Ramanujan–Sato series

Ramanujan–Sato series

In mathematics, a Ramanujan–Sato series generalizes Ramanujan's pi formulas such as,

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

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

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

Ramanujan made the enigmatic remark that there were "corresponding theories", but it was only in 2012 that H. H. Chan and S. Cooper found a general approach that used the underlying modular congruence subgroup ${\displaystyle \Gamma {0}(n)}$, while G. Almkvist has experimentally found numerous other examples also with a general method using differential operators.

Levels 1–4A were given by Ramanujan (1914), level 5 by H. H. Chan and S. Cooper (2012), 6A by Chan, Tanigawa, Yang, and Zudilin, 6B by Sato (2002), 6C by H. Chan, S. Chan, and Z. Liu (2004), 6D by H. Chan and H. Verrill (2009), level 7 by S. Cooper (2012), part of level 8 by Almkvist and Guillera (2012), 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 and Tn refers to the relevant McKay–Thompson series.

Level 1

Ramanujan–Sato series

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

${\displaystyle {\begin{aligned}j(\tau )&=\left({\frac {E{4}(\tau )}{\eta ^{8}(\tau )}}\right)^{3}={\frac {1}{q}}+744+196884q+21493760q^{2}+\cdots \\j^{}(\tau )&=432\,{\frac {{\sqrt {j(\tau )}}+{\sqrt {j(\tau )-1728}}}{{\sqrt {j(\tau )}}-{\sqrt {j(\tau )-1728}}}}={\frac {1}{q}}-120+10260q-901120q^{2}+\cdots \end{aligned}}}$

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 (OEIS: A007240) with a(0) = 744. Note that, as first noticed by J. McKay, the coefficient of the linear term of j(τ) almost equals 196883, which is the degree of the smallest nontrivial irreducible representation of the monster group, a relationship called monstrous moonshine. Similar phenomena will be observed in the other levels. Define

${\displaystyle s{1A}(k)={\binom {2k}{k}}{\binom {3k}{k}}{\binom {6k}{3k}}=1,120,83160,81681600,\ldots }$ (OEIS: A001421)

${\displaystyle s{1B}(k)=\sum {j=0}^{k}{\binom {2j}{j}}{\binom {3j}{j}}{\binom {6j}{3j}}{\binom {k+j}{k-j}}(-432)^{k-j}=1,-312,114264,-44196288,\ldots }$

Then the two modular functions and sequences are related by

${\displaystyle \sum {k=0}^{\infty }s{1A}(k)\,{\frac {1}{(j(\tau ))^{k+{\frac {1}{2}}}}}=\pm \sum {k=0}^{\infty }s{1B}(k)\,{\frac {1}{(j^{}(\tau ))^{k+{\frac {1}{2}}}}}}$

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

${\displaystyle {\frac {1}{\pi }}=12\,{\boldsymbol {i}}\,\sum {k=0}^{\infty }s{1A}(k)\,{\frac {163\cdot 3344418k+13591409}{\left(-640320^{3}\right)^{k+{\frac {1}{2}}}}},\quad j\left({\frac {1+{\sqrt {-163}}}{2}}\right)=-640320^{3}=-262537412640768000}$

${\displaystyle {\frac {1}{\pi }}=24\,{\boldsymbol {i}}\,\sum {k=0}^{\infty }s{1B}(k)\,{\frac {-3669+320{\sqrt {645}}\,\left(k+{\frac {1}{2}}\right)}{\left({-432}\,U{645}^{3}\right)^{k+{\frac {1}{2}}}}},\quad j^{}\left({\frac {1+{\sqrt {-43}}}{2}}\right)=-432\,U{645}^{3}=-432\left({\frac {127+5{\sqrt {645}}}{2}}\right)^{3}}$

where ${\displaystyle 645=43\times 15,}$ and ${\displaystyle U{n}}$ is a fundamental unit. The first belongs to a family of formulas that were rigorously proven by the Chudnovsky brothers in 1989 and later used to calculate 10 trillion digits of π in 2011. The second formula, and the ones for higher levels, were established by H.H. Chan and S. Cooper in 2012.

Level 2

Ramanujan–Sato series

Using Zagier's notation for the modular function of level 2,

${\displaystyle {\begin{aligned}j{2A}(\tau )&=\left(\left({\frac {\eta (\tau )}{\eta (2\tau )}}\right)^{12}+2^{6}\left({\frac {\eta (2\tau )}{\eta (\tau )}}\right)^{12}\right)^{2}={\frac {1}{q}}+104+4372q+96256q^{2}+1240002q^{3}+\cdots \\j{2B}(\tau )&=\left({\frac {\eta (\tau )}{\eta (2\tau )}}\right)^{24}={\frac {1}{q}}-24+276q-2048q^{2}+11202q^{3}-\cdots \end{aligned}}}$

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

${\displaystyle s{2A}(k)={\binom {2k}{k}}{\binom {2k}{k}}{\binom {4k}{2k}}=1,24,2520,369600,63063000,\ldots }$ (OEIS: A008977)

${\displaystyle s{2B}(k)=\sum {j=0}^{k}{\binom {2j}{j}}{\binom {2j}{j}}{\binom {4j}{2j}}{\binom {k+j}{k-j}}(-64)^{k-j}=1,-40,2008,-109120,6173656,\ldots }$

${\displaystyle \sum {k=0}^{\infty }s{2A}(k)\,{\frac {1}{(j{2A}(\tau ))^{k+{\frac {1}{2}}}}}=\pm \sum {k=0}^{\infty }s{2B}(k)\,{\frac {1}{(j{2B}(\tau ))^{k+{\frac {1}{2}}}}}}$

if the series converges and the sign chosen appropriately.

