Closed formula to $N:=\sum_{j=0}^{k/2}\left(\begin{array}{c} n \\ k-j \end{array}\right)\left(\begin{array}{c} k-j \\ j \end{array}\right) $

Question

I gave mysel the following

Problem: For $k$ even and $n\geq k$, can one find a closed formula for the sum $$ N:=\sum_{j=0}^{k/2}\left(\begin{array}{c} n \\ k-j \end{array}\right)\left(\begin{array}{c} k-j \\ j \end{array}\right)? $$

At first, I thought that some sort of Vandermonde's identity could solve this. Now I believe this is far from trivial, if possible at all. I want to make sure I am not mistaken.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Note that $\ds{{k - j \choose j} = 0}$ when $\ds{j > {k \over 2}}$: The sum can be performed $\ds{\mbox{" up to $\ds{+\infty"}$}}$: \begin{align} {\cal N}\pars{k,n} & \equiv \bbox[5px,#ffd]{% \sum_{j = 0}^{k/2}{n \choose k - j}{k - j \choose j}} \\[5mm] & = \sum_{j = -\infty}^{0}{n \choose k + j} {k + j \choose -j} \\[5mm] & = \sum_{j = -\infty}^{k}{n \choose j} {j \choose k - j} = \sum_{j = 0}^{\infty}{n \choose j}{j \choose k - j} \\[5mm] & = \sum_{j = 0}^{\infty}{n \choose j}\bracks{z^{k - j}} \pars{1 + z}^{j} \\[5mm] & = \bracks{z^{k}}\sum_{j = 0}^{\infty}{n \choose j} \bracks{z\pars{1 + z}}^{j} \\[5mm] & = \bracks{z^{k}}\pars{1 + z + z^{2}}^{n} \\[5mm] & = \bracks{z^{k}}{1 \over \bracks{1 -2\pars{\color{red}{-1/2}}z + z^{2}} ^{\color{red}{-n}}} \\[5mm] & = \sum_{j = 0}^{\infty}\on{C}_{j}^{\color{red}{-n}}\, \pars{\color{red}{-{1 \over 2}}}z^{j} \end{align}

$\ds{C_{j}^{\pars{\alpha}}}$ is a Gegenbauer Polynomial. $$ \implies \bbx{{\cal N}\pars{k,n} = C_{k}^{\pars{\color{red}{-n}}}\, \pars{\color{red}{-{1 \over 2}}}} \\ $$

• $\ds{{\cal N}\pars{k,n} = 0}$ when $\ds{k \geq 2n + 1}$.
• $\ds{{\cal N}\pars{k,n} = {\cal N}\pars{k,n - k}}$.

Plot of $\ds{{\cal N}\pars{k,10}}$: