Let $C\_n=\frac1{n+1}\binom{2n}n$ denote the Catalan numbers.


This question is motivated by the (unanswered) [MO post by Alexander Burstein](https://mathoverflow.net/questions/391533/a-bijective-proof-for-the-odd-companion-to-shapiros-catalan-convolution) and [my own (answered by Fedor Petrov) MO post](https://mathoverflow.net/questions/393119/looking-for-a-q-analogue-of-a-binomial-identity).
Specifically, Shapiro's convolution formula states that
$$\sum\_{k=0}^nC\_{2k}C\_{2n-2k}=4^nC\_n. \tag1$$



> 
> **QUESTION.** Is there a $q$-analogue to this "elegant" identity in equation (1)?
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Is there a $q$-analogue to Shapiro's convolution identity?
George E.Andrews gives a $q$-analog for Shapiro's convolution identity in the article [On Shapiroʼs Catalan convolution](https://doi.org/10.1016/j.aam.2010.07.003). The $q$-analog is


$$\sum\_{j=0}^n q^{2j} C\_{2j+1}(1,-q) C\_{2n+1-2j}(1,-q) = \frac{-q^{2n+1}(-q^2 : q^2)\_{n-1} C\_{n+1}(1,-q)}{(-q:q^2)\_{n+1}}$$
where
$$C\_n(\lambda,q) = \frac{q^{2n}(\frac{-\lambda}{q}:q^2)\_n}{(q^2:q^2)\_n}$$
and $\lim\_{q\to 1} C\_{n+1}(1,-q) = -2^{-2n-1} C\_n$.
Taking $q \to 1$ and multiplying by the needed power of $2$ in the above $q$-identity gives Shapiros Catalan convolution.


