# Proof of certain $q$-identity for $q$-Catalan numbers

Let us use the standard notation for $$q$$-integers, $$q$$-binomials, and the $$q$$-analog $$\operatorname{Cat}_q(n) := \frac{1}{[n+1]_q} \left[\matrix{2n \\ n}\right]_q.$$ I want to prove that for all integers $$n\geq 0$$, we have $$$$\operatorname{Cat}_q(n+2) = \sum_{0\leq j,k \leq n} q^{k(k+2) + j(n+2)} \left[\matrix{n \\ 2k}\right]_q \operatorname{Cat}_q(k) \frac{[n+4]_q}{[k+2]_q} \left[\matrix{n-2k \\ j}\right]_q.$$$$

I have tried quite a bit, but not succeeded. Using $$q$$-hypergeometric series, this is equivalent with proving $$\sum_{\substack{k\geq 0 \\ j \geq 0}} q^{k(k+2)+j(n+2)} \frac{ (q;q)_{n+4} }{ (q^{n+3};q)_{n+2} (q;q)_{j} } \frac{ (q^{n-2k+1};q)_{2k} (q^{n-2k-j+1};q)_{j} }{ (q;q)_{k} (q;q)_{k+2} } =1$$ which I have also not managed to prove. I believe that some WZ-method could solve this easily, but a human-friendly proof would be preferrable. Note that the identity above is very similar to a theorem by Andrews (see reference below). It states that $$$$\operatorname{Cat}_q(n+1) = \sum_{k \geq 0} q^{k(k+2)} \left[\matrix{n \\ 2k}\right]_q \operatorname{Cat}_q(k) \frac{(-q^{k+2};q)_{n-k}}{(-q;q)_k}.$$$$

UPDATE: I have managed to find a more general conjecture, which would imply the one above. It states that for integer $$n \geq 0$$, and general $$a,c$$, we have $$\sum_{s} \frac{ (-a q^n)^{s} q^{-\binom{s}{2}} (q^{-n};q)_{s} }{ (q;q)_{s} } {}_{2}\phi_{1}(cq^{s-1}/a,q^{-s};c;q,q) = \frac{ (ac ;q)_{n} }{ (c;q)_{n} }.$$

I use This book as my main reference for notation and identities.

Andrews, George E., (q)-Catalan identities, Alladi, Krishnaswami (ed.) et al., The legacy of Alladi Ramakrishnan in the mathematical sciences. New York, NY: Springer (ISBN 978-1-4419-6262-1/hbk; 978-1-4419-6263-8/ebook). 183-190 (2010). ZBL1322.11018.

• Is there a human-friendly proof for the non-$q$ analog of this identity? That might be a start. Jun 6, 2020 at 8:26
• Yes, there is a combinatorial interpretation, tusing triangulations. I'll get the reference later today. Jun 6, 2020 at 9:50
• If one takes $q=1$, makes the change of variables $n \to n-4$ and $k \to k-2$ and sum over all $j$, one obtains the formula in the article. The summands the number of triangulations of regular $(n+2)$-gons with $k$ ears (an ear is a triangle on the vertices $i$, $i+1$ and $i+2$, where the indices are taken modulo $n+2$. Jun 6, 2020 at 14:44
• Is it possible that the triangulation interpretation of the $q=1$ identity is compatible with the statistic on triangulations discussed here?: mathoverflow.net/questions/93136/… Jun 6, 2020 at 20:21
I managed to solve the problem, in the last general conjecture, one can apply the $$q$$-Chu-Vandermonde theorem. After some simplification, the resulting expression can be expressed as a $${}_2\phi_1$$ q-hypergeometric series, where one again can apply the $$q$$-Chu-Vandermonde theorem.