Interesting behavior of $\frac{n}{v_2(n!)+1}$. I've lately noticed some interesting behavior from the values of the function $f(n)=\frac{n}{v_2(n!)+1}$,
Where $v_p(n)$ is the $p$-adic valuation of $n$, and we also know that $v_p(n!)=\sum_{t=1}^\infty \left\lfloor \frac{n}{p^t} \right\rfloor$.
 I'll show the values of $f(n)$ for some values of $n$ below:
$(n,f(n))=(5,\frac{5}{4}),(10,\frac{10}{9}),(20,\frac{20}{19}),(30,\frac{10}{9}),(40,\frac{40}{39}),(50,\frac{25}{24}),(60,\frac{20}{19}),(100,\frac{50}{49}),...(900,\frac{300}{299}),(1000,\frac{200}{199})$
The first question that comes up is whether $f(n)>1$ for all $n>1$. To prove this, we need to prove $n>v_2(n!)+1$. 
The second question is whether all $f(n)$s are in the form of a super particular ratio$(\frac{a+1}{a})$ for all values of $n$. For this happening, we need $\frac{1}{f(n)-1}$ to be a natural number. After some calculations, we should get to prove $n-(v_2(n!)+1)|n$. This seems unlikely to be true for all $n$, but I would like knowing the values of $n$ for which it holds, as I don't have any idea to approach this problem! For the first problem i tried using the formula i mentioned earlier, and i got this:$n>1+\lfloor{\frac{n}{2}}\rfloor+\lfloor{\frac{n}{2^2}}\rfloor+\lfloor{\frac{n}{2^3}}\rfloor+...$ But i don't know where to get from here, since we've got to deal with an infinite number of floors in the RHS!
I would appriecate any help :)
Edit: As Gerry Myerson pointed out, It seems that $f(p)$ for some prime numbers $p$ is in the form of $\frac{a+2}{a}$ instead. It would be nice to see for which prime numbers $p$ this happens, but still i have no idea in approaching this problems.

For the first part of the question, there is a theorem by Legendre (1808).

Let $p$ be a prime and let $n = a_k p^k + a_{k-1} p_{k-1} + \ldots + a_1 p + a_0$ be the
  base-$p$ expansion of $n$, we have
  $$v_p(n!) = \frac{n - (a_k + a_{k-1} + \ldots + a_1 + a_0)}{p-1}$$

Apply this to the case $p = 2$, we have $v_2(n!) = n - \alpha(n)$ where $\alpha(n)$ is the number of set bits in binary expansion of $n$. This implies $n \ge v_2(n!) + 1$ for all $n$.
For a proof of the theorem, see 
this.
