TITLE: enumerative meaning of natural q-Catalan numbers
QUESTION [15 upvotes]: Define $[n]=(1-q^n)/(1-q)$ and $[n]!=[1][2][3] \cdots [n]$, so that $[2n]!/[n]![n+1]!$ is a polynomial in $q$ (the most algebraically natural $q$-analogue of the Catalan numbers); what enumerative interpretation(s) does it have, vis-a-vis the standard members of the "Catalan zoo"?  I would be especially interested in answers pertaining directly to triangulations of polygons, without any intervening bijections.

REPLY [3 votes]: NEWEST
Seem my random speculations below. It never hurts to look in the OEIS and it would appear that the coefficient of $q^k$ in $\frac{1}{[n+1]}\left[{2n\atop n}\right]_q$ is the number of Dyck words of length 2n having major index k . I have not had time to look into translating that into triangulations.
OLDER
We do know that the $q$-binomial coefficient $$\left[{m\atop k}\right]_q=\frac{[m]!}{[k]![m-k]!}$$ is a polynomial (with non-negative integer coefficients) equal to $\binom{m}{k}$ when $q=1$ and that the coeffcient of $q^d$ is the number of ordered lists of type $0^k1^{m-k}$ having $d$ inversions. Equivalently, the number of lattice paths from $(0,0)$ to $(k,m-k)$ which enclose an region of area $d$ (with the $x$-axis and $x=k$). 
Each of the lattice paths counted by $\binom{2n}{n}$ has an excedence $0 \le e \le n$ which is the number of horizontal edges above the diagonal. Each value of $e$ occurs $\frac{1}{n+1}\binom{2n}{n}$ times and the Catalan numbers count the paths of excedence $0.$
These $q$-Catalan polynomials $\frac{1}{[n+1]}\left[{2n\atop n}\right]_q$ again have non-negative integer coefficients and for $q=1$ equal the Catalan numbers which the excedence $0$ lattice paths from $(0,0)$ to $(2n,n)$. One could dream that there is an interpretation of the coefficient of $q^d$ but I certainly can't find one. A recurrence relation might be helpful in back constructing an interpretation.
The obvious greedy guess fails. For that one would have a recurrence $C_q(n+1)=\sum q^{(k+1)(n-k)}C_q(k)C_q(n-k).$ Those polynomials do not have a strong claim on being a natural q-analogue.