TITLE: An Identity Involving Narayana Numbers
QUESTION [5 upvotes]: Let $N(n,m)$ denote the Narayana number defined by $$N(n,m)=\frac{1}{n}{n\choose m}{n\choose m-1}.$$ Let $$A(n,k,\ell)=\sum_{\substack{i_0+i_1+\cdots+i_k=n\\ j_0+j_1+\cdots+j_k=\ell}}\,\prod_{t=0}^kN(i_t,j_t+1),$$ where the sum ranges over all compositions $(i_0,i_1,\ldots,i_k)$ of $n$ into exactly $k+1$ parts and all weak compositions $(j_0,j_1,\ldots,j_k)$ of $\ell$ into exactly $k+1$ parts. From numerical evidence, I have conjectured that $$A(n,k,\ell)=\frac{k+1}{n}{n\choose \ell}{n\choose \ell+k+1}.$$ This seems like a simple enough formula, but I am at a loss for how to prove it. Any help would be greatly appreciated.

REPLY [2 votes]: Suppose we have the Narayana number
$$N(n,m) = \frac{1}{n} {n\choose m} {n\choose m-1}$$
and let
$$A(n,k, l) =
{\Large \sum_{i_0+i_1+\cdots+i_k=n \atop j_0+j_1+\cdots+j_k=l}
\prod_{t=0}^k N(i_t, j_t+1)}$$
where the compositions for $n$ are regular and the ones for $l$
are weak and we seek to verify that
$$A(n,k,l) = \frac{k+1}{n}
{n\choose l} {n\choose l+k+1}.$$ 
Introducing 
$$G(z, u) = \sum_{p\ge 1} z^p \sum_{q\ge 0} u^q
\frac{1}{p} {p\choose q+1} {p\choose q}
\\ = \sum_{p\ge 1} \frac{1}{p} z^p \sum_{q\ge 0} u^q
{p\choose q+1} {p\choose q}$$
we have by inspection that
$$A(n,k,l) = [z^n] [u^l] G(z, u)^{k+1}.$$
To evaluate this introduce for the inner sum term
$${p\choose q+1} = {p\choose p-q-1} =
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{p-q}} (1+w)^p \; dw.$$
We get for the inner sum
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{p}} (1+w)^p 
\sum_{q\ge 0} {p\choose q} u^q w^q 
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{p}} (1+w)^p 
(1+uw)^p
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{p}}
(1+w(1+u+uw)))^p
\; dw.$$
Extracting the coefficient from this we get
$$[w^{p-1}] \sum_{q=0}^p {p\choose q}
w^q (1+u+uw)^q
\\ = \sum_{q=0}^{p-1} {p\choose q}
[w^{p-1-q}] (1+u+uw)^q
\\ = \sum_{q=0}^{p-1} {p\choose q}
{q\choose p-1-q} u^{p-1-q} (1+u)^{2q+1-p}.$$
This is
$$\sum_{q=0}^{p-1} {p\choose p-1-q}
{p-1-q\choose q} u^q (1+u)^{p-1-2q}
\\ = \sum_{q=0}^{p-1} {p\choose q+1}
{p-1-q\choose q} u^q (1+u)^{p-1-2q}.$$
Now observe that
$$\frac{1}{p} {p\choose q+1} {p-1-q\choose q}
= \frac{1}{q+1} {p-1\choose q} {p-1-q\choose q}
\\ = \frac{1}{q+1} {p-1\choose p-1-q} {p-1-q\choose q}
= \frac{1}{q+1} {p-1\choose 2q} {2q\choose q}.$$
where $$C_q = \frac{1}{q+1} {2q\choose q}$$ is a Catalan number.
