In mathematics , a Ramanujan–Sato series [ 1] [ 2] generalizes Ramanujan ’s pi formulas such as,
1
π
=
2
2
99
2
∑
k
=
0
∞
(
4
k
)
!
k
!
4
26390
k
+
1103
396
4
k
{\displaystyle {\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,
1
π
=
∑
k
=
0
∞
s
(
k
)
A
k
+
B
C
k
{\displaystyle {\frac {1}{\pi }}=\sum _{k=0}^{\infty }s(k){\frac {Ak+B}{C^{k}}}}
by using other well-defined sequences of integers
s
(
k
)
{\displaystyle s(k)}
obeying a certain recurrence relation , sequences which may be expressed in terms of binomial coefficients
(
n
k
)
{\displaystyle {\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
Γ
0
(
n
)
{\displaystyle \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 j n (τ ) is derived from Zagier[ 10] and T n 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
π
i
τ
{\displaystyle q=e^{2\pi i\tau }}
as in the rest of this article. Let,
j
(
τ
)
=
(
E
4
(
τ
)
η
8
(
τ
)
)
3
=
1
q
+
744
+
196884
q
+
21493760
q
2
+
…
j
∗
(
τ
)
=
432
j
(
τ
)
+
j
(
τ
)
−
1728
j
(
τ
)
−
j
(
τ
)
−
1728
=
1
q
−
120
+
10260
q
−
901120
q
2
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
with the j-function j (τ ), Eisenstein series E 4 , 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 ) is exceedingly close to
196883
{\displaystyle 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
1
A
(
k
)
=
(
2
k
k
)
(
3
k
k
)
(
6
k
3
k
)
=
1
,
120
,
83160
,
81681600
,
…
{\displaystyle s_{1A}(k)={\tbinom {2k}{k}}{\tbinom {3k}{k}}{\tbinom {6k}{3k}}=1,120,83160,81681600,\dots }
(OEIS : A001421 )
s
1
B
(
k
)
=
∑
j
=
0
k
(
2
j
j
)
(
3
j
j
)
(
6
j
3
j
)
(
k
+
j
k
−
j
)
(
−
432
)
k
−
j
=
1
,
−
312
,
114264
,
−
44196288
,
…
{\displaystyle 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,
∑
k
=
0
∞
s
1
A
(
k
)
1
(
j
(
τ
)
)
k
+
1
/
2
=
±
∑
k
=
0
∞
s
1
B
(
k
)
1
(
j
∗
(
τ
)
)
k
+
1
/
2
{\displaystyle \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:
1
π
=
12
i
∑
k
=
0
∞
s
1
A
(
k
)
163
⋅
3344418
k
+
13591409
(
−
640320
3
)
k
+
1
/
2
,
j
(
1
+
−
163
2
)
=
−
640320
3
{\displaystyle {\frac {1}{\pi }}=12\,{\boldsymbol {i}}\,\sum _{k=0}^{\infty }s_{1A}(k)\,{\frac {163\cdot 3344418k+13591409}{(-640320^{3})^{k+1/2}}},\quad j{\Big (}{\tfrac {1+{\sqrt {-163}}}{2}}{\Big )}=-640320^{3}}
1
π
=
24
i
∑
k
=
0
∞
s
1
B
(
k
)
−
3669
+
320
645
(
k
+
1
2
)
(
−
432
U
645
3
)
k
+
1
/
2
,
j
∗
(
1
+
−
43
2
)
=
−
432
(
127
+
5
645
2
)
3
=
−
432
U
645
3
{\displaystyle {\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
{\displaystyle 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,
j
2
A
(
τ
)
=
(
(
η
(
τ
)
η
(
2
τ
)
)
12
+
2
6
(
η
(
2
τ
)
η
(
τ
)
)
12
)
2
=
1
q
+
104
+
4372
q
+
96256
q
2
+
1240002
q
3
+
⋯
j
2
B
(
τ
)
=
(
η
(
τ
)
η
(
2
τ
)
)
24
=
1
q
−
24
+
276
q
−
2048
q
2
+
11202
q
3
−
⋯
{\displaystyle {\begin{aligned}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{aligned}}}
