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I am looking at $$f_{j,k}(x) = \sum_{n=0}^\infty \frac{x^n}{(k+j\cdot n)!}$$ where $j$ is a positive integer, and $k = 0,\ldots, j-1$. The case $j = 1$ admits the expression $$f_{1,k}(x) = e^x x^{-k} \frac{\Gamma(k) - \Gamma(k,x)}{\Gamma(k)}$$ in terms of respectively the Gamma function and the incomplete Gamma function. The problem emerges in the solution of stationary p.d.f. of simple chemical reaction networks Markov dynamics. I can share more insights on this if of help.

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    $\begingroup$ @LSpice: I think the question is implicit from the title- does $f_{j,k}(x)$ have an analytical expression similar to/extending the case $j=1$ given. But yes it would be good to write this explicitly. $\endgroup$ Jan 10, 2021 at 20:02
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    $\begingroup$ It is convenient to treat $g_{j,k}(x):=x^kf_{j,k}(x^j)$. Then $g_{j,k}$ solves a linear ode with constant coefficients, whence the usual analytic expression of solutions. (And $k=0$ sufficies, for the other are obtained by derivation) $\endgroup$ Jan 10, 2021 at 20:10
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    $\begingroup$ If I am not mistaken, the $g_{j,0}$ described by Pietro are linear combinations of exponentials: $\exp(\zeta^0x)+\cdots+\exp(\zeta^{j-1}x)=jg_{j,0}$ when $\zeta$ is a primitive $j$th root of unity. $\endgroup$
    – Pierre PC
    Jan 10, 2021 at 20:50
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    $\begingroup$ These are called Olivier functions. $\endgroup$
    – Ira Gessel
    Jan 11, 2021 at 0:42
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    $\begingroup$ See en.wikipedia.org/wiki/Series_multisection $\endgroup$ Jan 11, 2021 at 3:44

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It is convenient to treat $g_{j,k}(x):=x^kf_{j,k}(x^j)$. So $D g_{j,0}=g_{j,j-1}$ and $D g_{j,k}=g_{j,k-1}$ for $0<k<j$, so the $g_{j,0},\dots, g_{j,j-1}$ are the solutions of the linear homogeneous ode $(D^j-I)u=0$, with initial conditions $D^n g_{j,k}(0)=\delta_{n,k}$. whence the expressions $$g_{j,0}(x)=\frac 1j\sum_{n=0}^{j-1} \exp(e^{\frac{2in\pi}j} x)$$ and in general (also from the above, taking $k$ derivatives) $$g_{j,k}(x)=\frac 1j\sum_{n=0}^{j-1} e^{-\frac{2ink\pi}j} \exp(e^{\frac{2in\pi}j} x).$$

(Thanks to Pierre PC for the exact value of the coefficients).

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  • $\begingroup$ Thanks, very nice! $\endgroup$
    – tomate
    Jan 11, 2021 at 10:47

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