Ramanujan–Sato series

In mathematics, a Ramanujan–Sato series[1][2] generalizes Ramanujan’s pi formulas such as,

\frac{1}{\pi} = \frac{2 \sqrt 2}{99^2} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{26390k+1103}{396^{4k}}

to the form,

\frac{1}{\pi} = \sum_{k=0}^\infty s(k) \frac{Ak+B}{C^k}

by using other well-defined sequences of integers s(k) obeying a certain recurrence relation, sequences which may be expressed in terms of binomial coefficients \tbinom{n}{k}, and employing modular forms of higher levels.

Ramanujan made the enigmatic remark that there were "corresponding theories", but it was only recently that H.H. Chan and S. Cooper found a general approach that used the underlying modular congruence subgroup \Gamma_0(n),[3] while G. Almkvist has experimentally found numerous other examples also with a general method using differential operators.[4]

Levels 1–4A were given by Ramanujan (1917),[5] level 5 by H.H. Chan and S. Cooper (2012),[3] 6A by Chan, Tanigawa, Yang, and Zudilin,[6] 6B by Sato (2002),[7] 6C by H. Chan, S. Chan, and Z. Liu (2004),[1] 6D by H. Chan and H. Verrill (2009),[8] level 7 by S. Cooper (2012),[9] part of level 8 by Almkvist and Guillera (2012),[2] part of level 10 by Y. Yang, and the rest by H.H. Chan and S.Cooper.

The notation jn(τ) is derived from Zagier[10] and Tn refers to the relevant McKay–Thompson series.

Level 1

Examples for levels 1–4 were given by Ramanujan in his 1917 paper. Given q=e^{2\pi i \tau} as in the rest of this article. Let,

\begin{align} j(\tau) &= \Big(\tfrac{E_4(\tau)}{\eta^8(\tau)}\Big)^3 = \tfrac{1}{q} + 744 + 196884q + 21493760q^2 +\dots\\ j^*(\tau) &= 432\,\frac{\sqrt{j(\tau)}+ \sqrt{j(\tau)-1728}}{\sqrt{j(\tau)}- \sqrt{j(\tau)-1728}} = \tfrac{1}{q} - 120 + 10260q - 901120q^2 + \dots \end{align}

with the j-function j(τ), Eisenstein series E4, and Dedekind eta function η(τ). The first expansion is the McKay–Thompson series of class 1A ( A007240) with a(0) = 744. Note that, as first noticed by J. McKay, the coefficient of the linear term of j(τ) is exceedingly close to 196883 which is the smallest degree > 1 of the irreducible representations of the Monster group. Similar phenomena will be observed in the other levels. Define,

s_{1A}(k)=\tbinom{2k}{k}\tbinom{3k}{k}\tbinom{6k}{3k}=1, 120, 83160, 81681600,\dots ( A001421)
s_{1B}(k)=\sum_{j=0}^k\tbinom{2j}{j}\tbinom{3j}{j}\tbinom{6j}{3j}\tbinom{k+j}{k-j}(-432)^{k-j} =1, -312, 114264, -44196288,\dots

Then the two modular functions and sequences are related by,

\sum_{k=0}^\infty s_{1A}(k)\,\frac{1}{(j(\tau))^{k+1/2}}= \pm \sum_{k=0}^\infty s_{1B}(k)\,\frac{1}{(j^*(\tau))^{k+1/2}}

if the series converges and the sign chosen appropriately, though squaring both sides easily removes the ambiguity. Analogous relationships exist for the higher levels.

Examples:

\frac{1}{\pi} = 12\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{1A}(k)\,\frac{163\cdot3344418k+13591409}{(-640320^3)^{k+1/2}},\quad j\Big(\tfrac{1+\sqrt{-163}}{2}\Big)=-640320^3
\frac{1}{\pi} = 24\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{1B}(k)\,\frac{-3669+320\sqrt{645}\,(k+\tfrac{1}{2})}{\big(-432\,U_{645}^3\big)^{k+1/2}},\quad j^*\Big(\tfrac{1+\sqrt{-43}}{2}\Big)=-432\Big(\tfrac{127+5\sqrt{645}}{2}\Big)^{3}=-432\,U_{645}^{3}

and U_n is a fundamental unit. The first belongs to a family of formulas which were rigorously proven by the Chudnovsky brothers in 1989[11] and later used to calculate 10 trillion digits of π in 2011.[12] The second formula, and the ones for higher levels, was established by H.H. Chan and S. Cooper in 2012.[3]

