After thinking about how
$$\left(\sum_{k=0}^\infty \frac{1}{k!}\right)^{z} = \sum_{k=0}^\infty \frac{z^k}{k!}$$
I wondered about what kind of sequence $(a_n)_{n=0}^\infty$ satisfies
$$\left(\sum_{k=0}^\infty a_k\right)^{z} = \sum_{k=0}^\infty a_k z^k \tag{1}$$
for all $z \in \mathbb{C}$. For instance, any such sequence would satisfy
$$\left(\sum_{k=0}^\infty a_k z^k\right) \left(\sum_{k=0}^\infty a_k w^k\right) = \left(\sum_{k=0}^\infty a_k\right)^z \left(\sum_{k=0}^\infty a_k\right)^w = \left(\sum_{k=0}^\infty a_k\right)^{z+w} = \sum_{k=0}^\infty a_k (z+w)^k$$
I will use the notation $(a_n)$ to denote $(a_n)_{n=0}^\infty$ and $(a_n)^z$ to denote $(a_n z^n)$. I would like to find a product $(a_n) \cdot (b_n)$ of such sequences which satisfies
$$(a_n)^z \cdot (a_n)^w = (a_n)^{z+w} \tag{2}$$
for all $z,w \in \mathbb{C}$. If we only consider sequences which satisfy $a_n \binom{n}{k} = a_k a_{n-k}$ for all $k$, then a discrete convolution would be such a product since
$$(a_n)^z \cdot (a_n)^w = (c_n)$$
where
$$c_n = \sum_{k=0}^n a_k a_{n-k} z^k w^{n-k} = a_n \sum_{k=0}^n \binom{n}{k} z^k w^{n-k} = a_n (z+w)^n$$
as required.
Are there any other sequences which satisfy (2) where the product is a discrete convolution? Are there any other products which satisfy (2), for some subset of sequences satisfying (1)? And more importantly, is $(\frac{1}{n!})$ the only sequence satisfying (1)?
Best Answer
Suppose $a_k$ is a sequence of such that such that $\sum_{k=0}^\infty a_k$ converges. Write $c = \sum_{k=0}^\infty a_k$. Assume $c>0$, let $\log c$ be the natural logarithm of $c$. Then $$ \left(\sum_{k=0}^\infty a_k\right)^z = c^z = \exp(z\log c) = \sum_{k=0}^\infty \frac{(\log c)^k}{k!} z^k $$
So, in order to get the equation you request, there must be a number $c>0$ such that $a_k = \frac{(\log c)^k}{k!}$.
Of course the famous example is where $c=e$, and $\log c = 1$, so $(\log c)^k = 1$.
I suppose, in case $c > 0$ fails, then we have to inestigate choosing branches of the logarithm.
For example, we could define $(-1)^z = \exp(i \pi z)$ so that $$ a_k = \frac{i^k \pi^k}{k!} . $$