Ramanujan came up with this neat $\pi$ formula in Cambridge: $$\frac 1\pi=\frac {2\sqrt2}{99^2}\sum_{k=0}^\infty\frac {(4k)!}{k!^4}\frac {26390k+1103}{396^{4k}}\tag{1}$$ I am simply amazed by this formula, and would like to know if there are other similar types of formulas like this. $(1)$ is interesting in the way that it is efficient, and yet beautiful. It is also closely related to $640320^3$ in some way. (I'm not too sure though)

Question: Can $(1)$ be generalized and if so, what would be some examples?

  • $\begingroup$ It is way too complicated. See en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series $\endgroup$ – reuns Nov 6 '16 at 17:11
  • $\begingroup$ Why the downvote?? $\endgroup$ – Frank Nov 22 '16 at 15:34
  • $\begingroup$ BTW this formula was obtained by Ramanujan in India before he met G H Hardy. But it was published in a paper when he was in Cambridge. The handwritten Notebooks of Ramanujan contain this and many similar series for $1/\pi$. $\endgroup$ – Paramanand Singh May 23 '17 at 6:47

We have the sequence,

$$s(k)=\frac{(4k)!}{k!^4}=\tbinom{2k}{k}\tbinom{2k}{k}\tbinom{4k}{2k}=1, 24, 2520, 369600,\dots$$ where $\binom{n}{k}$ is the binomial coefficient. Given Ramanujan's formula $$\frac 1\pi=\frac {2\sqrt2}{99^2}\sum_{k=0}^\infty\frac {(4k)!}{k!^4}\frac {26390k+1103}{396^{4k}}\tag1$$ and multiplying everything by $\frac{11\times24^2}{11\times24^2}$ so the denominator $99^2$ becomes $(396^2)^{3/2}$, and corresponding changes to the numerator, then it can be expressed as $$\frac{1}{\pi} =\frac{192 \sqrt 2}{(396^2)^{3/2}} \sum_{k=0}^\infty \tbinom{2k}{k}\tbinom{2k}{k}\tbinom{4k}{2k}\frac{2\cdot58\cdot15015k+72798}{(396^4)^k}\tag2$$ In this form, its affinity is clear to the related Ramanujan-Sato series, $$\begin{aligned}\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2-8)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{2j}\tbinom{2k-4j}{k-2j}\tbinom{2j}{j}\frac{58\cdot15015k+(72798+37/2)}{(-396^2+8)^k}\\ \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+8)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{2j}\tbinom{2k-4j}{k-2j}\tbinom{2j}{j}\frac{58\cdot15015k+(72798-37/2)}{(396^2+8)^k}\end{aligned}\tag3$$ and, $$\begin{aligned}\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2-16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}\frac{58\cdot15015k+(72798+37)}{(-396^2+16)^k}\\ \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}\frac{58\cdot15015k+(72798-37)}{(396^2+16)^k}\end{aligned}\tag4$$ with the first two found by Almkvist (though not in this form). Notice the $396^2\pm8,\,$ $396^2\pm16$, and the use of the same integers. I mention this briefly in my blog, Ramanujan Once A Day (which I really should update).

  • $\begingroup$ Where and why do we have$$s(k)=\frac {(4k)!}{k!^4}=\binom{2k}{k}\binom{2k}{k}\binom{4k}{2k}$$And where did you get the sequence of numbers $1,24,2520,369600\ldots$? $\endgroup$ – Frank Nov 7 '16 at 19:18
  • $\begingroup$ @Frank: It is just a $s(0) = 1,\,s(1) = 24,\,s(2)=2520,\dots$ $\endgroup$ – Tito Piezas III Nov 8 '16 at 0:34
  • $\begingroup$ So looking at your entries in your Ramanujan Once A Day blog, I'm wondering how did you get $U_n=u+v\sqrt{489}$ and $$3\sqrt{3}(U_n^{1/2}-U_n^{-1/2})+18\\\sqrt{163}(U_n^{1/2}+U_n^{-1/2})\\ 3\sqrt{3}(U_n^{1/2}-U_n^{-1/2})+6$$in Entry $11$? I am very new to this so sorry if this seems unbelievably obvious... $\endgroup$ – Frank Nov 21 '16 at 4:13
  • $\begingroup$ @Frank: It's not obvious at all. It's not immediately apparent that there is a connection to the Pell equation $u^2-3\cdot163\,v^2=1$ and $e^{\pi\sqrt{163}} = 640320^3+743.9999999999995\dots$ The form $U_n^{1/m}\pm U_n^{-1/m}$ is found a lot in Ramanujan's works, and all I did to find the relations above was to use some assumptions. But it would take a short article to explain it. :) $\endgroup$ – Tito Piezas III Nov 21 '16 at 4:43
  • $\begingroup$ Could you post the work for $e^{\pi\sqrt{67}}$? I'm interested to see the work. So you're saying this is a bunch of guessing? $\endgroup$ – Frank Nov 21 '16 at 4:47

