In search of a $q$-analogue of a Catalan identity 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):
\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.

 A: 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.
$$\textrm{C}_n(\lambda,q)=\frac{q^{2n}(-\lambda/q; q^{2})_{n}}{(q^2;q^2)_{n}}$$
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).
A: Decided to make a cw post: it is sort of amusing.
Let $C_n(q)$ be defined by
$$\sum_{k=0}^n\binom{2n-2k}{n-k}_qC_k(q)q^{2n-2k}=\binom{2n+1}n_q,\qquad n=0,1,2,\dotsc.$$
Then
\begin{multline*}
C_n(q)=1+q+q^2+q^3+2 q^4+3 q^5+3 q^6+3 q^7+4 q^8+6 q^9+\dotsb\\\dotsb-7q^{(n+1)^2-6}-5q^{(n+1)^2-5}-3q^{(n+1)^2-4}-2q^{(n+1)^2-3}-q^{(n+1)^2-2}-q^{(n+1)^2-1}
\end{multline*}
where the “tail” is made from the partition numbers $1,1,2,3,5,7,11,15,22,30,42,\dotsc$ while the “head” satisfies
\begin{multline*}
1+q+q^2+q^3+2 q^4+3 q^5+3 q^6+3 q^7+4 q^8+6 q^9+7 q^{10}+6 q^{11}+6 q^{12}+8 q^{13}+\dotsb\\
=1/(1-q-q^4+q^6+q^{11}-q^{14}-q^{21}+q^{25}+q^{34}-q^{39}-q^{50}+q^{56}+\dotsb).
\end{multline*}
Cf.
\begin{align*}
&\qquad\qquad1+q-q^4-q^6+q^{11}+q^{14}-q^{21}-q^{25}+q^{34}+q^{39}-q^{50}-q^{56}+\dotsb\\
&=q^{-1}(1-\prod_{n\geqslant1}(1-q^n)).
\end{align*}
Have no idea how to prove these, or what happens in between ….
