TITLE: How can I simplify this sum any further?
QUESTION [11 upvotes]: Recently I was playing around with some numbers and I stumbled across the following formal power series:
$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)$$
I was able to "simplify" the above expression for $a=1$:
$$\sum_{k=0}^\infty\frac{x^k}{k!}\cdot2^k=e^{2x}$$
I also managed to simplify the expression for $a=2$ with the identity $\sum_{i=0}^\infty\frac{x^{2k}}{(2k)!}=\cosh(x)$:
$$\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\biggl(\sum_{l=0}^k\binom{2k}{2l}\biggr)=\mathbf[\cdots\mathbf]=\frac{1}{4}\cdot(e^{2x}+e^{-2x})+\frac{1}{2}=\frac{1}{2}\cdot(\cosh(2x)+1)$$
However, I couldn't come up with a general method for all $a\in\Bbb{N}$. I would be very thankful if someone could either guide me towards simplifying this expression or post his solution here.

REPLY [7 votes]: The explicit formula is as follows:
$$
S_a=\frac{1}{a^2}\left(\sum_{z^a=2^a}+2\sum_{p_a(z)=0}\right)e^{az}
$$
where the polynomials $p_a$ are given by A244608. For example,
\begin{align}
p_9(z)&= 1 - 13604 z^9 - 13359 z^{18} + 247 z^{27} + z^{36}\\
p_{10}(z)&=3125-383750 z^{10}-73749 z^{20}+502 z^{30}+z^{40}
\end{align}
whose roots in the complex plane look like follows:


The first few solutions are
\begin{align}
S_1&=e^{2 x}\\
S_2&=\frac{e^{-2 x}}{4}+\frac{e^{2 x}}{4}\\
S_3&=\frac{2 e^{-x}}{9}+\frac{e^{2 x}}{9}+\frac{2}{9} e^{\frac{x}{2}-\frac{1}{2} i \sqrt{3} x}+\frac{2}{9} e^{\frac{x}{2}+\frac{1}{2} i \sqrt{3} x}+\frac{1}{9} e^{-x-i \sqrt{3} x}+\frac{1}{9} e^{-x+i \sqrt{3} x}\\
S_4&=\frac{e^{-2 x}}{16}+\frac{1}{8} e^{(-1-i) x}+\frac{1}{8} e^{(-1+i) x}+\frac{1}{16} e^{-2 i x}+\frac{1}{16} e^{2 i x}+\frac{1}{8} e^{(1-i) x}+\frac{1}{8} e^{(1+i) x}+\frac{e^{2 x}}{16}
\end{align}
as given by
\begin{align}
p_1(z)&=0\\
p_2(z)&=0\\
p_3(z)&=1+z^3\\
p_4(z)&=4+z^4\\
p_5(z)&=-1+11z^5+z^{10}\\
p_6(z)&=-27+26z^6+z^{12}
\end{align}
etc. Quoting the OEIS entry, the coefficients are found as follows:

Let $\omega$ be a primitive $j$-th root of unity. Let $L(k)=\sum_{p=0}^{j-1} c(p)\omega^{kp}$ with $c(0)=2$ and $c(i)=C(j,i)$ if $i>0$. Then $p(j,X)=(X-L(1))(X-L(2))\dots(X-L([(n-1)/2]))$.

REPLY [2 votes]: Actually the true generalisation for your $a=1$, $a=2$ cases is this expression for your sum: $$\frac{1}{a^2}\sum_{s=0}^{a-1}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)}.$$ where $\omega$ is an $a$-th root of unity ($\omega^a=1$, $a \neq 1$). 
When $a=1$ we have $\omega=1$ and your sum is $e^{2x}$. For $a=2$ we have $\omega=-1$ and your sum is $\frac{1}{4}(e^{2x}+e^{-2x}+e^0+e^0)=\frac{1}{4}(e^{2x}+e^{-2x}+2)$. Both agreeing with your observations.
Proof:
Using the fact that $\sum_{l=0}^{a-1}\omega^{lr}=0$ unless $a|l$ we can show:
$$\sum_{l=0}^k\binom{ak}{al}=\frac{1}{a}\sum_{s=0}^{a-1}(1+\omega^s)^{ak}$$ and$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}=\frac{1}{a}\sum_{r=0}^{a-1}e^{\omega^r x}.$$
Substituting the above expressions into your sum we obtain:
$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)=\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\frac{1}{a}\sum_{s=0}^{a-1}(1+\omega^s)^{ak}=\frac{1}{a}\sum_{s=0}^{a-1}\sum_{k=0}^\infty\frac{(x(1+\omega^s))^{ak}}{(ak)!}$$$$=\frac{1}{a}\sum_{s=0}^{a-1}(\frac{1}{a}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)})=\frac{1}{a^2}\sum_{s=0}^{a-1}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)}.$$