Level 2

Using Zagier’s notation[10] for the modular function of level 2,

\begin{align} j_{2A}(\tau) &=\Big(\big(\tfrac{\eta(\tau)}{\eta(2\tau)}\big)^{12}+2^6 \big(\tfrac{\eta(2\tau)}{\eta(\tau)}\big)^{12}\Big)^2 = \tfrac{1}{q} + 104 + 4372q + 96256q^2 + 1240002q^3+\cdots \\ j_{2B}(\tau) &= \big(\tfrac{\eta(\tau)}{\eta(2\tau)}\big)^{24} = \tfrac{1}{q} - 24 + 276q - 2048q^2 + 11202q^3 - \cdots \end{align}

Note that the coefficient of the linear term of j2A(τ) is one more than 4371 which is the smallest degree > 1 of the irreducible representations of the Baby Monster group. Define,

s_{2A}(k)=\tbinom{2k}{k}\tbinom{2k}{k}\tbinom{4k}{2k}=1, 24, 2520, 369600, 63063000,\dots ( A008977)
s_{2B}(k)=\sum_{j=0}^k\tbinom{2j}{j}\tbinom{2j}{j}\tbinom{4j}{2j}\tbinom{k+j}{k-j}(-64)^{k-j}=1, -40, 2008, -109120, 6173656,\dots

Then,

\sum_{k=0}^\infty s_{2A}(k)\,\frac{1}{(j_{2A}(\tau))^{k+1/2}}= \pm \sum_{k=0}^\infty s_{2B}(k)\,\frac{1}{(j_{2B}(\tau))^{k+1/2}}

if the series converges and the sign chosen appropriately.

Examples:

\frac{1}{\pi} = 32\sqrt{2}\,\sum_{k=0}^\infty s_{2A}(k)\,\frac{58\cdot455k+1103}{(396^4)^{k+1/2}},\quad j_{2A}\Big(\tfrac{1}{2}\sqrt{-58}\Big)=396^4
\frac{1}{\pi} = 16\sqrt{2}\,\sum_{k=0}^\infty s_{2B}(k)\,\frac{-24184+9801\sqrt{29}\, (k+\tfrac{1}{2})}{(64\,U_{29}^{12})^{k+1/2}},\quad j_{2B}\Big(\tfrac{1}{2}\sqrt{-58}\Big)=64\Big(\tfrac{5+\sqrt{29}}{2}\Big)^{12}=64\,U_{29}^{12}

The first formula, found by Ramanujan and mentioned at the start of the article, belongs to a family proven by D. Bailey and the Borwein brothers in a 1989 paper.[13]

Level 3

Define,

\begin{align} j_{3A}(\tau) &=\Big(\big(\tfrac{\eta(\tau)}{\eta(3\tau)}\big)^{6}+3^3 \big(\tfrac{\eta(3\tau)}{\eta(\tau)}\big)^{6}\Big)^2 = \tfrac{1}{q} + 42 + 783q + 8672q^2 +65367q^3+\dots\\ j_{3B}(\tau) &= \big(\tfrac{\eta(\tau)}{\eta(3\tau)}\big)^{12} = \tfrac{1}{q} - 12 + 54q - 76q^2 - 243q^3 + 1188q^4 + \dots\\ \end{align}

where 782 is the smallest degree > 1 of the irreducible representations of the Fischer group Fi23 and,

s_{3A}(k)=\tbinom{2k}{k}\tbinom{2k}{k}\tbinom{3k}{k}=1, 12, 540, 33600, 2425500,\dots ( A184423)
s_{3B}(k)=\sum_{j=0}^k\tbinom{2j}{j}\tbinom{2j}{j}\tbinom{3j}{j}\tbinom{k+j}{k-j}(-27)^{k-j}=1, -15, 297, -6495, 149481,\dots

Examples:

\frac{1}{\pi} = 2\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{3A}(k)\,\frac{267\cdot53k+827}{(-300^3)^{k+1/2}},\quad j_{3A}\Big(\tfrac{3+\sqrt{-267}}{6}\Big) = -300^3
\frac{1}{\pi} = \boldsymbol{i}\,\sum_{k=0}^\infty s_{3B}(k)\,\frac{12497-3000\sqrt{89}\, (k+\tfrac{1}{2})}{(-27\,U_{89}^{2})^{k+1/2}},\quad j_{3B}\Big(\tfrac{3+\sqrt{-267}}{6}\Big)=-27\,\big(500+53\sqrt{89}\big)^2=-27\,U_{89}^{2}