${\displaystyle {\frac {1}{\pi }}=32{\sqrt {2}}\,\sum {k=0}^{\infty }s{2A}(k)\,{\frac {58\cdot 455k+1103}{\left(396^{4}\right)^{k+{\frac {1}{2}}}}},\quad j{2A}\left({\frac {\sqrt {-58}}{2}}\right)=396^{4}=24591257856}$

${\displaystyle {\frac {1}{\pi }}=16{\sqrt {2}}\,\sum {k=0}^{\infty }s{2B}(k)\,{\frac {-24184+9801{\sqrt {29}}\,\left(k+{\frac {1}{2}}\right)}{\left(64\,U{29}^{12}\right)^{k+{\frac {1}{2}}}}},\quad j{2B}\left({\frac {\sqrt {-58}}{2}}\right)=64\left({\frac {5+{\sqrt {29}}}{2}}\right)^{12}=64\,U{29}^{12}}$

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.

Level 3

Ramanujan–Sato series

${\displaystyle {\begin{aligned}j{3A}(\tau )&=\left(\left({\frac {\eta (\tau )}{\eta (3\tau )}}\right)^{6}+3^{3}\left({\frac {\eta (3\tau )}{\eta (\tau )}}\right)^{6}\right)^{2}={\frac {1}{q}}+42+783q+8672q^{2}+65367q^{3}+\cdots \\j{3B}(\tau )&=\left({\frac {\eta (\tau )}{\eta (3\tau )}}\right)^{12}={\frac {1}{q}}-12+54q-76q^{2}-243q^{3}+1188q^{4}+\cdots \\\end{aligned}}}$

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

${\displaystyle s{3A}(k)={\binom {2k}{k}}{\binom {2k}{k}}{\binom {3k}{k}}=1,12,540,33600,2425500,\ldots }$ (OEIS: A184423)

${\displaystyle s{3B}(k)=\sum {j=0}^{k}{\binom {2j}{j}}{\binom {2j}{j}}{\binom {3j}{j}}{\binom {k+j}{k-j}}(-27)^{k-j}=1,-15,297,-6495,149481,\ldots }$

${\displaystyle {\frac {1}{\pi }}=2\,{\boldsymbol {i}}\,\sum {k=0}^{\infty }s{3A}(k)\,{\frac {267\cdot 53k+827}{\left(-300^{3}\right)^{k+{\frac {1}{2}}}}},\quad j{3A}\left({\frac {3+{\sqrt {-267}}}{6}}\right)=-300^{3}=-27000000}$

${\displaystyle {\frac {1}{\pi }}={\boldsymbol {i}}\,\sum {k=0}^{\infty }s{3B}(k)\,{\frac {12497-3000{\sqrt {89}}\,\left(k+{\frac {1}{2}}\right)}{\left(-27\,U{89}^{2}\right)^{k+{\frac {1}{2}}}}},\quad j{3B}\left({\frac {3+{\sqrt {-267}}}{6}}\right)=-27\,\left(500+53{\sqrt {89}}\right)^{2}=-27\,U{89}^{2}}$

Level 4

Ramanujan–Sato series

${\displaystyle {\begin{aligned}j{4A}(\tau )&=\left(\left({\frac {\eta (\tau )}{\eta (4\tau )}}\right)^{4}+4^{2}\left({\frac {\eta (4\tau )}{\eta (\tau )}}\right)^{4}\right)^{2}=\left({\frac {\eta ^{2}(2\tau )}{\eta (\tau )\,\eta (4\tau )}}\right)^{24}=-\left({\frac {\eta \left({\frac {2\tau +3}{2}}\right)}{\eta (2\tau +3)}}\right)^{24}={\frac {1}{q}}+24+276q+2048q^{2}+11202q^{3}+\cdots \\j{4C}(\tau )&=\left({\frac {\eta (\tau )}{\eta (4\tau )}}\right)^{8}={\frac {1}{q}}-8+2q-62q^{3}+216q^{5}-641q^{7}+\ldots \\\end{aligned}}}$

where the first is the 24th power of the Weber modular function ${\displaystyle {\mathfrak {f}}(2\tau )}$. And,

${\displaystyle s{4A}(k)={\binom {2k}{k}}^{3}=1,8,216,8000,343000,\ldots }$ (OEIS: A002897)

${\displaystyle s{4C}(k)=\sum {j=0}^{k}{\binom {2j}{j}}^{3}{\binom {k+j}{k-j}}(-16)^{k-j}=(-1)^{k}\sum {j=0}^{k}{\binom {2j}{j}}^{2}{\binom {2k-2j}{k-j}}^{2}=1,-8,88,-1088,14296,\ldots }$ (OEIS: A036917)

${\displaystyle {\frac {1}{\pi }}=8\,{\boldsymbol {i}}\,\sum {k=0}^{\infty }s{4A}(k)\,{\frac {6k+1}{\left(-2^{9}\right)^{k+{\frac {1}{2}}}}},\quad j{4A}\left({\frac {1+{\sqrt {-4}}}{2}}\right)=-2^{9}=-512}$

${\displaystyle {\frac {1}{\pi }}=16\,{\boldsymbol {i}}\,\sum {k=0}^{\infty }s{4C}(k)\,{\frac {1-2{\sqrt {2}}\,\left(k+{\frac {1}{2}}\right)}{\left(-16\,U{2}^{4}\right)^{k+{\frac {1}{2}}}}},\quad j{4C}\left({\frac {1+{\sqrt {-4}}}{2}}\right)=-16\,\left(1+{\sqrt {2}}\right)^{4}=-16\,U{2}^{4}}$

Level 5

Ramanujan–Sato series

${\displaystyle {\begin{aligned}j{5A}(\tau )&=\left({\frac {\eta (\tau )}{\eta (5\tau )}}\right)^{6}+5^{3}\left({\frac {\eta (5\tau )}{\eta (\tau )}}\right)^{6}+22={\frac {1}{q}}+16+134q+760q^{2}+3345q^{3}+\cdots \\j{5B}(\tau )&=\left({\frac {\eta (\tau )}{\eta (5\tau )}}\right)^{6}={\frac {1}{q}}-6+9q+10q^{2}-30q^{3}+6q^{4}+\cdots \end{aligned}}}$