Note that the coefficient of the linear term of j 2A (τ ) is one more than
4371
{\displaystyle 4371}
which is the smallest degree > 1 of the irreducible representations of the Baby Monster group . Define,
s
2
A
(
k
)
=
(
2
k
k
)
(
2
k
k
)
(
4
k
2
k
)
=
1
,
24
,
2520
,
369600
,
63063000
,
…
{\displaystyle s_{2A}(k)={\tbinom {2k}{k}}{\tbinom {2k}{k}}{\tbinom {4k}{2k}}=1,24,2520,369600,63063000,\dots }
(OEIS : A008977 )
s
2
B
(
k
)
=
∑
j
=
0
k
(
2
j
j
)
(
2
j
j
)
(
4
j
2
j
)
(
k
+
j
k
−
j
)
(
−
64
)
k
−
j
=
1
,
−
40
,
2008
,
−
109120
,
6173656
,
…
{\displaystyle 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,
∑
k
=
0
∞
s
2
A
(
k
)
1
(
j
2
A
(
τ
)
)
k
+
1
/
2
=
±
∑
k
=
0
∞
s
2
B
(
k
)
1
(
j
2
B
(
τ
)
)
k
+
1
/
2
{\displaystyle \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:
1
π
=
32
2
∑
k
=
0
∞
s
2
A
(
k
)
58
⋅
455
k
+
1103
(
396
4
)
k
+
1
/
2
,
j
2
A
(
1
2
−
58
)
=
396
4
{\displaystyle {\frac {1}{\pi }}=32{\sqrt {2}}\,\sum _{k=0}^{\infty }s_{2A}(k)\,{\frac {58\cdot 455k+1103}{(396^{4})^{k+1/2}}},\quad j_{2A}{\Big (}{\tfrac {1}{2}}{\sqrt {-58}}{\Big )}=396^{4}}
1
π
=
16
2
∑
k
=
0
∞
s
2
B
(
k
)
−
24184
+
9801
29
(
k
+
1
2
)
(
64
U
29
12
)
k
+
1
/
2
,
j
2
B
(
1
2
−
58
)
=
64
(
5
+
29
2
)
12
=
64
U
29
12
{\displaystyle {\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,
j
3
A
(
τ
)
=
(
(
η
(
τ
)
η
(
3
τ
)
)
6
+
3
3
(
η
(
3
τ
)
η
(
τ
)
)
6
)
2
=
1
q
+
42
+
783
q
+
8672
q
2
+
65367
q
3
+
…
j
3
B
(
τ
)
=
(
η
(
τ
)
η
(
3
τ
)
)
12
=
1
q
−
12
+
54
q
−
76
q
2
−
243
q
3
+
1188
q
4
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
where
782
{\displaystyle 782}
is the smallest degree > 1 of the irreducible representations of the Fischer group Fi 23 and,
s
3
A
(
k
)
=
(
2
k
k
)
(
2
k
k
)
(
3
k
k
)
=
1
,
12
,
540
,
33600
,
2425500
,
…
{\displaystyle s_{3A}(k)={\tbinom {2k}{k}}{\tbinom {2k}{k}}{\tbinom {3k}{k}}=1,12,540,33600,2425500,\dots }
(OEIS : A184423 )
s
3
B
(
k
)
=
∑
j
=
0
k
(
2
j
j
)
(
2
j
j
)
(
3
j
j
)
(
k
+
j
k
−
j
)
(
−
27
)
k
−
j
=
1
,
−
15
,
297
,
−
6495
,
149481
,
…
{\displaystyle 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:
1
π
=
2
i
∑
k
=
0
∞
s
3
A
(
k
)
267
⋅
53
k
+
827
(
−
300
3
)
k
+
1
/
2
,
j
3
A
(
3
+
−
267
6
)
=
−
300
3
{\displaystyle {\frac {1}{\pi }}=2\,{\boldsymbol {i}}\,\sum _{k=0}^{\infty }s_{3A}(k)\,{\frac {267\cdot 53k+827}{(-300^{3})^{k+1/2}}},\quad j_{3A}{\Big (}{\tfrac {3+{\sqrt {-267}}}{6}}{\Big )}=-300^{3}}
1
π
=
i
∑
k
=
0
∞
s
3
B
(
k
)
12497
−
3000
89
(
k
+
1
2
)
(
−
27
U
89
2
)
k
+
1
/
2
,
j
3
B
(
3
+
−
267
6
)
=
−
27
(
500
+
53
89
)
2
=
−
27
U
89
2
{\displaystyle {\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,
j
4
A
(
τ
)
=
(
(
η
(
τ
)
η
(
4
τ
)
)
4
+
4
2
(
η
(
4
τ
)
η
(
τ
)
)
4
)
2
=
(
η
2
(
2
τ
)
η
(
τ
)
η
(
4
τ
)
)
24
=
1
q
+
24
+
276
q
+
2048
q
2
+
11202
q
3
+
…
j
4
C
(
τ
)
=
(
η
(
τ
)
η
(
4
τ
)
)
8
=
1
q