Level 4

Define,

\begin{align} j_{4A}(\tau)&=\Big(\big(\tfrac{\eta(\tau)}{\eta(4\tau)}\big)^{4}+4^2 \big(\tfrac{\eta(4\tau)}{\eta(\tau)}\big)^{4}\Big)^2 = \Big(\tfrac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \Big)^{24} = \tfrac{1}{q} + 24+ 276q + 2048q^2 +11202q^3+\dots\\ j_{4C}(\tau) &= \big(\tfrac{\eta(\tau)}{\eta(4\tau)}\big)^{8} = \tfrac{1}{q} -8 + 20q - 62q^3 + 216q^5 - 641q^7 + \dots\\ \end{align}

where the first is the 24th power of the Weber modular function \mathfrak{f}(\tau). And,

s_{4A}(k)=\tbinom{2k}{k}^3=1, 8, 216, 8000, 343000,\dots ( A002897)
s_{4C}(k)=\sum_{j=0}^k\tbinom{2j}{j}^3\tbinom{k+j}{k-j}(-16)^{k-j}= (-1)^k \sum_{j=0}^k\tbinom{2j}{j}^2\tbinom{2k-2j}{k-j}^2 =1, -8, 88, -1088, 14296,\dots ( A036917)

Examples:

\frac{1}{\pi} = 8\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{4A}(k)\,\frac{6k+1}{(-2^9)^{k+1/2}},\quad j_{4A}\Big(\tfrac{1+\sqrt{-4}}{2}\Big)=-2^9
\frac{1}{\pi} = 16\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{4C}(k)\,\frac{1-2\sqrt{2}\, (k+\tfrac{1}{2})}{(-16\,U_{2}^{4})^{k+1/2}},\quad j_{4C}\Big(\tfrac{1+\sqrt{-4}}{2}\Big) = -16\,\big(1+\sqrt{2}\big)^4=-16\,U_{2}^{4}

Level 5

Define,

\begin{align} j_{5A}(\tau)&=\big(\tfrac{\eta(\tau)}{\eta(5\tau)}\big)^{6}+5^3 \big(\tfrac{\eta(5\tau)}{\eta(\tau)}\big)^{6}+22 =\tfrac{1}{q} + 16 + 134q + 760q^2 +3345q^3+\dots\\ j_{5B}(\tau)&=\big(\tfrac{\eta(\tau)}{\eta(5\tau)}\big)^{6}= \tfrac{1}{q}- 6 + 9q + 10q^2 - 30q^3 + 6q^4 + \dots \end{align}

and,

s_{5A}(k)=\tbinom{2k}{k}\sum_{j=0}^k \tbinom{k}{j}^2\tbinom{k+j}{j} =1, 6, 114, 2940, 87570,\dots
s_{5B}(k)=\sum_{j=0}^k(-1)^{j+k}\tbinom{k}{j}^3\tbinom{4k-5j}{3k}=1, -5, 35, -275, 2275, -19255,\dots ( A229111)

where the first is the product of the central binomial coefficients and the Apéry numbers ( A005258)[9]

Examples:

\frac{1}{\pi} = \frac{5}{9}\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{5A}(k)\,\frac{682k+71}{(-15228)^{k+1/2}},\quad j_{5A}\Big(\tfrac{5+\sqrt{-5(47)}}{10}\Big)=-15228=-(18\sqrt{47})^2
\frac{1}{\pi} = \frac{6}{\sqrt{5}}\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{5B}(k)\,\frac{25\sqrt{5}-141(k+\tfrac{1}{2})}{(-5\sqrt{5}\,U_{5}^{15})^{k+1/2}},\quad j_{5B}\Big(\tfrac{5+\sqrt{-5(47)}}{10}\Big)=-5\sqrt{5}\,\big(\tfrac{1+\sqrt{5}}{2}\big)^{15}=-5\sqrt{5}\,U_{5}^{15}

Level 6

Modular functions

In 2002, Sato[7] established the first results for level > 4. Interestingly, it involved Apéry numbers which were first used to establish the irrationality of \zeta(3). First, define,

\begin{align}j_{6A}(\tau) &=j_{6B}(\tau)+\tfrac{1}{j_{6B}(\tau)}+2 = j_{6C}(\tau)+\tfrac{64}{j_{6C}(\tau)}+20 = j_{6D}(\tau)+\tfrac{81}{j_{6D}(\tau)}+18 =\tfrac{1}{q} + 14 + 79q + 352q^2 +\dots \end{align}
\begin{align}j_{6B}(\tau) &= \Big(\tfrac{\eta(2\tau)\eta(3\tau)}{\eta(\tau)\eta(6\tau)}\Big)^{12}=\tfrac{1}{q} + 12 + 78q + 364q^2 + 1365q^3+\dots \end{align}
\begin{align}j_{6C}(\tau) &= \Big(\tfrac{\eta(\tau)\eta(3\tau)}{\eta(2\tau)\eta(6\tau)}\Big)^{6}=\tfrac{1}{q} -6 + 15q -32q^2 + 87q^3-192q^4+\dots \end{align}
\begin{align}j_{6D}(\tau) &= \Big(\tfrac{\eta(\tau)\eta(2\tau)}{\eta(3\tau)\eta(6\tau)}\Big)^{4}=\tfrac{1}{q} -4 - 2q + 28q^2 - 27q^3 - 52q^4+\dots\end{align}
\begin{align}j_{6E}(\tau) &= \Big(\tfrac{\eta(2\tau)\eta^3(3\tau)}{\eta(\tau)\eta^3(6\tau)}\Big)^{3}=\tfrac{1}{q} +3 + 6q + 4q^2 - 3q^3 - 12q^4 +\dots\end{align}

