A combinatorial identity - Hockey Stick generalization

$$ \begin{align} \sum_{j=0}^m\binom{r+j}{j}\binom{s+j}{j} &=\sum_{j=0}^m\sum_{k=0}^r\binom{r}{k}\color{#C00}{\binom{j}{k}\binom{s+j}{j}}\tag1\\ &=\sum_{j=0}^m\sum_{k=0}^r\binom{r}{k}\color{#C00}{\binom{s+k}{k}\binom{s+j}{s+k}}\tag2\\ &=\sum_{k=0}^r\binom{r}{k}\binom{s+k}{k}\binom{s+m+1}{s+k+1}\tag3\\ &=\sum_{k=0}^r\sum_{j=0}^s\color{#C00}{\binom{r}{k}}\binom{s}{j}\color{#C00}{\binom{k}{j}}\binom{s+m+1}{s+k+1}\tag4\\ &=\sum_{k=0}^r\sum_{j=0}^s\color{#C00}{\binom{r}{j}}\binom{s}{j}\color{#C00}{\binom{r-j}{k-j}}\binom{s+m+1}{s+k+1}\tag5\\ &=\sum_{j=0}^s\binom{r}{j}\binom{s}{j}\binom{m+r+s+1-j}{r+s+1}\tag6\\ \end{align} $$ Explanation:
$(1)$: Vandermonde's Identity: $\binom{r+j}{j}=\sum_k\binom{r}{k}\binom{j}{j-k}$
$(2)$: expand red binomial coefficients as ratios of factorials
$(3)$: sum in $j$ using the Hockey-Stick Identity
$(4)$: Vandermonde's Identity: $\binom{s+k}{k}=\sum_j\binom{s}{j}\binom{k}{k-j}$
$(5)$: expand red binomial coefficients as ratios of factorials
$(6)$: $\binom{r-j}{k-j}=\binom{r-j}{r-k}$, then Vandermonde's Identity

With the convention $\binom{x}{n}=x^{\,\underline {\,n\,} } /n!$, in the RHS $r+s+1$ shall be a non-negative integer.
We can then apply symmetry of the binomial to rewrite our identity as $$ \sum\limits_{j = 0}^m {\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)} = \sum\limits_j^{} {\left( \matrix{ m - j \cr m - j \cr} \right)\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)} = \sum\limits_{j = 0}^s {\left( \matrix{ r \cr j \cr} \right)\left( \matrix{ s \cr j \cr} \right)\left( \matrix{ m + r + s + 1 - j \cr r + s + 1 \cr} \right)} = \sum\limits_j^{} {\left( \matrix{ r \cr j \cr} \right)\left( \matrix{ s \cr j \cr} \right)\left( \matrix{ m + r + s + 1 - j \cr m - j \cr} \right)} $$ i.e.: $$ \bbox[lightyellow] { \sum\limits_j^{} {\left( \matrix{ m - j \cr m - j \cr} \right)\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)} = \sum\limits_j^{} {\left( \matrix{ r \cr j \cr} \right)\left( \matrix{ s \cr j \cr} \right)\left( \matrix{ m + r + s + 1 - j \cr m - j \cr} \right)} \tag{1} }$$

A possible way to demonstrate it is by taking the ogf over $m$ as follows

For the LHS $$ \eqalign{ & G_{\,a} (z) = \sum\limits_{0\, \le \,m} {\sum\limits_{j = 0}^m {\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)z^{\;m} } } = \cr & = \sum\limits_{0\, \le \,m} {\sum\limits_j^{} {\left( \matrix{ m - j \cr m - j \cr} \right)\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)z^{\;m} } } = \cr & = \sum\limits_{0\, \le \,m} {\sum\limits_j^{} {\left( \matrix{ m - j \cr m - j \cr} \right)z^{\;m - j} \left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)z^{\;j} } } = \cr & = \left( {\sum\limits_{0\, \le \,m} {\left( \matrix{ m \cr m \cr} \right)z^{\;m} } } \right)\left( {\sum\limits_{0\, \le \,m} {\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)z^{\;j} } } \right) \cr & = {1 \over {1 - z}}\sum\limits_{0\, \le \,k} {\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)z^{\;j} } = \cr & = {1 \over {1 - z}}\sum\limits_{0\, \le \,k} {{{\left( {r + 1} \right)^{\,\overline {\,k\,} } \left( {s + 1} \right)^{\,\overline {\,k\,} } } \over {1^{\,\overline {\,k\,} } }}{{z^{\;k} } \over {k!}}} = \cr & = {1 \over {1 - z}}{}_2F_{\,1} \left( {\left. {\matrix{ {r + 1,s + 1} \cr 1 \cr } \;} \right|\;z} \right) \cr} $$

