Given the binomial function $\binom{n}{k}$.

1. Define the following sequences, $$\begin{aligned} u_1(k) &= \tbinom{2k}{k}\tbinom{3k}{k}\tbinom{6k}{3k} = 1, 120, 83160, 81681600,\dots \\ u_2(k) &= \tbinom{2k}{k}\sum_{j=0}^k (-3)^{k-3j} \tbinom{2j}{j}\tbinom{3j}{j}\tbinom{k}{3j} = 1, -6, 54, -420, 630,\dots\\ u_3(k) &= \tbinom{2k}{k}\sum_{j=0}^k (3)^{k-3j} \tbinom{2j}{j}\tbinom{3j}{j}\tbinom{k}{3j} = 1, 6, 54, 660, 10710, \dots \end{aligned}$$


$$\frac{1}{\pi} = 12\,\boldsymbol{i}\sum_{k=0}^\infty u_1(k) \frac{163\cdot 3344418k + 13591409}{(-640320^3)^{k+1/2}},\quad\text{(Chudnovsky)}\tag1$$

$$\frac{1}{\pi} = \frac{\boldsymbol{i}}{231}\,\sum_{k=0}^\infty u_2(k) \frac{163\cdot 4826 k + 58831}{(-640320-12)^{k+1/2}}\tag2$$

$$\color{red}{\frac{1}{\pi}} = \frac{\boldsymbol{i}}{53359}\,\sum_{k=0}^\infty u_3(k) \frac{163\cdot 1114806k + 13592857}{(-640320+12)^{k+1/2}}\tag3$$

Note that the cube power disappears from (2) and (3). In fact, the McKay-Thompson series $T_{3C}$, with the appropriate constant, gives the cube root of $j(\tau)$.

2. Furthermore, define,

$$\begin{aligned} v_1(k) &= \tbinom{2k}{k}\tbinom{2k}{k}\tbinom{4k}{2k} = 1, 24, 2520, 369600,\dots \\ v_2(k) &= \tbinom{2k}{k}\sum_{j=0}^k (4)^{k-2j} \tbinom{2j}{j}\tbinom{2j}{j}\tbinom{k}{2j} = 1, 8, 120, 2240, 47320, \dots \end{aligned}$$


$$\frac{1}{\pi} = 32\sqrt{2}\,\sum_{k=0}^\infty v_1(k) \frac{29\cdot\color{blue}{13\cdot70}\,k + 1103}{(396^4)^{k+1/2}},\quad\text{(Ramanujan)}\tag4$$

$$\color{red}{\frac{1}{\pi}} = \frac{\sqrt{-2}}{\color{blue}{70}}\,\sum_{k=0}^\infty v_2(k)\, \frac{58\cdot\color{blue}{13\cdot99}\,k + 6243}{(16-396^2)^{k+1/2}}\tag5$$

$$\frac{1}{\pi} = \frac{2\sqrt{2}}{\color{blue}{13}}\,\sum_{k=0}^\infty v_2(k) \frac{\color{blue}{70\cdot99}\,k + 579}{(16+396^2)^{k+1/2}}\tag6$$

Similar results can be found using other discriminants d. Only four of the above are in H.H.Chan and S. Cooper's paper "Rational analogues of Ramanujan's series for 1/π", but I found (3) and (5) (in red) serendipitously by assuming there might be some sort of "symmetry".

Question: Why did the assumption of symmetry work?

Note: This has been updated to reflect the connection to the Pell equations $\color{blue}{70}^2-29\cdot\color{blue}{13}^2 = -1$ and $\color{blue}{99}^4-29\cdot1820^2 = 1$, asked in this post.

P.S. For the context of these formulas, kindly see "Ramanujan-Sato series".


1 Answer 1


A general methods of construction of Ramanujan-type identities are outlined in http://arxiv.org/abs/1211.6563 (Some conjectured formulas for 1/Pi coming from polytopes, K3-surfaces and Moonshine, by Gert Almkvist), http://arxiv.org/abs/0712.1332 (Ramanujan-type formulae for 1/π: A second wind?, by Wadim Zudilin) and http://arxiv.org/abs/1302.0548 (Ramanujan-type formulae for 1/π: the art of translation, by Jesus Guillera and Wadim Zudilin). If your identities can be obtained by these methods, maybe the origin of why the assumption of symmetry had worked also will become clear.

See also a lovely historical review http://www.math.uiuc.edu/~berndt/articles/monthly567-587.pdf (Ramanujan's Series for 1/π: A Survey, by Nayandeep Deka Baruah, Bruce C. Berndt, and Heng Huat Chan).

  • $\begingroup$ Thanks. I recently learned about Almkvist's paper on "Some conjectured formulas for 1/Pi..." It has a wealth of information, but it would be nice if similar functions were grouped together. From what I've seen so far, $396^2+8$ and $396^2−8$ will work as well using different sequences. $\endgroup$ Apr 1, 2014 at 5:31

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