right\rfloor\ge 1.$$ Let us express this criterion using base-$2$ representation. If $j = \sum_{i \ge 0} a_i 2^i$, $n-j+1 = \sum_{i \ge 0} b_i 2^i$ and $n+1 = \sum_{i \ge 0} c_i 2^i$ then $$ \left\lfloor \frac{n+1}{2^k} \right\rfloor- \left\lfloor \frac{j}{2^k}\right\rfloor- \left\lfloor \frac{n-j+1}{2^k}\right\rfloor =\sum_{i \ge k} (c_i-a_i-b_i) 2^{i-k} $$ so that $$\begin{align} \sum_{2 \le 2^k \le n} \left\lfloor \frac{n+1}{2^k} \right\rfloor- \left\lfloor \frac{j}{2^k}\right\rfloor- \left\lfloor \frac{n-j+1}{2^k}\right\rfloor &= \sum_{i \ge 0} (c_i-a_i-b_i)(2^{i-1}+2^{i-2}+\ldots+1) \\ &= \sum_{i \ge 0} (c_i-a_i-b_i)(2^{i}-1)\\ &= n+1 - (j + (n+1-j)) - \sum_{i \ge 0} (c_i-a_i-b_i)\\ &= \sum_{i \ge 0} (a_i+b_i-c_i). \end{align}$$ Let $s_2(m)$ be the sum of digits of $m$ in base-$2$. We see that $p_{n,j}(1)\equiv 0$ if and only if $s_2(j) + s_2(n+1-j) > s_2(n+1)$. In general, $s_2(a)+s_2(b)-s_2(a+b)$ is the number of carries when adding $a$ and $b$ in base-$2$ (can be proved by induction, goes back to Kummer).

Is there a good reason why $a^{2b} + b^{2a} \le 1$ when $a+b=1$? The following problem is not from me, yet I find it a big challenge to give a nice (in contrast to 'heavy computation') proof. The motivation for me to post it lies in its concise content. If $a$ and $b$ are nonnegative real numbers such that $a+b=1$, show that $a^{2b} + b^{2a}\le 1$.

Fixed now. I spent some time looking for some clever trick but the most unimaginative way turned out to be the best. So, as I said before, the straightforward Taylor series expansion does it in no time. Assume that $a>b$. Put $t=a-b=1-2b$. Step 1: $$ \begin{aligned} a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right] \\ &\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right] \\& =1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right] \\ &=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right) \end{aligned} $$ (in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$) Step 2. We need the inequality $e^{ku}\ge (1+u)(1+u+\dots+u^{k-1})+\frac k{k+1}u^{k+1}$ for $u\ge 0$. For $k=1$ it is just $e^u\ge 1+u+\frac{u^2}{2}$. For $k\ge 2$, the Taylor coefficients on the left are $\frac{k^j}{j!}$ and on the right $1,2,2,\dots,2,1$ (up to the order $k$) and then $\frac{k}{k+1}$. Now it remains to note that $\frac{k^0}{0!}=1$, $\frac{k^j}{j!}\ge \frac {k^j}{j^{j-1}}\ge k\ge 2$ for $1\le j\le k$, and $\frac{k^{k+1}}{(k+1)!}\ge \frac{k}{k+1}$. Step 3: Let $u=\log\frac 1a$. We've seen in Step 1 that $a^{2b}\le 1-b(1-t\mu)$ where $\mu=u+(1+u)t$. In what follows, it'll be important that $\mu\le\frac 1a-1+\frac 1a t=1$ (we just used $\log\frac 1a\le \frac 1a-1$ here. We have $b^{2a}=b(a-t)^t$. Thus, to finish, it'll suffice to show that $(a-t)^t\le 1-t\mu$. Taking negative logarithm of both sides and recalling that $\frac 1a=e^u$, we get the inequality $$ tu+t\log(1-te^u)^{-1}\ge \log(1-t\mu)^{-1} $$ to prove. Now, note that, according to Step 2, $$ \begin{aligned} &\frac{e^{uk}}k\ge \frac{(1+u)(1+u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1} \ge\frac{(1+u)(\mu^{k-1}+\mu^{k-2}u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1} \\ &=\frac{\mu^k-u^k}{kt}+\frac{u^{k+1}}{k+1} \end{aligned} $$ Multiplying by $t^{k+1}$ and adding up, we get $$ t\log(1-te^u)^{-1}\ge -ut+\log(1-t\mu)^{-1} $$ which is exactly what we need. The end. P.S. If somebody is still interested, the bottom line is almost trivial once the top line is known. Assume again that $a>b$, $a+b=1$. Put $t=a-b$. $$ \begin{aligned} &\left(\frac{a^b}{2^b}+\frac{b^a}{2^a}\right)^2=(a^{2b}+b^{2a})(2^{-2b}+2^{-2a})-\left(\frac{a^b}{2^a}-\frac{b^a}{2^b}\right)^2 \\ &\le 1+\frac 14\{ [\sqrt 2(2^{t/2}-2^{-t/2})]^2-[(1+t)^b-(1-t)^a]^2\} \end{aligned} $$ Now it remains to note that $2^{t/2}-2^{-t/2}$ is convex on $[0,1]$, so, interpolating between the endpoints, we get $\sqrt 2(2^{t/2}-2^{-t/2})\le t$. Also, the function $x\mapsto (1+x)^b-(1-x)^a$ is convex on $[0,1]$ (the second derivative is $ab[(1-x)^{b-2}-(1+x)^{a-2}]$, which is clearly non-negative). But the derivative at $0$ is $a+b=1$, so $(1+x)^b-(1-x)^a\ge x$ on $[0,1]$. Plugging in $x=t$ finishes the story.

Exact Value of a Series It is very easy to show that the series $$\frac{1-1/2}{1\times2} - \frac{1-1/2+1/3}{2\times3} + \frac{1-1/2+1/3-1/4}{3\times4} - ...$$ i.e. $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}[1-\frac{1}{2} + \frac{1}{3} - ...+ \frac {(-1)^{n}}{n+1}]$$ is convergent. Can one find its exact value? Or is it unreasonable to hope for such a thing? Thank you for your answers.

Since $$\frac1{n(n+1)}=\frac1n-\frac1{n+1},$$ we have $$\begin{aligned} \sum_{n=1}^N\sum_{k=1}^{n+1}\frac{(-1)^{n+k}}{kn(n+1)} &=\sum_{n=1}^N\sum_{k=1}^{n+1}\frac{(-1)^{n+k}}{kn}+\sum_{n=2}^{N+1}\sum_{k=1}^n\frac{(-1)^{n+k}}{kn}\\\\ &=2\sum_{n=1}^N\sum_{k=1}^n\frac{(-1)^{n+k}}{kn}-\sum_{n=1}^N\frac1{n(n+1)}-1+\sum_{k=1}^{N+1}\frac{(-1)^{N+1+k}}{k(N+1)}\\\\ &=\sum_{n,k=1}^N\frac{(-1)^{n+k}}{kn}+\sum_{n=1}^N\frac1{n^2}-2+\frac1{N+1}+\frac{(-1)^{N+1}}{N+1}\sum_{k=1}^{N+1}\frac{(-1)^k}k\\\\ &=\left(\sum_{n=1}^N\frac{(-1)^{n}}n\right)^2+\sum_{n=1}^N\frac1{n^2}-2+\frac1{N+1}+\frac{(-1)^{N+1}}{N+1}\sum_{k=1}^{N+1}\frac{(-1)^k}k. \end{aligned}$$ As $\sum_{n=1}^\infty\frac{(-1)^n}n=-\log2$ and $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$, this implies $$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}\sum_{k=1}^{n+1}\frac{(-1)^{k+1}}k=(\log2)^2+\frac{\pi^2}6-2.$$ There may be a numerical error somewhere, but in principle the method should work.

A matrix diagonalization problem For matrices $X,Y\in [0,1]^{n\times m}$, for n > m, is there a square matrix $W\in R^{n\times n}$ so that $X^TWY$ is diagonal if and only if $Y = X$? Furthermore, $X$ and $Y$ are column normalized so that $X1_m = Y1_m = 1_m$, where $1_m$ is the m-length column vector with all entries equal to 1. I know that if $X$ and $Y$ are binary, then $W=I$.

UPDATE Sorry, previous version was wrong. For $n=m=2$, this computation shows that there is such a matrix. FURHTER UPDATE However, for $m=n=3$, it shows there isn't. Taking $n=m=2$, we see that $X$ and $Y$ are of the form $\begin{pmatrix} x & 1-x \\ 1-x & x \end{pmatrix}$ and $\begin{pmatrix} y & 1-y \\ 1-y & y \end{pmatrix}$. Set $W = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ Then $X^T W Y =: \begin{pmatrix} e & f \\ g & h \end{pmatrix}$ with $$f=c+(a-c)x+(d-c) y+(-a+b+c-d)xy$$ $$g=b+(a-b)x+(d-b) y+(-a+b+c-d)xy$$ Your desire is that we have $f(x,y) = g(x,y) =0$ iff $x=y$. Plugging in $x=y$, we deduce that we must have $b=c=a+d=0$. But then $f$ and $g$ vanish for all $x$ and $y$. And, indeed, $(a,b,c,d) = (1,0,0,-1)$ solves the problem for $m=n=2$. Now run the same analysis with $m=n=3$, thinking about $X$ and $Y$ of the form $\begin{pmatrix} x & 1-x & 0 \\ 1-x & x & 0 \\ 0 & 0 & 1 \end{pmatrix}$. We deduce that $w_{12} = w_{21}=0$ and $w_{11}=-w_{22}$. Similarly, $w_{22} = - w_{33}$, $w_{11} = - w_{33}$ and $w_{13}=w_{31} = w_{23} = w_{32}=0$. But the only solution to these linear equations is $W=0$..

When is the degree of this number 3? I am helping a friend of mine, that works in history of mathematics. She is studying the story of the solution of the cubic equation by Cardano. Sometimes she asks me some mathematical questions, that are very hard to motivate from a modern point of view, but that were interesting to Cardano. So please do not ask for motivations. The question is the following. Let $a$, $b$ be rational numbers, with $b$ not a square. Consider the number $$ t=\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}-\sqrt{a^2-b} $$ Under what conditions on $a$ and $b$ is the degree (over $\mathbb{Q}$) of $t$ equal to $3$? A sufficient condition can be found as follows. Let $P(x) = x^3+\alpha_2 x^2 + \alpha_1 x + \alpha_0$ be a rational polynomial. The general expression of the roots of $P$ is $$ \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \frac{\alpha_2}{3}, $$ where $$ q = \frac{2\alpha_2^3 - 9\alpha_2\alpha_1 + 27\alpha_0}{27} $$ and $$ p = \frac{3\alpha_1 - \alpha_2^2}{3}, $$ see here. So we can take $a = -\frac{q}{2}$ and $b = \frac{q^2}{4} + \frac{p^3}{27}$ and we need to force $\sqrt{a^2 -b} = \frac{\alpha_2}{3}$. This boils down to $\alpha_1 = \frac{\alpha_2^2 - 3 \sqrt[3]{3\alpha_2^2}}{3}$. We find that one of the solutions of $$ x^3 + \alpha_2 x^2+ \frac{\alpha_2^2 - 3 \sqrt[3]{3\alpha_2^2}}{3}x+\alpha_0=0 $$ has the required form (of course we need to assume that $\alpha_2$ is such that $\sqrt[3]{3 \alpha_2^2}$ is rational). In this case $a$ and $b$ are given by the above expressions. I suspect that if $t$ has degree $3$, then its minimal polynomial must be of this form and that $a$ and $b$ are as above, but I am not able to prove it. Note that the condition that $t$ has degree $3$ implies that it can be written as the sum of two cubic root and a rational number (because of the formula), but it is not completely clear that this way of writing $t$ is unique.

Edit: The takeaway is that there is another, exclusive way to generate $a$ and $b$ that given an equation of degree $3$. First, chose parameters $a$ and $s$ such that $2a/s(s^2-3)$ is not a perfect cube and $1-4/s^2(s^2-3)^2$ is not a perfect square. This is most such values of $s$. Then choose $b$ according to the formula: $b=a^2\left(1-\frac{4}{s^2(s^2-3)^2}\right)$ Then the number asked in the question is this degree $3$ number: $t=s\sqrt[3]{\frac{2a}{s(s^2-3)}}+ \frac{2a}{s(s^2-3)}$ and can not be written in the way described in the question. We get that from this analysis: This is equivalent to asking whether $a^2-b$ is the sixth power of a rational number, since $\sqrt[3]{3\alpha_2^2}=\sqrt[3]{27(a^2-b)}=3\sqrt[3]{a^2-b}$ is rational, and $\sqrt{a^2-b}=\alpha_2/3$ is also rational. Vice versa, if we know that $a^2-b=x^6$ we set: $\alpha_0=-2a+x^9-3x^5-(2/3)x^6$ $\alpha_1=3x^6-3x^2$ $\alpha_2=3x^3$ We can gain additional insight using Galois theory. Consider the extension $\mathbb Q(\sqrt{a^2-b},\sqrt[3]{a+\sqrt{b}},\sqrt[3]{a-\sqrt{b}},\sqrt{-3})/\mathbb Q(a,b)$. This is a Galois extension of degree $72$, given by adjoining three square roots and then two cube roots. It is easy to check that the Galois group is $S_3 \times S_3 \times S_2$, with the first factor permuting the conjugates of $\sqrt[3]{a^2-b}$, the second permuting the conjugates of $\sqrt[3]{\frac{a+\sqrt{b}}{a-\sqrt{b}}}+\sqrt[3]{\frac{a-\sqrt{b}}{a+\sqrt{b}}}$, and the third permuting the conjugates of $\sqrt{a^2-b}$. The stabilizer of the formula is Klein four group, generated by a transposition in each $S_3$, the first switching $\sqrt{-3}$ and $-\sqrt{-3}$ and the second switching $\sqrt{-3b}$ and $-\sqrt{-3b}$. Thus, the generic degree is $18$, as Igor Rivin's numerical evidence suggested. For a specific value of $a$ and $b$, the Galois group is a subgroup of this. To be degree $3$, the Galois group, mod its intersection with the Klein four group, must have order $3$, so the Galois group must have order $6$ or $12$. Whatever subgroup it is, the elements of the field fixed by it are rational numbers. For instance, since every possible subgroup fixes $\sqrt{a^2-b}$, it is always rational. Assume that $\sqrt{-3b}$ is irrational. Then the subgroup must have order $12$, since $\sqrt{b}$, $\sqrt{-3}$, and $\sqrt{-3b}$ are all irrational, the action on them implies the subgroup must have order a multiple of $4$, thus exactly $12$, so it contains the either Klein four group. There are two subgroups of order $12$ with this property, the one containing the entire first $S_3$ and the one containing the entire second $S_3$. For the one containing the second $S_3$, the invariants are generated by $\sqrt{a^2-b}$ and $\sqrt[3]{a^2-b}$. This is the case discussed in the question. We are asked to eliminate the other cases. For the one containing the first $S_3$, the invariants are $\sqrt{a^2-b}$ and $\sqrt[3]{\frac{a+\sqrt{b}}{a-\sqrt{b}}}+\sqrt[3]{\frac{a-\sqrt{b}}{a+\sqrt{b}}}$. Calling them $x$ and $y$, we have the formulas $y^3-3y=\frac{a+\sqrt{b}}{a-\sqrt{b}}+\frac{a-\sqrt{b}}{a+\sqrt{b}}=2\frac{a^2+b}{a^2-b}=4\frac{a^2}{a^2-b}-2$, so we have the nodal cubic curve: $y^3-3y+2=\left(\frac{2a}{x}\right)^2=z^2$ setting $z=2a/x$, we can parametrize the curve $(y-1)^2(y+2)=z^2$, so $y=s^2-2$, $z=s(s^2-3)$ provides a rational parameterization. (Thanks to Ricky for pointing out it is singular.) At this point we can just set $b=a^2-4a^2/z^2=a^2\left(1-\frac{4}{s^2(s^2-3)^2}\right)$. This gives the formula above.

square root of a certain matrix Hello, I'd like to know the square root of the following $n$ by $n$ matrix, for $n > 2$ and $r>0$: $R_{ii}=r+1$ for $i < n$ $R_{ij}=r$ otherwise The $2$ by $2$ case is given by $\sqrt{R}=\frac{1}{d} \left[\begin{array}{cc} 1+r+\sqrt{r} & r \\\ r & r+\sqrt{r}\end{array}\right]$ where $d=\sqrt{1+2r+2\sqrt{r}}$. Any thoughts? Many thanks.