−
8
+
20
q
−
62
q
3
+
216
q
5
−
641
q
7
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
where the first is the 24th power of the Weber modular function
f
(
τ
)
{\displaystyle {\mathfrak {f}}(\tau )}
. And,
s
4
A
(
k
)
=
(
2
k
k
)
3
=
1
,
8
,
216
,
8000
,
343000
,
…
{\displaystyle s_{4A}(k)={\tbinom {2k}{k}}^{3}=1,8,216,8000,343000,\dots }
(OEIS : A002897 )
s
4
C
(
k
)
=
∑
j
=
0
k
(
2
j
j
)
3
(
k
+
j
k
−
j
)
(
−
16
)
k
−
j
=
(
−
1
)
k
∑
j
=
0
k
(
2
j
j
)
2
(
2
k
−
2
j
k
−
j
)
2
=
1
,
−
8
,
88
,
−
1088
,
14296
,
…
{\displaystyle 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 }
(OEIS : A036917 )
Examples:
1
π
=
8
i
∑
k
=
0
∞
s
4
A
(
k
)
6
k
+
1
(
−
2
9
)
k
+
1
/
2
,
j
4
A
(
1
+
−
4
2
)
=
−
2
9
{\displaystyle {\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}}
1
π
=
16
i
∑
k
=
0
∞
s
4
C
(
k
)
1
−
2
2
(
k
+
1
2
)
(
−
16
U
2
4
)
k
+
1
/
2
,
j
4
C
(
1
+
−
4
2
)
=
−
16
(
1
+
2
)
4
=
−
16
U
2
4
{\displaystyle {\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,
j
5
A
(
τ
)
=
(
η
(
τ
)
η
(
5
τ
)
)
6
+
5
3
(
η
(
5
τ
)
η
(
τ
)
)
6
+
22
=
1
q
+
16
+
134
q
+
760
q
2
+
3345
q
3
+
…
j
5
B
(
τ
)
=
(
η
(
τ
)
η
(
5
τ
)
)
6
=
1
q
−
6
+
9
q
+
10
q
2
−
30
q
3
+
6
q
4
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
and,
s
5
A
(
k
)
=
(
2
k
k
)
∑
j
=
0
k
(
k
j
)
2
(
k
+
j
j
)
=
1
,
6
,
114
,
2940
,
87570
,
…
{\displaystyle 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
5
B
(
k
)
=
∑
j
=
0
k
(
−
1
)
j
+
k
(
k
j
)
3
(
4
k
−
5
j
3
k
)
=
1
,
−
5
,
35
,
−
275
,
2275
,
−
19255
,
…
{\displaystyle 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 }
(OEIS : A229111 )
where the first is the product of the central binomial coefficients and the Apéry numbers (OEIS : A005258 )[ 9]
Examples:
1
π
=
5
9
i
∑
k
=
0
∞
s
5
A
(
k
)
682
k
+
71
(
−
15228
)
k
+
1
/
2
,
j
5
A
(
5
+
−
5
(
47
)
10
)
=
−
15228
=
−
(
18
47
)
2
{\displaystyle {\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}}
1
π
=
6
5
i
∑
k
=
0
∞
s
5
B
(
k
)
25
5
−
141
(
k
+
1
2
)
(
−
5
5
U
5
15
)
k
+
1
/
2
,
j
5
B
(
5
+
−
5
(
47
)
10
)
=
−
5
5
(
1
+
5
2
)
15
=
−
5
5
U
5
15
{\displaystyle {\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
ζ
(
3
)
{\displaystyle \zeta (3)}
. First, define,
j
6
A
(
τ
)
=
j
6
B
(
τ
)
+
1
j
6
B
(
τ
)
+
2
=
j
6
C
(
τ
)
+
64
j
6
C
(
τ
)
+
20
=
j
6
D
(
τ
)
+
81
j
6
D
(
τ
)
+
18
=
1
q
+
14
+
79
q
+
352
q
2
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
j
6
B
(
τ
)
=
(
η
(
2
τ
)
η
(
3
τ
)
η
(
τ
)
η
(
6
τ
)
)
12
=
1
q
+
12
+
78
q
+
364
q
2
+
1365
q
3
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
j
6
C
(
τ
)
=
(
η
(
τ
)
η
(
3
τ
)
η
(
2
τ
)
η
(
6
τ
)
)
6
=
1
q
−
6
+
15
q
−
32
q
2
+
87
q
3
−
192
q
4
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
j
6
D
(
τ
)
=
(
η
(
τ
)
η
(
2
τ
)
η
(
3
τ
)
η
(
6
τ
)
)
4
=
1
q
−
4
−
2
q
+
28
q
2
−
27
q
3
−
52
q
4
