# Numerology with Ramanujan's pi formula

Given Ramanujan's famous $\frac1{\pi}$ formula $$\frac 1\pi=\frac {2\sqrt2}{99^2}\sum_{k=0}^\infty\frac {(4k)!}{k!^4}\frac {26390k+1103}{396^{4k}}$$

which is a level 2 Ramanujan-Sato series. It can also be expressed as $$\frac{1}{\pi} =\frac{192 \sqrt 2}{(396^2)^{3/2}} \sum_{k=0}^\infty \tbinom{2k}{k}\tbinom{2k}{k}\tbinom{4k}{2k}\frac{2\cdot58\cdot15015k+72798}{(396^4)^k}$$

where $\binom{n}{k}$ is the binomial coefficient. In this form, its affinity is clear to the following level 8 Ramanujan-Sato series,

\begin{aligned} \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+4)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_2(k)\,\frac{58\cdot15015k+(72798-37/4)}{(396^2+4)^k}\\ \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+8)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_3(k)\,\frac{58\cdot15015k+(72798-37/2)}{(396^2+8)^k}\\ \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_4(k)\,\frac{58\cdot15015k+(72798-37)}{(396^2+16)^k}\\ \frac{1}{\pi}&\overset{\color{red}?}=\frac{192\sqrt{2}}{(396^2+32)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_5(k)\,\frac{58\cdot15015k+(72798-2\cdot37)}{(396^2+32)^k}\\ \frac{1}{\pi}&\overset{\color{red}?}=\frac{192\sqrt{2}}{(396^2+64)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_6(k)\,\frac{58\cdot15015k+(72798-4\cdot37)}{(396^2+64)^k}\end{aligned}

and integer sequences $s_n(k)$ starting with $k=0$,

\begin{aligned} s_2(k)&=\sum_{j=0}^k\tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j}=1, 1, 5, 13, 61, 221,\dots\\ s_3(k)&=\sum_{j=0}^k\tbinom{k}{2j}\tbinom{2k-4j}{k-2j}\tbinom{2j}{j}=1, 2, 8, 32, 148, 712,\dots\\ s_4(k)&=\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}=1, 4, 20, 112, 676, 4304,\dots\\ s_5(k)&=1, 8, 68, 608, 5668, 54688, 542864,\dots\\ s_6(k)&=1, 16, 260, 4288, 71716, 1215296, 20848016,\dots \end{aligned}

Q: Are all the terms of $s_5(k)$ and $s_6(k)$ integers as well, and does it have a closed-form?

P.S. I have already checked the OEIS.

• I've re-indexed $s_n(k)$ to match the $396^2+2^n$ in the denominators, I hope this okay. Another reason is for me to ask: if we treat the very first formula of Ramanujan corresponding to $s_0(k)$ then what is the missing series that might be associated with $s_1(k)$? – T. Amdeberhan Mar 25 '17 at 18:10
• @T.Amdeberhan: I have also checked $396^2+2$ as a denominator. However, $s_1(k)$ seem to involve rationals already. (Notice that the second term of $s_n(k)$ for $n>1$ start with $1,2,4,8,16$.) – Tito Piezas III Mar 26 '17 at 0:06
• @T.Amdeberhan: I just found the general case. Kindly see below. – Tito Piezas III Mar 26 '17 at 3:15
• Because the title said Numerology I thought you might be asking something about the Freemasons or something. – user78249 Mar 26 '17 at 15:58

(Too long for a comment.) After staring hard at my question and recalling an old MSE post of mine, I made an inspired guess and found,

Level 8

$$\frac{1}{\pi}=\frac{192\sqrt{2}}{(396^2+4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\alpha^{k-2j} \tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j} \frac{58\cdot15015k+72798-37\color{blue}\alpha/4}{(396^2+4\color{blue}\alpha)^k}$$

for general real $$\alpha$$, so turns out it is unnecessary to restrict it to powers of $$2$$. (Thus, $$s_5(k)$$ and $$s_6(k)$$ do have a closed-form and are integer sequences.)

Level 9

Similarly, the Chudnovsky formula (a level 1 Ramanujan-Sato) yields,

$$\frac{1}{\pi}=\frac{12}{(640320-4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j} \frac{163A\,k+B+1448\color{blue}\alpha/3}{(-640320+4\color{blue}\alpha)^k}$$

where $$A=1114806, B=13591409$$.

(Added later.) Now that I know what to look for, one can find other families.

Level 6

For example, starting with H. H. Chan, W. Zudilin, et al's

$$\frac1{\pi}=\frac{192\sqrt3}{(2\cdot140^2)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}k\sum_{j=0}^k\tbinom{k}{j}^3\,\frac{140(561k+53)}{(2\cdot140^2)^k}$$

mentioned in H. H. Chan and S. Cooper's "Rational analogues of Ramanujan's series for 1/π", we find, $$\frac1{\pi}=\frac{192\sqrt3}{(2\cdot140^2+4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}k\sum_{j=0}^k \color{blue}\alpha^{k-j}\tbinom{k}{j}\sum_{m=0}^j\tbinom{j}{m}^3\,\frac{140(561k+53)-13\color{blue}\alpha/4}{(2\cdot140^2+4\color{blue}\alpha)^k}$$

Level 10

$$\frac1{\pi}=\frac{16\sqrt{5}}{\sqrt{19}(76^2+4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}k\sum_{j=0}^k \color{blue}\alpha^{k-j}\tbinom{k}{j}\tbinom{2j}{j}^{-1}\sum_{m=0}^j\tbinom{j}{m}^4\,\frac{19^2(408k+47)-157\color{blue}\alpha/4}{(76^2+4\color{blue}\alpha)^k}$$

P.S. However, I do not have a rigorous proof for these families and the relevant literature do not seem to address general $$\alpha$$.

