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Continued from this post

Ramanujan found this handy formula for $\pi$$$\frac 1\pi=\frac {\sqrt8}{99^2}\sum_{k=0}^{\infty}\binom{2k}k\binom{2k}k\binom{4k}{2k}\frac {26390k+1103}{396^{4k}}\tag1$$

Which is related to Heegner numbers.

Sometime after, the Chudnovsky brothers came up with another $\pi$ formula$$\frac 1\pi=\frac{12}{(640320)^{3/2}}\sum_{k=0}^\infty (-1)^k\frac {(6k)!}{(k!)^3(3k)!}\frac {545140134k+13591409}{640320^{3k}}\tag2$$ And according to Tito, $(2)$ has a total of $11$ other forms with integer denominators.

Question: What are all $11$ forms? And what $e$-transcendal number to they correspond to?

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Chudnovsky-type pi formulas use the j-function $j(\tau)$ as the denominator. For $\tau=\frac{1+\sqrt{-d}}{2}$ and $\tau = \frac{1}{2}\sqrt{-d}$, there are exactly 9+4=13 discriminants $d$ such that $j(\tau)$ is an integer. (However, two of these do not yield series that converge.)

The j-function can easily be calculated in WA as, for example

N[12^3KleinInvariantJ[(1+Sqrt[-163])/2],100]

For convenience, given the well-known sequence defined in three equivalent ways, $$h_k = \frac{(6k)!}{(3k)!\,k!^3}=\tbinom{2k}k\tbinom{3k}k\tbinom{6k}{3k}=\frac{12^{3k}\,\big(\tfrac{1}{2})_k\,\big(\tfrac{1}{6})_k\,\big(\tfrac{5}{6})_k}{k!^3}=1, 120, 83160, 81681600,\dots$$ with factorial $n!$, binomial coefficient $\binom nk$, and Pochhammer symbol $(a)_n$.

I. $\tau=\frac{1+\sqrt{-d}}{2}$

$$\begin{aligned} 1/π &= 3\sum_{k=0}^\infty (-1)^k\, h_k\, \frac{\color{blue}{7}\cdot 9k+8}{(15^3)^{k+1/2}}\\ 1/π &= 4 \sum_{k=0}^\infty (-1)^k\, h_k\, \frac{\color{blue}{11}\cdot 14k+15}{(32^3)^{k+1/2}}\\ 1/π &= 12 \sum_{k=0}^\infty (-1)^k\, h_k\, \frac{\color{blue}{19}\cdot 18k+25}{(96^3)^{k+1/2}}\\ 1/π &= \frac{4}{3} \sum_{k=0}^\infty (-1)^k\, h_k\, \frac{\color{red}{27}\cdot506k+837}{(3\cdot160^3)^{k+1/2}}\\ 1/π &= 12 \sum_{k=0}^\infty (-1)^k\, h_k\, \frac{\color{blue}{43}\cdot 378 k+789}{(960^3)^{k+1/2}}\\ 1/π &= 12 \sum_{k=0}^\infty (-1)^k\, h_k\, \frac{\color{blue}{67}\cdot3906 k+10177}{(5280^3)^{k+1/2}}\\ 1/π &= 12 \sum_{k=0}^\infty (-1)^k\, h_k\, \frac{\color{blue}{163}\cdot3344418 k+13591409}{(640320^3)^{k+1/2}}\end{aligned}$$

II. $\tau = \frac{1}{2}\sqrt{-d}$

$$\begin{aligned} 1/π &= 2\sum_{k=0}^\infty h_k\, \frac{\color{blue}{8}\cdot 14k+12}{(20^3)^{k+1/2}}\\ 1/π &= \frac{3}{2} \sum_{k=0}^\infty h_k\, \frac{\color{red}{12}\cdot 44k+48}{(2\cdot30^3)^{k+1/2}}\\ 1/π &= \frac{3}{\sqrt2} \sum_{k=0}^\infty h_k\, \frac{\color{red}{16}\cdot 63k+80}{(66^3)^{k+1/2}}\\ 1/π &= \frac{3}{2} \sum_{k=0}^\infty h_k\, \frac{\color{red}{28}\cdot513k+864}{(255^3)^{k+1/2}}\end{aligned}$$

with fundamental $d=3, 4, 7, 8, 11, 19, 43, 67, 163$ in blue and non-fundamental $d=12,16,27,28$ in red. However, the two smallest, namely $d=3,4,$ do not yield series that converges.

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  • $\begingroup$ Just wondering, but is there a quick way to calculate $e^{\pi\sqrt{d}}$ for any number $d$ that involves the q-series and j-functions? And $d$ doesn't have to be a Heegner number, it can also be $58$ or $22$. (Both discovered by Ramanujan I think) $\endgroup$ – Frank Nov 23 '16 at 13:50
  • $\begingroup$ @Frank: Yes, the j-function input for WA and Mathematica is the same. $\endgroup$ – Tito Piezas III Nov 23 '16 at 13:53
  • $\begingroup$ @Frank: I don't understand your first question. Can you clarify? $\endgroup$ – Tito Piezas III Nov 23 '16 at 13:56
  • $\begingroup$ How would you calculate $e^{\pi\sqrt{67}}$ using the j-functions? How about $e^{\pi\sqrt{58}}$? Wikipedia describes a procedure for only Heegner numbers. So it'll work in calculating $e^{\pi\sqrt{67}}$. But not $e^{\pi\sqrt{58}}$ since $d=58$ is not a Heegner number. $\endgroup$ – Frank Nov 23 '16 at 14:08
  • $\begingroup$ @Frank: There is something wrong with your question. This post is about a pi formula using the jfunction. You want a formula for $e^{\pi\sqrt{67}}$ using the jfunction, is that it? What for? $\endgroup$ – Tito Piezas III Nov 23 '16 at 14:29

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