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
j
6
E
(
τ
)
=
(
η
(
2
τ
)
η
3
(
3
τ
)
η
(
τ
)
η
3
(
6
τ
)
)
3
=
1
q
+
3
+
6
q
+
4
q
2
−
3
q
3
−
12
q
4
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
J. Conway and S. Norton showed[ 14] there are linear relations between the McKay–Thompson series T n , one of which was,
T
6
A
−
T
6
B
−
T
6
C
−
T
6
D
+
2
T
6
E
=
0
{\displaystyle T_{6A}-T_{6B}-T_{6C}-T_{6D}+2T_{6E}=0}
or using the above eta quotients j n ,
j
6
A
−
j
6
B
−
j
6
C
−
j
6
D
+
2
j
6
E
=
18
{\displaystyle j_{6A}-j_{6B}-j_{6C}-j_{6D}+2j_{6E}=18}
σ Sequences
For the modular function j 6A , 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
)
=
(
2
k
k
)
{\displaystyle c(k)={\tbinom {2k}{k}}}
and OEIS : A002893 as σ 1 (k ),
σ
1
(
k
)
=
(
2
k
k
)
∑
j
=
0
k
(
k
j
)
2
(
2
j
j
)
=
1
,
6
,
90
,
1860
,
44730
,
…
{\displaystyle \sigma _{1}(k)={\tbinom {2k}{k}}\sum _{j=0}^{k}{\tbinom {k}{j}}^{2}{\tbinom {2j}{j}}=1,6,90,1860,44730,\dots }
(OEIS : A002896 )
the product of c (k ) and (-1)^k OEIS : A093388 as σ 2 (k ),
σ
2
(
k
)
=
(
2
k
k
)
∑
j
=
0
k
(
k
j
)
(
−
8
)
k
−
j
∑
m
=
0
j
(
j
m
)
3
=
1
,
−
12
,
252
,
−
6240
,
167580
,
−
4726512
,
…
{\displaystyle \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 σ 3 (k ),
σ
3
(
k
)
=
(
2
k
k
)
∑
j
=
0
k
(
k
j
)
3
=
1
,
4
,
60
,
1120
,
24220
,
…
{\displaystyle \sigma _{3}(k)={\tbinom {2k}{k}}\sum _{j=0}^{k}{\tbinom {k}{j}}^{3}=1,4,60,1120,24220,\dots }
(OEIS : A181418 )
Each has a companion sequence. Respectively, these are the Apéry numbers,
s
6
B
(
k
)
=
∑
j
=
0
k
(
k
j
)
2
(
k
+
j
j
)
2
=
1
,
5
,
73
,
1445
,
33001
,
…
{\displaystyle s_{6B}(k)=\sum _{j=0}^{k}{\tbinom {k}{j}}^{2}{\tbinom {k+j}{j}}^{2}=1,5,73,1445,33001,\dots }
(OEIS : A005259 )
the Domb numbers (unsigned), or the number of 2n -step polygons on a diamond lattice,
s
6
C
(
k
)
=
(
−
1
)
k
∑
j
=
0
k
(
k
j
)
2
(
2
(
k
−
j
)
k
−
j
)
(
2
j
j
)
=
1
,
−
4
,
28
,
−
256
,
2716
,
…
{\displaystyle 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 }
(OEIS : A002895 )
and the Almkvist-Zudilin numbers,
s
6
D
(
k
)
=
∑
j
=
0
k
(
−
1
)
k
−
j
3
k
−
3
j
(
3
j
)
!
j
!
3
(
k
3
j
)
(
k
+
j
j
)
=
1
,
−
3
,
9
,
−
3
,
−
279
,
2997
,
…
{\displaystyle 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 }
(OEIS : A125143 )
Identities
The modular functions can be related as
P
=
Q
=
R
{\displaystyle P=Q=R}
where,
P
=
∑
k
=
0
∞
σ
1
(
k
)
1
(
j
6
A
(
τ
)
)
k
+
1
/
2
=
±
∑
k
=
0
∞
s
6
B
(
k
)
1
(
j
6
B
(
τ
)
)
k
+
1
/
2
Q
=
∑
k
=
0
∞
σ
2
(
k
)
1
(
j
6
A
(
τ
)
−
36
)
k
+
1
/
2
=
±
∑
k
=
0
∞
s
6
C
(
k
)
1
(
j
6
C
(
τ
)
)
k
+
1
/
2
R
=
∑
k
=
0
∞
σ
3
(
k
)
1
(
j
6
A
(
τ
)
−
4
)
k
+
1
/
2
=
±
∑
k
=
0
∞
s
6
D
(
k
)
1
(
j
6
D
(
τ
)
)
k
+
1
/
2
{\displaystyle {\begin{aligned}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{aligned}}}
if the series converges and the sign chosen appropriately.