Yes, there are similar formulas. For example this one was found by David and Gregory Chudnovsky, $$\frac 1\pi=\frac{12}{(640320)^{3/2}}\sum_{k=0}^\infty (-1)^k\frac {(6k)!}{(k!)^3(3k)!}\frac {545140134k+13591409}{640320^{3k}}$$ see this paper for an interesting survey on this fascinating subject.


(Per OP's request.)

Similar pi formulas can be found, rather surprisingly, from fundamental solutions to Pell equations for special $N$, $$u^2-3Nv^2=1\tag1$$

I. $N=163$

The WA command for the fundamental soln to $(1)$ is NumberFieldFundamentalUnits[Sqrt[3N]], hence,

$$U_n = u+v\sqrt{489}=7592629975 + 343350596 \sqrt{489}=\left(\tfrac{1}{18}(\color{brown}{640320}-6)\sqrt{3}+4826\sqrt{163}\right)^2$$ We have $$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+18 = \color{brown}{2^2\cdot3^3\cdot7^2\cdot11^2}$$ $$\sqrt{163}\big(U_n^{1/2}+U_n^{-1/2}\big) = \color{brown}{2^2\cdot19\cdot127\cdot163}$$ $$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+6 = \color{brown}{640320}$$ and we find these integers in the Chudnovsky algorithm $$12\sum_{k=0}^\infty (-1)^k \frac{(6k)!}{k!^3(3k)!} \frac{\color{brown}{2\cdot3^2\cdot7\cdot11\cdot19\cdot127\cdot163}\,k+13591409}{(\color{brown}{640320}^3)^{k+1/2}} = \frac{1}{\pi}$$

II. $N=67$

The one for $N=67$ has the same form, $$U_n = u+v\sqrt{201}=515095+ 36332\sqrt{201}= \left(\tfrac{1}{18}(\color{brown}{5280}-6)\sqrt{3}+62\sqrt{67}\right)^2$$ And, $$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+18 = \color{brown}{2^2\cdot3^3\cdot7^2}$$ $$\sqrt{67}\big(U_n^{1/2}+U_n^{-1/2}\big) = \color{brown}{2^2\cdot31\cdot67}$$ $$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+6 = \color{brown}{5280}$$ so $$12\sum_{k=0}^\infty (-1)^k \frac{(6k)!}{k!^3(3k)!} \frac{\color{brown}{2\cdot3^2\cdot7\cdot31\cdot67}\,k+10177}{(\color{brown}{5280}^3)^{k+1/2}} = \frac{1}{\pi}$$
and similarly for the large Heegner numbers $N=7,11,19,43,67,163.$

  • $\begingroup$ Does that mean that for $N=43$, the corresponding formula is$$\frac 1{\pi}=12\sum_{k=0}^{\infty}(-1)^k\frac {(6k)!}{k!^3(3k)!}\frac {5418k+789}{(960^3)^{k+1/2}}\tag{1}$$ $\endgroup$ – Frank Nov 22 '16 at 16:08
  • $\begingroup$ @Frank: Almost. It should be $3\times5418k=16254k$. I have the formulas in this article I made before. $\endgroup$ – Tito Piezas III Nov 22 '16 at 16:12
  • $\begingroup$ @Frank: There are exactly 11 Chudnovsky-type formulas with an integer denominator. Maybe you can ask a separate question for these so we can list them all down? $\endgroup$ – Tito Piezas III Nov 22 '16 at 16:20
  • $\begingroup$ What? I'll try to work out the remaining kinks later... :/ $\endgroup$ – Frank Nov 22 '16 at 17:00
  • $\begingroup$ Sure. math.stackexchange.com/questions/2026007/… $\endgroup$ – Frank Nov 22 '16 at 17:01

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