How can I evaluate $ \sum_{n=0}^{\infty}{\frac{x^{kn}}{(kn)!}} $ where $k$ is a natural number?

Question

I suddenly interested in the differential equation $$ f^{(k)}(x)=f(x) $$ So I tried to calculate for some $n$. When $ k=1 $, we know the solution $$ f(x)=A_0e^x=\sum_{n=0}^{\infty}{\frac{A_0x^n}{n!}} $$ Also, for $ k=2 $, $$ f(x)=Ae^x-Be^{-x}=\sum_{n=0}^{\infty}{(\frac{A_0x^{2n}}{(2n)!}+\frac{A_1x^{2n+1}}{(2n+1)!})} $$ where $ A_0=A+B $ and $ A_1=A-B $. Inductively, I could guess that the solution of the differential equation would be in the form $$ f(x)=\sum_{n=0}^{\infty}{\sum_{i=0}^{k-1}{\frac{A_ix^{kn+i}}{(kn+i)!}}} $$ But I could neither prove that it is the only solution nor get the explicit formula. How should I evaluate $ \sum_{n=0}^{\infty}{\frac{x^{kn}}{(kn)!}} $, cause if we know the answer for it, we can evaluate the original expression by differentiating it. Thanks to WolframAlpha, I know the answer for $ k=3 $, $$ \sum_{n=0}^{\infty}{\frac{x^{3n}}{(3n)!}}=\frac{1}{3}(2e^{-\frac {x}{2}}\cos{(\frac {\sqrt{3}}{2}x)}+e^{x}) $$ I think the answer might related to $ \sin $ and $ \cos $ of $ \frac {2\pi}{k} $.

Answer 1

This is a multisection of the exponential function.

Here is one of the many places this is discussed: https://en.wikipedia.org/wiki/Series_multisection

The main result (copied from there):

If $f(z) =\sum_{n=0}^{\infty} a_nz^n$ then $\sum_{m=0}^{\infty}a_{qm+p}z^{qm+p} =\dfrac1{q}\sum_{k=0}^{q-1} \omega^{-kp}f(\omega^kz) $ where $\omega =e^{\frac{2\pi i}{q}} $ is a primitive $q$-th root of unity.

Put $f(z) = e^z$.

Explicit formulas are given in the article above for the multisections of the exponential function. Here is the general one:

$\begin{array}\\ \sum_{m=0}^{\infty} \dfrac{z^{qm+p}}{(qm+p)!} &=\dfrac1{q}\sum_{k=0}^{q-1} e^{z\cos(2\pi k/q)} \cos\left(z \sin\left(\dfrac{2\pi k}{q}\right)-\dfrac{2\pi kp}{q}\right)\\ &=\dfrac1{q}\left(e^z+\sum_{k=1}^{q-1} e^{z\cos(2\pi k/q)} \cos\left(z \sin\left(\dfrac{2\pi k}{q}\right)-\dfrac{2\pi kp}{q}\right)\right)\\ \end{array} $