For the RHS $$ \eqalign{ & G_{\,b} (r,s,z) = \sum\limits_{0\, \le \,m} {\sum\limits_j^{} {\left( \matrix{ r \cr j \cr} \right)\left( \matrix{ s \cr j \cr} \right)\left( \matrix{ r + s + 1 + m - j \cr m - j \cr} \right)z^{\;m} } } = \cr & = \sum\limits_{0\, \le \,m} {\sum\limits_j^{} {\left( \matrix{ r \cr j \cr} \right)\left( \matrix{ s \cr j \cr} \right)z^{\;j} \left( \matrix{ r + s + 1 + m - j \cr m - j \cr} \right)z^{\;m - j} } } = \cr & = \left( {\sum\limits_{0\, \le \,k} {\left( \matrix{ r \cr k \cr} \right)\left( \matrix{ s \cr k \cr} \right)z^{\;k} } } \right)\sum\limits_{0\, \le \,k} {\left( \matrix{ r + s + 1 + k \cr k \cr} \right)z^{\;k} } = \cr & = \left( {\sum\limits_{0\, \le \,k} {\left( \matrix{ r \cr k \cr} \right)\left( \matrix{ s \cr k \cr} \right)z^{\;k} } } \right){1 \over {\left( {1 - z} \right)^{\;r + s + 2} }} = \cr & = {1 \over {\left( {1 - z} \right)^{\;r + s + 2} }}\sum\limits_{0\, \le \,k} {\left( \matrix{ r \cr k \cr} \right)\left( \matrix{ s \cr k \cr} \right)z^{\;k} } = \cr & = {1 \over {\left( {1 - z} \right)^{\;r + s + 2} }}\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\;k} \left( { - 1} \right)^{\;k} \left( \matrix{ k - r - 1 \cr k \cr} \right)\left( \matrix{ k - s - 1 \cr k \cr} \right)z^{\;k} = } \cr & = {1 \over {\left( {1 - z} \right)^{\;r + s + 2} }}\sum\limits_{0\, \le \,k} {{{\left( { - r} \right)^{\,\overline {\,k\,} } \left( { - s} \right)^{\,\overline {\,k\,} } } \over {1^{\,\overline {\,k\,} } }}{{z^{\;k} } \over {k!}}} = \cr & = {1 \over {\left( {1 - z} \right)^{\;r + s + 2} }}{}_2F_{\,1} \left( {\left. {\matrix{ { - r, - s} \cr 1 \cr } \;} \right|\;z} \right) \cr} $$

The Euler transformation for the Hypergeometric gives $$ {}_2F_{\,1} \left( {\left. {\matrix{ { - r, - s} \cr 1 \cr } \;} \right|\;z} \right) = \left( {1 - z} \right)^{\,1 + r + s} {}_2F_{\,1} \left( {\left. {\matrix{ {r + 1,s + 1} \cr 1 \cr } \;} \right|\;z} \right) $$ which completes the demonstration.

It is interesting to note that the sides in id. 1 are polynomials in $r,s$ of degree $m,m$.

Therefore the identity holds as well for real and even complex values of $r, \,s$.

