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.