Let $C_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, Lagrange Inversion: When and How): \begin{equation} \label1 \sum_{k=0}^n\binom{2n-2k}{n-k}C_k=\binom{2n+1}n \qquad \iff \qquad \sum_{i+j=n}\binom{2i}iC_j=\binom{2n+1}n. \tag1 \end{equation}

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QUESTION. Is there a $q$-analogue of \eqref{1}? Possibly, a combinatorial proof of \eqref{1} would shed some light into this.


2 Answers 2


This identity is known as Jonah's formula (special case with $n\rightarrow 2n$ and $r\rightarrow n$, see "Catalan Numbers with Applications" by Thomas Koshy, pg. 325-326 for a combinatorial proof)

$$\sum_{k=0}^r\binom{n-2k}{r-k}C_k=\binom{n+1}r$$ and a $q$-analogue was obtained by Andrews in "$q$-Catalan identities" in the book "The legacy of Alladi Ramakrishnan in the Mathematical Sciences". It's Theorem 3, pg. 186.

$$\frac{(1+q^{n-r+1})}{(1+q^{r+1})}\left[ {\begin{array}{c}n+1\\r\end{array} } \right]_{q^2}=-(-q\;;q)_{n+1}\sum_{k=0}^r\left[ {\begin{array}{c}n-2k\\r-k\end{array} } \right]_{q^2}\frac{\textrm{C}_{k+1}(-1;q)}{(-q\;;q)_{n-2k}}q^{-k-1}$$

where $\textrm{C}_n(\lambda,q)$ is a $q$-analogue of the Catalan numbers considered also by Andrews here.

$$\textrm{C}_n(\lambda,q)=\frac{q^{2n}(-\lambda/q; q^{2})_{n}}{(q^2;q^2)_{n}}$$

In the paper, he says that the general strategy is to go from a binomial coefficient identity to a generalized hypergeometric identity, and then we can look for a $q$-analogue of the latter. In this case, he used the Pfaff-Saalschütz summation formula and then he searched for a $q$-analogue of this one with the help of Bailey's and Gasper and Rahman's books. I can't help much more, I'm not familiar with these kind of hypergeometric identities.

If $n\rightarrow 2n$ and $r\rightarrow n$, the limit $q\rightarrow 1$ recovers the identity (1).


Decided to make a cw post: it is sort of amusing.

Let $C_n(q)$ be defined by $$\sum_{k=0}^n\binom{2n-2k}{n-k}_qC_k(q)q^{2n-2k}=\binom{2n+1}n_q,\qquad n=0,1,2,\dotsc.$$

Then \begin{multline*} C_n(q)=1+q+q^2+q^3+2 q^4+3 q^5+3 q^6+3 q^7+4 q^8+6 q^9+\dotsb\\\dotsb-7q^{(n+1)^2-6}-5q^{(n+1)^2-5}-3q^{(n+1)^2-4}-2q^{(n+1)^2-3}-q^{(n+1)^2-2}-q^{(n+1)^2-1} \end{multline*} where the “tail” is made from the partition numbers $1,1,2,3,5,7,11,15,22,30,42,\dotsc$ while the “head” satisfies \begin{multline*} 1+q+q^2+q^3+2 q^4+3 q^5+3 q^6+3 q^7+4 q^8+6 q^9+7 q^{10}+6 q^{11}+6 q^{12}+8 q^{13}+\dotsb\\ =1/(1-q-q^4+q^6+q^{11}-q^{14}-q^{21}+q^{25}+q^{34}-q^{39}-q^{50}+q^{56}+\dotsb). \end{multline*} Cf. \begin{align*} &\qquad\qquad1+q-q^4-q^6+q^{11}+q^{14}-q^{21}-q^{25}+q^{34}+q^{39}-q^{50}-q^{56}+\dotsb\\ &=q^{-1}(1-\prod_{n\geqslant1}(1-q^n)). \end{align*} Have no idea how to prove these, or what happens in between ….

  • $\begingroup$ I'm confused, shouldn't $C_k(q)$ depend on both $n$ and $k$? $\endgroup$ Aug 1, 2021 at 20:07
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    $\begingroup$ @SamHopkins Why? The "non-$q$" Catalan numbers only depend on one index. $\endgroup$ Aug 1, 2021 at 22:09
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    $\begingroup$ I see, good point. Then possibly your observations can be proved by taking the limit $n\to\infty$ of the sum, and considering the regimes $q < 1$ and $q > 1$. $\endgroup$ Aug 1, 2021 at 22:18
  • $\begingroup$ @SamHopkins Interesting idea. But I don't know how to proceed: leaving it as it is, the summands on the left become divisible by higher and higher powers of $q$, because of the $q^{2n-2k}$ term. While if one divides both sides by $q^{2n}$, then the right hand side acquires more and more negative powers of $q$. What is definitely significant is that both ${\binom{2n+1}n}_q$ and ${\binom{2n}n}_q$ tend to the partition generating function $1+q+2q^2+3q^3+5q^4+7q^5+...$ $\endgroup$ Aug 1, 2021 at 22:26

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