Let $P(a,b,c)$ be the $n\times n$ matrix where $$ P(a,b,c)_{ij} = \begin{cases} a & \text{ if } i,j < n \\ b & \text{ if } i < n \text{ and } j = n \\ b & \text{ if } i = n \text{ and } j < n \\ c & \text{ if } i = j = n. \end{cases} $$ If I understand correctly, you want the square root of $I+P(r,r,0)$. You can check that $$ (I+P(a,b,c))^2 = I + P((n-1)a^2+b^2+2a,(n-1)ab+bc+2b,(n-1)b^2+c^2+2c). $$ Thus, you just need to solve \begin{align*} (n-1)a^2+b^2+2a &= r \\\\ (n-1)ab+bc+2b &= r \\\\ (n-1)b^2+c^2+2c &= 0. \end{align*} Maple tells me that if we put \begin{align*} p &= \sqrt{1+(n-1)(r-r^2)} \\\\ q &= \sqrt{\frac{2p+2+(n-1)r}{(n-1)(n+3)}} \end{align*} then \begin{align*} a &= q + \frac{q-r-pq}{(n-1)r} \\\\ b &= q \\\\ c &= \frac{q-r-pq}{r}. \end{align*} UPDATE: The square root of $I+P(r,r,r-1)$ can be done similarly. If we put \begin{align*} s &= (1 - 2\sqrt{r} + nr)^{-1/2} \\\\ a &= rs + \frac{s - \sqrt{r}s - 1}{n-1} \\\\ b &= rs \\\\ c &= rs - \sqrt{r}s - 1 \end{align*} then $$ \sqrt{I + P(r,r,r-1)} = 1 + P(a,b,c). $$

Computing $\prod_p(\frac{p^2-1}{p^2+1})$ without the zeta function? We see that $$\frac{2}{5}=\frac{36}{90}=\frac{6^2}{90}=\frac{\zeta(4)}{\zeta(2)^2}=\prod_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\prod_p \left(\frac{(p^2-1)^2}{(p^2+1)(p^2-1)}\right)=\prod_p\left(\frac{p^2-1}{p^2+1}\right)$$ $$\implies \prod_p \left(\frac{p^2-1}{p^2+1}\right)=\frac{2}{5},$$ But is this the only way to compute this infinite product over primes? It seems like such a simple product, one that could be calculated without the zeta function. Note that $\prod_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\prod_p(\frac{p-1}{p-i})\prod_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\prod_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdots$.

An elementary proof of the identity $$ 2 \zeta(2)^2 = 5\zeta(4)$$ has been found by Don Zagier in the paper http://people.mpim-bonn.mpg.de/zagier/files/tex/ConsequencesCohomologySL/fulltext.pdf The idea seems similar to how David Speyer proves it in that he starts by defining the function $$ f(m,n) = \frac{2}{n^3m} + \frac{1}{n^2m^2} + \frac{2}{nm^3} $$ and then verifies that $$ f(m,n) -f(m,n+m) - f(m+n,n) = \frac{2}{m^2n^2}.$$ But since we also have $$ \sum_{n,m>0} f(m,n) - \sum_{n,m>0} f(m,n+m) - \sum_{m,n>0} f(m+n,n) = \sum_{n>0} f(n,n),$$ since only the diagonal terms survive, this gives the identity $$ 2 \zeta(2)^2 = 5\zeta(4).$$ The crucial function $f(m,n)$ seems to be part of a larger family of functions with ties to period polynomials, as explained in the paper.

How does this sequence grow Let $a(n)$ be the number of solutions of the equation $a^2+b^2\equiv -1 \pmod {p_n}$, where $p_n$ is the n-th prime and $0\le a \le b \le \frac{p_n-1}2$. Is the sequence $a(1),a(2),a(3),\dots$ non-decreasing? Data for the first thousand values of the sequence supports this conjecture. Here is an example for $n=5$: The fifth prime is 11. The equation $a^2+b^2 \equiv -1 \pmod {11}$ has just two solutions with the required conditions on $a$ and $b$, namely: $1^2+3^2=10$ and $4^2+4^2=32$. Here are the first fifty values of $a(n)$: 0,1,1,1,2,2,3,3,3,4,4,5,6,6,6,7,8,8,9,9,10,10,11,12,13,13,13,14,14,15,16,17,18,18,19,19,20,21,21,22,23,23,24,25,25,25,27,28,29,29

The answer is yes, and the number of solutions with a prime $p$ is $\lfloor \frac{p+5}{8} \rfloor$ when $p \not\equiv 1 \pmod{8}$ and is $\lfloor \frac{p+5}{8} \rfloor + 1$ when $p \equiv 1 \pmod{8}$. The equation $a^{2} + b^{2} + c^{2} = 0$ defines a conic in $\mathbb{P}^{2}/\mathbb{F}_{p}$. If $p > 2$ this conic has a point on it (by the standard pigeonhole argument that there is a solution to $a^{2} + b^{2} \equiv -1 \pmod{p}$), and so it is isomorphic to $\mathbb{P}^{1}$. Hence, there are $p+1$ points on this conic in $\mathbb{P}^{2}$. Every such point has the form $(a : b : 0)$ or $(a : b : 1)$. If $p \equiv 3 \pmod{4}$, there are no points of the form $(a : b : 0)$, while if $p \equiv 1 \pmod{4}$, then there is a solution to $x^{2} \equiv -1 \pmod{p}$ and $(1 : \pm x : 0)$ give two such points. Hence the number of solutions to $a^{2} + b^{2} \equiv -1 \pmod{p}$ with $0 \leq a \leq p-1$, $0 \leq b \leq p-1$ is $p+1$ or $p-1$ depending on what $p$ is mod $4$. Now it takes a bit more thought and some careful keeping track of solutions with $a$ or $b$ equal to zero, or $a = b$ to derive the formula.

Is $\lceil \frac{n}{\sqrt{3}} \rceil > \frac{n^2}{\sqrt{3n^2-5}}$ for all $n > 1$? An equivalent inequality for integers follows: $$(3n^2-5)\left\lceil n/\sqrt{3} \right\rceil^2 > n^4.$$ This has been checked for n = 2 to 60000. Perhaps there is some connection to the convergents to $\sqrt{3}$. $\lceil \frac{n}{\sqrt{3}} \rceil > \frac{n^2}{\sqrt{3n^2-5}}$

Equivalently, we want to know if $\mathrm{sqceiling}(n^2/3) > n^4/(3n^2-5) = n^2/3 + 5/9 + 25/(27n^2) + ...$ where $\mathrm{sqceiling}()$ is the function taking a real to the next exact square. This is equivalent to $\mathrm{sqceiling}(n^2/3) - n^2/3 > 5/9 + 25/(27n^2) + ...$ but since $n^2/3$ is never a square for $n>0$, and $n^2 \equiv 0\ \textrm{or}\ 1 (\textrm{mod}\ 3)$, we have $\mathrm{sqceiling}(n^2/3) - n^2/3 \ge 2/3$. Clearly for large enough $n$ 2/3 > 5/9 + 25/(27n^2) + ...

This inequality why can't solve it by now (Only four variables inequality)? I asked a question at Math.SE last year and later offered a bounty for it, only johannesvalks give Part of the answer; A few months ago, I asked the author(Pham kim Hung) in Facebook, he said that now there is no proof by hand.and use of software verification is correct,and I try it sometimes,and not succeed.Later asked a lot of people (such on AOPS 1,AOPS 2) have no proof interesting inequality: Let $a,b,c,d>0$, show that $$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}$$ In fact,we have $$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \underbrace{\sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}\ge \sqrt{\dfrac{a^2+b^2+c^2+d^2}{4}}}_{\text{Generalized mean}}$$ Now we only prove this not stronger inequality: $$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt{\dfrac{a^2+b^2+c^2+d^2}{4}}$$ Proof:By Holder inequality we have $$\left(\sum_{cyc}\dfrac{a^2}{b}\right)^2(a^2b^2+b^2c^2+c^2d^2+d^2a^2)\ge (a^2+b^2+c^2+d^2)^3$$ and Note $$a^2b^2+b^2c^2+c^2d^2+d^2a^2=(a^2+c^2)(b^2+d^2)\le\dfrac{(a^2+b^2+c^2+d^2)^2}{4}$$ Proof 2:(I hope following methods(creat is Mine) will usefull to solve my OP inequality,So I post it): \begin{align*}&\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^2-4(a^2+b^2+c^2+d^2)\\ &=\sum_{cyc}\dfrac{3a^4b^2d+5a^4c^3+24a^3cd^3+3a^2b^3c^2+10ab^3d^3+15bcd^5-60a^2bcd^3}{15a^2bcd}\\ &\ge 0 \end{align*} NoW I use computer $$\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^4-64(a^2+b^2+c^2+d^2)=\dfrac{a^{12}c^4d^4+4a^{10}b^3c^3d^4+4a^{10}bc^6d^3+4a^9bc^4d^6 +6a^8b^6c^2d^4+12a^8b^4c^5d^3+64a^8b^4c^4d^4+6a^8b^2c^8d^2+12a^7b^4c^3d^6+12a^7b^2c^6d^5+4a^6b^9cd^4+\cdots+4ab^4c^6d^9+b^4c^4d^{12}}{a^4b^4c^4d^4}$$

Here is a partial solution that reduces the problem to a (hopefully) simpler one. The inequality is homogeneous, so we may assume that the RHS equals one. Let $$ S=\left\{x\in\mathbb R^4;\frac14\sum_ix_i^4=1,x_i>0\right\} $$ and $$ f(a,b,c,d)=\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right). $$ Clearly $f$ and $S$ are smooth, $S$ is bounded and $f(x)\to\infty$ as $x$ approaches any boundary point, so it suffices to show that the inequality is satisfied at points given by Lagrange's multiplier theorem. We get the conditions $(a,b,c,d)\in S$ and \begin{eqnarray} && \frac14 \left( 2\frac ab-\left(\frac da\right)^2, 2\frac bc-\left(\frac ab\right)^2, 2\frac cd-\left(\frac bc\right)^2, 2\frac da-\left(\frac cd\right)^2 \right) \\&=& \lambda (a^3,b^3,c^3,d^3) \end{eqnarray} for some $\lambda\in\mathbb R$. Clearly we need to have $\lambda>0$. Taking the dot product with the vector $(a,b,c,d)$ we obtain $$ f(a,b,c,d) = 4\lambda $$ so it suffices to show that $\lambda\geq\frac14$. If this were not the case, we would have \begin{eqnarray} && \left( 2\frac ab-\left(\frac da\right)^2, 2\frac bc-\left(\frac ab\right)^2, 2\frac cd-\left(\frac bc\right)^2, 2\frac da-\left(\frac cd\right)^2 \right) \\&>& (a^3,b^3,c^3,d^3), \end{eqnarray} meaning inequality for each component. It would now suffice to show that this is impossible if $(a,b,c,d)\in S$.

Matrix Submodular Inequality Given $a,b,x > 0$ I know following the submodularity property holds: \begin{align} \frac{1}{a} - \frac{1}{a+x} \geq \frac{1}{a+b} - \frac{1}{a+b+x} \end{align} My question is, does this property hold for matrices? Precisely, for $A,B,X \succ 0$ is it the case that: \begin{align} A^{-1} - (A+X)^{-1} \succeq (A+B)^{-1} - (A+B+X)^{-1} \end{align} By '$\succeq$' I mean that if $A \succeq B$ then $A−B \succeq 0$, or, is positive semi-definite.

Consider $$ A = \pmatrix{1 & 0\cr 0 & 1\cr},\ X = \pmatrix{1 & 0\cr 0 & 0\cr},\ B = \pmatrix{1 & 1\cr 1 & 1\cr}$$ $$ \eqalign{A^{-1} &- (A+X)^{-1} = \pmatrix{1/2 & 0\cr 0 & 0\cr}\cr &\not\succeq (A+B)^{-1} - (A+X+B)^{-1} = \pmatrix{4/15 & -2/15 \cr -2/15 & 1/15}}$$ Yes, I know $B$ and $X$ are positive semidefinite rather than positive definite; add $\epsilon I$ for $\epsilon $ sufficiently small.

$2^n$-1 consisting only of small factors I've checked the factorization of $2^N - 1$ up through N = 120 for the largest prime factor, and it looks like the largest value of N where $2^N-1$ has a largest prime factor under 2500 is N = 60 (largest prime factor = 1321). As N gets larger, the largest prime factors get larger, even for the "abundant" numbers like 96, 108, and 120. Is there a way to prove that no value of N > 60 exists such that the largest prime factor of $2^N - 1$ is less than 2500?

It is true that if $N > 60$, then $2^{N} - 1$ has a prime factor $> 2500$. Here's another approach. First, observe that every prime factor of $2^{p} - 1$ is $\equiv 1 \pmod{p}$. Combining this with the observation that if $a | b$, then $2^{a} - 1 | 2^{b} - 1$, we see that if $2^{N} - 1$ has all prime factors $\leq 2500$, then all prime factors of $N$ are $< 2500$. Checking these primes, we see that $2^{p} - 1$ has a prime factor $> 2500$ unless $p = 2, 3, 5, 7, 11$ or $29$. Hence if all the prime factors of $2^{N} - 1$ are less than $2500$, then all prime divisors of $N$ are in the set $S = \{ 2, 3, 5, 7, 11, 29 \}$. We find that $65537$ is a prime factor of $2^{32} - 1$ and this means that $N$ cannot be a multiple of $32$ if $2^{N} - 1$ has all prime divisors $< 2500$. Similar arguments show that $N$ cannot be a multiple of $3^{3}$, $5^{3}$, $7^{3}$, $11^{2}$ or $29^{2}$. This implies that $N$ divides $56271600$, and checking all such divisors, we see that $N = 60$ is the largest possible.

