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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?

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  • $\begingroup$ It is not a deep question. The known formulas originate from the modular parametrization of the underlying functions and the fact that modular functions assume algebraic (sometimes rational) values at CM points. Proving that that a particular linear combination with algebraic coefficients of a hypergeometric function and its derivative at rational $z$ multiplied by $\pi$ is not a rational number is in general out of reach of existing methods in diophantine approximation. But a "belief" is that if there is no reason for an algebraic relation then there is no relation. $\endgroup$ – Wadim Zudilin Oct 28 '15 at 0:15
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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.
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