# The Chudnovsky pi formula $1/\pi$ revisited

Define the constants, $$A=163\cdot1114806\\B=13591409\\C=640320$$ Given the binomial coefficient $\binom{n}{k}$, then we have the pi formulas, $$\frac{1}{\pi} =\frac{12}{(C)^{3/2}}\sum^\infty_{k=0} \frac{(6k)!}{(3k)!\,k!^3} \frac{3Ak+ B}{(-C^3)^k}$$ and $$\frac{1}{\pi} =\frac{12}{(C+4)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^{k/3} (-1)^j\tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}\frac{Ak+B-\color{blue}{1448}/3}{(C+4)^k}$$ $$\frac{1}{\pi} =\frac{12}{(C-4)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^{k/3} (+1)^j\tbinom{k}{3j} \tbinom{2j}{j}\tbinom{3j}{j}\frac{Ak+B+\color{blue}{1448}/3}{(-C+4)^k}$$ and $$\frac{1}{\pi} =\frac{12}{(C+12)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k(-3)^{k-3j}\tbinom{k}{3j} \tbinom{2j}{j}\tbinom{3j}{j}\frac{Ak+B-\color{blue}{1448}}{(-C-12)^k}$$ $$\frac{1}{\pi}=\frac{12}{(C-12)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\,(+3)^{k-3j}\,\tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}\,\frac{Ak+B+\color{blue}{1448} }{(-C+12)^k}$$

The first is the Chudnovsky formula, while the rest are also Ramanujan-Sato series (of level 9?). One can give the general form of the Chudnovsky using Eisenstein series.

Q: But what yields the blue number $\beta$? (These are $\beta=4, 24, 76, 1448$ for $d=11,19,67,163$, respectively.)

P.S. A similar phenomenon happens for the Ramanujan pi formula which uses $d=58$. I discuss this briefly in my blog Ramanujan Once A Day.

• I wasn't aware of $S_{2}, S_{3}$. Do you have a reference for proof of these series? – Paramanand Singh Oct 4 '16 at 5:14
• @ParamanandSingh: $S_2$ and presumably $S_3$ are Ramanujan-Sato series of level 9. I believe they (or some version) are discussed in the paper by Chan and Cooper, or the one by Almkvist. I found these using the integer relations command of Mathematica (by assuming it had the above form). – Tito Piezas III Oct 4 '16 at 5:27