Furthermore (I realized just now going through my notes on binomial identities) it can be straightly deduced from this other basic identity $$ \bbox[lightyellow] { \eqalign{ & \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)\,} {( - 1)^{\,m - k} \left( \matrix{ x + y + 1 \cr m - k \cr} \right)\left( \matrix{ x + k \cr k \cr} \right)\left( \matrix{ y + k \cr k \cr} \right)} = \left( \matrix{ x \cr m \cr} \right)\left( \matrix{ y \cr m \cr} \right)\quad \Leftrightarrow \cr & \Leftrightarrow \quad \left( \matrix{ x + m \cr m \cr} \right)\left( \matrix{ y + m \cr m \cr} \right) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{ x \cr k \cr} \right)\left( \matrix{ y \cr k \cr} \right)\left( \matrix{ x + y + m - k \cr m - k \cr} \right)} \quad \left| \matrix{ \;{\rm integer}\,m \hfill \cr \,x,y \in C \hfill \cr} \right. \cr} \tag{2} }$$ that is called Suranyi's formula in this paper, and demonstrated therein.
It has also been dealt with and demonstrated in this related post.

In fact, from the above we get $$ \eqalign{ & \sum\limits_j^{} {\left( \matrix{ m - j \cr m - j \cr} \right)\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)} = \cr & = \sum\limits_j^{} {\sum\limits_k {\left( \matrix{ m - j \cr m - j \cr} \right)\left( \matrix{ r \cr k \cr} \right)\left( \matrix{ s \cr k \cr} \right)\left( \matrix{ r + s + j - k \cr j - k \cr} \right)} } = \cr & = \sum\limits_k^{} {\left( \matrix{ r \cr k \cr} \right)\left( \matrix{ s \cr k \cr} \right)\left( \matrix{ r + s + m + 1 - k \cr m - k \cr} \right)} \cr} $$

Using the symmetry $\binom{p}{q}=\binom{p}{p-q}$ of binomial coefficients, we want to show for non-negative integers $m$ and integers $0\leq r\leq s$: \begin{align*} \sum_{k=0}^m \binom{r+k}{r}\binom{s+k}{s} = \sum_{j=0}^s \binom{r}{j}\binom{s}{j} \binom{m+r+s-j+1}{r+s+1}\tag{1} \end{align*}

At first we look at the right-most binomial coefficient in (1) where we can apply the Hockey-stick identity. The summand $+1$ indicates a telescoping approach via $\binom{p+1}{q+1}-\binom{p}{q+1}=\binom{p}{q}$. Indeed, we obtain \begin{align*} \sum_{k=0}^m\binom{r+s-j+k}{r+s}&=\sum_{k=0}^m\left(\binom{r+s-j+k+1}{r+s+1}-\binom{r+s-j+k}{r+s+1}\right)\\ &=\binom{m+r+s-j+1}{r+s+1}\tag{2} \end{align*} where the telescoping sum permits cancellation of all terms besides the first and the last summand and the first summand vanishes also since it is $\binom{r+s-j}{r+s+1}=0$.

Using (2) we can now write the claim (1) in the form \begin{align*} \sum_{k=0}^m \binom{r+k}{r}\binom{s+k}{s} = \sum_{k=0}^m \sum_{j=0}^s \binom{r}{j}\binom{s}{j} \binom{m+r+s-j}{r+s} \tag{3} \end{align*}

In fact we can show that we have termwise equality in (3) for each $0\leq k\leq m$: \begin{align*} \binom{r+k}{r}\binom{s+k}{s} = \sum_{j=0}^s \binom{r}{j}\binom{s}{j} \binom{m+r+s-j}{r+s} \tag{4} \end{align*}

The identity (4) is named after Surányi (1955) which has already been mentioned by @GCab. We show the validity of (4) by closely following an approach given in Combinatorial identities by J. Riordan.

