# 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 $$d=19,43,67,163$$, respectively. (Note: Typo corrected.)

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? Oct 4, 2016 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). Oct 4, 2016 at 5:27
• Your $\beta_d$ seem related to the $A_N$ from this question: for example $1448\cdot 6 = 8688$. Aug 18, 2019 at 16:12
• @L.Miller: Amazing. you found the closed-form! I looked at my notes and realized I made a typo. Those $\beta_n = 4,24,76,1448$ are supposed to be for the 4 largest discriminants, namely $n=19,43,67,163$. Comparing it your $A_n = 24, 144, 456, 8688$, we find that $\color{red}{6\beta_n = A_n}$. Amazing! Aug 18, 2019 at 17:14
• @L.Miller: I'll give the formulas for $n=19,43,67,163$ later. Aug 18, 2019 at 17:29

This is mainly a re-post of my comment: In this question, I have defined

$$A_N:=\sqrt{-N}\cdot\frac{E_2(\tau_N)-\frac{3}{\pi\cdot Im(\tau_N)}}{\eta^4(\tau_N)}$$

where $$\eta$$ denotes the Dedekind $$\eta$$-Function and $$E_2$$ is the Eisenstein series of weight $$2$$, and $$\tau_N=\frac{N+\sqrt{-N}}{2}$$ is a quadratic irrationality with class number $$1$$.

For the terms $$\beta_N$$ of the question above, it holds $$\color{red}{e^{i\pi/3}\,6\beta_N =A_N}$$, or (with $$N=d$$):

$$\beta=\frac{\sqrt{-d}}{e^{i\pi/3}\,6}\cdot\frac{E_2(\tau_d)-\frac{3}{\pi\cdot Im(\tau_d)}}{\eta^4(\tau_d)}$$

A proof that the $$A_N$$ are algebraic integers of $$\mathbb Z$$ can be found here in Appendix B (which uses Appendix A).

• What is the background of $A_N$? (I mean, one does not just define a function randomly, it must have arisen from some context.) Aug 18, 2019 at 17:46
• Oh, I see it here in your earlier MO question. You were investigating the Chudnovsky formula Aug 18, 2019 at 17:48
• yes, and the results are in chapter 10 and appendices A/B of the above mentioned arxiv preprint arxiv.org/abs/1809.00533 Aug 18, 2019 at 18:03
• Made a minor change. I forgot that you affixed a cube root of unity to your MO question. Aug 19, 2019 at 11:15