Let $C\_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, [Lagrange Inversion: When and How](https://doi.org/10.1007/s10440-006-9077-7)):
\begin{equation}
\label1
\sum\_{k=0}^n\binom{2n-2k}{n-k}C\_k=\binom{2n+1}n \qquad \iff \qquad 
\sum\_{i+j=n}\binom{2i}iC\_j=\binom{2n+1}n. \tag1
\end{equation}


I like to ask



> 
> **QUESTION.** Is there a $q$-analogue of \eqref{1}? Possibly, a combinatorial proof of \eqref{1} would shed some light into this.
> 
> 
> 



In search of a $q$-analogue of a Catalan identity
This identity is known as Jonah's formula (special case with $n\rightarrow 2n$ and $r\rightarrow n$, see "Catalan Numbers with Applications" by Thomas Koshy, pg. 325-326 for a combinatorial proof)


$$\sum\_{k=0}^r\binom{n-2k}{r-k}C\_k=\binom{n+1}r$$
and a $q$-analogue was obtained by Andrews in "$q$-Catalan identities" in the book "The legacy of Alladi Ramakrishnan in the Mathematical Sciences". It's Theorem 3, pg. 186.


$$\frac{(1+q^{n-r+1})}{(1+q^{r+1})}\left[ {\begin{array}{c}n+1\\r\end{array} } \right]\_{q^2}=-(-q\;;q)\_{n+1}\sum\_{k=0}^r\left[ {\begin{array}{c}n-2k\\r-k\end{array} } \right]\_{q^2}\frac{\textrm{C}\_{k+1}(-1;q)}{(-q\;;q)\_{n-2k}}q^{-k-1}$$


where $\textrm{C}\_n(\lambda,q)$ is a $q$-analogue of the Catalan numbers considered also by Andrews [here](https://www.sciencedirect.com/science/article/pii/0097316587900331).


$$\textrm{C}\_n(\lambda,q)=\frac{q^{2n}(-\lambda/q; q^{2})\_{n}}{(q^2;q^2)\_{n}}$$


In the paper, he says that the general strategy is to go from a binomial coefficient identity to a generalized hypergeometric identity, and then we can look for a $q$-analogue of the latter. In this case, he used the Pfaff-Saalschütz summation formula and then he searched for a $q$-analogue of this one with the help of Bailey's and Gasper and Rahman's books. I can't help much more, I'm not familiar with these kind of hypergeometric identities.


If $n\rightarrow 2n$ and $r\rightarrow n$, the limit $q\rightarrow 1$ recovers the identity (1).


