I had completed a paper describing the $q$-Catalan numbers, which is the $q$-analog of the Catalan numbers.

The $n$-th Catalan numbers can be represented by:

$$C_n=\frac{1}{n+1}{2n \choose n}$$

and with the recurrence relation:

$$C_{n+1}=\sum^n_{i=0}C_i C_{n-i}\ \ \ \ \ \forall n\geq 0$$

Now, for the $q$-analog, I know the definition of that can be defined as:

$$\lim_{q\to 1}\frac{1-q^n}{1-q}=n$$

and we know that the definition of the $q$-analog, can be defined like this:

$$[n]_q=\frac{1-q^n}{1-q}=1+q+q^2+q^3+\cdots+q^{n-1}$$

which this is the $q$-analog of $n$.

and that for that $q$-analog of ${2n\choose n}$:

$$C_n(q)=\frac{1}{[n+1]_q}\begin{bmatrix}2n\\ n\end{bmatrix}_q$$

**So, everything up to this point I know what I'm doing, and I'm not sure if I did everything correct after this**

So, in order to generate the $q$-Catalan Numbers, I will need to use the Lagrange inversion formula.

And, then I got something like this:

$$G(X)=\sum^\infty_{i=0}C_i x^i$$

where $G(x)$ is the generating function, and that

$$G(x)=G_q(x)=\sum^\infty_{i=0}C_n(q)x^n=\sum^\infty_{i=0}C_nx^n=1+x+x^2(1+q)+\cdots$$

Since I know that for Catalan Numbers, it's true:

$$G(x)=(G(x))^2+1$$

So, the $q$-analog will just be:

$$G_q(x)=G(x)G_q(x)+1$$

So the recurrence relation for the $q$-analog Catalan Numbers:

$$C_{n+1}(q)=\sum^n_{i=0}C_i C_{n-1}q^i$$

**It just doesn't sound right here...**

Also, I don't have a clue that what does the $q$-Catalan Numbers count, can anyone help me with that or give me like a clue?

Help appreciated!