Product of a Finite Number of Matrices Related to Roots of Unity Does anyone have an idea how to prove the following identity? $$ \mathop{\mathrm{Tr}}\left(\prod_{j=0}^{n-1}\begin{pmatrix} x^{-2j} & -x^{2j+1} \\ 1 & 0 \end{pmatrix}\right)= \begin{cases} 2 & \text{if } n=0\pmod{6}\\ 1 & \text{if } n=1,5\pmod{6}\\ -1 & \text{if } n=2,4\pmod{6}\\ 4 & \text{if } n=3\pmod{6} \end{cases}, $$ where $x=e^{\frac{\pi i}{n}}$ and the product sign means usual matrix multiplication. I have tried induction but there are too many terms in all of four entries as $n$ grows. I think maybe using generating functions is the way?

UPDATE: As pointed out by Peter Mueller in the comments below, this reduction to continuants does not quite work as stated. I still believe that there is some connection with (possibly generalized) continuants and leave this answer as an unsuccessful attempt to demonstrate it (in hope to fix it later). Notice that $$\begin{pmatrix} x^{-2j} & -x^{2j+1} \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -x^{2j+1} & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} -x^{-4j-1} & 1 \\ 1 & 0 \end{pmatrix}.$$ Since diagonal matrices commute with others and $\prod_{j=0}^{n-1} (-x^{2j+1}) = (-1)^n x^{n^2} = 1$, we have $$\prod_{j=0}^{n-1} \begin{pmatrix} x^{-2j} & -x^{2j+1} \\ 1 & 0 \end{pmatrix} = \prod_{j=0}^{n-1} \begin{pmatrix} -x^{-4j-1} & 1 \\ 1 & 0 \end{pmatrix}.$$ The last product can be expressed in terms of continuants $K_m(a_1,\dots,a_m)$. Namely, for $j=0,\dots,n-1$, let $a_{j+1} = -x^{-4j-1}$. Then $$\prod_{j=0}^{n-1} \begin{pmatrix} -x^{-4j-1} & 1 \\ 1 & 0 \end{pmatrix}= \begin{pmatrix} K_n(a_1,\ldots,a_n) & K_{n-1}(a_1,\ldots,a_{n-1}) \\ K_{n-1}(a_2,\ldots,a_n) & K_{n-2}(a_2,\ldots,a_{n-1}) \end{pmatrix}. $$ So, it remains to compute values of these continuats. P.S. Fedor's answer provides a way to compute the continuants, using Euler's rule.

asymptotic for the number of involutions in GL(n,2) Is it known how the number of involutions in $GL_n(2)$, the group of $n\times n$ matrices over $\mathbb{Z}/2\mathbb{Z}$, behaves as $n\to\infty$ ? Equivalently, one may ask this for the number of $n\times n$ matrices $A$ over $\mathbb{Z}/2\mathbb{Z}$ satisfying $A^2=0$, as $(A+I)^2=I \mod 2$. Needless to say, there are $\lfloor n/2\rfloor$ conjugacy classes of involutions in $GL_n(2)$, and one can write down formula for the centraliser order for each class, $$\frac{\left|GL_n(2)\right|}{2^{k^2+2k(n-2k)}\left|GL_k(2)\right|\left|GL_{n-2k}(2)\right|},\quad 1\leq k\leq \left\lfloor \frac{n}{2} \right\rfloor,$$ but it's quite a mess to just sum them up.

We may write $|{\rm GL}(n,2)| = 2^{n^{2}} \prod_{j=1}^{n}( 1- \frac{1}{2^{j}}).$ As $n \to \infty$, the rightmost factor tends to $\left( \sum_{r=0 }^{\infty} \frac{p(r)}{2^{r}} \right)^{-1}$, where $p(r)$ is the number of partitions of $r$. Let us write $|{\rm GL}(n,2)| = 2^{n^{2}}f(n)$. Then we see that the number of involutions of ${\rm GL}(n,2)$ is given by $\sum_{k = 1}^{\lfloor \frac{n}{2} \rfloor} 2^{2k(n-k)} \frac{f(n)}{f(k)f(n-2k)}$ using the formula given in the question ( if $k = \frac{n}{2}$, we should interpret $|{\rm GL}(n-2k,2)|$ as $1$). Hence the number of involutions in ${\rm GL}(n,2)$ may be expressed as $2^{\frac{n^{2}}{2}} \left(\sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} 2^{- 2\left(\frac{n}{2}-k\right)^{2}}\left( \frac{\prod_{j=k+1}^{n}( 1- \frac{1}{2^{j}})}{ \prod_{m=1}^{n-2k}( 1- \frac{1}{2^{m}})}\right)\right)$ if $n$ is odd, and $2^{\frac{n^{2}}{2}} \left( \prod_{j=\frac{n}{2}+1}^{n}( 1- \frac{1}{2^{j}})+ \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor-1} 2^{- 2\left(\frac{n}{2}-k\right)^{2}}\left( \frac{\prod_{j=k+1}^{n}( 1- \frac{1}{2^{j}})}{ \prod_{m=1}^{n-2k}( 1- \frac{1}{2^{m}})}\right)\right)$ if $n$ is even. Hence when $n$ is even, the number of involutions is at most $2^{\frac{n^{2}}{2}} \prod_{j=\frac{n}{2}+1}^{n}( 1- \frac{1}{2^{j}}) \left( 1 + \frac{1}{\prod_{m=1}^{n-2}( 1- \frac{1}{2^{m}})} \left( \sum_{k=1}^{\frac{n}{2}} 2^{- 2(\frac{n}{2}-k)^{2}}\right)\right)$ and is at least $2^{\frac{n^{2}}{2}} \prod_{j=\frac{n}{2}+1}^{n}( 1- \frac{1}{2^{j}})$. As (even) $n \to \infty$, the first product appearing approaches $1$ from below and (the reciprocal of) the second product appearing tends to $\sum_{r=0}^{\infty} \frac{p(r)}{2^{r}}$, while the inner sum is at most $\sum_{j=0}^{\infty} \left(\frac{1}{4} \right)^{j^{2}}$. I omit the similar analysis when $n$ is odd.

Product of arithmetic progressions Let $(a_1,a_2\ldots,a_n)$ and $(b_1,b_2,\ldots,b_n)$ be two permutations of arithmetic progressions of natural numbers. For which $n$ is it possible that $(a_1b_1,a_2b_2,\dots,a_nb_n)$ is an arithmetic progression? The sequence is (trivially) an arithmetic progression when $n=1$ or $2$, and there are examples for $n=3,4,5,6$: $n=3$: $(11,5,8), (2,3,1)$ $n=4$: $(1,11,6,16), (10,4,13,7)$ $n=5$: $(8,6,4,7,5), (4,9,19,14,24)$ $n=6$: $(7,31,19,13,37,25), (35,11,23,41,17,29)$

First, it's easy to see that real solutions can be rescaled to be rational and therefore also integer. For $n=7$ there are no solutions. For $n=6$ there are essentially 4 primitive integral solutions, with only one consisting of natural numbers (as requested): $\big(-35+12(1,2,4,3,6,5)\big) \times \big(-5+2(5,6,1,2,3,4)\big) = -153+38(1,2,3,4,5,6)$ $\big(-19+6(1,2,5,6,3,4)\big) \times \big(-5+2(5,6,1,2,3,4)\big) = -81+16(1,2,3,4,5,6)$ $\big(-10+3(1,3,2,4,6,5)\big) \times \big(-3+1(3,1,2,6,4,5)\big) = -2+2(1,2,3,4,5,6)$ $\big(1+6(1,5,3,2,6,4)\big) \times \big(5+6(5,1,3,6,2,4)\big) = 149+96(1,2,3,4,5,6)$ Other sequences can be obtained by rescaling, switching $a$ and $b$, reversing the orders, and replacing permutations $a,b$ with $7-a,7-b$ while adjusting the other parameters accordingly. Sketch of proof. Using bold letters for permutations of of the numbers $1,2,\dots n$, we are looking for real solutions to this equation: $(a+d{\bf u})(b+e{\bf v})=c+f(1,2,\dots n)$ that is $(1,2,\dots n)=\frac{ab-c}{f}+\frac{bd}{f}{\bf u}+\frac{ae}{f}{\bf v}+\frac{de}{f}{\bf uv}$ Therefore, given $\bf u,v$, the coefficients $\frac{ab-c}{f},\frac{bd}{f},\frac{ae}{f},\frac{de}{f}$ can be computed by a standard linear regression with $(1,2,\dots n)$ as the dependent ($y$-)variable and intercept, $\bf u$, $\bf v$ and $\bf uv$ as the 4 independent ($x$-)variables. Once $\frac{bd}{f},\frac{ae}{f},\frac{de}{f}$ are known, then also $\frac{b}{e}$ and $\frac{a}{d}$ are, and these determine the coprime pairs $(a,d)$ and $(b,e)$, uniquely for $d,e>0$. There are $n!-2$ possibilities for each of $\bf u$ and $\bf v$ (excluding the trivial $(1,2,\dots n)$ and $(n,n-1,\dots 1)$). Moreover we can assume ${\bf v}\ge {\bf u}$ (lexicographically) and also, obviously, that whenever $u_j<u_i$ for $j>i$ then $\bf v$ must satisfy $v_j>v_i$. Last, we want to avoid the singular cases ${\bf u}+{\bf v}=(n+1,n+1,\dots n+1)$. Given all these restrictions there are only $14777$ pairs $({\bf u},{\bf v})$ left for $n=6$ and $328790$ left for $n=7$. I produced all these pairs with simple awk scripts and ran the $14777+328790$ regressions with awk+shell+Rscript, looking for the cases where both $\frac{de}{f}\ne 0$ and $\text{r-squared}=1$. The search produced 8 hits for $n=6$, that is $4$ pairs of related solutions, and no hits for $n=7$. I conjecture that there are no solutions for $n\ge 8$. The case $n=8$ is definitely within computational reach using the method above, but a better programmer than myself is needed for it. It would be easy to also find all the solutions for $n\le 5$, but I ran out of motivation.

Does the equation $(xy+1)(xy+x+2)=n^2$ have a positive integer solution? Does there exist a positive integral solution $(x, y, n)$ to $(xy+1)(xy+x+2)=n^2$? If there doesn't, how does one prove that?

It looks that Vieta jumping helps. For fixed positive integer $y$ choose a minimal positive integer $x$ for which $(xy+1)(xy+x+2)$ is a perfect square. Denote $4(xy+1)(xy+x+2)=4n^2=(2xy+x+3-z)^2$ for some integer $z=2n-2xy-x-3$, this yields $0<z<x+3$ and rewrites as $z^2-2z(2xy+x+3)+x^2+2x+1=0$. Note that $x$ must divide $z^2-6z+1$, for each $z\leqslant 5$ this gives several variants for $x$ for which it is straightforward to check that $y$ does not appear to be a positive integer. If $z\geqslant 6$, we may replace $x$ to $x'=(z^2-6z+1)/x>0$ (which is another root of the same quadratic equation in $x$.) This contradicts to the minimality since $z^2-6z+1<(z-3)^2<x^2$. Remark: for the new pair $(x',y)$ we have different value of $z$, as $2x'y+x'+3-z$ becomes negative, but it is still true that $(x'y+1)(x'y+x'+2)$ is a perfect square.

Number of integer partitions modulo 3 Motivated by Parity of number of partitions of $n!/6$ and $n!/2$, I asked my computer (and my FriCAS package for guessing) for an algebraic differential equation for the number of integer partitions modulo 3. This is what it answered: (1) -> s := [partition n for n in 1..] (1) [1, 2, 3, 5, 7, 11, 15, 22, 30, 42, ...] Type: Stream(Integer) (32) -> guessADE([s.i for i in 1..400], safety==290, maxDerivative==2)$GUESSF PF 3 (32) [ [ n [x ]f(x): 3 2 2 ,, 4 , 3 3 2 , 2 (x f(x) + 2 x f(x) + x)f (x) + x f (x) + (2 x f(x) + 2 x )f (x) + , 2 (2 x f(x) + 2)f (x) + 2 f(x) = 0 , 3 4 f(x) = 1 + 2 x + 2 x + O(x )] ] Of course, this is only a guess, but it seems fairly well tested. Only 110 terms were needed to guess the recurrence, all the other 290 were used to check it. My question is: is this known, and if not so, is this interesting?

Well, $f''f^2 +xf' ^3+2ff'^2 =0$ modulo 3 for $f=\prod(1-x^m)=\sum_{n\in \mathbb{Z} } (-1)^nx^{n(3n+1)/2}$ may be quickly seen as follows. Differentiating the power series for $f$ and expanding the brackets we see that we should prove that $$ \sum_{a(3a+1)/2+b(3b+1)/2+c(3c+1)/2=n} (-1)^{a+b+c} \left(a(a-2)-abc+2ab\right) $$ is divisible by 3. Multiplying the sum by 3 and using cycling shift of variables we reduce it to proving that the sum $$ \sum_{a(3a+1)/2+b(3b+1)/2+c(3c+1)/2=n} (-1)^{a+b+c} \left( a(a-2)+b(b-2)+c(c-2)-3abc+2ab+2bc+2ac\right) $$ is divisible by 9. The summand is $(a+b+c) (a+b+c-2)-3abc$. If $a=b=c$ this is $9a^2 - 3(a^3+2a)$, divisible by 9. Other triples partition by permuting the variables onto 3-tuples and 6-tuples, so the sum of $3abc$ is of course divisible by 9. As for $a+b+c$, it is congruent to $2n$ modulo 3, thus unless $n=3t+2$ the expression $(a+b+c) (a+b+c-2)$ is divisible by 3, as we need. It remains to show that for $n=3m+2$ the sum of $(-1)^{a+b+c}$ over our triples is divisible by 9. In other words, the coefficient of $x^{3m+2} $ in $f^3$ must be divisible by 9. We have $$ f^3=\prod (1-x^k)^3 =\prod (1-x^{3k}+3(x^k-x^{2k})). $$ Expanding the brackets and reducing modulo 9 we should prove that the expression $$ [x^{3m+2} ] 3 f(x^3) \sum_k \frac{x^k-x^{2k} } {1 - x^{3k} } $$ is divisible by 9. But in the latter sum $\sum_k (x^k-x^{2k}+x^{4k}-x^{5k}+\dots) $ all coefficients of powers $x^{3s+2} $ do cancel, since if $3s+2=kr$, the guys $x^{kr} $ and $x^{rk} $ go with different signs.

