Is there a closed form for $\sum_{|\gamma|=k} \gamma!$ Consider the multi-index $\gamma=(\gamma_1,\ldots, \gamma_n)$. Is there a closed form for the sum $\sum_{|\gamma|=k} \gamma!$ in terms of $n$ and $k$? Asymptotics, or good upper bounds are also very helpful.
Here is what I have tried. Let $f(x)=\sum_{i=0} i! x^{i+2}$. This generating function satisfies the ODE $x^2f'(x)=f(x)-x$. Then the sum $\sum_{|\gamma|=k} \gamma!$ is the coefficient of $x^{k+2n}$ in $f(x)^n$. However I don't have any means to find this coefficient.
 A: Dividing by $(\gamma_1+\cdots+\gamma_n)!$ you are trying to estimate $$\sum_{\gamma \vdash k}\binom{k}{\gamma_1,\dots,\gamma_n}^{-1}$$
One way to start would be to write it as an integral over the probability simplex (sum of generalized beta functions) or to try to find some information from the generating function. For $n=2$, for instance the integral representation is $$\sum_{i=0}^k \binom{k}{i}^{-1}=(k+1)\int_{0}^1 \frac{(1-t)^{k+1}-t^{k+1}}{1-2t}dt$$
and the generating function is $$\sum_{k=0}^{\infty} \left(\sum_{i=0}^k \binom{k}{i}^{-1}\right)x^k=\frac{2x}{1-x}-\frac{2x}{2-x}-\frac{2x \log(1-x)}{(2-x)^2}$$
from which it is not very hard to see that
$$\sum_{i=0}^k \binom{k}{i}^{-1} \sim 2+\frac{2}{k-1}$$ so $$\sum_{i=0}^k i!(k-i)!\sim k!\left(2+\frac{2}{k-1}\right)$$
A: Let's try $n=2$.  Maple says:
$$
\sum_{j = 0}^{k} j!(k - j)! = -\Biggl(\mathrm{LerchPhi} (2,1,k) k^{2} + 
\mathrm{LerchPhi} (2,1,k) k + \frac{i \pi k^{2}}{2^{k}} + \frac{i \pi k}{2^{k}} - 3 k - 1\Biggr) \frac{\Gamma (k)}{2}
$$
but doesn't know asymptotics for $\mathrm{LerchPhi}(2,1,k)$.
