Entry 33

In a similar manner to Entry 32, given the Chudnovsky algorithm $$\frac{1}{\pi} = \frac{12}{(640320)^{3/2}} \sum^\infty_{k=0} \frac{(6k)!}{(3k)!\,k!^3} \frac{3\cdot163\cdot1114806k + 13591409}{(-640320^3)^k}$$ we can use the cube root $\sqrt[3]{-640320^3}=-640320$ as the median, $$\begin{aligned}\frac{1}{\pi}&=\frac{12}{(640320+12)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k(-3)^{k-3j}\tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}\frac{163\cdot1114806k+(13591409-8\cdot181)}{(-640320-12)^k}\\ \frac{1}{\pi}&=\frac{12}{(640320-12)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\,3^{k-3j}\,\tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}\,\frac{163\cdot1114806k+(13591409+8\cdot181) }{(-640320+12)^k}\end{aligned}$$ where $\binom{n}{k}$ is the binomial coefficient. These are now level-9 Ramanujan-Sato series.