J. Conway and S. Norton showed[14] there are linear relations between the McKay–Thompson series Tn, one of which was,

T_{6A}-T_{6B}-T_{6C}-T_{6D}+2T_{6E} = 0

or using the above eta quotients jn,

j_{6A}-j_{6B}-j_{6C}-j_{6D}+2j_{6E} = 18

σ Sequences

For the modular function j6A, one can associate it with three different sequences. Define the number of 2n-step polygons on a cubic lattice, or the product of the central binomial coefficients c(k)=\tbinom{2k}{k} and  A002893 as σ1(k),

\sigma_1(k)=\tbinom{2k}{k}\sum_{j=0}^k \tbinom{k}{j}^2\tbinom{2j}{j} =1, 6, 90, 1860, 44730,\dots ( A002896)

the product of c(k) and (-1)^k  A093388 as σ2(k),

\sigma_2(k)=\tbinom{2k}{k}\sum_{j=0}^k \tbinom{k}{j}(-8)^{k-j}\sum_{m=0}^j\tbinom{j}{m}^3 =1, -12, 252, -6240, 167580, -4726512,\dots

and the product of c(k) and Franel numbers as σ3(k),

\sigma_3(k)=\tbinom{2k}{k}\sum_{j=0}^k \tbinom{k}{j}^3 =1, 4, 60, 1120, 24220,\dots ( A181418)

Each has a companion sequence. Respectively, these are the Apéry numbers,

s_{6B}(k)=\sum_{j=0}^k \tbinom{k}{j}^2\tbinom{k+j}{j}^2 =1, 5, 73, 1445, 33001,\dots ( A005259)

the Domb numbers (unsigned), or the number of 2n-step polygons on a diamond lattice,

s_{6C}(k)=(-1)^k \sum_{j=0}^k \tbinom{k}{j}^2 \tbinom{2(k-j)}{k-j} \tbinom{2j}{j} =1, -4, 28, -256, 2716,\dots ( A002895)

and the Almkvist-Zudilin numbers,

s_{6D}(k)=\sum_{j=0}^k (-1)^{k-j}\,3^{k-3j}\,\tfrac{(3j)!}{j!^3} \tbinom{k}{3j} \tbinom{k+j}{j} =1, -3, 9, -3, -279, 2997,\dots ( A125143)

Identities

The modular functions can be related as P=Q=R where,

\begin{align} P &= \sum_{k=0}^\infty \sigma_1(k)\,\frac{1}{(j_{6A}(\tau))^{k+1/2}} = \pm \sum_{k=0}^\infty s_{6B}(k)\,\frac{1}{(j_{6B}(\tau))^{k+1/2}}\\ Q &= \sum_{k=0}^\infty \sigma_2(k)\,\frac{1}{(j_{6A}(\tau)-36)^{k+1/2}} = \pm \sum_{k=0}^\infty s_{6C}(k)\,\frac{1}{(j_{6C}(\tau))^{k+1/2}}\\ R &= \sum_{k=0}^\infty \sigma_3(k)\,\frac{1}{(j_{6A}(\tau)-4)^{k+1/2}} = \pm \sum_{k=0}^\infty s_{6D}(k)\,\frac{1}{(j_{6D}(\tau))^{k+1/2}} \end{align}

if the series converges and the sign chosen appropriately.

Examples

One can use a value for j6A in three ways. For example, starting with,

j_{6A}\Big(\sqrt{\tfrac{-3}{6}}\Big)=10^2

then,

\begin{align} \frac{1}{\pi} &= \frac{\sqrt{3}}{5^2}\,\sum_{k=0}^\infty \sigma_1(k)\,\frac{16k+3}{(10^2)^{k}}\\ \frac{1}{\pi} &= \frac{\sqrt{3}}{2^3}\,\sum_{k=0}^\infty \sigma_2(k)\,\frac{10k+3}{(10^2-36)^{k}}\\ \frac{1}{\pi} &= \frac{1}{3\sqrt{2}}\,\sum_{k=0}^\infty \sigma_3(k)\,\frac{5k+1}{(10^2-4)^{k}}\\ \end{align}

