TITLE: When is a power series with coefficients in {0,1} algebraic?
QUESTION [7 upvotes]: If we have a series $F(x)=\sum a_n x^n$ where $a_n$ is in {0,1} for every integer $n$. Is $F$ algebraic over $\mathbb Q$ (set of rational numbers). If it is, under what conditions?
Thank you.

REPLY [15 votes]: There is a classical rational-transcendental dichotomy for power series with coefficients which don't grow too fast. Here are two famous results:

P. Fatou, "Series trigonometriques et series de Taylor", Acta Math. (1906), no. 30, 335–400.

Fatou proves that if a power series $F(x)\in \mathbb Z[[x]]$ converges inside the unit disk, then either $F(x)\in \mathbb Q(x)$ or $F(x)$ is transcendental over $\mathbb Q(x)$.

F. Carlson, "Uber Potenzreihen mit ganzzahligen Koeﬃzienten.", Math. Zeitschr. (1921), no. 9, 1–13.

Carlson proves that if $F(x)\in \mathbb Z[[x]]$ converges inside the unit disk, then either $F(x)$ is rational or $F(x)$ admits the unit circle as a natural boundary.
This tells you that your series if it is algebraic then it must be a rational function with rational coefficients. This happens if and only if the binary number $0.a_1a_2\dots$ is eventually periodic (to see this just plug in $x= \frac{1}{2}$).

REPLY [2 votes]: I think the answer to Asymptotics/growth for coefficients of algebraic power series will be helpful.