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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