For the other modular functions,

\frac{1}{\pi} = 8\sqrt{15}\,\sum_{k=0}^\infty s_{6B}(k)\,\Big(\tfrac{1}{2}-\tfrac{3\sqrt{5}}{20}+k\Big)\Big(\frac{1}{\phi^{12}}\Big)^{k+1/2}, \quad j_{6B}\Big(\sqrt{\tfrac{-5}{6}}\Big)=\Big(\tfrac{1+\sqrt{5}}{2}\Big)^{12}=\phi^{12}
\frac{1}{\pi} = \frac{1}{2}\,\sum_{k=0}^\infty s_{6C}(k)\,\frac{3k+1}{32^k}, \quad j_{6C}\Big(\sqrt{\tfrac{-1}{3}}\Big)=32
\frac{1}{\pi} = 2\sqrt{3}\,\sum_{k=0}^\infty s_{6D}(k)\,\frac{4k+1}{81^{k+1/2}}, \quad j_{6D}\Big(\sqrt{\tfrac{-1}{2}}\Big)=81

Level 7

Define

s_{7A}(k)=\sum_{j=0}^k \tbinom{k}{j}^2\tbinom{2j}{k}\tbinom{k+j}{j} =1, 4, 48, 760, 13840,\dots ( A183204)

and,

\begin{align} j_{7A}(\tau) &=\Big(\big(\tfrac{\eta(\tau)}{\eta(7\tau)}\big)^{2}+7 \big(\tfrac{\eta(7\tau)}{\eta(\tau)}\big)^{2}\Big)^2=\tfrac{1}{q} +10 + 51q + 204q^2 +681q^3+\dots\\ j_{7B}(\tau)&=\big(\tfrac{\eta(\tau)}{\eta(7\tau)}\big)^{4}= \tfrac{1}{q}- 4 + 2q + 8q^2 - 5q^3 - 4q^4 - 10q^5 + \dots \end{align}

Example:

\frac{1}{\pi} = \frac{3}{22^3}\,\sum_{k=0}^\infty s_{7A}(k)\, \frac{11895k+1286}{(-22^3)^{k}}, \quad j_{7A}\Big(\tfrac{7+\sqrt{-427}}{14}\Big) = -22^3+1 = -(39\sqrt{7})^2

No pi formula has yet been found using j7B.

Level 8

Define,

\begin{align} j_{4B}(\tau)&=\big(j_{2A}(2\tau)\big)^{1/2}=\tfrac{1}{q} + 52q + 834q^3 + 4760q^5 + 24703q^7+\dots\\ &= \Big(\big(\tfrac{\eta(\tau)\,\eta^2(4\tau)}{\eta^2(2\tau)\,\eta(8\tau)}\big)^{4}+4 \big(\tfrac{\eta^2(2\tau)\,\eta(8\tau)}{\eta(\tau)\,\eta^2(4\tau)}\big)^{4}\Big)^2 = \Big(\big(\tfrac{\eta(2\tau)\,\eta(4\tau)}{\eta(\tau)\,\eta(8\tau)}\big)^{4} - 4 \big(\tfrac{\eta(\tau)\,\eta(8\tau)}{\eta(2\tau)\,\eta^2(\tau)}\big)^{4}\Big)^2\\ j_{8A'}(\tau)&=\big(\tfrac{\eta(\tau)\,\eta^2(4\tau)}{\eta^2(2\tau)\,\eta(8\tau)}\big)^{8}=\tfrac{1}{q} - 8 + 36q - 128q^2 + 386q^3 -1024q^4+\dots\\ j_{8A}(\tau)&=\big(\tfrac{\eta(2\tau)\,\eta(4\tau)}{\eta(\tau)\,\eta(8\tau)}\big)^{8}=\tfrac{1}{q} + 8 + 36q + 128q^2 + 386q^3 +1024q^4+\dots.\\ \end{align}

The expansion of the first is the McKay–Thompson series of class 4B (and is a square root of another function) while the second, if unsigned, is that of class 8A given by the third. Let,

s_{4B}(k)=\tbinom{2k}{k}\sum_{j=0}^k 4^{k-2j}\tbinom{k}{2j}\tbinom{2j}{j}^2 =\tbinom{2k}{k}\sum_{j=0}^k \tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}=1, 8, 120, 2240, 47320,\dots
s_{8A'}(k)=\sum_{j=0}^k (-1)^{k}\tbinom{k}{j}^2\tbinom{2j}{k}^2 =1, -4, 40, -544, 8536,\dots

where the first is the product[2] of the central binomial coefficient and a sequence related to an arithmetic-geometric mean ( A081085),

Examples:

