For context, I recently learned about how $\exp(D)$ is the shift operator. Upon learning this, I started messing around with other functions of the derivative. Specifically, I started experimenting with "differential" equations (not sure if differential is the right word here) of the form $$f(x)=g(D)f(x)$$ Where $g$ is an analytic function (at 0) and $f$ is a smooth function. We find that for $g(x)=x$, $f(x) = e^x$; for $g(x)=e^x$ we find $f$ is a function with a period of one.
Then, I next asked the question "Is it possible for $g = f$?" Since $g$ is analytic, it follows that $f$ must be as well then. Thus Maclaurin expansions finds that $$\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n= \sum_{i=0}^\infty \frac{f^{(i)}(0)f^{(i)}(x)}{i!}$$
If we expand and apply linear independence, we can derive that $$f^{(n)}(0) = \sum_{i=0}^\infty \frac{f^{(i)}(0)f^{(n+i)}(0)}{i!}$$
Of course, most of this is unnecessary, we can simplify this to a question about sequences. Let $a_n = f^{(n)}(0)$. Then we have that sequence recursively defined as $$a_n = \sum_{i=0}^\infty \frac{a_ia_{n+i}}{i!}$$ This is the part that I'm struggling with. I don't know how I would go about finding such a sequence. Quite obviously $a_n = \delta_{n,0}$ works, but is overall a quite boring solution. I want to know whether there exist other sequences satisfying this recursive definition, and how do we find them?