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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!

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    $\begingroup$ There are various $q$-analogues of the Catalan numbers. You can pick a combinatorial description of the Catalan numbers, then pick an interesting parameter corresponding to that combinatorial description, and that gets you a $q$-analogue. Different $q$-analogues will satisfy different $q$-analogue-looking properties. $\endgroup$
    – Qiaochu Yuan
    Commented Apr 27, 2013 at 3:53
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    $\begingroup$ Actually I found that the recurrence relation for the $q$-Catalan Numbers is incorrect, but I'm not sure how to get it fixed ... also , the generating function is also incorrect. Although that $G(x)=(G(X))^2+1$, though however it doesn't work for the $q$-analog of the Catalan Numbers ... $\endgroup$
    – user67258
    Commented Apr 27, 2013 at 5:27

1 Answer 1

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From Mathoverflow:$\qquad\qquad\qquad$

https://mathoverflow.net/questions/93136/enumerative-meaning-of-natural-q-catalan-numbers

and as predicted in the comments, a $(q,t)$-interpretation. Specialize to get one of the $q$-interpretations

http://garsia.math.yorku.ca/MPWP/qtanalogs/catalan/combinterpret.html

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    $\begingroup$ Thanks for the comment. But I'm actually not doing the $q$-Catalan Numbers on the $(q,t)$-interpretation... Thanks for the information though, it's very helpful! $\endgroup$
    – user67258
    Commented Apr 27, 2013 at 5:00
  • $\begingroup$ The MO link has at least one straight $q$ interpretation, without $t$. $\endgroup$
    – zyx
    Commented Apr 27, 2013 at 5:33
  • $\begingroup$ Yes, I've notice that, thanks! $\endgroup$
    – user67258
    Commented Apr 27, 2013 at 5:34
  • $\begingroup$ For the other part of the question, I didn't understand what the question is. Are you asking what is the convolution formula for q-Catalan numbers? $\endgroup$
    – zyx
    Commented Apr 27, 2013 at 5:44
  • $\begingroup$ Oh, my bad, I was just trying to ask what does the $q$-Catalan numbers count (like statistically), comparing to the Catalan Numbers $\endgroup$
    – user67258
    Commented Apr 27, 2013 at 5:45

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