# Ramanujan's pi formulas with a twist

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}

Then,

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

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

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

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}

Then,

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

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

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

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".

• 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. – Tito Piezas III Apr 1 '14 at 5:31