Special sum of multinomial coefficients!

Question

I'm trying to find a explicit formula for the following sum in terms of $n$ and $k$. Any help would be appreciated.

\begin{equation} \sum_{a_1 + \dots + a_k = n} \frac{(2n)!}{(2 a_1)! \dots (2 a_k)!} = ? \end{equation}

Answer 1

Since $$(x_1+x_2+\ldots+x_k)^{2n} = \sum_{a_1+a_2+\ldots+a_k=2n}\frac{(2n)!}{a_1!a_2!\cdot\ldots\cdot a_k!} x_1^{a_1} x_2^{a_2}\cdot\ldots\cdot x_k^{a_k}$$ we may apply the following transformation to isolate in the RHS just the terms for which $a_1$ is even: $$ (x_1+x_2+\ldots+x_k)^{2n} \mapsto \frac{1}{2}\left[(x_1+x_2+\ldots+x_k)^{2n}+(-x_1+x_2+\ldots+x_k)^{2n}\right] $$ hence: $$ S=\sum_{\substack{a_1+a_2+\ldots+a_k=2n\\ a_i\text{ even}}}\frac{(2n)!}{a_1!a_2!\cdot\ldots\cdot a_k!}=\frac{1}{2^k}\sum_{e\in\{-1,+1\}^k}(e_1+e_2+\ldots+e_k)^{2n}$$ that leads to $$ S = \frac{1}{2^k}\sum_{j=0}^{k}\binom{k}{j}(k-2j)^{2n} $$ resembling (but not being equal) a Stirling number of the second kind. We may expand $(k-2j)^{2n}$ in terms of binomial coefficients and Stirling numbers of the second kind, then compute $$\sum_{j=0}^{k}\binom{k}{j}\binom{|k-2j|}{a} $$ to turn $S$ into a sum of powers of two multiplied by Stirling numbers of the second kind, but that still does not give a "nice" closed formula for $S$.

Answer 2

To cancel all of the multinomial coefficients with odd components, we can average $$ (\underbrace{\pm1\pm1\pm\dots\pm1}_{k\text{ terms}})^{2n} $$ which gives $$ \frac1{2^k}\sum_{j=0}^k\binom{k}{j}(k-2j)^{2n} $$ I have not yet been able to simplify this further.