How to simplify this combinatorial expression? Find 
\begin{eqnarray}
\sum_{j\in\mathbb{N}}(n-2j)^k\binom{n}{2j-m}
\end{eqnarray}
Note that this question is a generalization of this one. I tried to imitate the steps in the answer given in that post but without success. Any idea?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{j\ \in\ \mathbb{N}}\pars{n - 2j}^{\,k}\ {n \choose 2j - m} & =
\sum_{j = 0}^{\infty}\pars{n - j}^{\,k}\ {n \choose j - m}{1 + \pars{-1}^{\,j} \over 2} =
{\mrm{f}\pars{1} + \mrm{f}\pars{-1} \over 2}\label{1}\tag{1}
\\[2mm] &\ \mbox{where}\ \,\mrm{f}\pars{x} \equiv
\sum_{j = m}^{\infty}\pars{n - j}^{k}\ {n \choose j - m}x^{\,j}
\end{align}

\begin{align}
\mrm{f}\pars{x} & =
x^{m}\sum_{j = 0}^{\infty}{n \choose j}x^{\,j}\,\pars{n - m - j}^{\,k} =
x^{m}\sum_{j = 0}^{\infty}{n \choose j}x^{\,j}\ \overbrace{%
\bracks{k!\oint_{\verts{z} = 1}{\expo{\pars{n - m - j}z} \over z^{k + 1}}}
\,{\dd z \over 2\pi\ic}}^{\ds{\pars{n - m - j}^{\,k}}}
\\[5mm] & =
k!\,x^{m}\oint_{\verts{z} = 1}{\expo{\pars{n - m}z} \over z^{k + 1}}
\sum_{j = 0}^{\infty}{n \choose j}\pars{x\expo{-z}}^{\,j}\,{\dd z \over 2\pi\ic} =
k!\,x^{m}\oint_{\verts{z} = 1}{\expo{\pars{n - m}z} \over z^{k + 1}}
\pars{1 + x\expo{-z}}^{n}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
k!\,x^{m}\oint_{\verts{z} = 1}{\expo{-mz} \over z^{k + 1}}
\pars{\expo{z} + x}^{n}\,{\dd z \over 2\pi\ic}
\end{align}

$\ds{\Large\mrm{f}\pars{-1}:\ ?.}$
\begin{align}
\mrm{f}\pars{-1} & =
k!\,\pars{-1}^{m}\oint_{\verts{z} = 1}{\expo{-mz} \over z^{k + 1}}
\pars{\expo{z} - 1}^{n}\,{\dd z \over 2\pi\ic}
\end{align}
The integral over $\ds{z}$ can be evaluated by using an identity which involves the
Stirling Number of the Second Kind
$\ds{a \brace b}$. Namely,
\begin{equation}
\pars{\expo{z} - 1}^{s} = s!\sum_{j = 0}^{\infty}{j \brace s}{z^{\,j} \over j!}
\label{2}\tag{2}
\end{equation}
Then,
\begin{align}
\oint_{\verts{z} = 1}{\expo{-mz} \over z^{k + 1}}
\pars{\expo{z} - 1}^{s}\,{\dd z \over 2\pi\ic} & =
s!\sum_{j = 0}^{\infty}{j \brace s}{1 \over j!}
\oint_{\verts{z} = 1}{\expo{-mz} \over z^{k + 1 - j}}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
s!\sum_{j = 0}^{\infty}{j \brace s}{1 \over j!}\,
{\pars{-m}^{k - j} \over \pars{k - j}!}
\\[5mm] & =
\bracks{k \geq s}{s! \over k!}\,\pars{-1}^{k}\,m^{k}
\sum_{j = s}^{k}{j \brace s}{k \choose j}\,{\pars{-1}^{\,j} \over m^{\,j}}
\label{3}\tag{3}
\end{align}
\begin{align}
&\mbox{With this result,}\ \,\mrm{f}\pars{-1}\ \mbox{is given by}
\\
&\bbx{\mrm{f}\pars{-1} = \bracks{k \geq n}\pars{-1}^{m + k}\,m^{k}\,n!
\sum_{j = n}^{k}{j \brace n}{k \choose j}{\pars{-1}^{\,j} \over m^{\,j}}}
\label{4}\tag{4}
\end{align}

$\ds{\Large\mrm{f}\pars{1}:\ ?.}$
\begin{align}
\mrm{f}\pars{1} & =
k!\oint_{\verts{z} = 1}{\expo{-mz} \over z^{k + 1}}
\pars{\expo{z} + 1}^{n}\,{\dd z \over 2\pi\ic} =
k!\oint_{\verts{z} = 1}{\expo{-mz} \over z^{k + 1}}
2^{n}\pars{1 + {\expo{z} - 1 \over 2}}^{n}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
k!\sum_{\ell = 0}^{n}{n \choose \ell}
\oint_{\verts{z} = 1}{\expo{-mz} \over z^{k + 1}}
2^{n}\pars{\expo{z} - 1 \over 2}^{n\ell}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
k!\sum_{\ell = 0}^{n}{n \choose \ell}2^{\pars{1 - \ell}n}
\oint_{\verts{z} = 1}{\expo{-mz} \over z^{k + 1}}\pars{\expo{z} - 1}^{n\ell}
\,{\dd z \over 2\pi\ic}
\end{align}
With result \eqref{3}:
\begin{align}
&\mrm{f}\pars{1} =
k!\sum_{\ell = 0}^{n}{n \choose \ell}2^{\pars{1 - \ell}n}\braces{%
\bracks{k \geq n\ell}{\pars{n\ell}! \over k!}\,\pars{-1}^{k}\,m^{k}
\sum_{j = n\ell}^{k}{j \brace n\ell}{k \choose j}
\,{\pars{-1}^{\,j} \over m^{\,j}}}
\\[5mm] &
\bbx{\mrm{f}\pars{1} = \pars{-1}^{k}\,2^{n}\,m^{k}\sum_{\ell = 0}^{M}\sum_{j = n\ell}^{k}
{n \choose \ell}{j \brace n\ell}{k \choose j}\pars{-1}^{\,j}\,
{\pars{n\ell}! \over 2^{n\ell}m^{\,j}}}\label{5}\tag{5}
\\
&\mbox{where}\quad M \equiv
\min\braces{\left\lfloor\,{k \over n}\,\right\rfloor,n}
\end{align}


The final result is given by \eqref{1}, \eqref{4} and \eqref{5}. 

