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.


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 ….