Examples
One can use a value for j 6A in three ways. For example, starting with,
j
6
A
(
−
3
6
)
=
10
2
{\displaystyle j_{6A}{\Big (}{\sqrt {\tfrac {-3}{6}}}{\Big )}=10^{2}}
then,
1
π
=
3
5
2
∑
k
=
0
∞
σ
1
(
k
)
16
k
+
3
(
10
2
)
k
1
π
=
3
2
3
∑
k
=
0
∞
σ
2
(
k
)
10
k
+
3
(
10
2
−
36
)
k
1
π
=
1
3
2
∑
k
=
0
∞
σ
3
(
k
)
5
k
+
1
(
10
2
−
4
)
k
{\displaystyle {\begin{aligned}{\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{aligned}}}
For the other modular functions,
1
π
=
8
15
∑
k
=
0
∞
s
6
B
(
k
)
(
1
2
−
3
5
20
+
k
)
(
1
ϕ
12
)
k
+
1
/
2
,
j
6
B
(
−
5
6
)
=
(
1
+
5
2
)
12
=
ϕ
12
{\displaystyle {\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}}
1
π
=
1
2
∑
k
=
0
∞
s
6
C
(
k
)
3
k
+
1
32
k
,
j
6
C
(
−
1
3
)
=
32
{\displaystyle {\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}
1
π
=
2
3
∑
k
=
0
∞
s
6
D
(
k
)
4
k
+
1
81
k
+
1
/
2
,
j
6
D
(
−
1
2
)
=
81
{\displaystyle {\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
7
A
(
k
)
=
∑
j
=
0
k
(
k
j
)
2
(
2
j
k
)
(
k
+
j
j
)
=
1
,
4
,
48
,
760
,
13840
,
…
{\displaystyle s_{7A}(k)=\sum _{j=0}^{k}{\tbinom {k}{j}}^{2}{\tbinom {2j}{k}}{\tbinom {k+j}{j}}=1,4,48,760,13840,\dots }
(OEIS : A183204 )
and,
j
7
A
(
τ
)
=
(
(
η
(
τ
)
η
(
7
τ
)
)
2
+
7
(
η
(
7
τ
)
η
(
τ
)
)
2
)
2
=
1
q
+
10
+
51
q
+
204
q
2
+
681
q
3
+
…
j
7
B
(
τ
)
=
(
η
(
τ
)
η
(
7
τ
)
)
4
=
1
q
−
4
+
2
q
+
8
q
2
−
5
q
3
−
4
q
4
−
10
q
5
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
Example:
1
π
=
3
22
3
∑
k
=
0
∞
s
7
A
(
k
)
11895
k
+
1286
(
−
22
3
)
k
,
j
7
A
(
7
+
−
427
14
)
=
−
22
3
+
1
=
−
(
39
7
)
2
{\displaystyle {\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 j 7B .
Level 8
Define,
j
4
B
(
τ
)
=
(
j
2
A
(
2
τ
)
)
1
/
2
=
1
q
+
52
q
+
834
q
3
+
4760
q
5
+
24703
q
7
+
…
=
(
(
η
(
τ
)
η
2
(
4
τ
)
η
2
(
2
τ
)
η
(
8
τ
)
)
4
+
4
(
η
2
(
2
τ
)
η
(
8
τ
)
η
(
τ
)
η
2
(
4
τ
)
)
4
)
2
=
(
(
η
(
2
τ
)
η
(
4
τ
)
η
(
τ
)
η
(
8
τ
)
)
4
−
4
(
η
(
τ
)
η
(
8
τ
)
η
(
2
τ
)
η
2
(
τ
)
)
4
)
2
j
8
A
′
(
τ
)
=
(
η
(
τ
)
η
2
(
4
τ
)
η
2
(
2
τ
)
η
(
8
τ
)
)
8
=
1
q
−
8
+
36
q
−
128
q
2
+
386
q
3
−
1024
q
4
+
…
j
8
A
(
τ
)
=
(
η
(
2
τ
)
η
(
4
τ
)
η
(
τ
)
η
(
8
τ
)
)
8
=
1
q
+
8
+
36
q
+
128
q
2
+
386
q
3
+
1024
q
4
+
…
.