\frac{1}{\pi} = \frac{2\sqrt{2}}{13}\,\sum_{k=0}^\infty s_{4B}(k)\,\frac{70\cdot99\,k+579}{(16+396^2)^{k+1/2}},\quad j_{4B}\Big(\tfrac{1}{4}\sqrt{-58}\Big)=396^2
\frac{1}{\pi} = \frac{\sqrt{-2}}{70}\,\sum_{k=0}^\infty s_{4B}(k)\, \frac{58\cdot13\cdot99\,k + 6243}{(16-396^2)^{k+1/2}}
\frac{1}{\pi} = 2\sqrt{2}\,\sum_{k=0}^\infty s_{8A'}(k)\,\frac{-222+377\sqrt{2}\,(k+\tfrac{1}{2})}{\big(4(1+\sqrt{2})^{12}\big)^{k+1/2}},\quad j_{8A'}\Big(\tfrac{1}{4}\sqrt{-58}\Big)=4(1+\sqrt{2})^{12},\quad j_{8A}\Big(\tfrac{1}{4}\sqrt{-58}\Big)=4(99+13\sqrt{58})^{2}=4U_{58}^2

though no pi formula is yet known using j8A(τ).

Level 9

Define,

\begin{align} j_{3C}(\tau) &= \big(j(3\tau))^{1/3} =-6+\big(\tfrac{\eta^2(3\tau)}{\eta(\tau)\,\eta(9\tau)}\big)^6 -27 \big(\tfrac{\eta(\tau)\,\eta(9\tau)}{\eta^2(3\tau)}\big)^6=\tfrac{1}{q} + 248q^2 + 4124q^5 +34752q^8+\dots\\ j_{9A}(\tau) &= \big(\tfrac{\eta^2(3\tau)}{\eta(\tau)\,\eta(9\tau)}\big)^6 = \tfrac{1}{q} + 6 + 27q + 86q^2 + 243q^3 + 594q^4+\dots\\ \end{align}

The expansion of the first is the McKay–Thompson series of class 3C (and related to the cube root of the j-function), while the second is that of class 9A. Let,

s_{3C}(k)=\tbinom{2k}{k}\sum_{j=0}^k (-3)^{k-3j}\tbinom{k}{j}\tbinom{k-j}{j}\tbinom{k-2j}{j} =\tbinom{2k}{k}\sum_{j=0}^k(-3)^{k-3j}\tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j} = 1, -6, 54, -420, 630,\dots
s_{9A}(k)=\sum_{j=0}^k\tbinom{k}{j}^2\sum_{m=0}^j\tbinom{k}{m}\tbinom{j}{m}\tbinom{j+m}{k} =1, 3, 27, 309, 4059,\dots

where the first is the product of the central binomial coefficients and  A006077 (though with different signs).

Examples:

\frac{1}{\pi} = \frac{-\boldsymbol{i}}{9}\sum_{k=0}^\infty s_{3C}(k)\,\frac{602k+85}{(-960-12)^{k+1/2}},\quad j_{3C}\Big(\tfrac{3+\sqrt{-43}}{6}\Big)=-960
\frac{1}{\pi} = 6\,\boldsymbol{i}\,\sum_{k=0}^\infty s_{9A}(k)\,\frac{4-\sqrt{129}\,(k+\tfrac{1}{2})}{\big( -3\sqrt{3U_{129}}\big)^{k+1/2}},\quad j_{9A}\Big(\tfrac{3+\sqrt{-43}}{6}\Big)=-3\sqrt{3}\big(53\sqrt{3}+14\sqrt{43}\big) = -3\sqrt{3U_{129}}

Level 10

Modular functions

Define,

\begin{align}j_{10A}(\tau) &=j_{10B}(\tau)+\tfrac{16}{j_{10B}(\tau)}+8 = j_{10C}(\tau)+\tfrac{25}{j_{10C}(\tau)}+6 = j_{10D}(\tau)+\tfrac{1}{j_{10D}(\tau)}-2 =\tfrac{1}{q} + 4 + 22q + 56q^2 +\dots \end{align}
\begin{align}j_{10B}(\tau) &= \Big(\tfrac{\eta(\tau)\eta(5\tau)}{\eta(2\tau)\eta(10\tau)}\Big)^{4}=\tfrac{1}{q} - 4 + 6q - 8q^2 + 17q^3 - 32q^4 +\dots \end{align}
\begin{align}j_{10C}(\tau) &= \Big(\tfrac{\eta(\tau)\eta(2\tau)}{\eta(5\tau)\eta(10\tau)}\Big)^{2}=\tfrac{1}{q} - 2 - 3q + 6q^2 + 2q^3 + 2q^4+\dots\end{align}
\begin{align}j_{10D}(\tau) &= \Big(\tfrac{\eta(2\tau)\eta(5\tau)}{\eta(\tau)\eta(10\tau)}\Big)^{6}=\tfrac{1}{q} + 6 + 21q + 62q^2 + 162q^3 +\dots \end{align}
\begin{align}j_{10E}(\tau) &= \Big(\tfrac{\eta(2\tau)\eta^5(5\tau)}{\eta(\tau)\eta^5(10\tau)}\Big)=\tfrac{1}{q} + 1 + q + 2q^2 + 2q^3 - 2q^4 +\dots\end{align}

