A question on Ramanujan's $1/\pi$ formulas

Question

It is known that Ramanujan discovered a number of formulas for $1/\pi$. All of these formulas are of the form $$\frac{1}{\pi}=\sum_{n=0}^{\infty}\frac{(1/2)_n(s)_n(1-s)_n}{(1)_n^3}(a+bn)z^n,$$where $(a)_n$ is Pochhammer symbol and $-1\leq z< 1$$, a, b$ are all algebraic numbers.

Question: Fix $s$, for example, $s=1/4$. Suppose that $z$ is rational number. Are there finite many triples $(z,a,b)$ that satisfy Ramanujan's identity?

Answer 1

For a particular general formula, yes, there are only finitely many triples $(z,a,b)$ with rational $z$. For $\color{blue}{p = 2:}$

$$\begin{aligned} \frac{1}{\pi} &= \sum_{n=0}^\infty \frac{256^n \big(\tfrac{1}{2})_n \big(\tfrac{1}{4})_n \big(\tfrac{3}{4})_n}{n!^3} \frac{An+B}{C^{n+1/2}}\\ \small{\color{brown}{where,}}\;&\\ A &= \sqrt{C}\big(1-2\beta)\tfrac{1}{p}\sqrt{d}\\ B &= \frac{\sqrt{C}}{\pi}\Big(\,_2F_1\big(\tfrac{1}{4},\tfrac{3}{4};1;\beta\big)\Big)^{-2}-\tfrac{1}{2}\cdot\tfrac{1}{4}\cdot\tfrac{3}{4}\cdot\tfrac{1}{p}\cdot\frac{256\sqrt{d}}{\sqrt{C}}\,\frac{_2F_1\big(\tfrac{5}{4},\tfrac{7}{4};2;\beta\big)}{_2F_1\big(\tfrac{1}{4},\tfrac{3}{4};1;\beta\big)}\\ C &= r_2(\tau)\\ \beta &= \frac{1-\sqrt{1-\frac{256}{r_2(\tau)}}}{2}\\ \small{\color{brown}{and,}}\;&\\ r_2(\tau) &= \left( \Big(\tfrac{\eta(\tau)}{\eta(2\tau)}\Big)^{12}+ \Big(\tfrac{\sqrt{2}\,\eta(2\tau)}{\eta(\tau)}\Big)^{12}\right)^2\\ \small{\color{brown}{with,}}\;&\\ \tau &= \frac{1}{4}\sqrt{-d},\quad \text{or}\quad \tau = \frac{1}{4}\big(2+\sqrt{-d}\big) \end{aligned}$$

This formula generalizes the well-known one by Ramanujan and depends only on a single parameter, $\tau$, and ultimately on the discriminant $d$. Thus, it is easy to test various $d$ to see which yields rational $C$.

Examples:. Define,

$$h_2(n) = \frac{(4n)!}{n!^4} = \frac{256^n \big(\tfrac{1}{2})_n \big(\tfrac{1}{4})_n \big(\tfrac{3}{4})_n}{n!^3} = 1, 24, 2520, 369600, 63063000,\dots\tag1$$

which is sequence A008977.

1. Let $d = 148 = 4\times37$, with class number $h(-d) = 2$, and $\tau = \frac{2+\sqrt{-148}}{4}$. Note that given the prime-generating polynomial $F(n) = 2n^2-2n+19$, then $F(\tau) = 0$. Then,
   $$A = 85840i,\quad B = 4492i,\quad C = r_2(\tau) = -14112^2$$ $$\frac{1}{\pi} = 4\sum_{n=0}^\infty h_2(n) (-1)^n \frac{37\times580n+1123}{(14112^2)^{n+1/2}}$$

2. Let $d = 232 = 4\times58$, with class number $h(-d) = 2$, and $\tau = \frac{\sqrt{-232}}{4}$. Note that given the prime-generating polynomial $F(n) = 2n^2+29$, then $F(\tau) = 0$. Then,
   $$A = 844480\sqrt{2},\quad B = 35296\sqrt{2},\quad C = r_2(\tau) = 396^4$$ $$\frac{1}{\pi} = 32\sqrt{2}\sum_{n=0}^\infty h_2(n) \frac{58\times455n+1103}{(396^4)^{n+1/2}}$$

which is Ramanujan's famous formula alluded to earlier.

Remarks:

1. The formulas for $p=1,2,3,4$ are given in this article Ramanujan's pi formulas and the hypergeometric function. They are my own formulation, but you can re-derive them from the more complicated versions by the Borweins or by Guillera.
2. Using $p=2$, I count only $12$ formulas with rational $C$, but some of the $A,B$ involve a square root. They're in Pi Formulas and the Monster. This is an older work, so is not as stream-lined as the previous article.
3. Ramanujan's formulas have been generalized for $p>4$ by using other well-defined integer sequences similar to $(1)$. See Ramanujan-Sato series.