Q:

Smallest power of a prime whose factorizations don't have distinct sums of factors

Problem

Given a prime $p$, find smallest $n$ such that some unordered factorizations of $p^n$ have equal sums of factors.

Unordered factorizations are factorizations where order of factors is irrelevant and they do not include the trivial factor $1$. Notice that $n\gt 1$ for all primes $p$ because prime numbers have only one unordered factorization.

Examples
Prime $p=2$. It is trivial that $n=2$ for $p=2$ because $2+2=2\cdot 2$. That is, unordered factorizations of $2^2$ are $4$ and $2\cdot 2$, and they both have the same sum of factors $4 = 2+2$.
Prime $p=3$. But, $n=2$ is not a solution for $p=3$ because $9\ne 3+3$. Neither is $n=3$ because $27\ne 3+9 \ne 3+3+3$. Neither is $n=4$ because $81\ne 27 + 3\ne 9 + 9\ne 9 + 3 + 3\ne 3 + 3 + 3 + 3$. Eventually, we find that $n=12$ is the smallest one that fits, because then there exist the following duplicate sums of factors:
$$\begin{align}{}
3^{12}&=&27\cdot3^9&=&9^6 &\implies& 27+\sum_{i=1}^{9}3 &=& \sum_{i=1}^{6}9 &=& 54 \\
3^{12}&=&81\cdot9\cdot 3^6&=&27^4 &\implies& 81+9+\sum_{i=1}^{6}3 &=& \sum_{i=1}^{4}27 &=& 108
\end{align}$$
Notice that if $p^{n}$ or any number in general satisfies this property, then all multiples of that number also satisfy it.

Solution?
Prime $p\in\mathbb P$. Let $a(k)$ be the smallest such $n_k$ given the $k$th prime $p_k$. We have:
$$a(k) = 2, 12, 26, 34, 50, 58, 74, 82, \dots$$

Is it possible to find and prove a formula for this sequence?

I noticed the following appears to hold so far: $a(1)=2,a(2)=12,a(k)=4p_k+6,k\ge 3$.
This is because of the following unordered factorizations:
$$\begin{align}
p_k &\quad n &\quad \\
2 &\quad 2 &\quad  (2)(2) &=(2^2) \\
3 &\quad 12 &\quad (3)^9(3^3) &= (3^2)^6 &\quad (3)^6(3^2)(3^4) &= (3^3)^4 \\
5 &\quad 26 &\quad (5^2)^{11} (5^4) &= (5)^5(5^3)^7 \\
7 &\quad 34 &\quad (7)^{15}(7^4) &= (7)^7(7^3)^9 \\
11 &\quad 50 &\quad (11^2)^{23}(11^4) &= (11)^{11}(11^3)^{13}\\
13 &\quad 58 &\quad (13^2)^{27}(13^4) &= (13)^{13}(13^3)^{15}\\
17 &\quad 74 &\quad (17^2)^{35}(17^4) &= (17)^{17}(17^3)^{19}\\
19 &\quad 82 &\quad (19^2)^{39}(19^4) &= (19)^{19}(19^3)^{21}\\
\end{align}$$
Notice that the primes $p_k\ge 5$ follow the following pattern:
$$
(p^2)^{2p+1}(p^4) = (p)^{p}(p^3)^{p+2} \implies (p^2)\cdot(2p+1)+(p^4) = (p)\cdot p+(p^3)\cdot(p+2)
$$

This gives us an upper bound $a(k)\le 4p_k+6$ because the pattern holds for all natural numbers.

The equality was proven computationally for some small primes (as you can see above).
Can we prove that the equality always holds? I.e. can we prove $a(k)\ge 4p_k+6, k\ge 3$ ?

That is, it is left to prove that all unordered factorizations of numbers of the form
$$
p^{4p+5}
$$
have distinct sums of factors for all primes $p\ge 5$.

In other words, we need to prove that $\text{A001055}$$(p^{4p+5})$ $=$ $\text{A069016}$$(p^{4p+5})$.
Or maybe there exists a prime $p$ that is a counter-example? I.e. $p_k : a(k)\lt 4p_k+6$ ?

A:

Note: my solution is quite long and contains lots of cases, so some mistakes are unavoidable. Let me know if there is anything that needs explaining.

