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I have bumped into the following expressions involving $q$-binomial coefficients. $$ \sum_{s=0}^a (-1)^s q^{s^2-s} \left(\begin{array}{c}2b+1-2s\\2a-2s\end{array}\right)_q \left(\begin{array}{c}b\\s\end{array}\right)_{q^2} $$ The expressions depend on $a$ and $b$ and are zero unless $0\leq a\leq b$.

Have these expressions appeared before? What is known about them? Any references would be appreciated.

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This is a summable ${}_2\phi_1$. The standard reference is Gasper and Rahman, Basic Hypergeometric Series. It looks like your series is $$\frac{(q;q)_{2b+1}}{(q;q)_{2a}(q;q)_{2b+1-2a}}{}_2\phi_1\left(\begin{matrix}q^{-2a},q^{1-2a}\\q^{-2b-1}\end{matrix};q^2,q^{4a-2b-2}\right).$$ By the $q$-Gauss summation, this factors completely as $$\frac{(q;q)_{2b+1}}{(q;q)_{2a}(q;q)_{2b+1-2a}}\frac{(q^{-2b-2+2a};q^2)_a}{(q^{-2b-1};q^2)_a}.$$ If you compute your expression for some special values of $a$ and $b$ you should see this factorization; otherwise I made a mistake.

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    $\begingroup$ You could add that Askey's advice "whenever you see a sum involving binomial coefficients, write it in hypergeometric notation", is equally valid in the $q$-case. $\endgroup$
    – Hjalmar Rosengren
    Commented Jan 17, 2015 at 15:44
  • $\begingroup$ I think your formula is a bit off, but I will surely figure it out now. $\endgroup$
    – Lev Borisov
    Commented Jan 18, 2015 at 1:28
  • $\begingroup$ I am getting $q^{2a^2-a}(q^{2b+2-2a}-1)(q^{2b-4a+4};q^2)_{2a-1}/(q;q)_{2a}$... $\endgroup$
    – Lev Borisov
    Commented Jan 18, 2015 at 2:48
  • $\begingroup$ You have a minus sign wrong. My answer can be simplified to $q^{2a^2-a}(1-q^{2b+2-2a})(q^{2b-4a+4};q^2)_{2a-1}/(q;q)_{2a}$. To see this, rewrite the three factors involving $(q;q)_k$ using $(q;q)_{2k}=(q;q^2)_k(q^2;q^2)_k$, $(q;q)_{2k+1}=(q;q^2)_{k+1}(q^2;q^2)_k$ and the other two factors using $(a;q^2)_k=(-1)^kq^{k^2-k}(q^{2-2k}/a;q^2)_k$. $\endgroup$
    – Hjalmar Rosengren
    Commented Jan 18, 2015 at 6:03
  • $\begingroup$ Yup, got the sign wrong. $\endgroup$
    – Lev Borisov
    Commented Jan 18, 2015 at 12:19

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