At A conjecture concerning odd binomial coefficients I conjectured the following, which was proved by Fedor Petrov and Fedor Ushakov. Let the positive integer $n$ have binary expansion $2^{a_1}+\cdots + 2^{a_s}$. For $S\subseteq [s]=\{1,2,\dots,s\}$, let $k_S= \sum_{i\in S} 2^{a_i}$.
Theorem (Petrov, Ushakov). Let $T\subseteq [s]$. Then $$ \sum_{S\subseteq [s]} (-1)^{|S\cap T|}{n\choose k_S} $$ is divisible by $2^s$.
Donald Knuth asked me whether this result has a nice $q$-analogue. What seems to work best is to replace ${n\choose k_S}$ by $q^{k_S(k_S+1)/2}\boldsymbol{{n\choose k_S}}_q$, where $\boldsymbol{{n\choose k_S}}_q$ is a $q$-binomial coefficient. When $T=\emptyset$ (the "classical case"), it looks like the sum in the conjecture is a polynomial with nonnegative integer coefficients (this much is obvious) which is divisible by at least $s$ factors of the form $q^i+1$. For arbitrary $T$ this is no longer the case, e.g., $n=11$ (so $S=\{0,1,3\}$) and $T=\{1,2\}$. Is there a better $q$-analogue?
