# Algebraicity of a generating function and binomial numbers

It is an obvious fact that the sum $$\sum_{n\geq 0} \binom{2n}{n} x^n$$ defines an algebraic function. I am interested in the variation of this sum, namely

$$A(x)=\sum_{n\geq 0} \binom{2n}{n}^2 x^n$$

which is not an algebraic function due to the growth of the coefficients (see Enumerative Combinatorics, Vol. 2 from Stanley, for instance).

My question is the following: can we say something about the algebraicity/transcendence of $$A(x_0)$$, for any real $$x_0$$ in the region of convergence)? If $$x_0$$ is algebraic this is quite obvious I guess, but I do not know if there exists an argument saying something when $$x_0$$ is transcendent.

I assume that this is a very general question, and an answer for any transcendent number should be very difficult. But I would be happy to know if something it can be said for some particular transcendental number ($$\pi$$ or $$e$$, for instance).

Maybe the integral representation

$$A(x)=\sum_{n\geq 0} \binom{2n}{n}^2 x^n=\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{du}{(1-16x\cos^2 u)^{1/2}}$$

could be helpful, and something concerning elliptic integrals can be used here, but I do not know enough on this topic.

It is a straightforward computation to express $$A(x) = \sum_{n=0}^{\infty} \binom{2n}{n}^2x^n$$ in terms of hypergeometric functions, namely $$A(x) = {}_2F_1 (\tfrac12,\tfrac12;1;16x).$$ Then in turn one can express this function in terms of elliptic integrals, $${}_2F_1 (\tfrac12,\tfrac12;1;16x) = \frac{1}{\pi} \int_1^{\infty}\frac{du}{\sqrt{u(u-1)(u-16x)}}. \tag{1}\label{1}$$ The integral on the right is a period of the Legendre elliptic curve $$E_x:y^2 = u(u-1)(u-16x).$$ When $$x$$ is algebraic and $$x \neq 0$$, $$1$$, it is a result of Th. Schneider (1934) that such a period is transcendental over $$\overline{\mathbb{Q}}$$, and it was further proved by Schneider (1937) that the ratio in \eqref{1} is itself transcendental. So if $$x$$ is algebraic and $$0<|x|< \frac{1}{16}$$ (so that $$A(x)$$ converges), the value of $$A(x)$$ is transcendental over $$\overline{\mathbb{Q}}$$.