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](https://mathoverflow.net/questions/189126/).


**P.S.** For the context of these formulas, kindly see "*[Ramanujan-Sato series](http://en.wikipedia.org/wiki/Ramanujan-Sato_series)*".



Ramanujan's pi formulas with a twist
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).


