Consider the sequence (of rational numbers) given by
$$a\_n=\sum\_{k=1}^n\binom{n}k\frac{k}{n+k}.$$
Let $s(n)$ be the sum of binary digits of $n$, i.e. the total number of $1$'s.



> 
> **QUESTION.** Is it true that the $2$-adic valuation of the denominator of $a\_n$ equals $s(n)$?
> It seems so, experimentally.
> 
> 
> 



2-adic valuation of a certain binomial sum
First we notice that
\begin{split} 
a\_n & = n \int\_0^1 x^n (1+x)^{n-1}{\rm d}x \\
& = n \int\_0^1 (1-x)^n (2-x)^{n-1}{\rm d}x \\
& = n\sum\_{k=0}^{n-1} \binom{n-1}{k}2^k (-1)^{n-1-k} \int\_0^1 (1-x)^n x^{n-1-k}{\rm d}x \\
&= \sum\_{k=0}^{n-1} 2^k (-1)^{n-1-k} \binom{2n}{k} / \binom{2n}n. \\
\end{split}
Now, the numerator in the last expression is odd, and thus $\nu\_2(a\_n)=-\nu\_2(\tbinom{2n}n)=-s(n)$ by [Kummer's theorem](https://en.wikipedia.org/wiki/Kummer%27s_theorem). QED


