Unveiling math links by Ramanujan & Shimura in number theory
In mathematics, a Ramanujan–Sato series[1][2] 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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81af8209737989d6ac37d5181cb0d114435324a0)
to the form
![{\displaystyle {\frac {1}{\pi }}=\sum _{k=0}^{\infty }s(k){\frac {Ak+B}{C^{k}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af5576cb46d16d14b7e0c346cd174fe00ab65392)
by using other well-defined sequences of integers
obeying a certain recurrence relation, sequences which may be expressed in terms of binomial coefficients
, 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
,[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 (1914),[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[edit]
Examples for levels 1–4 were given by Ramanujan in his 1917 paper. Given
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfd1361b00ff8449334b0533e5c8654e4bde9054)
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. Similar phenomena will be observed in the other levels. Define
(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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/df299357554ed711cc65d4233cdf986bf1127608)
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}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eed18508fc44b9652b102ea150a55e6d34aeda79)
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:
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ee011ec4a47cdfecb3826da10d96dfdb7a5ebea)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be466f62af0b2b54e2382efe849e35fa2780a657)
where
and
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[edit]
Using Zagier's notation[10] 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/335aa6cb490d5e366ff1fa63729adec733a7e721)
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,
(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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd0f34ec5843b14dc49d69d5a3a8e2abe1f85941)
Then,
![{\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}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c420c4af7b226b7447096113865e2544976d92f)
if the series converges and the sign chosen appropriately.
Examples:
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a12809a25b8d8fc381a8c871e5194eb7b777bd09)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/382928b461d28e434610163726d124676bd8ab84)
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[edit]
Define,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f750ba0bdbe225b527509ab9544ffe5604404cfa)
where 782 is the smallest degree greater than 1 of the irreducible representations of the Fischer group Fi23 and,
(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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0681053e14707781cf92b54e7d1f797d2eac0ba)
Examples:
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff37cdd38fcd2447751d1446e3b7619dd75244d2)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e191b1a5523a220e14d735e6df02801834d5b5ff)
Level 4[edit]
Define,
![{\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+20q-62q^{3}+216q^{5}-641q^{7}+\ldots \\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cbfd5b201b105787f24f1b60d9bb468813b3a93)
where the first is the 24th power of the Weber modular function
. And,
(OEIS: A002897)
(OEIS: A036917)
Examples:
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/548f81527935caac257159d4846486aee3482639)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98e8bb8e961bb7cbe40c8511eaf7c858cfaea56a)
Level 5[edit]
Define,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2dcd434227976a03428e81390f53a4faf46a0ee6)
and,
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cf11bf325a6241c1db8c43001519c9cf42d13d3)
(OEIS: A229111)
where the first is the product of the central binomial coefficients and the Apéry numbers (OEIS: A005258)[9]
Examples:
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e964b9c6a064ba9e06c38c0b44edad91f4c144b)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63373673db4e2a4073b735b6326c7a21a0c9b5da)
Level 6[edit]
Modular functions[edit]
In 2002, Sato[7] established the first results for levels above 4. It involved Apéry numbers which were first used to establish the irrationality of
. 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e5171e838bc7c3ba2b0b6ec55e51d2a3ffdfa8a)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6515c78336ec8e29b27e78f24409c96f188b489)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73b5a925371ae2feb14686cec867744b7552f952)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22f7993f246bb34c2a7169c56d95e005efb57b88)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d6c98c31c768eaca30f141b5e1bfd95e21cb964)
The phenomenon of
being squares or a near-square of the other functions will also be manifested by
. Another similarity between levels 6 and 10 is J. Conway and S. Norton showed there are linear relations between the McKay–Thompson series Tn,[14] one of which was,
![{\displaystyle T_{6A}-T_{6B}-T_{6C}-T_{6D}+2T_{6E}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4d56e9e2d126d6c7445ef4701c89c62413f3ff0)
or using the above eta quotients jn,
![{\displaystyle j_{6A}-j_{6B}-j_{6C}-j_{6D}+2j_{6E}=22}](https://wikimedia.org/api/rest_v1/media/math/render/svg/846d090505a56b0387eb972baccb70a37b638da1)
A similar relation exists for level 10.
α Sequences[edit]
For the modular function j6A, one can associate it with three different sequences. (A similar situation happens for the level 10 function j10A.) Let,
(OEIS: A181418, labeled as s6 in Cooper's paper)
(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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/bde56d91949d71112b5b461693a26037da2ae306)
The three sequences involve the product of the central binomial coefficients
with: first, the Franel numbers
; second, OEIS: A002893, and third,
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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5bdd05df1de38c8d152a02350f6d755a53af533)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c9e1e4eeb72ee40ae9100bdd7e4321a40f4e277)
There are also associated sequences, namely the Apéry numbers,
(OEIS: A005259)
the Domb numbers (unsigned) or the number of 2n-step polygons on a diamond lattice,
(OEIS: A002895)
and the Almkvist-Zudilin numbers,
(OEIS: A125143)
where
![{\displaystyle {\frac {(3j)!}{j!^{3}}}={\binom {2j}{j}}{\binom {3j}{j}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d14a8fa37825e85a470db66b34ee071ee4f6647f)
Identities[edit]
The modular functions can be related as,
![{\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}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac6412b3449176afec1648200a42941c05e5abd2)
![{\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}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15bcf06d681d273ef340af2f239172b6287b6c58)
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}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3fa61dd15a88ed3159f1c7321c5f802a99a50d4a)
which implies,
![{\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}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/804ed5512fafe5c3153a9673ca42ba0d129ddcfc)
and similarly using α3 and α'3.
