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,

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

$$\color{red}{\frac{1}{\pi}} = \frac{\sqrt{-2}}{\color{blue}{70}}\,\sum_{k=0}^\infty v_2(k)\, (-1)^k \frac{58\cdot\color{blue}{13\cdot99}\,k + 6343}{(16-396^2)^{k+1/2}}\tag5$$

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

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][1].


**P.S.** For the context of these formulas, kindly see "*[Ramanujan-Sato series][2]*".


  [1]: http://mathoverflow.net/questions/189126/
  [2]: http://en.wikipedia.org/wiki/Ramanujan-Sato_series