I have a proof of the following fact related to ordinary generating functions, and I was curious if it was known, as it seems plausible it is classically known:


"Let $\lambda\_1,\ldots, \lambda\_k$ be algebraic numbers. Let $f(z)= \sum^\infty\_{n=0} c\_nz^n$ when each $c\_n\in \mathbb{Z}.$ Suppose $f$ analytically continues to $\mathbb{C}\setminus \{\lambda\_1,\ldots,\lambda\_k\}.$ Then, $f$ is a rational function."



Ordinary generating functions with finitely many singularities at algebraic numbers are rational
It is not necessary to assume that $\lambda\_i$ are algebraic. This is a special case of the result in G. Pólya, Mathematische Annalen (1928), Volume: 99, page 687-706, [page 704](https://gdz.sub.uni-goettingen.de/id/PPN235181684_0099?tify=%7B%22pages%22:%5B708%5D,%22view%22:%22help%22%7D) in particular:



> 
> Der Spezialfall, in dem $\mathfrak A$ abzählbar, also $\tau=0$ ist, liefert folgende Aussage: *Wenn die Koerffizieten $a\_0, a\_1, a\_2, \dots$ der Potenzreihe
> $$a\_0+a\_1z+a\_2z^2+\dots=F(z)$$
> ganze Zahlen sind, so muß für die Funktion $F(z)$ einer der folgenden drei Fälle zutreffen: Entweder ist $F(z)$ eine rationale Funktion, oder ist $F(z)$ eine mehrdeutige Funktion, oder besitzt $F(z)$ unabzählbar viele singuläre Punkte.*
> 
> 
> 


The paper actually works with the function $f(z)=F(1/z).$ This is defined near $\infty,$ and it's either rational, doesn't have a unique analytical continuation, or its analytic continuation is singular on an uncountable set $\mathfrak A.$ The quantity $\tau$ is the capacity of $\mathfrak A$ (Chebyshev constant, transfinite diameter), which is zero for compact countable sets.