${\displaystyle s{5A}(k)={\binom {2k}{k}}\sum {j=0}^{k}{\binom {k}{j}}^{2}{\binom {k+j}{j}}=1,6,114,2940,87570,\ldots }$

${\displaystyle s{5B}(k)=\sum {j=0}^{k}(-1)^{j+k}{\binom {k}{j}}^{3}{\binom {4k-5j}{3k}}=1,-5,35,-275,2275,-19255,\ldots }$ (OEIS: A229111)

where the first is the product of the central binomial coefficients and the Apéry numbers (OEIS: A005258)

${\displaystyle {\frac {1}{\pi }}={\frac {5}{9}}\,{\boldsymbol {i}}\,\sum {k=0}^{\infty }s{5A}(k)\,{\frac {682k+71}{(-15228)^{k+{\frac {1}{2}}}}},\quad j{5A}\left({\frac {5+{\sqrt {-5(47)}}}{10}}\right)=-15228=-(18{\sqrt {47}})^{2}}$

${\displaystyle {\frac {1}{\pi }}={\frac {6}{\sqrt {5}}}\,{\boldsymbol {i}}\,\sum {k=0}^{\infty }s{5B}(k)\,{\frac {25{\sqrt {5}}-141\left(k+{\frac {1}{2}}\right)}{\left(-5{\sqrt {5}}\,U{5}^{15}\right)^{k+{\frac {1}{2}}}}},\quad j{5B}\left({\frac {5+{\sqrt {-5(47)}}}{10}}\right)=-5{\sqrt {5}}\,\left({\frac {1+{\sqrt {5}}}{2}}\right)^{15}=-5{\sqrt {5}}\,U{5}^{15}}$

Level 6

Modular functions

In 2002, Takeshi Sato established the first results for levels above 4. It involved Apéry numbers which were first used to establish the irrationality of ${\displaystyle \zeta (3)}$. First, define,

${\displaystyle {\begin{aligned}j{6A}(\tau )&=\left({\sqrt {j{6B}(\tau )}}-{\frac {1}{\sqrt {j{6B}(\tau )}}}\right)^{2}=\left({\sqrt {j{6C}(\tau )}}+{\frac {8}{\sqrt {j{6C}(\tau )}}}\right)^{2}=\left({\sqrt {j{6D}(\tau )}}+{\frac {9}{\sqrt {j{6D}(\tau )}}}\right)^{2}-4={\frac {1}{q}}+10+79q+352q^{2}+\cdots \end{aligned}}}$

${\displaystyle {\begin{aligned}j{6B}(\tau )&=\left({\frac {\eta (2\tau )\eta (3\tau )}{\eta (\tau )\eta (6\tau )}}\right)^{12}={\frac {1}{q}}+12+78q+364q^{2}+1365q^{3}+\cdots \end{aligned}}}$

${\displaystyle {\begin{aligned}j{6C}(\tau )&=\left({\frac {\eta (\tau )\eta (3\tau )}{\eta (2\tau )\eta (6\tau )}}\right)^{6}={\frac {1}{q}}-6+15q-32q^{2}+87q^{3}-192q^{4}+\cdots \end{aligned}}}$

${\displaystyle {\begin{aligned}j{6D}(\tau )&=\left({\frac {\eta (\tau )\eta (2\tau )}{\eta (3\tau )\eta (6\tau )}}\right)^{4}={\frac {1}{q}}-4-2q+28q^{2}-27q^{3}-52q^{4}+\cdots \end{aligned}}}$

${\displaystyle {\begin{aligned}j{6E}(\tau )&=\left({\frac {\eta (2\tau )\eta ^{3}(3\tau )}{\eta (\tau )\eta ^{3}(6\tau )}}\right)^{3}={\frac {1}{q}}+3+6q+4q^{2}-3q^{3}-12q^{4}+\cdots \end{aligned}}}$

The phenomenon of ${\displaystyle j{6A}}$ being squares or a near-square of the other functions will also be manifested by ${\displaystyle j{10A}}$. Another similarity between levels 6 and 10 is J. Conway and S. Norton showed there are linear relations between the McKay–Thompson series Tn, one of which was,

${\displaystyle T{6A}-T{6B}-T{6C}-T{6D}+2T{6E}=0}$

${\displaystyle j{6A}-j{6B}-j{6C}-j{6D}+2j{6E}=22}$

α Sequences

For the modular function j6A, one can associate it with three different sequences. (A similar situation happens for the level 10 function j10A.) Let,

${\displaystyle \alpha {1}(k)={\binom {2k}{k}}\sum {j=0}^{k}{\binom {k}{j}}^{3}=1,4,60,1120,24220,\ldots }$ (OEIS: A181418, labeled as s6 in Cooper's paper)

${\displaystyle \alpha {2}(k)={\binom {2k}{k}}\sum {j=0}^{k}{\binom {k}{j}}\sum {m=0}^{j}{\binom {j}{m}}^{3}={\binom {2k}{k}}\sum {j=0}^{k}{\binom {k}{j}}^{2}{\binom {2j}{j}}=1,6,90,1860,44730,\ldots }$ (OEIS: A002896)

${\displaystyle \alpha {3}(k)={\binom {2k}{k}}\sum {j=0}^{k}{\binom {k}{j}}(-8)^{k-j}\sum {m=0}^{j}{\binom {j}{m}}^{3}=1,-12,252,-6240,167580,-4726512,\ldots }$

