Let $\lambda>0$ and $m \ge 1, r\ge 1$ be integers. Can one calculate the sum of the series $$ \sum_{n \ge 0}\frac{\lambda^{mn+r}}{(mn+r)!}e^{-\lambda}? $$
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Obviously the factor $e^{-\lambda}$ is quite irrelevant here. For any $r\in[0,m-1)$ it is possible to compute the closed form for
$$S=\sum_{n\geq 0}\frac{\lambda^{mn+r}}{(mn+r)!}=\sum_{n\equiv r\!\pmod{\!\!m}}\frac{\lambda^n}{n!}\tag{1}$$ since it is possible to write the indicator function of the intergers $\equiv r\pmod{m}$ in terms of the $m$-th roots of unity (discrete Fourier transform). For instance, given that $\zeta_m=\exp\left(\frac{2\pi i}{m}\right)$ we have:
$$\frac{1}{m}\sum_{h=0}^{m-1}\zeta_m^{hn} = \mathbb{1}_{n\equiv 0\!\pmod{\!\!m}}\tag{2}$$ hence for $r=0$ we have: $$ S = \frac{1}{m}\sum_{n\geq 0}\frac{(1+\zeta_m^n+\zeta_m^{2n}+\ldots+\zeta_m^{(m-1)n})\,\lambda^n}{n!}\\=\frac{1}{m}\left(e^\lambda+e^{\lambda\zeta_m}+e^{\lambda\zeta_m^2}+\ldots+e^{\lambda\zeta_m^{m-1}}\right)\tag{3}$$ and it is not difficult to do the same for different values of $r$.
Jack D'Aurizio
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