An intermediate step deduces Jacobi’s triple product identity by taking the $q$-binomial theorem $$ \prod_{i=1}^{m-1}(1+xq^i)=\sum_{j=0}^m\binom{m}{j}_q q^{\binom{j}{2}}x^j $$ and deducing $$ \prod_{i=1}^s(1+x^{-1}q^i)\prod_{i=0}^{t-1}(1+xq^i)=\sum_{j=-s}^t\binom{s+t}{s+j}_q q^{\binom{j}{2}}x^j $$ and then letting $s\to\infty$ and $t\to\infty$. I don’t follow the intermediate deduction, what’s the way to see it? (Thank you Colin McQuillan for pointing this out.) Much later edit: By letting $s\to\infty$ […]

For real $0<q<1$, integer $n >0 $ and integer $k\ge 0$, define $$[k, n]_q \equiv -\sum_{m=1}^{n} q^{m(k+1)} (q^{-n}; q)_m = -\sum_{m=1}^{n} q^{m(k+1)} \prod_{l=0}^{m-1} (1-q^{l-n})$$ where $(\cdot\; ; q)_n$ is a $q$-Pochhammer symbol. These functions express exact occupation numbers of $k$-th energy level in an ideal Fermi gas with equidistant spectrum and exactly $n$ fermions. (For […]

It is a standard exercise in combinatorics to show that the binomial coefficient satisfies the reciprocity law $\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$ for $n, k \geqslant 0$, which is the multiset coefficient up to sign. Does the $q$-binomial coefficient $\binom{-n}{k}_{q}$ satisfy a similar reciprocity law with a nice combinatorial interpretation (and proof), which specializes to the […]

I read in some papers that the Stirling numbers (of the second kind) have a natural $q$-analog $S_q(n,k)$, which satisfy the recurrence $$ S_q(n,k)=(k)_qS_q(n-1,k)+q^{k-1}S_q(n-1,k-1) $$ with the conditions that $S_q(0,k)=\delta_{0,k}$ and $S_q(n,0)=\delta_{n,0}$. How is this recurrence arrived at? Even if this recurrence is taken as definition, there must be some motivation for it. Thank you.

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