Polynomial expression for $\frac 1{2^n} \sum_{i=0}^n \binom{n}{i}(2i-n)^{2k}$ Let
$$F (n,k)=\frac {1}{2^n}\sum_{i=0}^n \binom{n}{i}(2i-n)^{2k},$$
where $n,k$ are non-negative integers.
By numerical tests the expression is an integer polynomial in $n $ of order $k $:
$$ F(n,0)=1;
F (n,1)=n;
F (n,2)=n (3n-2),$$
and so on.
Is there a simple general expression for the polynomial?
 A: It  is convenient to use the  coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
\begin{align*}
n![z^n]e^{jz}=j^n\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\frac{1}{2^n}}&\color{blue}{\sum_{j=0}^n\binom{n}{j}(2j-n)^{2k}}\\
&=\frac{1}{2^n}\sum_{j=0}^n\binom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}\tag{2}\\
&=\frac{(2k)!}{2^n}[z^{2k}]e^{-nz}\sum_{j=0}^n\binom{n}{j}\left(e^{2z}\right)^j\tag{3}\\
&=\frac{(2k)!}{2^n}[z^{2k}]e^{-nz}\left(1+e^{2z}\right)^n\tag{4}\\
&=(2k)![z^{2k}]\left(\frac{e^{z}+e^{-z}}{2}\right)^n\tag{5}\\
&\,\,\color{blue}{=(2k)![z^{2k}]\left(\cosh z\right)^n}
\end{align*}
We see OPs formula is essentially the coefficient of $z^{2k}$ of $\left(\cosh z\right)^n$ which does not have a closed formula as far as I know.

Comment:


*

*In (2) we apply the coefficient of operator according to (1).

*In (3) use the linearity of the coefficient of operator.

*In (4) we apply the binomial theorem.

*In (5) we write the expression somewhat more conveniently.
A: Let $\omega $ be a $n$-dimensional vector with binary components $\omega_i=\pm1$ and $\Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $\binom {n}{i}$. Thus
$$
F (n,k)=\frac {1}{2^n}\sum_{\omega\in\Omega_n } \left (\sum_{i=1}^n\omega_i \right)^{2k}
=\frac {1}{2^n}\sum_{\omega\in\Omega_n } \sum_{p_i\ge0}^{\sum_i p_i=2k}\binom {2k} {p_1,p_2,\dots,p_n}\prod_i \omega_i^{p_i}  
=\sum_{p_i\ge0}^{\sum_i p_i=2k}\binom {2k} {p_1,p_2,\dots,p_n}\left (\frac {1}{2^n}\sum_{\omega\in\Omega_n }\prod_i \omega_i^{p_i}\right)
=\sum_{p_i\ge0,\;p_i\,\text {mod}\,2=0}^{\sum_i p_i=2k}\binom {2k} {p_1,p_2,\dots,p_n}.
$$
To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
$$
F (n,k)=\sum_{l=1}^n T (k,l)n^\underline{l},\tag {1}
$$
where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size, and $n^\underline{l}$ is falling factorial. $T(k,l)$ can be recognized as the OEIS sequence A156289 with known close-form and recurrence expressions.

Note added: by numerical evidence  the polynomial (1) can be expressed in the terms of usual powers as:
$$
F (n,k)=\sum_{l=1}^n A (k,l)n^l,\tag {2}
$$
with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact  the so called Omega polynomial.
A: I'll play with the cosh and
see if I get 
anything other than
the original problem.
$\begin{array}\\
\cosh^n(x)
&=\frac1{2^n}(e^x+e^{-x})^n\\
&=\frac1{2^n}\sum_{k=0}^n \binom{n}{k}e^ke^{(n-k)x}\\
&=\frac1{2^n}e^{nx}\sum_{k=0}^n \binom{n}{k}e^{-2kx}\\
&=\frac1{2^n}\sum_{i=0}^{\infty} \dfrac{(nx)^i}{i!}\sum_{k=0}^n \binom{n}{k}\sum_{j=0}^{\infty} \dfrac{(-2kx)^j}{j!}\\
&=\frac1{2^n}\sum_{k=0}^n \binom{n}{k}\sum_{i=0}^{\infty} \dfrac{(nx)^i}{i!}\sum_{j=0}^{\infty} \dfrac{(-2kx)^j}{j!}\\
&=\frac1{2^n}\sum_{k=0}^n \binom{n}{k}\sum_{m=0}^{\infty}\sum_{i=0}^{m} \dfrac{(nx)^i}{i!} \dfrac{(-2kx)^{m-i}}{(m-i)!}\\
&=\frac1{2^n}\sum_{k=0}^n \binom{n}{k}\sum_{m=0}^{\infty}x^m\sum_{i=0}^{m} \dfrac{(n)^i}{i!} \dfrac{(-2k)^{m-i}}{(m-i)!}\\
&=\frac1{2^n}\sum_{m=0}^{\infty}x^m\sum_{k=0}^n \binom{n}{k}\sum_{i=0}^{m} \dfrac{(n)^i}{i!} \dfrac{(-2k)^{m-i}}{(m-i)!}\\
&=\frac1{2^n}\sum_{m=0}^{\infty}x^m\sum_{i=0}^{m}\dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}\sum_{k=0}^n \binom{n}{k}  k^{m-i}\\
&=\frac1{2^n}\sum_{m=0}^{\infty}\dfrac{x^m}{m!}\sum_{i=0}^{m}\binom{m}{i}(-2)^{m-i}n^i\sum_{k=0}^n \binom{n}{k}  k^{m-i}\\
&=\frac1{2^n}\sum_{m=0}^{\infty}\dfrac{(-2)^mx^m}{m!}\sum_{i=0}^{m}\binom{m}{i}(-n/2)^i\sum_{k=0}^n \binom{n}{k}  k^{m-i}\\
&=\frac1{2^n}\sum_{m=0}^{\infty}\dfrac{(-2)^mx^m}{m!}\sum_{k=0}^n \binom{n}{k}\sum_{i=0}^{m}\binom{m}{i}(-n/2)^i  k^{m-i}\\
&=\frac1{2^n}\sum_{m=0}^{\infty}\dfrac{(-2)^mx^m}{m!}\sum_{k=0}^n \binom{n}{k}(-n/2+k)^m\\
&=\frac1{2^n}\sum_{m=0}^{\infty}\dfrac{x^m}{m!}\sum_{k=0}^n \binom{n}{k}(2k-n)^m\\
\end{array}
$
And this is the OP.
Oh well.