The three sequences involve the product of the central binomial coefficients ${\displaystyle c(k)={\tbinom {2k}{k}}}$ with: first, the Franel numbers ${\displaystyle \textstyle \sum {j=0}^{k}{\tbinom {k}{j}}^{3}}$; second, OEIS: A002893, and third, ${\displaystyle (-1)^{k}}$ OEIS: A093388. Note that the second sequence, α2(k) is also the number of 2n-step polygons on a cubic lattice. Their complements,

${\displaystyle \alpha '{2}(k)={\binom {2k}{k}}\sum {j=0}^{k}{\binom {k}{j}}(-1)^{k-j}\sum {m=0}^{j}{\binom {j}{m}}^{3}=1,2,42,620,12250,\ldots }$

${\displaystyle \alpha '{3}(k)={\binom {2k}{k}}\sum {j=0}^{k}{\binom {k}{j}}(8)^{k-j}\sum {m=0}^{j}{\binom {j}{m}}^{3}=1,20,636,23840,991900,\ldots }$

There are also associated sequences, namely the Apéry numbers,

${\displaystyle s{6B}(k)=\sum {j=0}^{k}{\binom {k}{j}}^{2}{\binom {k+j}{j}}^{2}=1,5,73,1445,33001,\ldots }$ (OEIS: A005259)

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

${\displaystyle s{6C}(k)=(-1)^{k}\sum {j=0}^{k}{\binom {k}{j}}^{2}{\binom {2(k-j)}{k-j}}{\binom {2j}{j}}=1,-4,28,-256,2716,\ldots }$ (OEIS: A002895)

${\displaystyle s{6D}(k)=\sum {j=0}^{k}(-1)^{k-j}\,3^{k-3j}\,{\frac {(3j)!}{j!^{3}}}{\binom {k}{3j}}{\binom {k+j}{j}}=1,-3,9,-3,-279,2997,\ldots }$ (OEIS: A125143)

${\displaystyle {\frac {(3j)!}{j!^{3}}}={\binom {2j}{j}}{\binom {3j}{j}}}$

Identities

${\displaystyle P=\sum {k=0}^{\infty }\alpha {1}(k)\,{\frac {1}{\left(j{6A}(\tau )\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }\alpha {2}(k)\,{\frac {1}{\left(j{6A}(\tau )+4\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }\alpha {3}(k)\,{\frac {1}{\left(j{6A}(\tau )-32\right)^{k+{\frac {1}{2}}}}}}$

${\displaystyle Q=\sum {k=0}^{\infty }s{6B}(k)\,{\frac {1}{\left(j{6B}(\tau )\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }s{6C}(k)\,{\frac {1}{\left(j{6C}(\tau )\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }s{6D}(k)\,{\frac {1}{\left(j{6D}(\tau )\right)^{k+{\frac {1}{2}}}}}}$

if the series converges and the sign chosen appropriately. It can also be observed that,

${\displaystyle P=Q=\sum {k=0}^{\infty }\alpha '{2}(k)\,{\frac {1}{\left(j{6A}(\tau )-4\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }\alpha '{3}(k)\,{\frac {1}{\left(j{6A}(\tau )+32\right)^{k+{\frac {1}{2}}}}}}$

${\displaystyle \sum {k=0}^{\infty }\alpha {2}(k)\,{\frac {1}{\left(j{6A}(\tau )+4\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }\alpha '{2}(k)\,{\frac {1}{\left(j{6A}(\tau )-4\right)^{k+{\frac {1}{2}}}}}}$

Examples

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

${\displaystyle \Delta =j{6A}\left({\sqrt {\frac {-17}{6}}}\right)=198^{2}-4=\left(140{\sqrt {2}}\right)^{2}=39200}$

and noting that ${\displaystyle 3\cdot 17=51}$ then,

${\displaystyle {\begin{aligned}{\frac {1}{\pi }}&={\frac {24{\sqrt {3}}}{35}}\,\sum {k=0}^{\infty }\alpha {1}(k)\,{\frac {51\cdot 11k+53}{(\Delta )^{k+{\frac {1}{2}}}}}\\{\frac {1}{\pi }}&={\frac {4{\sqrt {3}}}{99}}\,\sum {k=0}^{\infty }\alpha {2}(k)\,{\frac {17\cdot 560k+899}{(\Delta +4)^{k+{\frac {1}{2}}}}}\\{\frac {1}{\pi }}&={\frac {\sqrt {3}}{2}}\,\sum {k=0}^{\infty }\alpha {3}(k)\,{\frac {770k+73}{(\Delta -32)^{k+{\frac {1}{2}}}}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {1}{\pi }}&={\frac {12{\sqrt {3}}}{9799}}\,\sum {k=0}^{\infty }\alpha '{2}(k)\,{\frac {11\cdot 51\cdot 560k+29693}{(\Delta -4)^{k+{\frac {1}{2}}}}}\\{\frac {1}{\pi }}&={\frac {6{\sqrt {3}}}{613}}\,\sum {k=0}^{\infty }\alpha '{3}(k)\,{\frac {51\cdot 770k+3697}{(\Delta +32)^{k+{\frac {1}{2}}}}}\\\end{aligned}}}$

though the formulas using the complements apparently do not yet have a rigorous proof. For the other modular functions,

${\displaystyle {\frac {1}{\pi }}=8{\sqrt {15}}\,\sum {k=0}^{\infty }s{6B}(k)\,\left({\frac {1}{2}}-{\frac {3{\sqrt {5}}}{20}}+k\right)\left({\frac {1}{\phi ^{12}}}\right)^{k+{\frac {1}{2}}},\quad j{6B}\left({\sqrt {\frac {-5}{6}}}\right)=\left({\frac {1+{\sqrt {5}}}{2}}\right)^{12}=\phi ^{12}}$

