Guidance on the function $\sum_{k=1}^{n}\binom{2n}{n-k}k^{2j}$ I am writing this post to gather whether or not anyone has some knowledge or information about the function
\begin{equation}
\sum_{k=0}^{n-1}\binom{2n}{k}(n-k)^{2j} = \sum_{k=1}^{n}\binom{2n}{n-k}k^{2j},
\end{equation}
for $n,j\in\mathbb{N}_{0}$. This function arises in the $\textit{Taylor}$ series for $\cos(t)^{2n}$, which is a recreational problem I am trying to pursue. I must embarrassingly admit: I have absolutely no idea where to even begin with tackling this. However, I am not so much interested in finding an exact expression for this function, which should hopefully make my task a bit easier. For instance, using Wolfram Mathematica and plugging in several values of $j>0$ suggests that
\begin{equation}
\sum_{k=1}^{n}\binom{2n}{n-k}k^{2j} = 2^{2n-j-1}\left((2j-1)!!\cdot n^{j} + P_{j-1}(n)\right),
\end{equation}
where $n!!$ is the double factorial function, and $P_{k}(n)$ denotes a polynomial in $n$ of degree at most $k$. Proving the above statement (if it is true in the first place) would be sufficient. I am curious if this function or similar sorts have been well-studied by mathematicians? From the small amount of looking around that I have done, it seems that there is very little known about
\begin{equation}
\sum_{k=0}^{n}\binom{n}{k}k^{s},
\end{equation}
which surprises me. To clarify, I am not asking for a direct proof or solution to my conjectured statement, but rather any insights that can be offered (via your responses, recommended texts and/or online resources, etc.) which could potentially further this investigation, as I would greatly appreciate these sorts of things. If I make any progress on this, I will make sure to post updates. Thank you for your time.
 A: Concerning
\begin{equation}
f_s=\sum_{k=0}^{n}\binom{n}{k}k^{s}
\end{equation}
we know the solution in terms of hypergeometric functions. For example
$$f_4=n \, _4F_3(2,2,2,1-n;1,1,1;-1)$$
$$f_5=n \, _5F_4(2,2,2,2,1-n;1,1,1,1;-1)$$
$$f_6=n \, _6F_5(2,2,2,2,2,1-n;1,1,1,1,1;-1)$$ which clearly shows the pattern.
Beside the well known $f_0=2^n$ and $f_1=n 2^{n-1}$, we have
$$f_s= 2^{n-s}\,n\,P_{s-1}(n)$$
There is other patterns : if $s$ is odd, $P_{s-1}(n)=n\,Q_{s-2}(n)$ and
if $s$ is even, $P_{s-1}(n)=(n+1)\,R_{s-2}(n)$
where the first polynomials are
$$\left(
\begin{array}{cc}
s & Q_{s-2}(n) \\
 3 & n+3 \\
 5 & n^3+10 n^2+15 n-10 \\
 7 & n^5+21 n^4+105 n^3+35 n^2-210 n+112 \\
 9 & n^7+36 n^6+378 n^5+1008 n^4-1575 n^3-2436 n^2+5292 n-2448
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
s & R_{s-2}(n) \\
 2 & 1 \\
 4 & n^2+5 n-2 \\
 6 & n^4+14 n^3+31 n^2-46 n+16 \\
 8 & n^6+27 n^5+183 n^4+97 n^3-832 n^2+860 n-272 
\end{array}
\right)$$ where we can also find patterns for the coefficients (try with $OEIS$ to find some of them).
A: When looking at
\begin{align*}
\sum_{k=0}^n\binom{n}{k}k^s\tag{1}
\end{align*}
we can think of the series expansion of the exponential function
\begin{align*}
e^{kz}=1+kz+\frac{(kz)^2}{2}+\cdots = \sum_{j=0}^{\infty}\frac{(kz)^j}{j!}
=\sum_{j=0}^\infty \color{blue}{k^j}\frac{z^j}{j!}
\end{align*}
Denoting with $[z^s]$ the coefficient of $z^s$ of a series we see (assuming $s$ being a non-negative integer)
\begin{align*}
k^s=s![z^s]e^{kz}\tag{2}
\end{align*}
With the relation (2) we can apply the binomial theorem in (1) as follows:

We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\binom{n}{k}k^s}
&=\sum_{k=0}^n\binom{n}{k}s![z^s]e^{kz}\\
&=s![z^s]\sum_{k=0}^n\binom{n}{k}\left(e^z\right)^k\\
&\,\,\color{blue}{=s![z^s]\left(1+e^z\right)^n}
\end{align*}
and find a representation of (1) using a generating function.

