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One more time, let us see how else the Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$ can be generalized. First recall the generating function $$C(x):=\frac{1-\sqrt{1-4x}}x=\sum_{n\geq0}C_n\,x^n.$$ Adopting multivariate conventions: let $\pmb{a}=(a_1,\dots,a_k)\in\mathbb{Z}_{\geq0}^k$ and $\pmb{x}=(x_1,\dots,x_k)$ be indeterminates, $\pmb{x}^{\pmb{a}}=x_1^{a_1}\cdots x_k^{a_k}$.

Definition. The multi-indexed Catalan numbers, denoted by $C(T_{\pmb{a}})$ or $C(T_{a_1,\dots,a_k})$, is defined by $$\sum_{\pmb{a}\geq\pmb{0}}C(T_{\pmb{a}})\,\pmb{x}^{\pmb{a}} =\frac{\prod_{j=1}^kC(x_j)}{1-\sum_{j=1}^kx_j\,C(x_j)}.$$

Question. What do the numbers $C(T_{\pmb{a}})$ mean?

To get some insight, you may take a look at page 7 of this paper by A. Postnikov describing $T_{a_1,a_2,a_3}$ as the graph that has a central node with $3$ attached chains of $a_1, a_2$ and $a_3$ vertices.

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  • $\begingroup$ Possibly related: sciencedirect.com/science/article/pii/S0195669806000278. I would look into generalized tree hook length formulas, since this is what followed from Postnikov's paper. $\endgroup$
    – Mosquite
    Commented May 31, 2017 at 18:19
  • $\begingroup$ Thank you, but this does not help nor is it related to the problem at hand. $\endgroup$ Commented May 31, 2017 at 18:36

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