Just like the level 6, there are also linear relations between these,

T_{10A}-T_{10B}-T_{10C}-T_{10D}+2T_{10E} = 0

or using the above eta quotients jn,

j_{10A}-j_{10B}-j_{10C}-j_{10D}+2j_{10E} = 6

Sequences

Let,

v_{1}(k)=\sum_{j=0}^k \tbinom{k}{j}^4 =1, 2, 18, 164, 1810,\dots ( A005260)
v_{2}(k)=\tbinom{2k}{k}\sum_{j=0}^k \tbinom{2j}{j}^{-1}\tbinom{k}{j}\sum_{m=0}^j \tbinom{j}{m}^4 =1, 4, 36, 424, 5716,\dots
v_{3}(k)=\tbinom{2k}{k}\sum_{j=0}^k \tbinom{2j}{j}^{-1}\tbinom{k}{j} (-4)^{k-j}\sum_{m=0}^j \tbinom{j}{m}^4 =1, -6, 66, -876, 12786,\dots

their complements,

v_{2}'(k)=\tbinom{2k}{k}\sum_{j=0}^k \tbinom{2j}{j}^{-1}\tbinom{k}{j} (-1)^{k-j}\sum_{m=0}^j \tbinom{j}{m}^4 =1, 0, 12, 24, 564, 2784,\dots
v_{3}'(k)=\tbinom{2k}{k}\sum_{j=0}^k \tbinom{2j}{j}^{-1}\tbinom{k}{j} (4)^{k-j}\sum_{m=0}^j \tbinom{j}{m}^4 =1, 10, 162, 3124, 66994,\dots

and,

s_{10B}(k)=1, -2, 10, -68, 514, -4100, 33940,\dots
s_{10C}(k)=1, -1, 1, -1, 1, 23, -263, 1343, -2303,\dots
s_{10D}(k)=1, 3, 25, 267, 3249, 42795, 594145,\dots

though closed-forms are not yet known for the last three sequences.

Identities

The modular functions can be related as,[15]

U = \sum_{k=0}^\infty v_1(k)\,\frac{1}{(j_{10A}(\tau))^{k+1/2}} = \sum_{k=0}^\infty v_2(k)\,\frac{1}{(j_{10A}(\tau)+4)^{k+1/2}} = \sum_{k=0}^\infty v_3(k)\,\frac{1}{(j_{10A}(\tau)-16)^{k+1/2}}

if the series converges. It can also be observed that U=V=W where,

V =\sum_{k=0}^\infty v_2'(k)\,\frac{1}{(j_{10A}(\tau)-4)^{k+1/2}} = \sum_{k=0}^\infty v_3'(k)\,\frac{1}{(j_{10A}(\tau)+16)^{k+1/2}}
W = \sum_{k=0}^\infty s_{10B}(k)\,\frac{1}{(j_{10B}(\tau))^{k+1/2}} = \sum_{k=0}^\infty s_{10C}(k)\,\frac{1}{(j_{10C}(\tau))^{k+1/2}} = \sum_{k=0}^\infty s_{10D}(k)\,\frac{1}{(j_{10D}(\tau))^{k+1/2}}

Since the exponent has a fractional part, the sign of the square root must be chosen appropriately though it is less an issue when jn is positive.

Examples

Starting with,

j_{10A}\Big(\sqrt{\tfrac{-19}{10}}\Big) = 76^2

then,

\begin{align} \frac{1}{\pi} &= \frac{5}{\sqrt{95}}\,\sum_{k=0}^\infty v_1(k)\,\frac{408k+47}{(76^2)^{k+1/2}}\\ \frac{1}{\pi} &= \frac{1}{17\sqrt{95}}\,\sum_{k=0}^\infty v_2(k)\,\frac{19\cdot 1824k+3983}{(76^2+4)^{k+1/2}}\\ \frac{1}{\pi} &= \frac{5}{481\sqrt{95}}\,\sum_{k=0}^\infty v_2'(k)\,\frac{19\cdot 10336k+22675}{(76^2-4)^{k+1/2}}\\ \frac{1}{\pi} &= \frac{1}{6\sqrt{95}}\,\,\sum_{k=0}^\infty v_3(k)\,\,\frac{19\cdot 646k+1427}{(76^2-16)^{k+1/2}}\\ \frac{1}{\pi} &= \frac{5}{181\sqrt{95}}\,\sum_{k=0}^\infty v_3'(k)\,\frac{19\cdot 3876k+8405}{(76^2+16)^{k+1/2}} \end{align}

though the ones using the complements do not yet have a rigorous proof. A conjectured formula using one of the last three sequences is,

\frac{1}{\pi} = \frac{\boldsymbol{i}}{\sqrt{5}}\,\sum_{k=0}^\infty s_{10C}(k)\frac{10k+3}{(-25)^{k+1/2}},\quad j_{10C}\Big(\tfrac{1+\,\boldsymbol{i}}{2}\Big) = -25

which implies there might be examples for all sequences of level 10.