Examples[edit]
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/544925ce266fbfa611bcde0a9bb9b3d752e1b89b)
and noting that
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8e990e21abcb3fb7b1a80d8c86aa4afa274003e)
as well as,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a511338af8a5774add411dfbaaccbcafca8a37be)
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22f60bf10d1ccf9cfbf32903937af7ca7c30e70c)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/361a4b375af0d839c2de6b1c63ae8b2c4df8b645)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd0be49590a751625402f0543447d89016312ab3)
Level 7[edit]
Define
(OEIS: A183204)
and,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37a109682cb89895604c0c26a3f8dad3692fec41)
Example:
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a6997908d6edc678a11a703fe0646a6a699f070)
No pi formula has yet been found using j7B.
Level 8[edit]
Modular functions[edit]
Levels
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d160a6c7ebfa341bf9405c01a98004a1dc81651)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d428666c4763b87b1998bf7947bbf71e6d1f759)
But this is not one of the nine Conway-Norton-Atkin linear dependencies since
is not a moonshine function. However, it is related to one as,
![{\displaystyle j_{8A'}(\tau )=-j_{8A}{\Big (}\tau +{\tfrac {1}{2}}{\Big )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ae9b2bff4e58f5135de8f1bcd4bbeca3ba0f255)
Sequences[edit]
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ee035942af72d80bafd08361ef5d362b7d0346a)
![{\displaystyle s_{4D}(k)={\binom {2k}{k}}^{3}=1,8,216,8000,343000,\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/308ec6683b03d725c36e8b1774ce4034128dab45)
(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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9a96e7b200bfc86860811916dd5716d2b5f24f1)
where the first is the product[2] of the central binomial coefficient and a sequence related to an arithmetic-geometric mean (OEIS: A081085).
Identities[edit]
The modular functions can be related as,
![{\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}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91367a79baebc2a70c7c812f218342feb9425347)
if the series converges and signs chosen appropriately. Note also the different exponent of
from the others.
Examples[edit]
Recall that
while
. 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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/952f5d566c21743e1d3cd2a2a42f7f0936ccded6)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3ac9f78d34defcb291b85cebed12a6f1a50f544)
![{\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},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00f5b9db163b3aeb74bbb331490fe60c73c059da)
For another level 8 example,
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5d6dde7b9978fc5c1e28af8b98daf5ea678c030)
Level 9[edit]
Define,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f361967c04cbabf4b837e0665ee369e1761311e3)
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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/dae08daf068c55ad2dc3c51e8ae6fac37a9b6d20)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e28179928ea6b1d6864d63cfb655c6a9a8db97de)
where the first is the product of the central binomial coefficients and OEIS: A006077 (though with different signs).
Examples:
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b4ca0e6f5ea68f24388cdb6961fed1c661fcd44)
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b181cf98aec67fbd89138d7957e5c658cf15f87)
Level 10[edit]
Modular functions[edit]
Define,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3ee3904fa9b3c2c27a9a6e6b1621db7f6c8eb8e)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05c994b4b006a29a512b6f781694fe0e3f415abf)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39f92ab6401be75eff0d9b766456b37f7f48ed3b)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f55cf81f13e40637100ea51140752fc1bb412941)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be5ae9b856d9757fa39d155a791f91cc29a4aade)
Just like
, the function
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a86398a17af2354295bd8b67ada1ce9c1902f4cf)
or using the above eta quotients jn,
![{\displaystyle j_{10A}-j_{10B}-j_{10C}-j_{10D}+2j_{10E}=6}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4ee0a8b928ea081e46126889c73e343bb1a5ffd)
β sequences[edit]
Let,
(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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/022b17d4eedf073a093a47f79e45fb83efba9265)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/52e6b866b608b0da8005443062adef0e96ccb1cd)
their complements,
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5900b0be57a177f9cc7901750fd2a4af4bd0855)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5421ad33b45c3347a9dc2b8c59aa6064c1b0d67)
and,
![{\displaystyle s_{10B}(k)=1,-2,10,-68,514,-4100,33940,\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/745ec62f2b7533b81bddb95b698c2b48d31a3aa1)
![{\displaystyle s_{10C}(k)=1,-1,1,-1,1,23,-263,1343,-2303,\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fcb477bfcd47b53427b4c9117f0308b975f939a)
![{\displaystyle s_{10D}(k)=1,3,25,267,3249,42795,594145,\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/13a46d4189170f2f02c64224e99af51b2d33c1d3)
though closed forms are not yet known for the last three sequences.
Identities[edit]
The modular functions can be related as,[15]
![{\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}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5f49029f3ee7e91bce4f5096aa5a7ea3252819c)
![{\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}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a0473e714be2803dcb754220c47b04b272882e4)
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}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75ccc64c1e5a0c5ba3b22058f91e8f66bbe190e5)
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[edit]
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc7859cd6b64386c7230e227b04791519ce9f4e4)
and noting that
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/707dcf1637a904c0a0a872af04de0746f3ebd54f)
as well as,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5c79f6b4365357fd8e6892a39d49b47bc2fb52b)
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2306784c9627253fcd578400adfb9b16e410a40)
which implies there might be examples for all sequences of level 10.
Level 11[edit]
Define the McKay–Thompson series of class 11A,
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/82c4dbf4692688b050c8bc3d5059644ca4a17804)
or sequence (OEIS: A128525) and where,
![{\displaystyle F={\frac {\eta (3\tau )\,\eta (33\tau )}{\eta (\tau )\,\eta (11\tau )}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5943d3a9ed0f913072d4debe0b3c93a3c8450bdd)
and,