A conjecture concerning odd binomial coefficients

This is true.

As usual, the notation $\nu_2(x)=m$ for a non-zero integer $x$ and a non-negative integer $m$ means that $x=2^m r$ for an odd $r$. $[x^a]f(x)$ denotes the coefficient of $x^a$ in a polynomial or Laurent polynomial $f(x)$. We assume, without loss of generality, that $a_1<\dots<a_s$.

We use the following

Lemma. Let $a_1<\dots<a_k$ be non-negative integers, and $t_1,\dots,t_k$ be non-zero integers such that $|t_i|\le 2^{a_i}$. Put $\sigma_i=t_1+\dots+t_i$. Assume that $\sigma_k=0$, but $\sigma_i\ne 0$ for $i<k$. Then $(k-1)+\nu_2(t_1\dots t_k)\leqslant a_1+\dots+a_k$.

Proof. (provided by Fedor Ushakov)

Let $p_i=\nu_2(\sigma_i)$ for $i=1,\ldots,k-1$ and $p_k=a_k$. It suffices to check $$a_i-\nu_2(t_i)\geqslant p_i-p_{i-1}\tag{$\heartsuit$}$$ for all $i=2,\ldots,k$, then sum up $(\heartsuit)$ and the obvious inequality $a_1\geqslant \nu_2(t_1)$ to get $$a_1+\ldots+a_k-\nu_2(t_1\ldots t_k)\geqslant p_k-p_1=a_k-p_1\geqslant a_k-a_1\geqslant k-1.$$ Well, $(\heartsuit)$ is obvious when the RHS is non-positive. If it is positive, then $\nu_2(t_i)=\nu_2(\sigma_i-\sigma_{i-1})=\nu_2(\sigma_{i-1})=p_{i-1}$, and $(\heartsuit)$ reads as $a_i\geqslant p_i$ that holds for $i=k$ by definition and for $i<k$ since $0<|\sigma_i|<2^{a_i+1}$.

Now denote $\varepsilon_i=-1$ for $i\in T$ and $\varepsilon_i=+1$ otherwise. Then your sum equals $$ [x^0](1+x)^n\prod_{i=1}^s (1+\varepsilon_i x^{-2^{a_i}})= [x^0]\prod_{i=1}^s f_i(x)=\sum_{\substack{t_1,\ldots,t_s\\\sum t_i=0}}\prod_{i=1}^s[x^{t_i}]f_i(x),\tag{$\clubsuit$} $$ where $f_i(x)=(1+\varepsilon_i x^{-2^{a_i}})(1+x)^{2^{a_i}}= (1+x)^{2^{a_i}}+\varepsilon_i (1+1/x)^{2^{a_i}}$. We prove that this sum is divisible by $2^s$ using induction on $s$ (the base case $s=0$ is clear).

Note that $[x^0]f_i(x)$ is either 0 or 2. This allows to exclude the terms in the RHS of $(\clubsuit)$ for which some $t_i$ vanish, since this reduces to the induction hypothesis.

Then we have the sum over sequences $(t_1,\ldots,t_s)$ of non-zero integers which sum up to 0 (further simply $s$-sequences) of the expression $\prod_{i=1}^s \delta_i{2^{a_i}\choose |t_i|}$, where $\delta_i=-1$ if simultaneously $\varepsilon_i=-1$ and $t_i<0$, otherwise $\delta_i=1$. For an $s$-sequence $\tau=(t_1,\ldots,t_s)$ we consider the following partition of $\{1,\ldots,s\}$ into intervals: $\Delta_1$ is the shortest initial interval of $\{1,\ldots,s\}$ for which $\sum_{i\in \Delta_1} t_i=0$, $\Delta_2$ is the shortest initial interval of $\{1,\ldots,s\}\setminus \Delta_1$ for which $\sum_{i\in \Delta_2} t_i=0$, etc. Let the last interval be $\Delta_N$. We call two $s$-sequences friends, if one is achieved from the other by changing the signs of all $t_i$ in some interval $\Delta_r$, or by several such operations. Note that $\tau$ has exactly $2^N$ friends, all these friends are mutual friends, and have the same partition. As for the corresponding terms in the RHS of $(\clubsuit)$, they are either all equal, or can be partitioned onto annihilating pairs. Thus, it suffices to prove that $\prod_i {2^{a_i}\choose |t_i|}$ is divisible by $2^{s-N}$. Since ${2^a\choose t}=\frac{2^a}{t}{2^a-1\choose t-1}$ for $t>0$, this follows from the Lemma applied to each of the $N$ intervals.