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Let $a_n := \frac{1}{2n+1}\binom{3n}{n}$. Then it is known that (one can find references in the OEIS for this.) $$ a_n = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n-1} } a_i a_j a_k. $$

Is there a natural q-analog of this recursion for the corresponding q-binomials? That is, define the q-analog $b_n := \frac{1}{[2n+1]_q}\binom{3n}{n}_q$. Is there some suitable function $f$ (depending on $i,j,k$) such that the following holds: $$ b_n = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n-1} }q^{f(i,j,k)} b_i b_j b_k. $$

EDIT: With the connection with Catalan numbers, the analogous questions is, with $c_n := \frac{1}{n+1}\binom{2n}{n}$ is there a q-analog of the classical recursion $$ c_n = \sum_{\substack{i,j \geq 0 \\ i+j=n-1} } c_i c_j ? $$ And, this is as far as I know, there is no q-analog of this recursion, compatible with the q-analog of $c_n$.

Essentially: This amounts to choosing between the Carlitz (q-analog of recursion), or the q-binomial version of q-Catalan numbers.

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  • $\begingroup$ These are (the first case of) Fuss-Catalan numbers, right? "q-Fuss-Catalan" gets some hits on Google... $\endgroup$ Jul 10, 2021 at 15:10
  • $\begingroup$ As mentioned, it's in the OEIS-- oeis.org/A001764 $\endgroup$ Jul 10, 2021 at 17:11
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    $\begingroup$ @SamHopkins Ah, yes. A good reference seem to be doi.org/10.1007/BF03024089 where the recursion is (0.1), for p=3. $\endgroup$ Jul 10, 2021 at 17:17

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