${\displaystyle {\frac {1}{\pi }}={\frac {1}{2}}\,\sum {k=0}^{\infty }s{6C}(k)\,{\frac {3k+1}{32^{k}}},\quad j{6C}\left({\sqrt {\frac {-1}{3}}}\right)=32}$

${\displaystyle {\frac {1}{\pi }}=2{\sqrt {3}}\,\sum {k=0}^{\infty }s{6D}(k)\,{\frac {4k+1}{81^{k+{\frac {1}{2}}}}},\quad j{6D}\left({\sqrt {\frac {-1}{2}}}\right)=81}$

Level 7

Ramanujan–Sato series

${\displaystyle s{7A}(k)=\sum {j=0}^{k}{\binom {k}{j}}^{2}{\binom {2j}{k}}{\binom {k+j}{j}}=1,4,48,760,13840,\ldots }$ (OEIS: A183204)

${\displaystyle {\begin{aligned}j{7A}(\tau )&=\left(\left({\frac {\eta (\tau )}{\eta (7\tau )}}\right)^{2}+7\left({\frac {\eta (7\tau )}{\eta (\tau )}}\right)^{2}\right)^{2}={\frac {1}{q}}+10+51q+204q^{2}+681q^{3}+\cdots \\j{7B}(\tau )&=\left({\frac {\eta (\tau )}{\eta (7\tau )}}\right)^{4}={\frac {1}{q}}-4+2q+8q^{2}-5q^{3}-4q^{4}-10q^{5}+\cdots \end{aligned}}}$

${\displaystyle {\frac {1}{\pi }}={\frac {\sqrt {7}}{22^{3}}}\,\sum {k=0}^{\infty }s{7A}(k)\,{\frac {11895k+1286}{\left(-22^{3}\right)^{k}}},\quad j{7A}\left({\frac {7+{\sqrt {-427}}}{14}}\right)=-22^{3}+1=-\left(39{\sqrt {7}}\right)^{2}=-10647}$

Level 8

Modular functions

Levels ${\displaystyle 2,4,8}$ are related since they are just powers of the same prime. Define,

${\displaystyle {\begin{aligned}j{4B}(\tau )&={\sqrt {j{2A}(2\tau )}}=\left({\sqrt {j{4D}(\tau )}}+{\frac {8}{\sqrt {j{4D}(\tau )}}}\right)^{2}-16=\left({\sqrt {j{8A}(\tau )}}-{\frac {4}{\sqrt {j{8A}(\tau )}}}\right)^{2}=\left({\sqrt {j{8A'}(\tau )}}+{\frac {4}{\sqrt {j{8A'}(\tau )}}}\right)^{2}\\&=\left({\frac {\eta (2\tau )}{\eta (4\tau )}}\right)^{12}+2^{6}\left({\frac {\eta (4\tau )}{\eta (2\tau )}}\right)^{12}={\frac {1}{q}}+52q+834q^{3}+4760q^{5}+24703q^{7}+\cdots \\j{4D}(\tau )&=\left({\frac {\eta (2\tau )}{\eta (4\tau )}}\right)^{12}={\frac {1}{q}}-12q+66q^{3}-232q^{5}+639q^{7}-1596q^{9}+\cdots \\j{8A}(\tau )&=\left({\frac {\eta (2\tau )\,\eta (4\tau )}{\eta (\tau )\,\eta (8\tau )}}\right)^{8}={\frac {1}{q}}+8+36q+128q^{2}+386q^{3}+1024q^{4}+\cdots \\j{8A'}(\tau )&=\left({\frac {\eta (\tau )\,\eta ^{2}(4\tau )}{\eta ^{2}(2\tau )\,\eta (8\tau )}}\right)^{8}={\frac {1}{q}}-8+36q-128q^{2}+386q^{3}-1024q^{4}+\cdots \\j{8B}(\tau )&=\left({\frac {\eta ^{2}(4\tau )}{\eta (2\tau )\,\eta (8\tau )}}\right)^{12}={\sqrt {j{4A}(2\tau )}}={\frac {1}{q}}+12q+66q^{3}+232q^{5}+639q^{7}+\cdots \\j{8E}(\tau )&=\left({\frac {\eta ^{3}(4\tau )}{\eta (2\tau )\,\eta ^{2}(8\tau )}}\right)^{4}={\frac {1}{q}}+4q+2q^{3}-8q^{5}-q^{7}+20q^{9}-2q^{11}-40q^{13}+\cdots \end{aligned}}}$

Just like for level 6, five of these functions have a linear relationship,

${\displaystyle j{4B}-j{4D}-j{8A}-j{8A'}+2j{8E}=0}$

But this is not one of the nine Conway-Norton-Atkin linear dependencies since ${\displaystyle j{8A'}}$ is not a moonshine function. However, it is related to one as,

${\displaystyle j{8A'}(\tau )=-j{8A}{\Big (}\tau +{\tfrac {1}{2}}{\Big )}}$

Sequences

${\displaystyle s{4B}(k)={\binom {2k}{k}}\sum {j=0}^{k}4^{k-2j}{\binom {k}{2j}}{\binom {2j}{j}}^{2}={\binom {2k}{k}}\sum {j=0}^{k}{\binom {k}{j}}{\binom {2k-2j}{k-j}}{\binom {2j}{j}}=1,8,120,2240,47320,\ldots }$

${\displaystyle s{4D}(k)={\binom {2k}{k}}^{3}=1,8,216,8000,343000,\ldots }$

${\displaystyle s{8A}(k)=\sum {j=0}^{k}{\binom {k}{j}}^{2}{\binom {2j}{k}}^{2}=1,4,40,544,8536,\ldots }$ (OEIS: A290575)

${\displaystyle s{8B}(k)=\sum {j=0}^{k}{\binom {2j}{j}}^{3}{\binom {2k-4j}{k-2j}}=1,2,14,36,334,\ldots }$

where the first is the product of the central binomial coefficient and a sequence related to an arithmetic-geometric mean (OEIS: A081085).