How can one integrate over the unit cube, subject to certain (quantum-information-theoretic) constraints? To begin, we have two constraints \begin{equation} C1=x>0\land z>0\land y>0\land x+2 y+3 z<1 \end{equation} and \begin{equation} C2=x>0\land y>0\land x+2 y+3 z<1\land x^2+x (3 z-2 y)+(y+3 z)^2<3 z. \end{equation} $C1$ ensures the nonnegative-definiteness of a class of $9 \times 9$ ("two-qutrit") density matrices ($\rho$). $C2$ also ensures this, as well as the nonnegative-definiteness of the "partial transpose" ($\rho^{PT}$) of $\rho$. The integration--subject to $C1$--of the value 36 over the unit cube $\{x,y,z\} \in [0,1]^3$ yields 1. The integration--subject to $C2$--of the value 36 over the unit cube yields (the Hilbert-Schmidt positive partial transpose probability) $\frac{8 \pi}{27 \sqrt{3}} \approx 0.537422$. Now, we are interested in similarly enforcing both $C1 \land C3$ (yielding an "entanglement probability") and $C2 \land C3$ (yielding a "bound-entanglement probability"), where, the entanglement constraint $C3$ is \begin{equation} b \left(-x \left(a^2-a (b+2)+2 b+1\right)+2 y (a (-a+b+2)+b-1)-3 z (a-b-1) (a+b-1)+(a-1)^2\right)<0, \end{equation} with its three parameters $a,b,c$ subject to \begin{equation} C4= b>0\land c>0\land 0<a<1\land a+b+c=2\land (a-1)^2=b c. \end{equation} Now, I suspect these last two problems are too difficult to resolve in their full generality (leaving $a,b,c$ unspecified). But, to begin, if we take \begin{equation} \{a,b,c\}=\left\{\frac{1}{4} \left(3-\sqrt{5}\right),\frac{1}{2},\frac{1}{4} \left(3+\sqrt{5}\right)\right\}, \end{equation} the integration of 36 over the unit cube subject to $C1 \land C3$ yields $\frac{5}{132} \left(5+\sqrt{5}\right) \approx 0.274093$. Alternatively, for \begin{equation} \{a,b,c\} = \left\{\frac{1}{3},\frac{1}{3},\frac{4}{3}\right\}, \end{equation} the integration of 36 over the unit cube subject to $C1 \land C3$ yields $\frac{125}{486} \approx 0.257202$. However, I have not been so far able to obtain the counterparts for these last two results for $C2 \land C3$. (Using numerical integration, we get the much lower values of 0.001497721920258410 and 0.003256122941383665, respectively.) A set of values of $a,b,c$ which satisfy $C4$ are \begin{equation} \left\{\frac{2}{3} (\cos (\alpha )+1),\frac{2}{3} \left(-\frac{1}{2} \sqrt{3} \sin (\alpha )-\frac{\cos (\alpha )}{2}+1\right),\frac{2}{3} \left(\frac{1}{2} \sqrt{3} \sin (\alpha )-\frac{\cos (\alpha )}{2}+1\right)\right\}, \end{equation} for $\frac{\pi}{3} \leq \alpha \leq \frac{5 \pi}{3}$. So, I would like to obtain results of integration of the value 36 over $[0,1]^3$ of $C1 \land C3$ and $C2 \land C3$ for either specific values of $a,b,c$, satisfying $C4$, or even without particular values being specified.

Well, maybe the second comment of LSpice made me fully realize that the constraint $C4$ means that there is only one degree of freedom between $a,b,c$. So, as a start I took $a=\frac{1}{3}$. Then, using the Mathematica GenericCylindricalDecomposition command and choosing the order of integration of $x,y,z$ to select the simplest (as measured by LeafCount) output, I was able to obtain an entanglement probability of $\frac{125}{486}=\frac{5^3}{2 \cdot 3^5} \approx 0.257202$. Additionally, using the same approach, the bound-entanglement probability proved to be \begin{equation} \frac{-204+56 \sqrt{3} \pi +7 \log (7)-336 \sqrt{3} \csc ^{-1}\left(2 \sqrt{7}\right)}{1134} \approx 0.00325612. \end{equation} So, I'll now try to use other values of $a$--rather than $\frac{1}{3}$ (and extend this answer, if I so succeed, and the results are of interest). I thought that the problem of getting results for general $0 \leq a \leq 1$ would be much more formidable, as the resulting forms that $C1 \land C3$ and $C2 \land C3$ take seemed quite involved. However, much to my surprise, the calculation for $C1 \land C3$ for the entanglement probability as a function of $a$ greatly simplified, yielding \begin{equation} -\frac{(a-2)^3}{9 a^2-30 a+27}. \end{equation} For $a=\frac{1}{3}$, the function gives the above-reported $\frac{125}{486}$. Further, Nicholas Tessore was able to find the formula for $C2 \land C3$, giving the bound-entanglement probability. It took the form \begin{equation} \label{Tessore} -\frac{A+B}{54 (4-3 a)^{3/2} (2 a-3)} , \end{equation} where \begin{equation} A=8 \sqrt{12-9 a} \left(6 a^2-17 a+12\right) \cos ^{-1}\left(\frac{a (3 a-8)+6}{6-4 a}\right) \end{equation} and \begin{equation} B=3 \sqrt{a} \left(2 \left(9 a^3-57 a^2+108 a-64\right)+3 (3-2 a) a \log (9-6 a)\right). \end{equation} So, a complete answer has now been successfully given to the question put.

how to prove an equation involving sums of Kronecker symbol Let $p\equiv 8 \mod 9$ be a prime, I find the following equation: $$2\sum_{\substack{0<x<p\\ 2|x}}\sum_{r|p^2-x^2}\left(\frac{-3}{r}\right)=p+1.$$ where $\left(\frac{-3}{r}\right)$ is the Kronecker symbol. I checked it for many $p$ using computer. Does anyone have ideal how to prove it?

The identity can be rewritten as $$\sum_{\substack{|x|<p\\ 2|x}}\sum_{r|p^2-x^2}\left(\frac{-3}{r}\right)=p+2,$$ because for $x=0$ the inner sum is $1-1+1=1$. Writing $x=2c$, the identity becomes $$\sum_{|c|<p/2}\,\sum_{r|p^2-4c^2}\left(\frac{-3}{r}\right)=p+2.$$ The inner sum counts the number of integral representations $p^2-4c^2=a^2+ab+b^2$ divided by $6$, hence the identity is equivalent to the statement that the number of integral representations of $p^2$ by the quadratic form $a^2+ab+b^2+4c^2$ equals $6(p+2)$. We shall verify this by Siegel's mass formula, as it appears in Über die analytische Theorie der quadratischen Formen, Ann. of Math. 36 (1935), 527-606. As the class of $a^2+ab+b^2+4c^2$ is alone in its genus, the number of representations of $p^2$ can be calculated as a product of local densities: $$r(p^2)=\alpha_\infty\alpha_2\alpha_3\alpha_p\prod_{q\nmid 6p}\alpha_q.$$ By Hilfssatz 26 and (71) and the line below (59) in Siegel's paper, $$\alpha_\infty=\frac{p}{\sqrt{3}}\cdot\frac{\pi^{3/2}}{\Gamma(3/2)}=\frac{2\pi }{\sqrt{3}}p.$$ By Hilfssatz 13 in Siegel's paper, $$\alpha_2=\frac{3}{2}\qquad\text{and}\qquad\alpha_3=\frac{4}{3}.$$ By Hilfssatz 16 in Siegel's paper, $$\alpha_p=\left(1-p^{-2}\right)\left(1+\frac{p^{-1}}{1+p^{-1}}\right)=(1-p^{-1})(1+2p^{-1}).$$ Finally, by Hilfssatz 12 in Siegel's paper, $$\prod_{q\nmid 6p}\alpha_q=\prod_{q\nmid 6p}(1+\chi(q)q^{-1})=\frac{2}{1-p^{-1}}\prod_{q\neq 3}(1+\chi(q)q^{-1}),$$ where $\chi$ denotes the nontrivial quadratic character modulo $3$. Therefore, $$r(p^2)=(p+2)\frac{8\pi}{\sqrt{3}}\prod_{q\neq 3}(1+\chi(q)q^{-1}).$$ We can identify the product over $q\neq 3$ as $$\prod_{q\neq 3}(1+\chi(q)q^{-1})=\prod_{q\neq 3}\frac{1-q^{-2}}{1-\chi(q)q^{-1}}=\frac{9}{8}\cdot\frac{6}{\pi^2}L(1,\chi)=\frac{3\sqrt{3}}{4\pi},$$ hence in the end $$r(p^2)=(p+2)\frac{8\pi}{\sqrt{3}}\cdot\frac{3\sqrt{3}}{4\pi}=6(p+2).$$ The proof is complete.

How to determine the coefficient of the main term of $S_{k}(x)$? Let $k\geqslant 2$ be an integer, suppose that $p_1,p_2,\dotsc,p_k$ are primes not exceeding $x$. Write $$ S_{k}(x) = \sum_{p_1 \leqslant x} \dotsb \sum_{p_k \leqslant x} \frac{1}{p_1+\dotsb +p_k}. $$ By AM-GM inequality, $p_{1}+\dotsb + p_{k} \geqslant k \sqrt[k]{p_{1}\dotsm p_{k}}$, we have $$ S_{k}(x) \leqslant \frac{1}{k} \sum_{p_{1}\leqslant x}\dotsb \sum_{p_{k} \leqslant x} \frac{1}{\sqrt[k]{p_{1}\dotsm p_{k}}} = \frac{1}{k} \left( \sum_{p \leqslant x} p^{-\frac{1}{k}} \right)^{k}. $$ By Prime Number Theorem and summation by parts we see that $$ \sum_{p \leqslant x} p^{-\frac{1}{k}} = \mathrm{Li}\big( x^{1-\frac{1}{k}} \big) + O \left( x^{1-\frac{1}{k}}\mathrm{e}^{-c\sqrt{\log x}} \right), $$ Here $\mathrm{Li}(x)$ is the logarithmic integral, and $\mathrm{Li}(x)\sim x/\log x$. Hence $$ S_{k}(x) \leqslant \left( \frac{k^{k-1}}{(k-1)^{k}} +o(1) \right) \frac{x^{k-1}}{\log^{k} x}. $$ On the other hand, $p_{1}+\dotsb +p_{k} \leqslant kx$, we have $$ S_{k}(x) \geqslant \frac{1}{kx} \sum_{p_{1} \leqslant x} \dotsb \sum_{p_{k} \leqslant x} 1 = \frac{1}{kx} \left( \sum_{p \leqslant x} 1 \right)^{k} = \frac{\pi^{k}(x)}{kx} = \frac{(1+o(1))}{k} \frac{x^{k-1}}{ \log^{k} x}. $$ My question is how to determine the coefficient of the main term of $S_{k}(x)$? Thanks!

Thank you, Mr. Petrov, but you made a little mistake. A detailed calculation of $c$ is as follows: Write $g(x)=(1-\mathrm{e}^{-x})^k= \sum\limits_{j=0}^{k} \binom{k}{j} (-1)^{j} \mathrm{e}^{-jx}$, integrating by parts we get \begin{align} \int_{0}^{\infty} g(x) x^{-k} \,\mathrm{d} x & = \int_{0}^{\infty} g(x) \,\mathrm{d} \left( \frac{x^{-k+1}}{-k+1} \right) \nonumber \\ & = \left. \frac{g(x)}{(-k+1)x^{k-1}} \right|_{0}^{\infty} + \frac{1}{k-1} \int_{0}^{\infty} \frac{g'(x)}{x^{k-1}} \mathrm{d} x, \end{align} since $\lim\limits_{x\to 0} \dfrac{g(x)}{x^{k-1}} = \lim\limits_{x\to +\infty} \dfrac{g(x)}{x^{k-1}} = 0$, so that \begin{align*} \frac{1}{k-1} \int_{0}^{\infty} \frac{g'(x)}{x^{k-1}} \mathrm{d} x & = \frac{1}{k-1} \int_{0}^{\infty} g'(x) \, \mathrm{d} \left( \frac{x^{-k+2}}{-k+2} \right) \\ & = - \left. \frac{g'(x)}{(k-1)(k-2)x^{k-2}} \right|_{0}^{\infty} + \frac{1}{(k-1)(k-2)} \int_{0}^{\infty} \frac{g''(x)}{x^{k-2}} \mathrm{d} x, \end{align*} where $g'(x)=k(1-\mathrm{e}^{-x})^{k-1}\cdot \mathrm{e}^{-x}$ and $\lim\limits_{x\to 0} \dfrac{-g'(x)}{(k-1)(k-2)x^{k-2}}= \lim\limits_{x\to +\infty} \dfrac{-g'(x)}{(k-1)(k-2)x^{k-2}}=0$. Hence, integrating by parts $k-1$ times gives \begin{align} \int_{0}^{\infty} \frac{\sum\limits_{j=0}^{k} \binom{k}{j} (-1)^{j}\mathrm{e}^{-jx}}{x^k} \, \mathrm{d} x & =\frac{1}{(k-1)!}\int_{0}^{\infty} \frac{\sum\limits_{j=0}^{k} \binom{k}{j} (-1)^j(-j)^{k-1} \mathrm{e}^{-jx}}{x} \,\mathrm{d} x \nonumber \\ & =\frac{1}{(k-1)!}\int_{0}^{\infty} \sum\limits_{j=1}^{k} \binom{k}{j} (-1)^{k+j-1}j^{k-1} \frac{\mathrm{e}^{-jx}}{x} \, \mathrm{d} x. \quad (\ast) \end{align} Notice that $(-1)^{k+j-1}=(-1)^{k+j+1}=-(-1)^{k-j}$, and consider the Stirling number of the second kind, we get \begin{align} \frac{1}{(k-1)!} \sum_{j=1}^{k} (-1)^{k+j-1} \binom{k}{j} j^{k-1} & = -k \cdot \frac{1}{k!} \sum_{j=1}^{k} (-1)^{k-j} \binom{k}{j} j^{k-1} \\ & = -k\cdot S(k-1,k)=0. \end{align} Set $\displaystyle a_{j} = \frac{(-1)^{k+j-1}j^{k-1}}{(k-1)!} \binom{k}{j}$, then $\sum\limits_{j=1}^{k} a_{j}=0$. Using the Frullani's integral formula $\int_{0}^{\infty} \frac{\mathrm{e}^{-jx}- \mathrm{e}^{-Ax}}{x} \mathrm{d} x = \log A - \log j$ with $0<j<A$. Write $(\ast)$ as \begin{align*} \int_{0}^{\infty} \sum_{j=1}^{k} a_{j} \frac{\mathrm{e}^{-jx}}{x} \mathrm{d} x & = \lim_{A\to + \infty} \int_{0}^{\infty} \sum_{j=1}^{k} a_{j} \frac{\mathrm{e}^{-jx}- \mathrm{e}^{-Ax}}{x} \mathrm{d} x \\ & = \lim_{A\to +\infty} \sum_{j=1}^{k} a_{j} (\log A - \log j) = - \sum_{j=1}^{k} a_{j} \log j, \end{align*} where $\lim\limits_{A\to +\infty} \sum\limits_{j=1}^{k} a_{j} \log A =0$. We obtain $$ \int_{0}^{\infty} \left(\frac{1-\mathrm{e}^{-x}}{x}\right)^k \,\mathrm{d}x = c = \frac{1}{(k-1)!} \sum_{j=2}^{k} (-1)^{k+j} j^{k-1} \binom{k}{j} \log j. $$

Integrality of a binomial sum The following sequence appears to be always an integer, experimentally. QUESTION. Let $n\in\mathbb{Z}^{+}$. Are these indeed integers? $$\sum_{k=1}^n\frac{(4k - 1)4^{2k - 1}\binom{2n}n^2}{k^2\binom{2k}k^2}.$$ POSTSCRIPT. After Carlo's cute response and several useful comments, I like to ask this: is there a combinatorial proof?

