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) $ 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.

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