Identities

${\displaystyle \pm \sum {k=0}^{\infty }s{4B}(k)\,{\frac {1}{\left(j{4B}(\tau )+16\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }s{4D}(k)\,{\frac {1}{\left(j{4D}(\tau )\right)^{2k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }s{8A}(k)\,{\frac {1}{\left(j{8A}(\tau )\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }(-1)^{k}s{8A}(k)\,{\frac {1}{\left(j{8A'}(\tau )\right)^{k+{\frac {1}{2}}}}}}$

if the series converges and signs chosen appropriately. Note also the different exponent of ${\displaystyle \left(j{4D}(\tau )\right)^{2k+{\frac {1}{2}}}}$ from the others.

Examples

Recall that ${\displaystyle j{2A}\left({\tfrac {\sqrt {-58}}{2}}\right)=396^{4},}$ while ${\displaystyle j{4B}\left({\tfrac {\sqrt {-58}}{4}}\right)=396^{2}}$. Hence,

${\displaystyle {\frac {1}{\pi }}={\frac {2{\sqrt {2}}}{13}}\,\sum {k=0}^{\infty }s{4B}(k)\,{\frac {70\cdot 99\,k+579}{\left(396^{2}+16\right)^{k+{\frac {1}{2}}}}},\qquad j{4B}\left({\frac {\sqrt {-58}}{4}}\right)=396^{2}}$

${\displaystyle {\frac {1}{\pi }}=2{\sqrt {2}}\,\sum {k=0}^{\infty }s{8A}(k)\,{\frac {-222+70{\sqrt {58}}\,\left(k+{\frac {1}{2}}\right)}{\left(4\left(99+13{\sqrt {58}}\right)^{2}\right)^{k+{\frac {1}{2}}}}},\qquad j{8A}\left({\frac {\sqrt {-58}}{4}}\right)=4\left(99+13{\sqrt {58}}\right)^{2}=4U{58}^{2}}$

${\displaystyle {\frac {1}{\pi }}=2\,\sum {k=0}^{\infty }(-1)^{k}s{8A}(k)\,{\frac {-222{\sqrt {2}}+13\times 58\,\left(k+{\frac {1}{2}}\right)}{\left(4\left(1+{\sqrt {2}}\right)^{12}\right)^{k+{\frac {1}{2}}}}},\qquad j{8A'}\left({\frac {\sqrt {-58}}{4}}\right)=4\left(1+{\sqrt {2}}\right)^{12}=4U{2}^{12},}$

${\displaystyle {\frac {1}{\pi }}={\frac {1}{16}}{\sqrt {\frac {3}{5}}}\,\sum {k=0}^{\infty }s{8B}(k)\,{\frac {210k+43}{(64)^{k+{\frac {1}{2}}}}},\qquad j{8B}\left({\frac {\sqrt {-7}}{4}}\right)=2^{6}=64}$

Level 9

Ramanujan–Sato series

${\displaystyle {\begin{aligned}j{3C}(\tau )&=\left(j(3\tau )\right)^{\frac {1}{3}}=-6+\left({\frac {\eta ^{2}(3\tau )}{\eta (\tau )\,\eta (9\tau )}}\right)^{6}-27\left({\frac {\eta (\tau )\,\eta (9\tau )}{\eta ^{2}(3\tau )}}\right)^{6}={\frac {1}{q}}+248q^{2}+4124q^{5}+34752q^{8}+\cdots \\j{9A}(\tau )&=\left({\frac {\eta ^{2}(3\tau )}{\eta (\tau )\,\eta (9\tau )}}\right)^{6}={\frac {1}{q}}+6+27q+86q^{2}+243q^{3}+594q^{4}+\cdots \\\end{aligned}}}$

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,

${\displaystyle s{3C}(k)={\binom {2k}{k}}\sum {j=0}^{k}(-3)^{k-3j}{\binom {k}{j}}{\binom {k-j}{j}}{\binom {k-2j}{j}}={\binom {2k}{k}}\sum {j=0}^{k}(-3)^{k-3j}{\binom {k}{3j}}{\binom {2j}{j}}{\binom {3j}{j}}=1,-6,54,-420,630,\ldots }$

${\displaystyle s{9A}(k)=\sum {j=0}^{k}{\binom {k}{j}}^{2}\sum {m=0}^{j}{\binom {k}{m}}{\binom {j}{m}}{\binom {j+m}{k}}=1,3,27,309,4059,\ldots }$

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

${\displaystyle {\frac {1}{\pi }}={\frac {-{\boldsymbol {i}}}{9}}\sum {k=0}^{\infty }s{3C}(k)\,{\frac {602k+85}{\left(-960-12\right)^{k+{\frac {1}{2}}}}},\quad j{3C}\left({\frac {3+{\sqrt {-43}}}{6}}\right)=-960}$

${\displaystyle {\frac {1}{\pi }}=6\,{\boldsymbol {i}}\,\sum {k=0}^{\infty }s{9A}(k)\,{\frac {4-{\sqrt {129}}\,\left(k+{\frac {1}{2}}\right)}{\left(-3{\sqrt {3U{129}}}\right)^{k+{\frac {1}{2}}}}},\quad j{9A}\left({\frac {3+{\sqrt {-43}}}{6}}\right)=-3{\sqrt {3}}\left(53{\sqrt {3}}+14{\sqrt {43}}\right)=-3{\sqrt {3U{129}}}}$

Level 10

Modular functions

${\displaystyle {\begin{aligned}j{10A}(\tau )&=\left({\sqrt {j{10D}(\tau )}}-{\frac {1}{\sqrt {j{10D}(\tau )}}}\right)^{2}=\left({\sqrt {j{6B}(\tau )}}+{\frac {4}{\sqrt {j{10B}(\tau )}}}\right)^{2}=\left({\sqrt {j{10C}(\tau )}}+{\frac {5}{\sqrt {j{10C}(\tau )}}}\right)^{2}-4={\frac {1}{q}}+4+22q+56q^{2}+\cdots \end{aligned}}}$