There is a way to prove Zhi-Wei Sun's identity as well as Carlo Beenakker's identity. Of course, both can be treated in accord with Fedor Petrov's induction. Let's focus on Sun's identity. Divide through by $\binom{2n}n\binom{3n}n$ to write $$A_n:=\sum_{k=1}^n\frac{(9k-2)27^{k-1}}{k^2\binom{2k}k\binom{3k}k}=\frac{27^n}{3\binom{2n}n\binom{3n}n}-\frac13. \tag1$$ so that $$A_n-A_{n-1}=\frac{(9n-2)27^{n-1}}{n^2\binom{2n}n\binom{3n}n}.$$ Let $a_n=\binom{2n}n\binom{3n}nA_n$ (which is exactly Sun's LHS) to get the recursive equation $$n^2a_n-3(3n-1)(3n-2)a_{n-1}=(9n-2)27^{n-1}.\tag2$$ First, we find a solution to the homogeneous equation $n^2a_n-3(3n-1)(3n-2)a_{n-1}=0$ as follows $$a_n^{(h)}=\binom{2n}n\binom{3n}n. \tag4$$ A particular solution to the non-homogeneous equation (2) can be determined by mimicking the RHS as $a_n^{(p)}=(bn+c)27^n$. Now, plug this back in (2) to solve for $b$ and $c$: \begin{align*} n^2(bn+c)27^n-3(3n-1)(3n-2)(bn-b+c)27^{n-1}&=(9n-2)27^{n-1} \\ \iff 27n^2(bn+c)-3(3n-1)(3n-2)(bn-b+c)&=9n-2 \\ \iff \qquad b=0 \qquad \text{and} \qquad c=\frac13. \end{align*} Therefore, the general solution takes the form $$a_n=a_n^{(p)}+\beta\,a_n^{(h)}=\frac{27^n}3+\alpha\binom{2n}n\binom{3n}n.$$ Since $a_0=A_0=0$, we compute $\beta=-\frac13$ and hence $$a_n=\frac{27^n}3-\frac13\binom{2n}n\binom{3n}n=\frac{27^n}3-\binom{2n}n\binom{3n-1}{n-1}. \qquad \square$$

Alternative proofs sought after for a certain identity Here is an identity for which I outlined two different arguments. I'm collecting further alternative proofs, so QUESTION. can you provide another verification for the problem below? Problem. Prove that $$\sum_{k=1}^n\binom{n}k\frac1k=\sum_{k=1}^n\frac{2^k-1}k.$$ Proof 1. (Induction). The case $n=1$ is evident. Assume the identity holds for $n-1$. Then, \begin{align*} \sum_{k=1}^{n+1}\binom{n+1}k\frac1k-\sum_{k=1}^n\binom{n}k\frac1k &=\frac1{n+1}+\sum_{k=1}^n\left[\binom{n+1}k-\binom{n}k\right]\frac1k \\ &=\frac1{n+1}+\sum_{k=1}^n\binom{n}{k-1}\frac1k \\ &=\frac1{n+1}+\frac1{n+1}\sum_{k=1}^n\binom{n+1}k \\ &=\frac1{n+1}\sum_{k=1}^{n+1}\binom{n+1}k=\frac{2^{n+1}-1}{n+1}. \end{align*} It follows, by induction assumption, that $$\sum_{k=1}^{n+1}\binom{n+1}k\frac1k=\sum_{k=1}^n\binom{n}k\frac1k+\frac{2^{n+1}-1}{n+1}=\sum_{k=1}^n\frac{2^k-1}k+\frac{2^{n+1}-1}{n+1} =\sum_{k=1}^{n+1}\frac{2^k-1}k.$$ The proof is complete. Proof 2. (Generating functions) Start with $\sum_{k=1}^n\binom{n}kx^{k-1}=\frac{(x+1)^n-1}x$ and integrate both sides: the left-hand side gives $\sum_{k=1}^n\binom{n}k\frac1k$. For the right-hand side, let $f_n=\int_0^1\frac{(x+1)^n-1}x\,dx$ and denote the generating function $G(q)=\sum_{n\geq0}f_nq^n$ so that \begin{align*} G(q)&=\sum_{n\geq0}\int_0^1\frac{(x+1)^n-1}x\,dx\,q^n =\int_0^1\sum_{n\geq0}\frac{(x+1)^nq^n-q^n}x\,dx \\ &=\int_0^1\frac1x\left[\frac1{1-(x+1)q}-\frac1{1-q}\right]dx=\frac{q}{1-q}\int_0^1\frac{dx}{1-(x+1)q} \\ &=\frac{q}{1-q}\left[\frac{\log(1-(1+x)q)}{-q}\right]_0^1=\frac{\log(1-q)-\log(1-2q)}{1-q} \\ &=\frac1{1-q}\left[-\sum_{m=1}^{\infty}\frac1mq^m+\sum_{m=1}^{\infty}\frac{2^m}mq^m\right]=\frac1{1-q}\sum_{m=1}^{\infty}\frac{2^m-1}m\,q^m \\ &=\sum_{n\geq1}\sum_{k=1}^n\frac{2^k-1}k\,q^n. \end{align*} Extracting coefficients we get $f_n=\sum_{k=1}^n\frac{2^k-1}k$ and hence the argument is complete.

Here's a sketch of a proof of a generalization: \begin{multline} \quad \sum_{k=1}^n\binom nk \frac{t^k}{k+a}\\ =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}. \quad \tag {$*$} \end{multline} (This is a generalization of Terry Tao's generalization, which is the case $a=0$.) We start with the identity $$\sum_{k=0}^n \binom nk \frac{t^k}{k+a} = \frac {1}{a\binom{a+n}{n}}\sum_{k=0}^n \binom{a+k-1}{k} (1+t)^k.$$ This is a special case of a well-known linear transformation for the hypergeometric series, the case $b=a+1$ of $${}_2F_1(-n,a; b\mid -t) =\frac{(b-a)_n}{(b)_n}\,_2F_1(-n,a; 1-n-b+a\mid 1+t),$$ where $(u)_n = u(u+1)\cdots (u+n-1)$, which can be proved easily in several ways. Since $\frac{1}{a}\binom{a+k-1}{k} = \frac {1}{k}\binom{a+k-1}{k-1}$ for $k\ge 1$, we have $$\sum_{k=1}^n\binom nk \frac{t^k}{k+a} =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}+C$$ where $C$ is a constant (as a polynomial in $t$). But $C=0$ since each summand has no constant term in $t$, and $(*)$ follows.

Is the solution to this trig function known to be algebraic or transcendental? This largest solution to this gorgeous equation is the first local extremum on a function related to the Fibonacci sequence: $$x^2 \cdot \sin \left(\frac{2\pi}{x+1} \right) \cdot \left(3+2 \cos \left(\frac{2\pi}{x} \right) \right) = (x+1)^2 \cdot \sin \left(\frac{2\pi}{x} \right) \cdot \left(3+2 \cos \left(\frac{2\pi}{x+1} \right) \right)$$ This is as simplified as I could get it. The largest solution to this equation is around $x = 2.1392.$ It appears there is no closed-form solution for this; is there any way to prove if the solution is algebraic or transcendental? P.S. Can anyone approximate this constant to more decimal places? ANSWERED

It should be possible to show that $x$ is irrational using Theorem 7 of Trigonometric diophantine equations (On vanishing sums of roots of unity) by J. H. Conway and A. J. Jones, Acta Arithmetica 30 (1976), 229–240, although I have not carried out the full calculation. By letting $\alpha = 2\pi/(x+1)$ and $\beta = 2\pi/x$ and using standard trig identities, we can rewrite the given equation as $$3x^2\sin \alpha -3(x+1)^2\sin\beta - (2x+1)\sin(\alpha+\beta) + (2x^2+2x+1)\sin(\alpha-\beta) = 0.$$ We can convert from sines to cosines via $\sin \gamma \equiv \cos(\pi/2 - \gamma)$. Then Theorem 7 of Conway and Jones tells us that there are only a few "primitive" ways of getting a rational linear combination of four cosines of rational multiples of $\pi$ to vanish: $$\eqalign{{1\over2} &= \cos{\pi\over 3}\cr 0 &= -\cos\phi + \cos\biggl({\pi\over 3}-\phi\biggr) + \cos\biggl({\pi\over 3}+\phi\biggr)\cr {1\over2} &= \cos{\pi\over 5} - \cos{2\pi\over 5}\cr {1\over2} &= \cos{\pi\over 7} - \cos{2\pi\over 7} + \cos{3\pi \over 7}\cr {1\over2} &= \cos{\pi\over 5} - \cos{\pi\over 15} + \cos{4\pi\over 15}\cr {1\over2} &= -\cos{2\pi\over 5} + \cos{2\pi\over 15}-\cos{7\pi\over 15}\cr {1\over2} &= \cos{\pi\over 7} + \cos{3\pi\over 7} - \cos{\pi\over 21} + \cos{8\pi\over21}\cr {1\over2} &= \cos{\pi\over 7} -\cos{2\pi\over7}+\cos{2\pi\over21}-\cos{5\pi\over 21}\cr {1\over2} &= -\cos{2\pi\over 7} + \cos{3\pi\over 7}+ \cos{4\pi\over21} + \cos{10\pi\over 21}\cr {1\over2} &= -\cos{\pi\over 15}+\cos{2\pi\over15}+\cos{4\pi\over 15}-\cos{7\pi\over 15}\cr }$$ One should be able to go through this list and check case by case that no rational value of $x$ can yield the desired equation. I do not know how to show that $x$ must be transcendental.

How can I find all integer solutions of $3^n - x^2 = 11$ I know that $n$ can't be even because of the following argument: Let $n = 2p$. Then we can use the difference of two squares and it becomes like this : $(3^p + x)(3^p - x) = 11; 3^p + x = 11 , 3^p - x = 1$. $3^p = 6$ which is not possible if $p$ is an integer. I also found out that $x$ has to be an even number. I think that the only solution is $3^3 - 4^2 = 11$ but how can I find the true answer?

$$3^{n} - x^2 = 11$$ According to Silverman's answer, we take the three cases $n=3a, n=3a+1,$ and $n=3a+2.$ The problem can be reduced to finding the integer points on elliptic curves as follows. $\bullet n=3a$ Let $X = 3^a, Y=x$, then we get $Y^2 = X^3 - 11.$ According to LMFDB, this elliptic curve has integer points $(X,Y)=(3,\pm 4), (15,\pm 58)$ with rank $2.$ Hence $(X,Y)= (3,\pm 4) \implies (n,x)=(3,\pm4).$ $\bullet n=3a+1$ Let $X = 3^{a+1}, Y=3x$, then we get $Y^2 = X^3 - 99.$ This elliptic curve has rank $0$ and has no integer point, so there is no integer solution $(n,x).$ $\bullet n=3a+2$ Let $X = 3^{a+2}, Y=9x$, then we get $Y^2 = X^3 - 891.$ This elliptic curve has integer points $(X,Y)=(31,\pm 170)$ with rank $1.$ Hence there is no integer solution $(n,x).$ Thus, there are only integer solutions $(n,x)=(3,\pm4).$

Integers representable as binary quadratic forms It is known that odd prime $p$ can be represented as $p=x^2+y^2$ if and only if $p \equiv 1$ mod $4$, represented as $p=x^2+2y^2$ if and only if $p \equiv 1$ or $3$ mod $8$, represented as $p=x^2+3y^2$ if and only if $p \equiv 0$ or $1$ mod $3$, etc. The situation with representation as $p=x^2+ny^2$ becomes more complicated at $n=11$: if $p=x^2+11y^2$ then $p=0,1,3,4,5$ or $9$ mod $11$, (that is, $p=11$ or is the qudratic residue mod $11$) but the converse statement is not true. In fact, primes in these residues classes are represented as either $p=x^2+11y^2$ or $p=3x^2+2xy+4y^2$, and these sets of primes are disjoint. Example $2^2 + 11 \cdot 1^2 = 15 = 3 \cdot 5$ shows that all prime factors of an integer representable as $x^2+11y^2$ can be of the form $3x^2+2xy+4y^2$. My question is whether the opposite can be true: do there exists any integer representable $3x^2+2xy+4y^2$ that has all its prime factors in the form $x^2+11y^2$? Update inspired by Will Jagy's answer. The answer mentions polynomial $f(z)=z^3+z^2-z+1$. Ok, if $p$ is in the form $x^2+11y^2$ then $f(z)=0$ is solvable (in fact has $3$ solutions if $p\neq 11$) modulo $p$. If $p_1$ and $p_2$ are two distinct primes of this form, it follows that $f(z)=0$ is solvable modulo $m=p_1p_2$, and so on. But why $z^3+z^2-z+1=0$ cannot be solvable modulo a (not necessarily prime) integer $m$ representable as $m=3x^2+2xy+4y^2$?

you don't seem to be mentioning Gauss composition. You have a genus of forms, equivalent to $\langle 1,8,27 \rangle,$ then $\langle 3,8,29 \rangle,$ then $\langle 9,8,3 \rangle.$ These are convenient for Dirichlet's description of composition. There is a cancellation for principal forms: if $x^2 + 8xy + 27 y^2$ represents both $p$ and $np,$ it also represents $n.$ You mention 15, $$ \left( 3x^2 + 8xy + 9 y^2 \right) \left( 9z^2 + 8zw + 3 w^2 \right) = \color{magenta}{ u^2 + 8 uv + 27 v^2,} $$ where $u = 3xw + 9 yz +8yw$ and $v=xz-yw.$ Lots more...A prime $p \neq 2,11$ with Legendre symbol $(-44|p) = 1$ is represented by $x^2 + 8xy + 27 y^2$ if and only if the polynomial $z^3 + z^2 - z + 1$ factors into three distinct linear factors $\pmod p.$ Cubic because class number $h(-44) = 3$ Oh, Dirichlet composition is available everywhere, I copied from D. A. Cox, Primes of the Form $x^2 + n y^2$ there is always more

What are the integer solutions to $z^2-y^2z+x^3=0$? The question is to describe ALL integer solutions to the equation in the title. Of course, polynomial parametrization of all solutions would be ideal, but answers in many other formats are possible. For example, answer to famous Markoff equation $x^2+y^2+z^2=3xyz$ is given by Markoff tree. See also this previous question Solve in integers: $y(x^2+1)=z^2+1$ for some other examples of formats of acceptable answers. In general, just give as nice description of the integer solution set as you can. If we consider the equation as quadratic in $z$ and its solutions are $z_1,z_2$, then $z_1+z_2=y^2$ while $z_1z_2=x^3$, so the question is equivalent to describing all pairs of integers such that their sum is a perfect square while their product is a perfect cube. An additional motivation is that, together with a similar equation $xz^2-y^2z+x^2=0$, this equation is the smallest $3$-monomial equation for which I do not know how to describe all integer solutions. Here, the "smallest" is in the sense of question What is the smallest unsolved Diophantine equation?, see also Can you solve the listed smallest open Diophantine equations? .

