Closed form or estimate for $m-$fold sum of binomial coefficients

Question

The coefficient of $x^j$ in the polynomial $$ (1+x+x^2+\cdots+x^{k-1})^m $$ is, I believe, $$ \large{ C(k,m;j)\stackrel{\triangle}{=}\sum_{s=1,\ldots,j} ~~\sum_{\stackrel{f_1+\cdots+f_s=j}{f_i\geq 1}} \binom{m}{f_1} \binom{m}{f_2} \cdots \binom{m}{f_l}. } $$ Are there reasonable estimates for this quantity? I am interested in $k$ and $m$ going to infinity together, perhaps $k=\lambda m$ for some $\lambda>0,$ and in how close the probability distribution obtained by normalizing $$\{C(k,m;j): 0\leq j\leq (m-1)k\}$$supported on $\{0,1,\ldots,(k-1)m\}$ is to the corresponding binomial distribution $\mathrm{Bin}((k-1)m,1/2)$ on the same support.

Answer 1

Since $(1+x+x^2+\cdots+x^{k-1})^m=\left(\dfrac{1-x^k}{1-x}\right)^m$, you can expand as follows:

$(1+x+x^2+\cdots+x^{k-1})^m=\left(\dfrac{1-x^k}{1-x}\right)^m=\sum\limits_{i=0}^n\dbinom{m}{i}(-x)^{ki}\sum\limits_{j\geq0}\dbinom{m+j-1}{j}x^j$.

So, the coefficient of $x^j$ is

$\sum\limits_{i=0}^{\left\lfloor j/k\right\rfloor}(-1)^i\dbinom{m}{i}\dbinom{m+j-ki-1}{m-1}$.

This should provide you with better estimates.