# In search of a $q$-analogue of a Catalan identity

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): $$\begin{equation} \label1 \sum_{k=0}^n\binom{2n-2k}{n-k}C_k=\binom{2n+1}n \qquad \iff \qquad \sum_{i+j=n}\binom{2i}iC_j=\binom{2n+1}n. \tag1 \end{equation}$$

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

• I can only say that substituting $q$-binomial coefficients for binomial coefficients and solving for $C_k$ gives something awful:$$C_0=1$$$$C_1=q^2$$$$C_2=q^6+q^5+q^4-q^2$$$$C_3=q^{12}+q^{11}+2q^{10}+2q^9+3q^8+q^7-q^6-2q^5-2q^4$$$$\cdots$$ Aug 1, 2021 at 16:39
• @მამუკაჯიბლაძე: often there are "hidden" powers of $q$ that make these $q$-analogues work. Aug 1, 2021 at 16:46
• – Nemo
Aug 1, 2021 at 17:52
• This article sciencedirect.com/science/article/pii/0097316585900895 gives q-analog of Catalan convolution for Carlitz's q-analog of Catalan numbers.
– Nemo
Aug 1, 2021 at 17:59
• Also see my answer to OP's another post, mathoverflow.net/questions/263164/generating-q-catalan-numbers/…
– Nemo
Aug 1, 2021 at 18:04

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 ….

• I'm confused, shouldn't $C_k(q)$ depend on both $n$ and $k$? Aug 1, 2021 at 20:07
• @SamHopkins Why? The "non-$q$" Catalan numbers only depend on one index. Aug 1, 2021 at 22:09
• 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 $q > 1$. Aug 1, 2021 at 22:18
• @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 $1+q+2q^2+3q^3+5q^4+7q^5+...$ Aug 1, 2021 at 22:26