${\displaystyle {\begin{aligned}j{10B}(\tau )&=\left({\frac {\eta (\tau )\eta (5\tau )}{\eta (2\tau )\eta (10\tau )}}\right)^{4}={\frac {1}{q}}-4+6q-8q^{2}+17q^{3}-32q^{4}+\cdots \end{aligned}}}$

${\displaystyle {\begin{aligned}j{10C}(\tau )&=\left({\frac {\eta (\tau )\eta (2\tau )}{\eta (5\tau )\eta (10\tau )}}\right)^{2}={\frac {1}{q}}-2-3q+6q^{2}+2q^{3}+2q^{4}+\cdots \end{aligned}}}$

${\displaystyle {\begin{aligned}j{10D}(\tau )&=\left({\frac {\eta (2\tau )\eta (5\tau )}{\eta (\tau )\eta (10\tau )}}\right)^{6}={\frac {1}{q}}+6+21q+62q^{2}+162q^{3}+\cdots \end{aligned}}}$

${\displaystyle {\begin{aligned}j{10E}(\tau )&=\left({\frac {\eta (2\tau )\eta ^{5}(5\tau )}{\eta (\tau )\eta ^{5}(10\tau )}}\right)={\frac {1}{q}}+1+q+2q^{2}+2q^{3}-2q^{4}+\cdots \end{aligned}}}$

Just like ${\displaystyle j{6A}}$, the function ${\displaystyle j{10A}}$ is a square or a near-square of the others. Furthermore, there are also linear relations between these,

${\displaystyle T{10A}-T{10B}-T{10C}-T{10D}+2T{10E}=0}$

${\displaystyle j{10A}-j{10B}-j{10C}-j{10D}+2j{10E}=6}$

β sequences

${\displaystyle \beta {1}(k)=\sum {j=0}^{k}{\binom {k}{j}}^{4}=1,2,18,164,1810,\ldots }$ (OEIS: A005260, labeled as s10 in Cooper's paper)

${\displaystyle \beta {2}(k)={\binom {2k}{k}}\sum {j=0}^{k}{\binom {2j}{j}}^{-1}{\binom {k}{j}}\sum {m=0}^{j}{\binom {j}{m}}^{4}=1,4,36,424,5716,\ldots }$

${\displaystyle \beta {3}(k)={\binom {2k}{k}}\sum {j=0}^{k}{\binom {2j}{j}}^{-1}{\binom {k}{j}}(-4)^{k-j}\sum {m=0}^{j}{\binom {j}{m}}^{4}=1,-6,66,-876,12786,\ldots }$

${\displaystyle \beta {2}'(k)={\binom {2k}{k}}\sum {j=0}^{k}{\binom {2j}{j}}^{-1}{\binom {k}{j}}(-1)^{k-j}\sum {m=0}^{j}{\binom {j}{m}}^{4}=1,0,12,24,564,2784,\ldots }$

${\displaystyle \beta {3}'(k)={\binom {2k}{k}}\sum {j=0}^{k}{\binom {2j}{j}}^{-1}{\binom {k}{j}}(4)^{k-j}\sum {m=0}^{j}{\binom {j}{m}}^{4}=1,10,162,3124,66994,\ldots }$

${\displaystyle s{10B}(k)=1,-2,10,-68,514,-4100,33940,\ldots }$

${\displaystyle s{10C}(k)=1,-1,1,-1,1,23,-263,1343,-2303,\ldots }$

${\displaystyle s{10D}(k)=1,3,25,267,3249,42795,594145,\ldots }$

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

Identities

${\displaystyle U=\sum {k=0}^{\infty }\beta {1}(k)\,{\frac {1}{\left(j{10A}(\tau )\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }\beta {2}(k)\,{\frac {1}{\left(j{10A}(\tau )+4\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }\beta {3}(k)\,{\frac {1}{\left(j{10A}(\tau )-16\right)^{k+{\frac {1}{2}}}}}}$

${\displaystyle V=\sum {k=0}^{\infty }s{10B}(k)\,{\frac {1}{\left(j{10B}(\tau )\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }s{10C}(k)\,{\frac {1}{\left(j{10C}(\tau )\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }s{10D}(k)\,{\frac {1}{\left(j{10D}(\tau )\right)^{k+{\frac {1}{2}}}}}}$

if the series converges. In fact, it can also be observed that,

${\displaystyle U=V=\sum {k=0}^{\infty }\beta {2}'(k)\,{\frac {1}{\left(j{10A}(\tau )-4\right)^{k+{\frac {1}{2}}}}}=\sum {k=0}^{\infty }\beta {3}'(k)\,{\frac {1}{\left(j{10A}(\tau )+16\right)^{k+{\frac {1}{2}}}}}}$

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

Just like level 6, the level 10 function j10A can be used in three ways. Starting with,