{\displaystyle {\begin{aligned}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{aligned}}}
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
4
B
(
k
)
=
(
2
k
k
)
∑
j
=
0
k
4
k
−
2
j
(
k
2
j
)
(
2
j
j
)
2
=
(
2
k
k
)
∑
j
=
0
k
(
k
j
)
(
2
k
−
2
j
k
−
j
)
(
2
j
j
)
=
1
,
8
,
120
,
2240
,
47320
,
…
{\displaystyle 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
8
A
′
(
k
)
=
∑
j
=
0
k
(
−
1
)
k
(
k
j
)
2
(
2
j
k
)
2
=
1
,
−
4
,
40
,
−
544
,
8536
,
…
{\displaystyle 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 (OEIS : A081085 ),
Examples:
1
π
=
2
2
13
∑
k
=
0
∞
s
4
B
(
k
)
70
⋅
99
k
+
579
(
16
+
396
2
)
k
+
1
/
2
,
j
4
B
(
1
4
−
58
)
=
396
2
{\displaystyle {\frac {1}{\pi }}={\frac {2{\sqrt {2}}}{13}}\,\sum _{k=0}^{\infty }s_{4B}(k)\,{\frac {70\cdot 99\,k+579}{(16+396^{2})^{k+1/2}}},\quad j_{4B}{\Big (}{\tfrac {1}{4}}{\sqrt {-58}}{\Big )}=396^{2}}
1
π
=
−
2
70
∑
k
=
0
∞
s
4
B
(
k
)
58
⋅
13
⋅
99
k
+
6243
(
16
−
396
2
)
k
+
1
/
2
{\displaystyle {\frac {1}{\pi }}={\frac {\sqrt {-2}}{70}}\,\sum _{k=0}^{\infty }s_{4B}(k)\,{\frac {58\cdot 13\cdot 99\,k+6243}{(16-396^{2})^{k+1/2}}}}
1
π
=
2
2
∑
k
=
0
∞
s
8
A
′
(
k
)
−
222
+
377
2
(
k
+
1
2
)
(
4
(
1
+
2
)
12
)
k
+
1
/
2
,
j
8
A
′
(
1
4
−
58
)
=
4
(
1
+
2
)
12
,
j
8
A
(
1
4
−
58
)
=
4
(
99
+
13
58
)
2
=
4
U
58
2
{\displaystyle {\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 j 8A (τ ).
Level 9
Define,
j
3
C
(
τ
)
=
(
j
(
3
τ
)
)
1
/
3
=
−
6
+
(
η
2
(
3
τ
)
η
(
τ
)
η
(
9
τ
)
)
6
−
27
(
η
(
τ
)
η
(
9
τ
)
η
2
(
3
τ
)
)
6
=
1
q
+
248
q
2
+
4124
q
5
+
34752
q
8
+
…
j
9
A
(
τ
)
=
(
η
2
(
3
τ
)
η
(
τ
)
η
(
9
τ
)
)
6
=
1
q
+
6
+
27
q
+
86
q
2
+
243
q
3
+
594
q
4
+
…
{\displaystyle {\begin{aligned}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{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,
s
3
C
(
k
)
=
(
2
k
k
)
∑
j
=
0
k
(
−
3
)
k
−
3
j
(
k
j
)
(
k
−
j
j
)
(
k
−
2
j
j
)
=
(
2
k
k
)
∑
j
=
0
k
(
−
3
)
k
−
3
j
(
k
3
j
)
(
2
j
j
)
(
3
j
j
)
=
1
,
−
6
,
54
,
−
420
,
630
,
…
{\displaystyle 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
9
A
(
k
)
=
∑
j
=
0
k
(
k
j
)
2
∑
m
=
0
j
(
k
m
)
(
j
m
)
(
j
+
m
k
)
=
1
,
3
,
27
,
309
,
4059
,
…
{\displaystyle 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 OEIS : A006077 (though with different signs).
