# 2-adic valuation of a certain binomial sum

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.

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. QED
• I inserted $\binom{n-1}{k}$ in the third line. Jun 15, 2021 at 23:32
• Btw, integers $a_n\cdot \binom{2n}{n}$ are listed in OEIS A240721. Jun 16, 2021 at 19:39
Just for the record, I noticed that if we take a common denominator for $$a_n$$ as coming out of the terms $$n+k$$ then one would get $$\prod_{j=1}^n(n+j)=\frac{(2n)!}{n!}$$. Hence, we may rewrite the given sequence as follows: \begin{align*} \sum_{k=1}^n\binom{n}k\frac{k}{n+k} &=\frac{n!}{(2n)!}\sum_{k=1}^nk\binom{n}k\frac{\prod_{j=1}^n(n+j)}{n+k} \\ &=\binom{2n}n^{-1}\sum_{k=1}^n\frac{(n+1)\cdots(n+k-1)}{(k-1)!}\cdot\frac{(n+k+1)\cdots(2n)}{(n-k)!} \\ &=\binom{2n}n^{-1}\sum_{k=1}^n\binom{n+k-1}{k-1}\binom{2n}{n+k} \end{align*} and the numerators (evidently integers) actually agree with what Max Alekseyev's referral to OEIS A240721. As an aside, we gather the identity that $$\sum_{k=1}^n\binom{n+k-1}{k-1}\binom{2n}{n+k}= \sum_{k=0}^{n-1}\binom{2n}k2^k(-1)^{n-1-k}.$$