We start with the left-hand side of (4) and obtain \begin{align*} \color{blue}{\binom{r+k}{r}}&\color{blue}{\binom{s+k}{s}}\\ &=\sum_{j=0}^r\binom{r}{r-j}\binom{k}{j}\binom{s+k}{s}\tag{5}\\ &=\sum_{j=0}^r\binom{r}{r-j}\binom{s+j}{j}\binom{s+k}{s+j}\tag{6}\\ &=\sum_{j=0}^r\binom{r}{r-j}\binom{s+j}{j}\sum_{l=0}^{r-j}(-1)^{r-j-l}\binom{s+k+l}{s+r}\binom{r-j}{l}\tag{7}\\ &=\sum_{l=0}^r\binom{s+k+l}{s+r}\sum_{j=0}^{r-l}(-1)^{r-j-l}\binom{s+j}{j}\binom{r}{r-j}\binom{r-j}{l}\tag{8}\\ &=\sum_{l=0}^r\binom{s+k+l}{s+r}\binom{r}{l}\sum_{j=0}^{r-l}(-1)^{-j-l}\binom{s+j}{j}\binom{r-l}{r-l-j}\tag{9}\\ &=\sum_{l=0}^r\binom{s+k+l}{s+r}\binom{r}{l}(-1)^{r-l}\sum_{j=0}^{r-l}\binom{-s-1}{j}\binom{r-l}{r-l-j}\tag{10}\\ &=\sum_{l=0}^r\binom{s+k+l}{s+r}\binom{r}{l}(-1)^{r-l}\binom{-s-1+r-l}{r-l}\tag{11}\\ &=\sum_{l=0}^r\binom{s+k+l}{s+r}\binom{r}{l}\binom{s}{r-l}\tag{12}\\ &\,\,\color{blue}{=\sum_{l=0}^r\binom{r}{l}\binom{s}{l}\binom{r+s+k-l}{r+s}}\tag{13} \end{align*} and the claim (4) follows for $0\leq k\leq m$ and so also the claim (1).

Comment:

• In (5) we apply Vandermonde's identity to $\binom{r+k}{r}$.

• In (6) we use the binomial identity $\binom{k}{j}\binom{s+k}{s}=\binom{s+j}{j}\binom{s+k}{s+j}$.

• In (7) we use the binomial identity $\binom{n}{m}=\sum_{l=0}^r(-1)^{r-l}\binom{n+l}{m+r}\binom{r}{l}$ which is shown at the end of this post.

• In (8) we exchange the sums.

• In (9) we use the binomial identity $\binom{r}{r-j}\binom{r-j}{l}=\binom{r}{l}\binom{r-l}{r-l-j}$.

• In (10) we apply the binomial identity $\binom{-p}{q}=(-1)^q\binom{p+q-1}{q}$ to $\binom{s+j}{j}$.

• In (11) we apply again Vandermonde's identity.

• In (12) we use again $\binom{-p}{q}=(-1)^q\binom{p+q-1}{q}$.

• In (13) we change the order of summation $l\to r-l$.

Proof of (7): \begin{align*} \binom{n}{m}=\sum_{l=0}^r(-1)^{r-l}\binom{n+l}{m+r}\binom{r}{l}\tag{14} \end{align*}

We obtain \begin{align*} \color{blue}{\binom{n}{m}}&=(-1)^m\binom{-n+m-1}{m}\tag{12}\\ &=(-1)^m\sum_{l}\binom{-n+m-1-r}{l}\binom{r}{m-l}\tag{11}\\ &=\sum_{l}(-1)^{m+l}\binom{n-m+r+l}{l}\binom{r}{m-l}\tag{12}\\ &=\sum_{l}(-1)^l\binom{n+r-l}{m-l}\binom{r}{l}\tag{15}\\ &=\sum_{l}(-1)^l\binom{n+r-l}{n-m-l}\binom{r}{l}\tag{16}\\ &=\sum_{l}(-1)^l\binom{n+r-l}{m+r}\binom{r}{l}\tag{17}\\ &\,\,\color{blue}{=\sum_{l}(-1)^{r-l}\binom{n+l}{m+r}\binom{r}{l}}\tag{13} \end{align*} and the claim (14) follows.

Comment:

• In (15) we change the order of summation $l\to m-l$.

• In (16) we set $m\to n-m$, since $\binom{n}{m}=\binom{n}{n-m}$.

• In (17) we use $\binom{p}{q}=\binom{p}{p-q}$.