${\displaystyle j{10A}\left({\sqrt {\frac {-19}{10}}}\right)=76^{2}=5776}$

and noting that ${\displaystyle 5\cdot 19=95}$ then,

${\displaystyle {\begin{aligned}{\frac {1}{\pi }}&={\frac {5}{\sqrt {95}}}\,\sum {k=0}^{\infty }\beta {1}(k)\,{\frac {408k+47}{\left(76^{2}\right)^{k+{\frac {1}{2}}}}}\\{\frac {1}{\pi }}&={\frac {1}{17{\sqrt {95}}}}\,\sum {k=0}^{\infty }\beta {2}(k)\,{\frac {19\cdot 1824k+3983}{\left(76^{2}+4\right)^{k+{\frac {1}{2}}}}}\\{\frac {1}{\pi }}&={\frac {1}{6{\sqrt {95}}}}\,\,\sum {k=0}^{\infty }\beta {3}(k)\,\,{\frac {19\cdot 646k+1427}{\left(76^{2}-16\right)^{k+{\frac {1}{2}}}}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {1}{\pi }}&={\frac {5}{481{\sqrt {95}}}}\,\sum {k=0}^{\infty }\beta {2}'(k)\,{\frac {19\cdot 10336k+22675}{\left(76^{2}-4\right)^{k+{\frac {1}{2}}}}}\\{\frac {1}{\pi }}&={\frac {5}{181{\sqrt {95}}}}\,\sum {k=0}^{\infty }\beta {3}'(k)\,{\frac {19\cdot 3876k+8405}{\left(76^{2}+16\right)^{k+{\frac {1}{2}}}}}\end{aligned}}}$

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

${\displaystyle {\frac {1}{\pi }}={\frac {\boldsymbol {i}}{\sqrt {5}}}\,\sum {k=0}^{\infty }s{10C}(k){\frac {10k+3}{\left(-5^{2}\right)^{k+{\frac {1}{2}}}}},\quad j{10C}\left({\frac {1+\,{\boldsymbol {i}}}{2}}\right)=-5^{2}}$

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

Level 11

Ramanujan–Sato series

${\displaystyle j{11A}(\tau )=(1+3F)^{3}+\left({\frac {1}{\sqrt {F}}}+3{\sqrt {F}}\right)^{2}={\frac {1}{q}}+6+17q+46q^{2}+116q^{3}+\cdots }$

or sequence (OEIS: A128525) and where,

${\displaystyle F={\frac {\eta (3\tau )\,\eta (33\tau )}{\eta (\tau )\,\eta (11\tau )}}}$

${\displaystyle s{11A}(k)=1,4,28,268,3004,36784,476476,\ldots }$ (OEIS: A284756)

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

${\displaystyle (k+1)^{3}s{k+1}=2(2k+1)\left(5k^{2}+5k+2\right)s{k}-8k\left(7k^{2}+1\right)s{k-1}+22k(k-1)(2k-1)s{k-2}}$

${\displaystyle {\frac {1}{\pi }}={\frac {\boldsymbol {i}}{22}}\sum {k=0}^{\infty }s{11A}(k)\,{\frac {221k+67}{(-44)^{k+{\frac {1}{2}}}}},\quad j{11A}\left({\frac {1+{\sqrt {\frac {-17}{11}}}}{2}}\right)=-44}$

Higher levels

Ramanujan–Sato series

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

Similar series

Ramanujan–Sato series

R. Steiner found examples using Catalan numbers ${\displaystyle C{k}}$,

${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k-n}\right)^{2}{\frac {(4z)k+\left(4^{2n-3}-(4n-3)z\right)}{16^{k}}}\qquad z\in \mathbb {Z} ,\quad n\geq 2,\quad n\in \mathbb {N} }$

and for this a modular form with a second periodic for k exists:

${\displaystyle k={\frac {(-20-12{\boldsymbol {i}})+16n}{16}},\qquad k={\frac {(-20+12{\boldsymbol {i}})+16n}{16}}}$

${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k-2}\right)^{2}{\frac {3k+{\frac {1}{4}}}{16^{k}}}}$ ${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k-1}\right)^{2}{\frac {(4z+1)k-z}{16^{k}}}\qquad z\in \mathbb {Z} }$ ${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k-1}\right)^{2}{\frac {-1k+{\frac {1}{2}}}{16^{k}}}}$ ${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k-1}\right)^{2}{\frac {0k+{\frac {1}{4}}}{16^{k}}}}$ ${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k-1}\right)^{2}{\frac {{\frac {k}{5}}+{\frac {1}{5}}}{16^{k}}}}$ ${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k-1}\right)^{2}{\frac {{\frac {k}{3}}+{\frac {1}{6}}}{16^{k}}}}$ ${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k-1}\right)^{2}{\frac {{\frac {k}{2}}+{\frac {1}{8}}}{16^{k}}}}$ ${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k-1}\right)^{2}{\frac {2k-{\frac {1}{4}}}{16^{k}}}}$ ${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k-1}\right)^{2}{\frac {3k-{\frac {1}{2}}}{16^{k}}}}$ ${\displaystyle {\frac {1}{\pi }}=\sum {k=0}^{\infty }\left(2C{k}\right)^{2}{\frac {{\frac {k}{16}}+{\frac {1}{16}}}{16^{k}}}}$

with the last (comments in OEIS: A013709) found by using a linear combination of higher parts of Wallis-Lambert series for ${\displaystyle {\tfrac {4}{\pi }}}$ and Euler series for the circumference of an ellipse.

Using the definition of Catalan numbers with the gamma function the first and last for example give the identities

${\displaystyle {\frac {1}{4}}=\sum {k=0}^{\infty }{\left({\frac {\Gamma ({\frac {1}{2}}+k)}{\Gamma (2+k)}}\right)}^{2}\left(4zk-(4n-3)z+4^{2n-3}\right)\qquad z\in \mathbb {Z} ,\quad n\geq 2,\quad n\in \mathbb {N} }$

${\displaystyle 4=\sum {k=0}^{\infty }{\left({\frac {\Gamma ({\frac {1}{2}}+k)}{\Gamma (2+k)}}\right)}^{2}(k+1)}$.

${\displaystyle {\frac {1}{\pi }}={\frac {1}{4}}\sum {k=0}^{\infty }{\frac {{\binom {2k}{k}}^{2}}{k+1}}\,{\frac {1}{16^{k}}}}$

${\displaystyle \lim {k\rightarrow \infty }{\frac {16^{k}}{k{\binom {2k}{k}}^{2}}}=\pi }$

which is a consequence of Stirling's approximation.