Since the equation is NOT homogeneous, it is trivial to find infinite families of solutions with $g=\gcd(x,y)>1$. For instance, choose any integers $a$ and $b$, set $c=ab^2-a^2$, so multiplying by $c^8$ gives $(ac^4)^2-ac^4(bc^2)^2+c^9=0$, giving the trivial solutions $(x,y,z)=(c^3,bc^2,ac^4)$, but there are many other ways to find ``trivial'' solutions. If you assume $g=\gcd(x,y)=1$, the solution is pretty standard: since we have a monic second degree equation in $z$, there exist integer solutions if and only if the discriminant is a square, giving the auxiliary equation $y^4-4x^3=d^2$, rewritten as $((y^2-d)/2)((y^2+d)/2)=x^3$, and since the factors are coprime, they are both cubes, say $a^3$ and $b^3$, so $x=ab$, $y^2=a^3+b^3$, hence $z=b^3$. The equation in $y$ is a standard superFermat equation of elliptic type, which is entirely parametrized by the following three parametrizations (see for instance my GTM 240 chapter 14 for a proof), where $s$ and $t$ are coprime integers, and any solution belongs to one and only one parametrization: $$(a,b,y)=(s(s+2t)(s^2-2ts+4t^2),-4t(s-t)(s^2+ts+t^2),\pm(s^2-2ts-2t^2))$$ with $s$ odd and $s\not\equiv t\pmod3$, $$(a,b,y)=(s^4-4ts^3-6t^2s^2-4t^3s+t^4,2(s^4+2ts^3+2t^3s+t^4),3(s^2-t^2)(s^4+2s^3t+\ 6s^2t^2+2st^3+t^4))$$ with $s\not\equiv t\pmod{2}$ and $s\not\equiv t\pmod 3$, $$(a,b,y)=(-3s^4+6t^2s^2+t^4,3s^4+6t^2s^2-t^4,6st(3s^4+t^4))$$ with $s\not\equiv t\pmod{2}$ and $3\nmid t$.

$3^n - 2^m = \pm 41$ is not possible. How to prove it? $3^n - 2^m = \pm 41$ is not possible for integers $n$ and $m$. How to prove it?

As a valuable hint for solving the problem, I consider the following extract from my lectures on elementary number theory. Theorem ($\approx1320$; Levi ben Gerson 1288--1344). The equations $$ (1) \quad 3^p-2^q=1 $$ and $$ (2) \quad 2^p-3^q=1 $$ have no solutions in integers $p,q>1$, except the solution $p=2$, $q=3$ to equation (1). Proof. (1) If $p=2k+1$, then $$ 2^q=3^p-1=3\cdot9^k-1\equiv2\pmod4, $$ which is impossible for $q>1$. If $p=2k$, then $2^q=3^p-1=(3^k-1)(3^k+1)$ implying $3^k-1=2^u$ and $3^k+1=2^v$. Since $2^v-2^u=(3^k+1)-(3^k-1)=2$, we have $v=2$ and $u=1$. This corresponds to the unique solution $q=u+v=3$ and $p=2$. (2) If $q\ge1$, then $3^q+1$ is not divisible by~$8$. Indeed, if $q=2k$, then $3^q+1=9^k+1\equiv2\pmod8$; and if $q=2k+1$, then $3^q+1=3\cdot9^k+1\equiv4\pmod8$. Therefore $p\le2$, hence $p=2$. The latter implies $q=1$ which does not correspond to a solution.

Sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ Are there known sharp upper bounds (in terms of $k$ or $\omega(k)$, the number of distinct prime divisors of $k$) for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ for $k > 1$ subject to the constraint $\sum_{p \mid k} \frac{1}{p+1} < 1$? (This would be a special case of the general result of O. Izhboldin and L. Kurliandchik referenced in the comment.) A related question: Suppose $(p_i)$ is a set of $n$ consecutive primes which minimizes $1 - \sum_{i = 1}^n \frac{1}{p_i+1} > 0$ for a given $n > 1$. Are there known bounds for $1 - \sum_{i = 1}^{n} \frac{1}{p_i + 1}$ from below in terms of $n$, e.g., $n^{-\delta n}$ for some fixed $\delta > 0$? Thanks!

If we are allowed to consider somewhat weaker bounds, both answers depend only on the number of unitary divisors of $k$, which is $2^{\omega(k)}$. By the quoted result of O. Izhboldin and L. Kurliandchik (see Fedor's and Myerson's comments), for any set of $n$ positive integers {$a_{1}, \dots, a_{n}$} such that $\sum_{i = 1}^{n} \frac{1}{a_{i}} <1$, we have \begin{eqnarray} \sum_{i = 1}^{n} \frac{1}{a_{i}} \leq \sum_{i = 1}^{n} \frac{1}{d_{i}} = \frac{d_{n+1} - 2}{d_{n+1} -1} < 1, \end{eqnarray} where $d_{i}$ is the $i^{\text{th}}$-Euler number, which satisfies the quadratic recurrence $d_{i} = d_{1} \cdots d_{i-1} + 1$ with $d_{1} = 2$. The first few terms of the sequence are 2, 3, 7, 43, 1807.... (A000058). It is relatively straightforward to show that the aforementioned recurrence is equivalent to $d_{i} = d_{i-1}(d_{i-1}-1) + 1$, and it is known from the recurrence that $d_{i} = \lfloor \theta^{2^{i}} + \frac{1}{2} \rfloor$, where $\theta \approx 1.2640$.... Thus, \begin{eqnarray} \sum_{p \mid k} \frac{1}{p + 1} \leq \sum_{i = 1}^{\omega(k)} \frac{1}{d_{i}} = \frac{d_{\omega(k) + 1} - 2}{d_{\omega(k)+1} - 1} = \frac{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 2}{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 1}. \end{eqnarray} One can also show that $d_{i} - 1 = \lfloor \vartheta^{2^{i-1}} - \tfrac{1}{2} \rfloor$, where where $\vartheta \approx 1.5979$...., so we have \begin{eqnarray} 1 - \sum_{p \mid k} \frac{1}{p + 1} \geq 1 - \frac{d_{\omega(k)+1} - 2}{d_{\omega(k)+1} - 1} = \frac{1}{d_{\omega(k)+1} - 1} = \lfloor \vartheta^{2^{\omega(k)}} - \tfrac{1}{2} \rfloor^{-1} > 0. \end{eqnarray} Remark E. Deutsche points out that the sequence {$d_{i} - 1$} (A007018) counts the number of ordered rooted trees with out-going degree up to 2 with all leaves at top level.

How to decide this function takes integer values? I am working a combinatorial question. I need to decide when the following function $$\frac{m(7m^2 - 22m +7)}{ \sqrt{ (m^2- 2m + 9)(9m^2 - 2m +1) } } $$takes integer values with following assumption: 1) m takes all positive integer values; 2) $(m^2- 2m + 9)(9m^2 - 2m +1)$ is a perfect square (= $n^2$ for some integer $n$). Limited computation for $m < 20000$ shows $m = 1, 5$. Many thanks, Jianmin

An elementary approach is enough here. If the given ratio is an integer, then any prime $p$ which divides $m^2 -2m+9$ divides either $m$ or $7m^2-22m+7$. If $p$ divides $m$ then clearly $p =3$. If $p$ divides $7m^2-22m+7$ then $p$ divides $$[7m^2-14m+63 - (7m^2-22m+7)] = 8m+56. $$ Hence either $p=2$ or $p$ divides $m+7.$ In the second case, $p$ divides $$(m^2-2m+9) -(m+7) = m^2-3m+2 = (m-1)(m-2).$$ Now $p$ divides $8$ if $m \equiv 1$ (mod p) and $p$ divides $9$ if $m \equiv 2$ (mod p). Hence $(m-1)^2 +8$ has the form $2^a 3^b$. The greatest common divisor of $7m^2 -22m+7$ and $9m^2-2m+1$ is also very restricted. If $p$ is an odd prime which divides both $7m^2-22m +7$ and $9m^2-2m+1$, then (subtracting the first from the second and dividing by $2$), $p$ divides $m^2 +10m-3$. Then (subtracting the first equation from $7$ times the new equation), $p$ divides $92m-28 = 4(23m-7)$ and hence $p$ divides $7m^2 +m = 7m^2-22m +7 + (23m-7).$ Since $p$ does not divide $m$, $p$ divides $7m+1$, and thus divides $(23m-7)-3(7m+1) = 2(m-5).$ But now $p$ also divides $7(m-5) + 36$, so $p$ divides $36$ and $p=3$. Hence we see that $m^2 -2m+9$ has the form $2^a 3^b$ and $9m^2 - 2m+1$ has the form $2^e 3^f$ for integers $a,b,e,f$. But now notice that the greatest common divisor $d$ of $m^2 -2m+9$ and $9m^2 - 2m+1$ divides $16m-80$, which is the latter equation minus 9 times the former. Also, $16$ does not divide $d$, since` $(m-1)^2 +8$ is not divisible by 16 (and neither is $8m^2 +(m-1)^2$). If $h$ is the odd part of $d$, then $m \equiv 5$ (mod h), and $h$ divides $24$. Hence $d$ divides $24$. If $m$ is divisible by $3$, then $9m^2 - 2m+1$ must be a power of $2$, but is not divisible by $16$, yielding $m=0$ under current assumptions. Hence we may suppose that $m$ is not divisible by $3$. If $9$ divides $m^2-2m +9$, then $ m \equiv 2 $ (mod 9) so $9m^2 - 2m+1 \in \{3,12,24\}$, a contradiction. Hence $m^2-2m +9 = 24$ and $m= 5$ or $m^2-2m +9 = 8$ and $m= 1$ are the only remaining possibilities (for positive $m$ not divisible by 3, as we are currently assuming). Thus $m=1 $ and $m=5$ are the only positive integers which yield an integral value. (Later remark: Note that the additional solution $m=-1$ mentioned in the comment of Noam Elkies corresponds to the case $9m^2 - 2m+1 = 12 = m^2 -2m +9$, which does not occur for positive $m$).

Building a genus-$n$ torus from cubes I wonder if this has been studied: What is the fewest number of unit cubes from which one can build an $n$-toroid? The cubes must be glued face-to-face, and the boundary of the resulting object should be topologically equivalent to an $n$-torus, by which I mean a genus-$n$ handlebody in $\mathbb{R}^3$ (as per Kevin Walker's terminological correction). For example, 8 cubes are needed to form a 1-toroid:            And it seems that 13 cubes are needed for a 2-toroid:            I know how intricate is the analogous question for minimizing the number of triangles from which one can build a torus (cf. Császár's Torus), but I am hoping that my much easier question has an answer for arbitrary $n$. Thanks for ideas and/or pointers! Addendum. Here is Steve Huntsman's 20-cube candidate for genus-5:           

Theorem. Let $c(g)$ be the minimum number of cubes such that the boundary of some configuration of $c(g)$ cubes is a genus $g$ surface. Then $c(g)/g \to 2$ as $g \to \infty$. Proof. We write $\chi(X)$ for the compactly supported Euler characteristic of $X$, i.e., $\chi(X) = \sum (-1)^i \dim H^i_c(X, \mathbb{Q})$. Note this is not a homotopy invariant: the compactly supported Euler characteristic of $\mathbb{R}^n$ is $(-1)^n$. It does however have the property that for reasonable finite disjoint union decompositions, $X = \coprod X_i$, we have $\chi(X) = \sum \chi(X_i)$. Let $H_g$ be a closed genus $g$ handle-body. Then $$\chi(H_g) = 1-g.$$ On the other hand, let $K$ be a configuration of cubes. We write $K^0, K^1, K^2, K^3$ for the sets of vertices, edges, squares, and cubes, respectively, and $k^0, k^1, k^2, k^3$ for their cardinalities. Then $$\chi(K) = k^0 - k^1 + k^2 - k^3$$ We will count $k^0, k^1, k^2, k^3$ by looking at the interior of the $2 \times 2$ cube around each vertex. That is, abutting some vertex $v$, there is $1$ vertex, $6$ edges, $12$ faces, and $8$ cubes. More to the point, each $i$-dimensional face abuts $2^i$ vertices. We write $K_v$ to mean the configuration localized at $v$; we write $K^i_v$ for the set of $i$-dimensional faces which abut the vertex $v$, and $k^i_v$ for its cardinality. Thus: $$\chi(K) = \sum_{v \in \mathbb{Z}^3} \sum_{i=0}^3 (-2)^{-i} \cdot k^i_v $$ so we estimate $$\frac{\chi(K)}{k^3} = \frac{\sum_{v \in \mathbb{Z}^3} \sum_{i=0}^3 (-2)^{-i} \cdot k^i_v }{\sum_{v \in \mathbb{Z}^3} 2^{-3} \cdot k^3_v } \ge \min_{v \in \mathbb{Z}^3} \sum_i \frac{(-2)^{-i} \cdot k^i_v }{2^{-3} \cdot k^3_v }$$ The above inequality comes from the following fact: a (weighted) average is greater than the minimum term being averaged. Thus for any $a_i, b_i$ with $b_i > 0$, we have $$\frac{\sum_i a_i}{\sum_i b_i} = \sum_i \frac{a_i}{b_i} \cdot \frac{b_i}{\sum b_i} \ge \min_i \frac{a_i}{b_i}$$ Now let us analyze the possibilities for the right hand quantity $$\tau(K_v) := \frac{8}{k_v^3} \cdot \left(k_v^0 - \frac{k_v^1}{2} + \frac{k_v^2}{4} - \frac{k_v^3}{8} \right) $$ In fact, in the $2\times 2$ cube around a vertex, there are, up to symmetry, only 9 possible configurations of cubes whose boundary is (locally at that vertex) topologically a manifold: one each for every number of boxes other than 4, and for 4 boxes, the square, and the tripod configuration where, if say $v = (0,0,0)$, the cubes are the ones with most negative coordinates $(-1,-1,-1)$, $(-1, -1, 0)$, $(-1, 0, -1)$, and $(0, -1, -1)$. Note that the neighborhood of every point in the interior of a very porous solid is a tripod configuration. It remains to compute in each of these cases the above quantity. For example, in the configuration $C_1$ when there is one cube, there is one vertex abutting the central one, three edges, three faces, and one cube. Thus this contributes $$ \tau(C_1) = \frac{8}{1} \cdot \left(1 - \frac{3}{2} + \frac{3}{4} - \frac{1}{8} \right) = 1$$ We tabulate the remaining cases: $$ \tau(C_2) = \frac{8}{2} \cdot \left(1 - \frac{4}{2} + \frac{5}{4} - \frac{2}{8} \right) = 0$$ $$ \tau(C_3) = \frac{8}{3} \cdot \left(1 - \frac{5}{2} + \frac{7}{4} - \frac{3}{8} \right) = -\frac{1}{3}$$ $$ \tau(C_4) = \frac{8}{4} \cdot \left(1 - \frac{5}{2} + \frac{8}{4} - \frac{4}{8} \right) = 0$$ $$ \tau(C_4') = \frac{8}{4} \cdot \left(1 - \frac{6}{2} + \frac{9}{4} - \frac{4}{8} \right) = -\frac{1}{2}$$ $$ \tau(C_5) = \frac{8}{5} \cdot \left(1 - \frac{6}{2} + \frac{10}{4} - \frac{5}{8} \right) = -\frac{1}{5}$$ $$ \tau(C_6) = \frac{8}{6} \cdot \left(1 - \frac{6}{2} + \frac{11}{4} - \frac{6}{8} \right) = 0$$ $$ \tau(C_7) = \frac{8}{7} \cdot \left(1 - \frac{6}{2} + \frac{12}{4} - \frac{7}{8} \right) = \frac{1}{7}$$ $$ \tau(C_8) = \frac{8}{8} \cdot \left(1 - \frac{6}{2} + \frac{12}{4} - \frac{8}{8} \right) = 0$$ Here, $C_4'$ is the tripod configuration. We conclude that $$\frac{1-g(K)}{\# K^3} = \frac{\chi(K)}{\# K^3} \ge -\frac{1}{2}$$ hence $\# K^3 \ge 2g(K) - 2$. On the other hand, for any family of configurations in which the fraction of $v$ with $K_v \sim C_4'$ -- i.e., the probability that (the neighborhood of a given cube is a very porous solid) -- tends to 1, the estimates above are sharp and $\# K^3 /g(K) \to 2$. (An explicit calculation for the very porous cube appears in the comments above.) $\blacksquare$

