# Generating function for the products of pairs of Narayana numbers

The Narayana numbers OEIS sequence A001263 are given by: $$\operatorname{N}(n, k) = \frac{1}{n} {n \choose k} {n \choose k-1},$$ and have the generating function: $$G(z,t) = \sum_{n=1}^\infty \sum_{k=1}^n \operatorname{N}(n, k) z^n t^{k-1} = \frac{1-z(t+1) - \sqrt{1-2z(t+1)+z^2(t-1)^2}}{2tz}.$$ I would instead like to calculate the generating function of the product of pairs of Narayana numbers:

$$S(z,t,u) = \sum_{n=1}^\infty \sum_{k,m=1}^n \operatorname{N}(n, k)\operatorname{N}(n, m) z^n t^{k-1} u^{m-1}$$ In the special case $$S(z,1,1)$$, I believe this reduces to the generating function of the squares of Catalan numbers, which is known in terms of elliptic integrals and was the subject of my previous question. Unfortunately I am not sure how to generalize the approaches to derive that generating function to this case.

Even if $$S(z,t,u)$$ cannot be written in terms of (presumably) elliptic integrals or hypergeometric functions, can anything be said about its convergence?

• I suppose that the result would just be an infinite sum of the product of two hypergeometric functions Commented Jul 9 at 8:25
• @ClaudeLeibovici Can you possibly elaborate on this? Why do you expect this? Do you know how one would go about calculating on explicit form? Many thanks. Commented Jul 9 at 22:03
• You and I had good intuitions Commented Jul 10 at 7:55

Start writing $$\operatorname{N}(n, k) = \frac{1}{n} {n \choose k} {n \choose k-1}=\frac{n \, \big(\Gamma (n)\big)^2}{\Gamma (k)\, \Gamma(k+1)\, \Gamma (n+1-k)\, \Gamma(n+2-k)}$$ and the same for $$\operatorname{N}(n, m)$$ and consider with these expressions

$$S(z,t,u) = \sum_{n=1}^\infty z^n \sum_{k=1}^n t^{k-1}\sum_{m=1}^n \operatorname{N}(n, k)\operatorname{N}(n, m)\,\, u^{m-1}$$

The summation over $$m$$ gives $$\frac{n \, \big(\Gamma (n)\big)^2}{\Gamma (k) \Gamma(k+1) \Gamma (n+1-k) \Gamma(n+2-k)}\,\, _2F_1(1-n,-n;2;u)$$ The summation over $$k$$ gives $$\, _2F_1(1-n,-n;2;t) \,\,\,_2F_1(1-n,-n;2;u)$$

Finally

$$\large\color{blue}{S(z,t,u) = \sum_{n=1}^\infty \, _2F_1(1-n,-n;2;t) \,\,\,_2F_1(1-n,-n;2;u)\,\, z^n}$$

For the case where $$t=u=1$$, this gives $$S(z,1,1)=\sum_{n=1}^\infty \,\Big(\, _2F_1(1-n,-n;2;1){}\Big)^2\,\, z^n$$ As you expected $$\, _2F_1(1-n,-n;2;1){}$$ corresponds to Catalan numbers.

$$S(z,1,1)=\sum_{n=1}^\infty \,\Bigg(\frac{4^n \Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi } \Gamma (n+2)}\Bigg)^2\, z^n$$ which is

$$S(z,1,1)=\frac{\, _2F_1\left(-\frac{1}{2},-\frac{1}{2};1;16 z\right)}{4 z}-\frac{1}{4 z}-1$$ This is also $$S(z,1,1)=\frac 1 {2\pi z} \Big((16 z-1)\, K(16 z)+2\, E(16 z)-\frac \pi 2\Big)$$

Edit

What could be interesting (at least to me) is to find the inverse of $$S(z,1,1)=k$$ if $$k>1$$

A decent approximation is given by the $$[n,n]$$ Padé approximant $$P_n$$ such as

$$P_2=\frac{1-\frac{29 }{3}z+11z^2}{1-\frac{32 }{3}z+\frac{53}{3}z^2}$$ giving $$z \sim \frac{(29-32k)+ \sqrt{388 k^2-824 k+445}}{2 (33-53 k) }$$ Using this value and expanding around $$k=1$$ wold give $$k=1+(k-1)+115 (k-1)^5+O\left((k-1)^6\right)$$ Since $$k \leq \frac{16}{\pi }-4$$ the quadratic term is really small.

• Many thanks. I guess naively I was hoping that the sum over $z$ could be performed also, so that I could better understand the behaviour of S, (as it corresponds to a physical observable I wish to understand), but I guess from your form it is pretty clear that such a hope is in vain, and the hypergeometric form is certainly simpler to deal with than the original version. Commented Jul 10 at 18:25