A combinatorial identity related to Chebyshev differential equation Let $m,k$ be an positive integers with $k\le m$. Does anyone have a proof that $$\sum_{j=k}^m {2m+1\choose 2j+1}{j\choose k}=\frac{2^{2(m-k)}(2m-k)!}{(2m-2k)!k!}?$$
This is related to Chebyshev differential equation.
 A: Consider the generating function
$$
\begin{align}
&\sum_{m=k}^\infty\sum_{j=k}^m\binom{2m+1}{2j+1}\binom{j}{k}x^{2m}\\
&=\sum_{m=k}^\infty\sum_{j=k}^m\binom{2m+1}{2m-2j}\binom{j}{k}x^{2m}\\
&=\sum_{j=k}^\infty\sum_{m=0}^\infty\binom{-2j-2}{2m}\binom{j}{k}x^{2m+2j}\\
&=\frac12\sum_{j=k}^\infty\left((1+x)^{-2j-2}+(1-x)^{-2j-2}\right)\binom{j}{k}x^{2j}\\
&=\frac12\sum_{j=k}^\infty\left(\frac{x^{2j}}{(1+x)^{2j+2}}+\frac{x^{2j}}{(1-x)^{2j+2}}\right)(-1)^{j-k}\binom{-k-1}{j-k}\\
&=\frac12\sum_{j=0}^\infty\left(\frac{x^{2j+2k}}{(1+x)^{2j+2k+2}}+\frac{x^{2j+2k}}{(1-x)^{2j+2k+2}}\right)(-1)^j\binom{-k-1}{j}\\
&=\frac12\left(\frac{x^{2k}}{(1+x)^{2k+2}}\left(1-\frac{x^2}{(1+x)^2}\right)^{-k-1}+\frac{x^{2k}}{(1-x)^{2k+2}}\left(1-\frac{x^2}{(1-x)^2}\right)^{-k-1}\right)\\
&=\frac12\left(x^{2k}(1+2x)^{-k-1}+x^{2k}(1-2x)^{-k-1}\right)\\
&=\sum_{m=0}^\infty\binom{-k-1}{2m}2^{2m}x^{2k+2m}\\
&=\sum_{m=k}^\infty\binom{-k-1}{2m-2k}2^{2m-2k}x^{2m}\\
&=\sum_{m=k}^\infty\binom{2m-k}{2m-2k}2^{2m-2k}x^{2m}\\
\end{align}
$$
Therefore, equating powers of $x$, we get
$$
\sum_{j=k}^m\binom{2m+1}{2j+1}\binom{j}{k}=\binom{2m-k}{2m-2k}2^{2m-2k}
$$