When is the degree of this number 3? I am helping a friend of mine, that works in history of mathematics. She is studying the story of the solution of the cubic equation by Cardano. Sometimes she asks me some mathematical questions, that are very hard to motivate from a modern point of view, but that were interesting to Cardano. So please do not ask for motivations. The question is the following. Let $a$, $b$ be rational numbers, with $b$ not a square. Consider the number $$ t=\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}-\sqrt{a^2-b} $$ Under what conditions on $a$ and $b$ is the degree (over $\mathbb{Q}$) of $t$ equal to $3$? A sufficient condition can be found as follows. Let $P(x) = x^3+\alpha_2 x^2 + \alpha_1 x + \alpha_0$ be a rational polynomial. The general expression of the roots of $P$ is $$ \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \frac{\alpha_2}{3}, $$ where $$ q = \frac{2\alpha_2^3 - 9\alpha_2\alpha_1 + 27\alpha_0}{27} $$ and $$ p = \frac{3\alpha_1 - \alpha_2^2}{3}, $$ see here. So we can take $a = -\frac{q}{2}$ and $b = \frac{q^2}{4} + \frac{p^3}{27}$ and we need to force $\sqrt{a^2 -b} = \frac{\alpha_2}{3}$. This boils down to $\alpha_1 = \frac{\alpha_2^2 - 3 \sqrt[3]{3\alpha_2^2}}{3}$. We find that one of the solutions of $$ x^3 + \alpha_2 x^2+ \frac{\alpha_2^2 - 3 \sqrt[3]{3\alpha_2^2}}{3}x+\alpha_0=0 $$ has the required form (of course we need to assume that $\alpha_2$ is such that $\sqrt[3]{3 \alpha_2^2}$ is rational). In this case $a$ and $b$ are given by the above expressions. I suspect that if $t$ has degree $3$, then its minimal polynomial must be of this form and that $a$ and $b$ are as above, but I am not able to prove it. Note that the condition that $t$ has degree $3$ implies that it can be written as the sum of two cubic root and a rational number (because of the formula), but it is not completely clear that this way of writing $t$ is unique.

$$t:= \sqrt[3]{d(1 + x^6) + \sqrt{d^2 (1 - x^6)^2}} + \sqrt[3]{d(1 + x^6) - \sqrt{d^2 (1 - x^6)^2}} + \sqrt{d^2(1 +x^6)^2 - d^2(1 - x^6)^2}$$ $$= \sqrt[3]{2 d} + \sqrt[3]{2 d x^6} + \sqrt{4 d^2 x^6}$$ $$ = (1 + x^3) \sqrt[3]{2d} + 2 d x^3;$$ $$\frac{p^3}{27} = \left(\frac{q^2}{4} + \frac{p^3}{27}\right) - \left(\frac{q}{2}\right)^2 =^{?} b - a^2 = d^2 (1 - x^6)^2 - d^2(1 + x^6)^2 = - 4d^2 x^6 \Rightarrow 2d \in \mathbf{Q}^3$$

Is $\lceil \frac{n}{\sqrt{3}} \rceil > \frac{n^2}{\sqrt{3n^2-5}}$ for all $n > 1$? An equivalent inequality for integers follows: $$(3n^2-5)\left\lceil n/\sqrt{3} \right\rceil^2 > n^4.$$ This has been checked for n = 2 to 60000. Perhaps there is some connection to the convergents to $\sqrt{3}$. $\lceil \frac{n}{\sqrt{3}} \rceil > \frac{n^2}{\sqrt{3n^2-5}}$

A tight result is $$\lceil \frac{n}{\sqrt{3}}\rceil \ge \frac{n^2}{\sqrt{3n^2-6+\frac{12}{n^2+2}}}$$ This means that the desired inequality holds for $n \ge 4.$ The cases $n=2,3$ can be checked separately. As noted elsewhere, for $m=\lceil \frac{n}{\sqrt{3}}\rceil$ we have $n^2 \le 3m^2-2$ with equality exactly for the cases $n=1,5,19,71,265,989,\cdots$ A001834 with $\frac{n}{m} \lt \sqrt{3}$ a convergent to that square root. A bit of manipulation then gives the result above and the information that equality holds for exactly those values of $n.$ Details: Change the desired inequality to $$m^2 \ge \frac{n^4}{3n^2-k}$$ with $k$ to be determined. Since $n^2=3m^2-j$ with $j \ge 2,$ we have $m^2=\frac{n^2+j}{3}.$ This yields $$k \le \frac{3jn^2}{n^2+j} =3j-\frac{3j^2}{n^2+j}.$$ Since we do sometime have $j=2,$ The bound $k \le 6-\frac{12}{n^2+2}$ is sometime an equality. It is not hard to show that $\frac{3jn^2}{n^2+j}$ increases when $j \ge 2$ does.

This inequality why can't solve it by now (Only four variables inequality)? I asked a question at Math.SE last year and later offered a bounty for it, only johannesvalks give Part of the answer; A few months ago, I asked the author(Pham kim Hung) in Facebook, he said that now there is no proof by hand.and use of software verification is correct,and I try it sometimes,and not succeed.Later asked a lot of people (such on AOPS 1,AOPS 2) have no proof interesting inequality: Let $a,b,c,d>0$, show that $$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}$$ In fact,we have $$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \underbrace{\sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}\ge \sqrt{\dfrac{a^2+b^2+c^2+d^2}{4}}}_{\text{Generalized mean}}$$ Now we only prove this not stronger inequality: $$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt{\dfrac{a^2+b^2+c^2+d^2}{4}}$$ Proof:By Holder inequality we have $$\left(\sum_{cyc}\dfrac{a^2}{b}\right)^2(a^2b^2+b^2c^2+c^2d^2+d^2a^2)\ge (a^2+b^2+c^2+d^2)^3$$ and Note $$a^2b^2+b^2c^2+c^2d^2+d^2a^2=(a^2+c^2)(b^2+d^2)\le\dfrac{(a^2+b^2+c^2+d^2)^2}{4}$$ Proof 2:(I hope following methods(creat is Mine) will usefull to solve my OP inequality,So I post it): \begin{align*}&\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^2-4(a^2+b^2+c^2+d^2)\\ &=\sum_{cyc}\dfrac{3a^4b^2d+5a^4c^3+24a^3cd^3+3a^2b^3c^2+10ab^3d^3+15bcd^5-60a^2bcd^3}{15a^2bcd}\\ &\ge 0 \end{align*} NoW I use computer $$\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^4-64(a^2+b^2+c^2+d^2)=\dfrac{a^{12}c^4d^4+4a^{10}b^3c^3d^4+4a^{10}bc^6d^3+4a^9bc^4d^6 +6a^8b^6c^2d^4+12a^8b^4c^5d^3+64a^8b^4c^4d^4+6a^8b^2c^8d^2+12a^7b^4c^3d^6+12a^7b^2c^6d^5+4a^6b^9cd^4+\cdots+4ab^4c^6d^9+b^4c^4d^{12}}{a^4b^4c^4d^4}$$

As to why the question is hard - one could reformulate it as $$\frac{x+y+z+t}{4}\stackrel{?}{\ge} \sqrt[4]{\frac{(x^8 y^4 z^2 t)^{4/15}+(y^8 z^4 t^2 x)^{4/15}+(z^8 t^4 x^2 y)^{4/15}+(t^8 x^4 y^2 z)^{4/15}}{4}}$$ where $x=a^2/b$, $y=b^2/c$, $z=c^2/d$ and $t=d^2/a$. Then there is a tug of war between $$\frac{x+y+z+t}{4}\le \sqrt[4]{\frac{x^4+y^4+z^4+t^4}{4}} $$ and $$\frac{x+y+z+t}{4}\ge \frac{(x^8 y^4 z^2 t)^{1/15}+(y^8 z^4 t^2 x)^{1/15}+(z^8 t^4 x^2 y)^{1/15}+(t^8 x^4 y^2 z)^{1/15}}{4}$$ with the latter apparently winning.

Infinitely many $k$ such that $[a_k,a_{k+1}]>ck^2$ Let $a_n\in \mathbf{N}$ be an infinite sequence such that $\forall i\neq j, a_i\neq a_j$. I have the following theorem: For $0<c<\frac{3}{2}$, there are infinitely many $k$ for which $[a_k,a_{k+1}]>ck$, where $[\cdots]$ denotes least common multiple. Idea of proof: By contradiction. Suppose there is an $M$ such that for all $k>M$, $[a_k,a_{k+1}]\leq ck$. On the one hand, $\displaystyle \sum_{k=1}^{n}\frac{1}{[a_k,a_{k+1}]}\geq \sum_{k=M+1}^{n}\frac{1}{[a_k,a_{k+1}]}\geq \sum_{k=M+1}^{n}\frac{1}{ck}$ On the other hand, $\displaystyle \frac{1}{[a_k,a_{k+1}]}$ $\displaystyle =\frac{(a_k,a_{k+1})}{a_ka_{k+1}}$ $\displaystyle =(a_k,a_{k+1})\frac{1}{a_k+a_{k+1}}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$ $\displaystyle =\frac{1}{\frac{a_k}{(a_k,a_{k+1})}+\frac{a_{k+1}}{(a_k,a_{k+1})}}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$ $\displaystyle \leq \frac{1}{3}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$ $\displaystyle \sum_{k=1}^{n}\frac{1}{[a_k,a_{k+1}]}\leq \frac{1}{3}\sum_{k=1}^{n}(\frac{1}{a_k}+\frac{1}{a_{k+1}})\leq \frac{2}{3}\sum_{k=1}^{n}\frac{1}{k}$ And I make the following conjecture: $\exists c>0$, there are infinitely many $k$ for which $[a_k,a_{k+1}]>ck^2$. But I cannot prove or disprove it.

In fact there are sequences $\{a_k\}$ of pairwise distinct positive integers such that $[a_k, a_{k+1}] \ll k^{1+\epsilon}$ for all $\epsilon > 0$. We first exhibit a sequence with $[a_k, a_{k+1}] \ll k^{3/2} \log^3 k$ for all $k>1$, which already disproves the conjecture that $[a_k, a_{k+1}] \gg k^2$ infinitely often. The sequence will consist of all numbers of the form $p_m p_n$ with $m$ odd and $n$ even, listed in the following order: $$ \begin{array}{cccccccc} 17\cdot 3 & \rightarrow & 17\cdot 7 & \rightarrow & 17\cdot 13 & \rightarrow & 17\cdot 19 & \cr \uparrow & & & & & & \downarrow & \cr 11\cdot 3 & \leftarrow & 11\cdot 7 & \leftarrow & 11\cdot 13 & & 11\cdot 19 & \cr & & & & \uparrow & & \downarrow & \cr 5\cdot 3 & \rightarrow & 5\cdot 7 & & 5\cdot 13 & & 5\cdot 19 & \cr \uparrow & & \downarrow & & \uparrow & & \downarrow & \cr 2\cdot 3 & & 2\cdot 7 & \rightarrow & 2\cdot 13 & & 2\cdot 19 & \rightarrow \end{array} $$ Then each $a_k$ is $p_m p_n$ with $m,n \ll \sqrt k$, so $p_m, p_n \ll \sqrt k \log k$; and each $[a_k,a_{k+1}]$ is the product of three such primes, so $\ll k^{3/2} \log^3 k$, as claimed. Likewise, for each $M>1$ we can use products of $M$ primes to obtain a sequence $\{a_k\}$ of pairwise distinct positive integers such that $[a_k, a_{k+1}] \ll k^{(M+1)/M} \log^M k \ll k^{1+\epsilon}$ for all $\epsilon > 1/M$. Finally, by concatenating ever-longer initial segments of the sequences for $M=2$, $M=3$, $M=4$, etc. we construct a sequence $\{a_k\}$ of pairwise distinct positive integers such that $[a_k, a_{k+1}] \ll k^{1+\epsilon}$ for all $\epsilon > 0$.

odd length Chevalley relations (in rank two) The unipotent radicals $\text{N}$ of the Borel subgroups of the complex algebraic groups of type $A_2$, $B_2$, and $G_2$ can each be abstractly presented using two one-parameter subgroups $x_1, x_2: \Bbb{C} \longrightarrow \text{N}$ subject to the following defining relations, namely: \begin{equation} \begin{array}{llc} x_1(a)\, x_2(b) \, x_1(c) &= \ x_2 \Bigg( \displaystyle {bc \over {a+c}} \Bigg) \, x_1 \Big(a + b\Big) \, x_2 \Bigg( {ab \over {a+c}}\Bigg) &\text{for type $A_2$} \\ \\ x_1(a) \, x_2(b) \, x_1(c) \, x_2(d) &= \ x_2 \Bigg(\displaystyle {bc^2d \over {\pi_2}} \Bigg) \, x_1 \Bigg( {\pi_2 \over {\pi_1}} \Bigg) \, x_2 \Bigg( {\pi_1^2 \over \pi_2} \Bigg) \, x_1 \Bigg( {abc \over \pi_1} \Bigg)&\text{for type $B_2$} \\ \\ &\text{where} \ \pi_1 = ab + \big(a +c \big)d & \\ &\text{and} \ \pi_2 = a^2b + \big(a +c \big)^2d & \end{array} \end{equation} For $G_2$ a six term relation is valid which converts a factorisation $x_1(a)x_2(b)x_1(c)x_2(d)x_1(e)x_2(f)$ into a product of the form $x_2(A)x_1(B)x_2(C)x_1(D)x_2(E)x_1(F)$ where $A$, $B$, $C$, $D$, $E$, and $F$ are subtraction-free rational expressions in the complex parameters $a$, $b$, $c$, $d$, $e$, and $f$ --- see the paper "Total positivity in Schubert varieties" by Arkady Berenstein and Andrei Zelevinsky. Using a truncated version of the $\text{SL}_2 \Big( \Bbb{C} \Big)$ loop-group, namely $\text{SL}_2 \Big( \mathcal{L_d} \Big)$ where $\mathcal{L_d}$ is the Auslander ring $\Bbb{C}\big[ t \big] \Big/ \Big( t^d = 0 \Big)$ one can check the existence of similar relations of length $2d$ for $d \geq 1$; the cases of $d=1$ and $d=2$ correspond to types $B_2$ and $G_2$ respectively. Hint: interpret the abstract one-parameter groups $x_1$ an $x_2$ as the $\mathcal{L}_d$-valued $2 \times 2$ matrices \begin{equation} x_1(a) = \begin{pmatrix} 1 & a \\ 0 & 1\end{pmatrix} \qquad x_2(a) = \begin{pmatrix} 1 & 0 \\ at & 1 \end{pmatrix} \end{equation} These relations correspond to coxeter relations of length $2d$ within the group of dihedral symmetries of a regular $2d$-gon. Question: Is there a length five relation of this kind ? If so, is there a nice representation of the corresponding nilpotent group ? best, A. Leverkühn

