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 \begin{equation} \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. \end{equation}

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 \begin{equation} \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}. \end{equation}

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.

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    $\begingroup$ Is there a human-friendly proof for the non-$q$ analog of this identity? That might be a start. $\endgroup$ – Alexander Burstein Jun 6 '20 at 8:26
  • $\begingroup$ Yes, there is a combinatorial interpretation, tusing triangulations. I'll get the reference later today. $\endgroup$ – Per Alexandersson Jun 6 '20 at 9:50
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    $\begingroup$ Here is the combinatorial interpretation Per is talking about: core.ac.uk/download/pdf/82091698.pdf $\endgroup$ – Joakim Uhlin Jun 6 '20 at 14:29
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    $\begingroup$ 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 $n+2$. $\endgroup$ – Joakim Uhlin Jun 6 '20 at 14:44
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    $\begingroup$ 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/… $\endgroup$ – Sam Hopkins Jun 6 '20 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.

Skipping lots of details the proof still requires a few pages - it will be included in an upcoming paper, and once in arxiv, I'll add a link.


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