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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 and my own (answered by Fedor Petrov) MO post. 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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  • $\begingroup$ There are several such convolutions involving the factor of $4$. For example, for $f(x)=A025227(x)$, we have $(f(x)-f(-x))^2=(xr)\circ(4x^2)$, where $r(x)$ is the ogf for the large Schröder numbers, and $(A185010(x))^2=m(4x)$, where $m(x)$ is the ogf for the Motzkin numbers. $\endgroup$
    – Alexander Burstein
    Commented May 27, 2021 at 7:26

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George E.Andrews gives a $q$-analog for Shapiro's convolution identity in the article On Shapiroʼs Catalan convolution. 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.

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  • $\begingroup$ Thank you for this info! $\endgroup$
    – T. Amdeberhan
    Commented May 19, 2021 at 21:29

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