Entry 10

The Chudnovsky algorithm is $$\frac{1}{\pi} = 12 \sum^\infty_{k=0} (-1)^k\frac{ (6k)!}{k!^3\,(3k)!}\frac{ (163 \times 3344418 k + 13591409)}{ (640320^3)^{k + 1/2}}\tag1$$ Recall the famous $$e^{\pi\sqrt{163}}=640320^3+743.99999999999925\dots$$ The formula $(1)$ was inspired by Ramanujan's work and is used to calculate world records for the digits of pi. Now why didn't he discover this? Actually, he almost did. In his list of 17 formulas, there were two that belong to this family, one of which is $$\frac{1}{\pi} = \frac{18\sqrt{3}}{85\sqrt{85}} \sum_{n=0}^{\infty} \frac{\big(\tfrac{1}{2}\big)_n\big(\tfrac{1}{6}\big)_n\big(\tfrac{5}{6}\big)_n}{n!^3}\, (133n+8)\,\Big(\frac{4}{85}\Big)^{3n}\tag2$$ where $(a)_n$ is a Pochhammer symbol. However, this can be translated into the form of $(1)$ namely $$\frac{1}{\pi} = 162 \sum^\infty_{k=0} \frac{ (6k)!}{k!^3\,(3k)!}\frac{ (133 k + 8)}{ (255^3)^{k + 1/2}}\tag3$$ Similarly, $$e^{\pi\sqrt{28}}=255^3-744.01\dots$$ where the "excess" $744$ indicates that the j-function is involved. In fact, in Ramanujan's Lost Notebook, he had calculations involving Eisenstein series using the primes $d=11,19,43,67,163$ which is precisely the family which $(1)$ belongs to. Thus, if only Ramanujan lived longer, he would almost have surely found the Chudnovsky algorithm.