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.

  • $\begingroup$ I wasn't aware of $S_{2}, S_{3}$. Do you have a reference for proof of these series? $\endgroup$
    – Paramanand Singh
    Oct 4, 2016 at 5:14
  • $\begingroup$ @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). $\endgroup$ Oct 4, 2016 at 5:27
  • $\begingroup$ Your $\beta_d$ seem related to the $A_N$ from this question: for example $1448\cdot 6 = 8688$. $\endgroup$
    – L. Milla
    Aug 18, 2019 at 16:12
  • $\begingroup$ @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! $\endgroup$ Aug 18, 2019 at 17:14
  • $\begingroup$ @L.Miller: I'll give the formulas for $n=19,43,67,163$ later. $\endgroup$ Aug 18, 2019 at 17:29

1 Answer 1


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).

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

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