Lieber Señor Leverkühn, Instead of using the Auslander ring $\Bbb{C}\big[ t \big] \Big/ \Big( t^d = 0 \Big)$ use a two variable version $\mathcal{L}_{2+3} = \Bbb{C}\big[s,t \big]\Big/ \Big(s^3t^2 = s^2t^3 = 0 \Big)$ and form the truncated "double" loop group $\text{SL}_2 \Big( \mathcal{L}_{2+3} \Big)$. As in your construction let's define two one-parameter subgroups, namely \begin{equation} x_1(a) := \begin{pmatrix} 1 & as \\ 0 & 1 \end{pmatrix}\quad \text{and} \quad x_2(a) := \begin{pmatrix} 1 & 0 \\ at & 1 \end{pmatrix}. \end{equation} Now we compute the two relevant length five products: \begin{equation} \begin{pmatrix} 1 & as \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ bt & 1 \end{pmatrix}\begin{pmatrix} 1 & cs \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ dt & 1 \end{pmatrix}\begin{pmatrix} 1 & es \\ 0 & 1 \end{pmatrix} \ = \end{equation} \begin{equation} \begin{pmatrix} 1 + \big(ab + ad + cd \big)st + \big(abcd \big)s^2t^2 & \big(a + c + e \big)s + \big(ade + abe + cde + abc\big)s^2t \\ \big(b + d \big)t + \big(bcd \big)st^2 & 1 + \big( bc + be + de \big)st + \big( bcde \big)s^2t^2 \end{pmatrix} \end{equation} \begin{equation} \cdots \ \ \text{and} \ \ \cdots \end{equation} \begin{equation} \begin{pmatrix} 1 & 0 \\ At & 1 \end{pmatrix}\begin{pmatrix} 1 & Bs \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ Ct & 1 \end{pmatrix}\begin{pmatrix} 1 & Ds \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ Et & 1 \end{pmatrix}\ = \end{equation} \begin{equation} \begin{matrix} 1 + \big( BC + BE + DE \big)st + \big( BCDE \big)s^2t^2 & \big(B + D \big)s + \big(BCD \big)s^2t \\ \big(A + C + E \big)t + \big(ADE + ABE + CDE + ABC \big)st^2 & 1 + \big(AB + AD + CD \big)st + \big( ABCD \big)s^2t^2 \end{matrix} \end{equation} Now solve for $A$, $B$, $C$, $D$, and $E$ to obtain the conjectured "dihedral" relation. yours always, Ines. Ok, Here's a possible remedy which may avoid the inconsistency that you've discovered. Instead of $\mathcal{L}_{2+3}$ consider \begin{equation} \mathcal{L}_{4} \ = \ \Bbb{C}\big[t \big]\Big/ \Big(t^4 = 0 \Big) \end{equation} together with the one-parameter subgroups \begin{equation} x_1(a) := \begin{pmatrix} 1 & at \\ 0 & 1 \end{pmatrix}\quad \text{and} \quad x_2(a) := \begin{pmatrix} 1 & 0 \\ at & 1 \end{pmatrix}. \end{equation} The length five products are then: \begin{equation} \begin{pmatrix} 1 & at \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ bt & 1 \end{pmatrix}\begin{pmatrix} 1 & ct \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ dt & 1 \end{pmatrix}\begin{pmatrix} 1 & et \\ 0 & 1 \end{pmatrix} \ = \end{equation} \begin{equation} \begin{pmatrix} 1 + \big(ab + ad + cd \big)t^2 & \big(a + c + e \big)t + \big(ade + abe + cde + abc\big)t^3 \\ \big(b + d \big)t + \big(bcd \big)t^3 & 1 + \big( bc + be + de \big)t^2 \end{pmatrix} \end{equation} \begin{equation} ----- \ \text{and} \ ----- \end{equation} \begin{equation} \begin{pmatrix} 1 & 0 \\ At & 1 \end{pmatrix}\begin{pmatrix} 1 & Bt \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ Ct & 1 \end{pmatrix}\begin{pmatrix} 1 & Dt \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ Et & 1 \end{pmatrix}\ = \end{equation} \begin{equation} \begin{pmatrix} 1 + \big( BC + BE + DE \big)t^2 & \big(B + D \big)t + \big(BCD \big)t^3 \\ \big(A + C + E \big)t + \big(ADE + ABE + CDE + ABC \big)t^3 & 1 + \big(AB + AD + CD \big)t^2 \end{pmatrix} \end{equation} yours, Ines

Prove this conjecture inequality This following problem is from my Conjecture many years ago, Question : Let $a,b>0,n\in N^{+},n\ge 3$,such $$a^n+b^n+(2n+2)(ab)^n\le 2n$$ Conjecture: then $a+b\le 2$ or $a+b>2.a>0.b>0,n\ge 3$,then we prove $$a^n+b^n+(2n+2)a^nb^n-2n>0$$ or it suffuce to prove $$a^n+b^n+(2n+2)a^nb^n-2n>0\rm{when}~ a+b=2$$ This inequality is from this conjecture $$\sqrt[n]{\dfrac{x}{1+y}}+\sqrt[n]{\dfrac{y}{1+x}}\le 2,\rm{where}~ x+y\le 2n,n\ge 3,n\in N^{+}$$ it is not hard to prove this case $n=3,4$,but for $n\ge 5$, use this $$\sqrt[n]{\dfrac{x}{1+y}}=a,\sqrt[n]{\dfrac{y}{1+x}}=b$$ then $$x=\dfrac{a^n(b^n+1)}{1-(ab)^n},y=\dfrac{b^n(a^n+1)}{1-(ab)^n}$$take $x+y\le 2n\Longrightarrow a^n+b^n+(2n+2)(ab)^n\le 2n$

Take $a=1+x,b=1-x$, $|x|\leqslant 1$. We have to prove $(1+x)^n+(1-x)^n+(2n+2)(1-x^2)^n\geqslant 2n$. By Bernoulli inequality we have $(1-x^2)^n\geqslant 1-nx^2$ and by binomial expansion $(1+x)^n+(1-x)^n=2+n(n-1)x^2+2\binom{n}4 x^4+\dots$ So, it suffices to prove that $$2+n(n-1)x^2+2\binom{n}4 x^4+2n+2-n(2n+2)x^2\geqslant 2n,$$ this is quadratic in $x^2$, discriminant equals $(n^2+3n)^2-32\binom{n}4$, this is negative for large enough $n$ (about $n\geqslant 42$, I think). Well, this is not quite satisfactory, but at least it works for all but finitely many $n$. Next trick is generalized Bernoulli $(1-x^2)^n\geqslant 1-nx^2+\binom{n}2 x^4-\binom{n}3x^6$ (this follows from your favourite remainder term in Taylor's expansion for $(1-t)^n$). Assume that $n$ satisfies $2\binom{n}6\geqslant (2n+2)\binom{n}3$, this is true for $n\geqslant 17$. Then it suffices to prove $$4-(n^2+3n)x^2+\left(2\binom{n}4+(2n+2)\binom{n}2\right)x^4\geqslant 0.$$ For $n\geqslant 17$ this is true, so we have already proved your inequality for $n\geqslant 17$. Well, for $n=14,15,16$ the term with $x^8$ allows to beat negative term with $x^6$ (that is, after we subtract $(2-\frac{n^2+3n}4x)^2$ from our expression and divide by $x^4$, we get a positive quadratic trinomial in $x^2$.) Computer plots confirm that for $3\leqslant n\leqslant 13$ conjecture also holds, but this is probably not what you need.

Another question about the golden ratio and other numbers This is an extension of "A question about the golden ratio and other numbers." Given $r$, suppose that $$c_0+c_1x+c_2x^2+ \cdots = \frac{1} {\lfloor{r}\rfloor+\lfloor{2r}\rfloor x+\lfloor{3r}\rfloor x^2+ \cdots}.$$ Let $L(r) = \lim_{i\to\infty} \frac{c_{i+1}}{c_i}$. Can someone prove that the limit $L(\phi)$ exists, where $\phi = \frac{1+ \sqrt{5}}{2}$? It appears that $$L(\phi) = -1.688924110769165206686359\ldots$$ $$(c_0,c_1,c_2,\ldots) = (1,−3,5,−9,17,−30,52,−90,154,−262,446,−758,1285,\ldots).$$ Also, it appears that $L(F_{k+1}/F_{k})$ exists, for $k \ge 5$, where $F_k$ denotes the $k$th Fibonacci number; e.g., $$\begin{eqnarray} L(8/5) &=&-1.69562076\ldots \newline L(13/8) &=& -1.76404686\ldots \newline L(21/13) &=& -1.68892398\ldots \newline L(34/21) &=& -1.68880982\ldots \end{eqnarray}$$

Some observations, but not a solution yet. Let $t_k=\frac{F_{k+1}}{F_k}$ where $F_k$ is the $k^{th}$-Fibonacci number. The convention here for $F_k$ is that $F_3=2, F_4=3, F_5=5, \dots$. Denote the RHS in the above series by $$\Psi_k(x)=\left(\sum_{n=1}^{\infty}\lfloor nt_k\rfloor x^{n-1}\right)^{-1}.$$ Then, it seems that $$\Psi_k(x)=\frac{(1-x)(1-x^{F_k})}{P_k(x)}$$ for some polynomial $P_k(x)$ with the following rather curious properties: (1) it has degree $F_k-1$; (2) its coefficients are either $1$ or $2$ (although not clear which is which); (3) $P_k(1)=F_k$; (4) $L(t_k)=$ the smallest real root of $P_k(x)$; (5) there might be a way to relate $(1-x)P_k(x)$ to other $(1-x)P_{\ell}(x)$ for $\ell<k$, recursively. A few examples: \begin{align} \Psi_4(x)&=\frac{(1-x)(1-x^3)}{1+2x+2x^3} \\ \Psi_5(x)&=\frac{(1-x)(1-x^5)}{1+2x+x^2+2x^3+2x^4} \\ \Psi_6(x)&=\frac{(1-x)(1-x^8)}{1+2x+x^2+2x^3+2x^4+x^5+2x^6+2x^7} \\ \Psi_7(x)&=\frac{(1-x)(1-x^{12})}{1+2x+x^2+2x^3+2x^4+x^5+2x^6+x^7+2x^8+2x^9+x^{10}+2x^{11}+2x^{12}}. \end{align}

The average of reciprocal binomials This question is motivated by the MO problem here. Perhaps it is not that difficult. Question. Here is an cute formula. $$\frac1n\sum_{k=0}^{n-1}\frac1{\binom{n-1}k}=\sum_{k=1}^n\frac1{k2^{n-k}}.$$ I've one justification along the lines of Wilf-Zeilberger (see below). Can you provide an alternative proof? Or, any reference? The claim amounts to $a_n=b_n$ where $$a_n:=\sum_{k=0}^{n-1}\frac{2^n}{n\binom{n-1}k} \qquad \text{and} \qquad b_n:=\sum_{k=1}^n\frac{2^k}k.$$ Define $F(n,k):=\frac{2^n}{n\binom{n-1}k}$ and $\,G(n,k)=-\frac{2^n}{(n+1)\binom{n}k}$. Then, it is routinely checked that $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k),\tag1$$ for instance by dividing through with $F(n,k)$ and simplifying. Summing (1) over $0\leq k\leq n-1$: \begin{align} \sum_{k=0}^{n-1}F(n+1,k)-\sum_{k=0}^{n-1}F(n,k) &=a_{n+1}-\frac{2^{n+1}}{n+1}-a_n, \\ \sum_{k=0}^{n-1}G(n,k+1)-\sum_{k=0}^{n-1}G(n,k) &=-\sum_{k=1}^n\frac{2^n}{(n+1)\binom{n}k}+\sum_{k=0}^{n-1}\frac{2^n}{(n+1)\binom{n}k}=0. \end{align} Therefore, $a_{n+1}-a_n=\frac{2^{n+1}}{n+1}$. But, it is evident that $b_{n+1}-b_n=\frac{2^{n+1}}{n+1}$. Since $a_1=b_1$, it follows $a_n=b_n$ for all $n\in\mathbb{N}$. ===================================== Here is an alternative approach to Fedor's answer below in his elaboration of Fry's comment. With $\frac1{n+1}\binom{n}k=\int_0^1x^{n-k}(1-x)^kdx$, we get $a_{n+1}=2^{n+1}\int_0^1\sum_{k=0}^nx^{n-k}(1-x)^kdx$. So, \begin{align} 2^{n+1}\int_0^1 dx\sum_{k=0}^nx^{n-k}(1-x)^k &=2^{n+1}\int_0^1x^ndx\sum_{k=0}^n\left(\frac{1-x}x\right)^k \\ &=2^{n+1}\int_0^1x^n\frac{\left(\frac{1-x}x\right)^{n+1}-1}{\frac{1-x}x-1}dx \\ &=\int_0^1\frac{(2-2x)^{n+1}-(2x)^{n+1}}{1-2x}\,dx:=c_{n+1}. \end{align} Let's take successive difference of the newly-minted sequence $c_{n+1}$: \begin{align} c_{n+1}-c_n &=\int_0^1\frac{(2-2x)^{n+1}-(2-2x)^n+(2x)^n-(2x)^{n+1}}{1-2x}\,dx \\ &=\int_0^1\left[(2-2x)^n+(2x)^n\right]dx=2^{n+1}\int_0^1x^ndx=\frac{2^{n+1}}{n+1}. \end{align} But, $b_{n+1}-b_n=\frac{2^{n+1}}{n+1}$ and hence $a_n=b_n$.

Let me elaborate on Fry's suggestion and your forthcoming comment. $$\frac1{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}k=\frac1{2^{n+1}}\int_0^2(1+x+\dots+x^n)dx=\frac1{2^{n+1}}\int_0^2\frac{1-x^{n+1}}{1-x}dx=\\2\int_{0}^1\frac{(1/2)^{n+1}-t^{n+1}}{1-2t}dt=2\int_{0}^1\frac{(1-s)^{n+1}-(1/2)^{n+1}}{1-2s}ds.$$ We used change of variables $x=2t$, $s=1-t$. Now take a half-sum of two last expressions (identifying $t$ and $s$), you get $$ \int_{0}^1\frac{(1-t)^{n+1}-t^{n+1}}{1-2t}dt=\sum_{k=0}^n\int_0^1(1-t)^kt^{n-k}dt=\sum_{k=0}^n\frac1{(n+1)\binom{n}k}. $$