Examples:
1
π
=
−
i
9
∑
k
=
0
∞
s
3
C
(
k
)
602
k
+
85
(
−
960
−
12
)
k
+
1
/
2
,
j
3
C
(
3
+
−
43
6
)
=
−
960
{\displaystyle {\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}
1
π
=
6
i
∑
k
=
0
∞
s
9
A
(
k
)
4
−
129
(
k
+
1
2
)
(
−
3
3
U
129
)
k
+
1
/
2
,
j
9
A
(
3
+
−
43
6
)
=
−
3
3
(
53
3
+
14
43
)
=
−
3
3
U
129
{\displaystyle {\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,
j
10
A
(
τ
)
=
j
10
B
(
τ
)
+
16
j
10
B
(
τ
)
+
8
=
j
10
C
(
τ
)
+
25
j
10
C
(
τ
)
+
6
=
j
10
D
(
τ
)
+
1
j
10
D
(
τ
)
−
2
=
1
q
+
4
+
22
q
+
56
q
2
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
j
10
B
(
τ
)
=
(
η
(
τ
)
η
(
5
τ
)
η
(
2
τ
)
η
(
10
τ
)
)
4
=
1
q
−
4
+
6
q
−
8
q
2
+
17
q
3
−
32
q
4
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
j
10
C
(
τ
)
=
(
η
(
τ
)
η
(
2
τ
)
η
(
5
τ
)
η
(
10
τ
)
)
2
=
1
q
−
2
−
3
q
+
6
q
2
+
2
q
3
+
2
q
4
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
j
10
D
(
τ
)
=
(
η
(
2
τ
)
η
(
5
τ
)
η
(
τ
)
η
(
10
τ
)
)
6
=
1
q
+
6
+
21
q
+
62
q
2
+
162
q
3
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
j
10
E
(
τ
)
=
(
η
(
2
τ
)
η
5
(
5
τ
)
η
(
τ
)
η
5
(
10
τ
)
)
=
1
q
+
1
+
q
+
2
q
2
+
2
q
3
−
2
q
4
+
…
{\displaystyle {\begin{aligned}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{aligned}}}
Just like the level 6, there are also linear relations between these,
T
10
A
−
T
10
B
−
T
10
C
−
T
10
D
+
2
T
10
E
=
0
{\displaystyle T_{10A}-T_{10B}-T_{10C}-T_{10D}+2T_{10E}=0}
or using the above eta quotients j n ,
j
10
A
−
j
10
B
−
j
10
C
−
j
10
D
+
2
j
10
E
=
6
{\displaystyle j_{10A}-j_{10B}-j_{10C}-j_{10D}+2j_{10E}=6}
Sequences
Let,
v
1
(
k
)
=
∑
j
=
0
k
(
k
j
)
4
=
1
,
2
,
18
,
164
,
1810
,
…
{\displaystyle v_{1}(k)=\sum _{j=0}^{k}{\tbinom {k}{j}}^{4}=1,2,18,164,1810,\dots }
(OEIS : A005260 )
v
2
(
k
)
=
(
2
k
k
)
∑
j
=
0
k
(
2
j
j
)
−
1
(
k
j
)
∑
m
=
0
j
(
j
m
)
4
=
1
,
4
,
36
,
424
,
5716
,
…
{\displaystyle 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
)
=
(
2
k
k
)
∑
j
=
0
k
(
2
j
j
)
−
1
(
k
j
)
(
−
4
)
k
−
j
∑
m
=
0
j
(
j
m
)
4
=
1
,
−
6
,
66
,
−
876
,
12786
,
…
{\displaystyle 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
)
=
(
2
k
k
)
∑
j
=
0
k
(
2
j
j
)
−
1
(
k
j
)
(
−
1
)
k
−
j
∑
m
=
0
j
(
j
m
)
4
=
1
,
0
,
12
,
24
,
564
,
2784
,
…
{\displaystyle 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
)
=
(
2
k
k
)
∑
j
=
0
k
(
2
j
j
)
−
1
(
k
j
)
(
4
)
k
−
j
∑
m
=
0
j
(
j
m
)
4
=
1
,
10
,
162
,
3124
,
66994
,
…
{\displaystyle 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
10
B
(
k
)
=
1
,
−
2
,
10
,
−
68
,
514
,
−
4100
,
33940
,
…
{\displaystyle s_{10B}(k)=1,-2,10,-68,514,-4100,33940,\dots }
s
10
C
(
k
)
=
1
,
−
1
,
1
,
−
1
,
1
,
23
,
−
263
,
1343
,
−
2303
,
…
{\displaystyle s_{10C}(k)=1,-1,1,-1,1,23,-263,1343,-2303,\dots }
s
10
D
(
k
)
=
1
,
3
,
25
,
267
,
3249
,
42795
,
594145
,
…
{\displaystyle 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
=
∑
k
=
0
∞
v
1
(
k
)
1
(
j
10
A
(
τ
)
)
k
+
1
/
2
=
∑
k
=
0
∞
v
2
(
k
)
1
(
j
10
A
(
τ
)
+
4
)
k
+
1
/
2
=
∑
k
=
0
∞
v
3
(
k
)
1
(
j
10
A
(
τ
)
−
16
)
k
+
1
/
2
{\displaystyle 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}}}}
V
=
∑
k
=
0
∞
s
10
B
(
k
)
1
(
j
10
B
(
τ
)
)
k
+
1
/
2
=
∑
k
=
0
∞
s
10
C
(
k
)
1
(
j
10
C
(
τ
)
)
k
+
1
/
2
=
∑
k
=
0
∞
s
10
D
(
k
)
1
(
j
10
D
(
τ
)
)
k
+
1
/
2
{\displaystyle V=\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}}}}
if the series converges. In fact, it can also be observed that,
U
=
V
=
∑
k
=
0
∞
v
2
′
(
k
)
1
(
j
10
A
(
τ
)
−
4
)
k
+
1
/
2
=
∑
k
=
0
∞
v
3
′
(
k
)
1
(
j
10
A
(
τ
)
+
16
)
k
+
1
/
2
{\displaystyle U=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}}}}
Since the exponent has a fractional part, the sign of the square root must be chosen appropriately though it is less an issue when j n is positive.