Level 11

Define the McKay–Thompson series of class 11A,

j_{11A}(\tau)= (1+3F)^3+(\tfrac{1}{\sqrt{F}}+3\sqrt{F})^2=\tfrac{1}{q} + 6 + 17q + 46q^2 + 116q^3 +\dots

where,

F = \tfrac{\eta(3\tau)\,\eta(33\tau)}{\eta(\tau)\,\eta(11\tau)}

and,

s_{11A}(k) = 1,\, 4,\, 28,\, 268,\, 3004,\, 36784,\, 476476,\dots

No closed-form in terms of binomial coefficients is yet known for the sequence but it obeys the recurrence relation,

(k + 1)^3 s_{k + 1} = 2(2k + 1)(5k^2 + 5k + 2)s_k\,-\,8k(7k^2 + 1)s_{k - 1}\,+\,22k(k - 1)(2k - 1)s_{k - 2}

with initial conditions s(0) = 1, s(1) = 4.

Example:[16]

\frac{1}{\pi}=\frac{\boldsymbol{i}}{22}\sum_{k=0}^\infty s_{11A}(k)\,\frac{221k+67}{(-44)^{k+1/2}},\quad j_{11A}\Big(\tfrac{1+\sqrt{-17/11}}{2}\Big)=-44

Higher levels

As pointed out by Cooper,[16] there are analogous sequences for certain higher levels.

See also

References

  1. 1 2 Heng Huat Chan, Song Heng Chan, and Zhiguo Liu, "Domb's numbers and Ramanujan–Sato type series for 1/Pi" (2004)
  2. 1 2 3 Gert Almkvist and Jesus Guillera, Ramanujan–Sato Like Series (2012)
  3. 1 2 3 H.H. Chan and S. Cooper, "Rational analogues of Ramanujan's series for 1/π", Mathematical Proceedings of the Cambridge Philosophical Society / Volume 153 / Issue 02 / September 2012, pp. 361–383
  4. ↑ G. Almkvist, Some conjectured formulas for 1/Pi coming from polytopes, K3-surfaces and Moonshine, http://arxiv.org/abs/1211.6563
  5. ↑ S. Ramanujan, "Modular equations and approximations to pi", Quart. J. Math. (Oxford) 45 (1914)
  6. ↑ Chan, Tanigawa, Yang, and Zudilin, "New analogues of Clausen’s identities arising from the theory of modular forms" (2011)
  7. 1 2 T. Sato, "Apéry numbers and Ramanujan's series for 1/π", Abstract of a talk presented at the Annual meeting of the Mathematical Society of Japan, 2002
  8. ↑ H. Chan and H. Verrill, "The Apéry numbers, the Almkvist–Zudilin Numbers, and new series for 1/π", Advances in Mathematics, Vol 186, 2004
  9. 1 2 S. Cooper, "Sporadic sequences, modular forms and new series for 1/π", Ramanujan Journal 2012
  10. 1 2 D. Zagier, "Traces of Singular Moduli", (p.15-16), http://people.mpim-bonn.mpg.de/zagier/files/tex/TracesSingModuli/fulltext.pdf
  11. ↑ Chudnovsky, David V.; Chudnovsky, Gregory V. (1989), "The Computation of Classical Constants", Proceedings of the National Academy of Sciences of the United States of America 86 (21): 8178–8182, doi:10.1073/pnas.86.21.8178, ISSN 0027-8424, JSTOR 34831, PMC 298242, PMID 16594075.
  12. ↑ Yee, Alexander; Kondo, Shigeru (2011), 10 Trillion Digits of Pi: A Case Study of summing Hypergeometric Series to high precision on Multicore Systems, Technical Report, Computer Science Department, University of Illinois.
  13. ↑ J.M. Borwein, P.B. Borwein and D.H. Bailey, "Ramanujan, modular equations, and approximations to pi; Or how to compute one billion digits of pi", Amer. Math. Monthly, 96 (1989) 201–219
  14. ↑ J. Conway and S. Norton, "Monstrous Moonshine", p.319, http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.103.3704&rep=rep1&type=pdf
  15. ↑ S. Cooper, "Level 10 analogues of Ramanujan’s series for 1/π", Theorem 4.3, p.85, J. Ramanujan Math. Soc. 27, No.1 (2012)
  16. 1 2 S. Cooper, "Ramanujan’s theories of elliptic functions to alternative bases, and beyond", Askey 80 Conference, Dec 2013, http://www.math.umn.edu/~stant001/ASKEYABS/Shaun_Cooper.pdf

External links

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