We thus get for the sum
$$\sum_{p\ge 1} z^p
\sum_{q=0}^{p-1} {p-1\choose 2q} C_q
u^q (1+u)^{p-1-2q}
\\ = z \sum_{p\ge 0} z^p
\sum_{q=0}^{p} {p\choose 2q} C_q
u^q (1+u)^{p-2q}
\\ = z \sum_{q\ge 0} C_q u^q (1+u)^{-2q}
\sum_{p\ge q} {p\choose 2q} z^p (1+u)^p
\\ = z \sum_{q\ge 0} C_q u^q (1+u)^{-2q}
\sum_{p\ge 2q} {p\choose 2q} z^p (1+u)^p
\\ = z \sum_{q\ge 0} C_q u^q (1+u)^{-2q} (1+u)^{2q} z^{2q}
\sum_{p\ge 0} {p+2q\choose 2q} z^p (1+u)^p
\\ = z \sum_{q\ge 0} C_q u^q z^{2q} \frac{1}{(1-z(1+u))^{2q+1}}.$$
Using the generating function of the Catalan numbers
$$Q(w) = \sum_{q\ge 0} C_q w^q = \frac{1-\sqrt{1-4w}}{2w}$$
which has functional equation
$$Q(w) = 1 + w Q(w)^2$$ 
we obtain
$$Q\left(\frac{uz^2}{(1-z(1+u))^2}\right)
= 1 + \frac{uz^2}{(1-z(1+u))^2}
Q\left(\frac{uz^2}{(1-z(1+u))^2}\right)^2$$
which is
$$G(z, u) \frac{1-z(1+u)}{z}
= 1 + u G(z, u)^2.$$
Extract the coefficient in $z$ first. We get from the functional equation
$$z = \frac{G(z,u)}{u G(z,u)^2 + (1+u) G(z,u) + 1}.$$
The coefficient extractor integral is
$$[z^n] G(z, u) =
\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
G(z, u)^{k+1} \; dz.$$
which becomes with $G(z, u) = v$
$$\frac{1}{2\pi i} 
\int_{|v|=\epsilon} 
\frac{(uv^2+(1+u)v+1)^{n+1}}{v^{n+1}} 
\\ \times v^{k+1} \left(\frac{1}{uv^2+(1+u)v+1}
- \frac{v}{(uv^2+(1+u)v+1)^2}(2uv+(1+u))\right) \; dv
\\ = \frac{1}{2\pi i} 
\int_{|v|=\epsilon} 
\frac{(uv^2+(1+u)v+1)^{n-1}}{v^{n-k}}
(1 - uv^2) \; dv.$$
This is
$$\frac{1}{2\pi i} 
\int_{|v|=\epsilon} 
\frac{(1+v)^{n-1} (1+uv)^{n-1}}{v^{n-k}}
(1 - uv^2) \; dv.$$
Extracting the coefficient on $[u^l]$ we get two pieces
which are, first piece $A$
$${n-1\choose l} \frac{1}{2\pi i} 
\int_{|v|=\epsilon} 
\frac{(1+v)^{n-1} v^l}{v^{n-k}} \; dv
= {n-1\choose l} {n-1\choose n-k-l-1}$$
which is $${n-1\choose l} {n-1\choose k+l}
= {n-1\choose l} \frac{k+l+1}{n} {n\choose k+l+1}
\\ =  (n-l)\frac{k+l+1}{n^2} {n\choose l} {n\choose k+l+1}.$$
and piece $B$ which is
$$- {n-1\choose l-1} \frac{1}{2\pi i} 
\int_{|v|=\epsilon} 
\frac{(1+v)^{n-1} v^{l-1}}{v^{n-k}} \; v^2 dv
= -{n-1\choose l-1} {n-1\choose n-k-l-2}$$
which is
$$- {n-1\choose l-1} {n-1\choose k+l+1}
= - {n-1\choose l-1} \frac{n-k-l-1}{n} {n\choose k+l+1}
\\ = - l \frac{n-k-l-1}{n^2} {n\choose l} {n\choose k+l+1}.$$
Collecting the two pieces we finally obtain
$$\left((n-l)\frac{k+l+1}{n^2} + l \frac{-n+k+l+1}{n^2} \right)
{n\choose l} {n\choose k+l+1}
\\ = \left(n\frac{k+l+1}{n^2} + l \frac{-n}{n^2} \right)
{n\choose l} {n\choose k+l+1}
\\ = \frac{k+1}{n} {n\choose l} {n\choose k+l+1}$$
as claimed, QED.
Remark. The closed form of $G(z,u)$ can be computed as follows:
$$\frac{z}{1-z(1+u)}
\frac{1-\sqrt{1-4uz^2/(1-z(1+u))^2}}{2 uz^2/(1-z(1+u))^2}
\\ = \frac{z}{(1-z(1+u))^2}
\frac{1-z(1+u)-\sqrt{1-2z(1+u)+z^2(1+u)^2-4uz^2}}{2 uz^2/(1-z(1+u))^2}
\\ = \frac{1-z(1+u)-\sqrt{1-2z(1+u)+z^2(1+u)^2-4uz^2}}{2 uz}.$$
The above material incorporates data from
OEIS A055151 and from 
OEIS A001263 on Narayana numbers.