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Is there an explicit formula for the following quantity?

$$f_m(a_1,\ldots,a_n):=\sum_{\substack{k_1+\ldots+k_n=m \\ k_1,\ldots,k_n\in \mathbb{N}}} k_1^{a_1}\ldots k_n^{a_n}\ ,\hspace{1cm} m,a_1,\ldots,a_n\in \mathbb{N}$$

(for instance $f_m(0,\ldots,0)$ is simply the number of compositions of m into n parts, $f_m(1,1)=\frac{m(m+1)(m-1)}{6}$ and so on). I would like an answer both for the case where the $k_i$'s can and cannot attain the value $0$, if possible.

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I'm not sure if there is anything simpler than $f_m(a_1,\dots,a_n)$ being the coefficient of $x^m$ in the product of polylogarithms: $$\mathrm{Li}_{-a_1}(x)\cdot \mathrm{Li}_{-a_2}(x)\cdots \mathrm{Li}_{-a_n}(x).$$

In particular, since $\mathrm{Li}_{-1}(x) = x\cdot (1-x)^{-2}$, we have $$f_m(\underbrace{1,1,\dots,1}_n) = \mathrm{Coeff}_{x^m} x^n (1-x)^{-2n} = (-1)^{m-n}\cdot \binom{-2n}{m-n} = \binom{m+n-1}{2n-1}.$$

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  • $\begingroup$ Thank you very much for your answer. Actually that is what I started with. To provide some motivation, I was trying to compute, in a completely formal way, the product $\delta_+^{(a_1)}(x)\ldots\delta_+^{(a_n)}(x)$ as a linear combination of $\delta_+(x)$ and its derivatives, where $\delta_+=\sum_{k\geq 0}e^{ikx}$, or more precisely, the positive-Fourier-modes part of the Dirac delta distribution. Such distribution can of course be expressed as polylogarithms, as you suggest, and my question becomes: how to express your product of poylogs as a linear combination of other polylogs? $\endgroup$
    – issoroloap
    Commented Dec 3, 2014 at 23:58
  • $\begingroup$ Let me elaborate a little. I meant $i^s \mathrm{Li}_{-s}(e^{ix}) = \sum_{k>0} (ik)^s e^{ikx} = \left(\frac{1}{1-e^{ix}} \right)^{(s)} = \delta_+^{(s)}(x)$. $\endgroup$
    – issoroloap
    Commented Dec 4, 2014 at 0:06
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    $\begingroup$ $\mathrm{Li}_{-a}(x)$ is a rational function vanishing at $\infty$ whose only pole is at $x=1$, so this argument shows that for fixed $a_1,\ldots,a_n$, $f_m(a_1,\ldots,a_n)$ is a polynomial in $m$. $\endgroup$
    – Julian Rosen
    Commented Dec 4, 2014 at 0:09
  • $\begingroup$ actually the same argument, together with the property that $\mathrm{Li}_{-a}(x)=(-1)^{a+1}\mathrm{Li}_{-a}(\frac{1}{x})$ shows that $f_m(a_1,\ldots,a_n)$ is a $\mathbb{Z}_2$-homogeneous polynomial! $\endgroup$
    – issoroloap
    Commented Feb 7, 2015 at 18:10

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