Entry 32

It is well-known that$$\frac{1}{\pi} =\frac{2 \sqrt 2}{99^2} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot70\cdot13\,k+1103}{(396^4)^k}$$But it turns out we can also use the square root \(\sqrt{396^4}=\pm396^2\) as a median point for two other pi formulas. First express the above as$$\frac{1}{\pi} =\frac{192 \sqrt 2}{(396^2)^{3/2}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{2\cdot58\cdot15015k+72798}{(396^4)^k}$$then$$\begin{aligned}\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2-16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}\frac{58\cdot15015k+(72798+37)}{(-396^2+16)^k}\\ \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}\frac{58\cdot15015k+(72798-37)}{(396^2+16)^k}\end{aligned}$$ where \(\binom{n}{k}\) is the binomial coefficient. Note that they have a beautifully symmetric form and how the same integers (which figure in Pell equations as discussed in Entry 1) appear in all three formulas. These two are level-8 Ramanujan-Sato series.