Let $m$ and $n$ be two non-negative integers. I am interested in finding a closed formula or an asymptotic approximation of the following sum $$ S_{n,m} = \sum_{k_1+\dots+k_m = n}\max(k_1,\dots,k_m)\binom{n}{k_1,\dots ,k_m} $$ Any hint would be much appreciated. Thanks.
A good asymptotic approximation with respect to $n$ (respectively $m$) while $m$ (respectively $n$) is fixed is also fine.
Edit: I found out that this question has been asked on mathoverflow more than 3 years ago but has only been answered to partially. This may mean that there is a chance a solution is already lurking somewhere or the problem itself might be quite hard.
A partial answer with a simple lower bound: $$\begin{eqnarray*}\sum_{k_1+\ldots+k_m=n}\max(k_1,\ldots,k_m)\binom{n}{k_1,\ldots,k_m}&\color{red}{\geq}& \frac{1}{m}\sum_{k_1+\ldots+k_m=n}(k_1+\ldots+k_m)\binom{n}{k_1,\ldots,k_m}\\ &=&\frac{1}{m}\,{\large\nabla\cdot}\!\!\!\left.\sum_{k_1+\ldots+k_m=n}x_1^{k_1}\cdots x_m^{k_m}\binom{n}{k_1,\ldots,k_m}\right|_{\bar{x}=1}\\&=&\frac{1}{m}\,\left.{\nabla\cdot}(x_1+\ldots+x_m)^n\right|_{\bar{x}=1}\\&=&\color{red}{n\,m^{n-1}}.\end{eqnarray*}$$ Here $\nabla\cdot$ is the divergence operator.