Let $n$ be the smallest number such that $p^n$ have two unordered factorizations with equal sum of factors. We will assume that $n \le 4p+5$ and derive a contradiction.
Denote by $A=(p^{a_1})^{n_1}\dots (p^{a_k})^{n_k}$ and $B=(p^{b_1})^{m_1}\dots(p^{b_l})^{m_l}$ the two unordered factorizations, with $a_1 > \dots > a_k$ and $b_1 > \dots > b_l$. Without loss of generality, assume that $A$ has the higher power of $p$, i.e. that $a_1 \ge b_1$.
Observation 1: $\{a_1,\dots,a_k\} \cap \{b_1,\dots,b_l\}=\emptyset$.
This is because if $a_i=b_j$ for some $i,j$, then we can subtract $p^{a_i}$ from both factorizations to obtain two unordered factorizations with equal sum for $p^{n-a_i}$, contradicting minimality of $n$.
Observation 2: $a_1 \le 5$.
This comes from considering the equations $$\begin{equation}\label{eqn1} b_1m_1+\dots+b_lm_l=n \qquad (1)\end{equation}$$ and $$ \begin{equation} m_1p^{b_1}+\dots+m_lp^{b_l}=n_1p^{a_1}+\dots+n_kp^{a_k}\qquad (2)\end{equation}$$ If you solve for the maximum of the LHS of (2) given (1) (and with $m_j \in \mathbb{R}$ instead), then the maximum is attained at $m_1=n/b_1$ and $m_j=0$ for all $j \ge 2$, where the maximum value is $\frac{n}{b_1}p^{b_1}$. On the other hand, the RHS of (2) gives the lower bound of $p^{a_1}$, hence we must have $$ \frac{n}{b_1} p^{b_1} \ge p^{a_1} \iff n \ge b_1 p^{a_1-b_1} .$$ Since $n \le 4p+5 \le p^2$ for $p \ge 5$, we have that $$ a_1-b_1=2, b_1=1 \qquad \text{or} \qquad a_1-b_1=1, b_1 \le 4$$ and in both cases $a_1 \le 5$.
We now consider the various cases $a_1 \in \{2,3,4,5\}$, with $a_1=4$ being the hardest.

If $a_1=2$, then $b_1=1$ and we get two factorizations $(p^2)^{n/2}$ and $p^n$. They do not have equal sum of factors for $p>2$.

If $a_1=3$, then we have the following options:

$a_2=2$, $b_1=1$. We get two factorizations $(p^3)^{n_1}(p^2)^{n_2}$ and $p^n$. They do not have equal sum since $$np \le (4p+5)p < p^3$$ for $p \ge 5$.
$a_2=1$, $b_1=2$. We get two factorizations $(p^3)^{n_1}p^{n_2}$ and $(p^2)^{n/2}$. Since $$\frac{n}{2}p^2 \le \frac{4p+5}{2}p^2 < 3p^3$$ for $p \ge 5$, we have that $n_1 \in \{1,2\}$. Either values don't lead to equal sum.
$b_1=2$, $b_2=1$. We get two factorizations $(p^3)^{n/3}$ and $(p^2)^{m_1}p^{m_2}$. Then $$ m_1 p^2+m_2p \le \frac{n}{2}p^2 < \frac{n}{3}p^3. $$
$b_1=2$. We get $(p^3)^{n/3}$ and $(p^2)^{n/2}$, whose sums are not equal.
$b_1=1$. We get $(p^3)^{n/3}$ and $p^n$, whose sums are not equal.

If $a_1=5$, then $b_1=4$ as we have argued above. By the same line of argument, we can see that $$ m_1p^4+\dots +m_l p^{b_l} < 2p^5 $$ for all choices of $m_1,\dots,m_l$, so $n_1=1$, and $$ m_1p^4+\dots +m_l p^{b_l} \ge p^5 $$ only when $m_1 \ge p$. Therefore, we have that $$ A = (p^5)\cdot(\text{unordered factorization of }p^{n-5}) \quad \text{and} \quad B= (p^4)^p\cdot(\text{unordered factorization of }p^{n-4p}).$$ In particular, this implies that $n \in \{4p, \dots, 4p+5\}$. Some more checking shows that for such $n$, no factorization of $p^{n-5}$ has the same sum of factors as any factorization of $p^{n-4p}$.

If $a_1=4$, then $b_1=3$ from the argument in Observation 2. Using the same argument as in the above case, we conclude that $n_1=1$ and $m_1 \ge p$. Therefore, we have $$ A = (p^4)\cdot(\text{unordered factorization of }p^{n-4}) \quad \text{and} \quad B= (p^3)^p\cdot(\text{unordered factorization of }p^{n-3p}).$$ It remains to consider the following options for $a_2,\dots,a_k$:

$a_2=1$. Then if $A=(p^4)p^{n-4}$ and $B=(p^3)^{m_1}(p^2)^{m_2}$ have equal sum, we must have $$ (n-4)p=(m_1-p)p^3+m_2p^2 \ge \frac{n}{2} p^2,$$ which is false for $p \ge 5$ and $n \le 4p+5$.
$a_2=2$, $a_3=1$. Then $A=(p^4)(p^2)^{n_2}p^{n_3}$ and $B=(p^3)^{n/3}$. If their sums are equal, then $$ (n/3-p)p^3=n_2p^2+n_3p$$ so in particular $p \mid n_3$. Since $n_3 \le n-4 \le 4p+1$ we must have $n_3 \in \{p,2p,3p,4p\}$. Plugging each value of $n_3$ and $n_2=n-4-n_3$ into the equation, we find that none yields an integer value for $n$.
$a_2=2$. Then $A=(p^4)(p^2)^{(n-4)/2}$ and $B=(p^3)^{m_1}p^{m_2}$. If their sum are equal, then $$\frac{n-4}{2}p^2=(m_1-p)p^3+m_2p$$ so $p \mid m_2$, and similar to above we get $m_2 \in \{p,2p,3p\}$. Again, plugging each value of $m_2$ and $m_1=n-m_2$ into the equation, we find that none yields an integer value for $n \le 4p+5$.