Note: The level $$\color{red}m$$ is given by an eta quotient $$\frac{\eta(\tau)}{\eta(\color{red}m\,\tau)}$$. For example, define

$$\lambda(\tau) =-6+\big(\tfrac{\eta^2(3\tau)}{\eta(\tau)\,\eta(\color{red}9\tau)}\big)^6 -27 \big(\tfrac{\eta(\tau)\,\eta(\color{red}9\tau)}{\eta^2(3\tau)}\big)^6$$

so $$\lambda\Big(\tfrac{3+\sqrt{-163}}{6}\Big)=-640320$$ is level $$9$$.

• This is a very naive comment, but have you tried to derive the RHS with respect to $\alpha$? – Sylvain JULIEN Mar 26 '17 at 13:32
• @SylvainJULIEN: I wouldn't know where to start. It was luck and persistence that I detected a pattern. – Tito Piezas III Mar 26 '17 at 13:42
• By the way, when you say 'for general $\alpha$ ', is $\alpha$ supposed to be an integer or any non negative real number ? Cause if it has to be an integer, maybe one should check $1$-periodicity and all the Fourier related stuff. – Sylvain JULIEN Mar 26 '17 at 13:48
• @SylvainJULIEN: For any non-zero real number $\alpha$, positive or negative, as long as the denominator does not vanish, or the denominator becomes so small that the series no longer converges – Tito Piezas III Mar 26 '17 at 13:54
• FWIW, if we denote by $s_k(\alpha)$ the sum $\sum_{j=0}^k\alpha^{k-2j}\binom{k}{2j}\binom{2j}{j}\binom{2j}{j}$ in your formula, the Wilf-Zeilberger methods seem to produce a recurrence $\alpha(\alpha^2-16)(k+1)(k+2)s_k(\alpha)-(3\alpha^2-16)(k+2)^2s_{k+1}(\alpha)+\alpha(3k^2+15k+19)s_{k+2}(\alpha)-(k+3)^2s_{k+3}(\alpha)=0$ – Vladimir Dotsenko Mar 26 '17 at 15:06

(Per S. Cooper's request.)

I. Table relating level $1$ with level $9$.

The general form apparently is,

$$\frac{1}{\pi}=\frac{12}{(C)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k} \tbinom{3k}{k}\tbinom{6k}{3k} \frac{\color{red}3A\,k+B}{(-C^3)^k}$$ and, $$\frac{1}{\pi}=\frac{12}{(C-4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j} \frac{A\,k+B+N\color{blue}\alpha/3}{(-C+4\color{blue}\alpha)^k}$$

for general real $\color{blue}\alpha$ and where,

$$\begin{array}{|c|c|c|c|c|} \hline d&A&B&C&N\\ \hline 11&154/9&5&32&4/3\\ 19&114&25&96&4\\ 43 &5418 &789 &960 &24\\ 67 &87234 &10177 &5280 &76\\ 163 &181713378 &13591409 &640320 &1448\\ \hline \end{array}$$

The variables $A,B,C$ are known to have closed-form expressions in terms of $d$. Presumably $N$ should have as well.

P.S. Note also the equivalent forms,

$$s_k(\color{blue}\alpha)=\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}=\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{j}\tbinom{k-j}{j}\tbinom{k-2j}{j}$$

where the latter form is used in H. Chan and S. Cooper's paper "Rational analogues of Ramanujan's series for 1/π". The case $\alpha=-3$,

$$s_k(-3) = 1, -3, 9, -21, 9, 297, -2421$$

is one of the six sporadic sequences studied by Zagier and Cooper.

II. Table relating level $2$ with level $8$.

The general form apparently is,

$$\frac{1}{\pi}=\frac{192\sqrt{2}}{(C)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k} \tbinom{2k}{k}\tbinom{4k}{2k} \frac{\color{red}2A\,k+B}{(C^2)^k}$$ and, $$\frac{1}{\pi}=\frac{192\sqrt{2}}{(C+4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\alpha^{k-2j} \tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j} \frac{A\,k+B-M\color{blue}\alpha/4}{(C+4\color{blue}\alpha)^k}$$

for general real $\color{blue}\alpha$ and where,

$$\begin{array}{|c|c|c|c|c|} \hline d&A&B&C&M\\ \hline 6&\sqrt2&\sqrt2/4&(4\sqrt3)^2&\sqrt2/12\\ 10&10&2&12^2&1/3\\ 18 &70\sqrt6 &21\sqrt6/2 &28^2 &\sqrt6/2\\ 22 &385\sqrt2 &209\sqrt2/4 &(12\sqrt{11})^2 &17\sqrt2/12\\ 58 &870870&72798&396^2&37\\ \hline \end{array}$$