Examples
Starting with,
j
10
A
(
−
19
10
)
=
76
2
{\displaystyle j_{10A}{\Big (}{\sqrt {\tfrac {-19}{10}}}{\Big )}=76^{2}}
then,
1
π
=
5
95
∑
k
=
0
∞
v
1
(
k
)
408
k
+
47
(
76
2
)
k
+
1
/
2
1
π
=
1
17
95
∑
k
=
0
∞
v
2
(
k
)
19
⋅
1824
k
+
3983
(
76
2
+
4
)
k
+
1
/
2
1
π
=
5
481
95
∑
k
=
0
∞
v
2
′
(
k
)
19
⋅
10336
k
+
22675
(
76
2
−
4
)
k
+
1
/
2
1
π
=
1
6
95
∑
k
=
0
∞
v
3
(
k
)
19
⋅
646
k
+
1427
(
76
2
−
16
)
k
+
1
/
2
1
π
=
5
181
95
∑
k
=
0
∞
v
3
′
(
k
)
19
⋅
3876
k
+
8405
(
76
2
+
16
)
k
+
1
/
2
{\displaystyle {\begin{aligned}{\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{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,
1
π
=
i
5
∑
k
=
0
∞
s
10
C
(
k
)
10
k
+
3
(
−
25
)
k
+
1
/
2
,
j
10
C
(
1
+
i
2
)
=
−
25
{\displaystyle {\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
11
A
(
τ
)
=
(
1
+
3
F
)
3
+
(
1
F
+
3
F
)
2
=
1
q
+
6
+
17
q
+
46
q
2
+
116
q
3
+
…
{\displaystyle 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
=
η
(
3
τ
)
η
(
33
τ
)
η
(
τ
)
η
(
11
τ
)
{\displaystyle F={\tfrac {\eta (3\tau )\,\eta (33\tau )}{\eta (\tau )\,\eta (11\tau )}}}
and,
s
11
A
(
k
)
=
1
,
4
,
28
,
268
,
3004
,
36784
,
476476
,
…
{\displaystyle 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
(
2
k
+
1
)
(
5
k
2
+
5
k
+
2
)
s
k
−
8
k
(
7
k
2
+
1
)
s
k
−
1
+
22
k
(
k
−
1
)
(
2
k
−
1
)
s
k
−
2
{\displaystyle (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]
1
π
=
i
22
∑
k
=
0
∞
s
11
A
(
k
)
221
k
+
67
(
−
44
)
k
+
1
/
2
,
j
11
A
(
1
+
−
17
/
11
2
)
=
−
44
{\displaystyle {\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
^ a b Heng Huat Chan, Song Heng Chan, and Zhiguo Liu, "Domb's numbers and Ramanujan–Sato type series for 1/Pi" (2004)
^ a b c Gert Almkvist and Jesus Guillera, Ramanujan–Sato Like Series (2012)
^ a b c 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
^ G. Almkvist, Some conjectured formulas for 1/Pi coming from polytopes, K3-surfaces and Moonshine, http://arxiv.org/abs/1211.6563
^ S. Ramanujan, "Modular equations and approximations to pi", Quart. J. Math. (Oxford) 45 (1914)
^ Chan, Tanigawa, Yang, and Zudilin, "New analogues of Clausen’s identities arising from the theory of modular forms" (2011)
^ a b 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
^ H. Chan and H. Verrill, "The Apéry numbers, the Almkvist–Zudilin Numbers, and new series for 1/π", Advances in Mathematics, Vol 186, 2004
^ a b S. Cooper, "Sporadic sequences, modular forms and new series for 1/π", Ramanujan Journal 2012
^ a b D. Zagier, "Traces of Singular Moduli", (p.15-16), http://people.mpim-bonn.mpg.de/zagier/files/tex/TracesSingModuli/fulltext.pdf
^ 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 .
^ 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 .
^ 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
^ 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
^ 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)
^ a b 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