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 A , B , C {\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 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 (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 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 )&=\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 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 ) 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
s 1 A ( k ) = ( 2 k k ) ( 3 k k ) ( 6 k 3 k ) = 1 , 120 , 83160 , 81681600 , … {\displaystyle s_{1A}(k)={\binom {2k}{k}}{\binom {3k}{k}}{\binom {6k}{3k}}=1,120,83160,81681600,\ldots } (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}{\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
∑ 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+{\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.
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 = − 262537412640768000 {\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} 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 U 645 3 = − 432 ( 127 + 5 645 2 ) 3 {\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 645 = 43 × 15 , {\displaystyle 645=43\times 15,} 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 )&=\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 j 2A (τ ) is one more than 4371 which is the smallest degree greater than 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)={\binom {2k}{k}}{\binom {2k}{k}}{\binom {4k}{2k}}=1,24,2520,369600,63063000,\ldots } (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}{\binom {2j}{j}}{\binom {2j}{j}}{\binom {4j}{2j}}{\binom {k+j}{k-j}}(-64)^{k-j}=1,-40,2008,-109120,6173656,\ldots } 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+{\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.
Examples:
1 π = 32 2 ∑ k = 0 ∞ s 2 A ( k ) 58 ⋅ 455 k + 1103 ( 396 4 ) k + 1 2 , j 2 A ( − 58 2 ) = 396 4 = 24591257856 {\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} 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 ( − 58 2 ) = 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}}\,\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.[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 )&=\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 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)={\binom {2k}{k}}{\binom {2k}{k}}{\binom {3k}{k}}=1,12,540,33600,2425500,\ldots } (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}{\binom {2j}{j}}{\binom {2j}{j}}{\binom {3j}{j}}{\binom {k+j}{k-j}}(-27)^{k-j}=1,-15,297,-6495,149481,\ldots } 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 = − 27000000 {\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} 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}}\,\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 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 = − ( η ( 2 τ + 3 2 ) η ( 2 τ + 3 ) ) 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 )&=\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}}} where the first is the 24th power of the Weber modular function f ( 2 τ ) {\displaystyle {\mathfrak {f}}(2\tau )} . And,
s 4 A ( k ) = ( 2 k k ) 3 = 1 , 8 , 216 , 8000 , 343000 , … {\displaystyle s_{4A}(k)={\binom {2k}{k}}^{3}=1,8,216,8000,343000,\ldots } (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}{\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 )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 = − 512 {\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} 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}}\,\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 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 )&=\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}}} 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)={\binom {2k}{k}}\sum _{j=0}^{k}{\binom {k}{j}}^{2}{\binom {k+j}{j}}=1,6,114,2940,87570,\ldots } 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}{\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 )[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+{\frac {1}{2}}}}},\quad j_{5A}\left({\frac {5+{\sqrt {-5(47)}}}{10}}\right)=-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\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[7] established the first results for levels above 4. 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 ) + 8 j 6 C ( τ ) ) 2 = ( j 6 D ( τ ) + 9 j 6 D ( τ ) ) 2 − 4 = 1 q + 10 + 79 q + 352 q 2 + ⋯ {\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}}} 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 )&=\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}}} 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 )&=\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}}} 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 )&=\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}}} 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 )&=\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 j 6 A {\displaystyle j_{6A}} being squares or a near-square of the other functions will also be manifested by j 10 A {\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 T n ,[14] 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 = 22 {\displaystyle j_{6A}-j_{6B}-j_{6C}-j_{6D}+2j_{6E}=22} A similar relation exists for level 10.
α Sequences For the modular function j 6A , one can associate it with three different sequences. (A similar situation happens for the level 10 function j 10A .) Let,
α 1 ( k ) = ( 2 k k ) ∑ j = 0 k ( k j ) 3 = 1 , 4 , 60 , 1120 , 24220 , … {\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 s 6 in Cooper's paper) α 2 ( k ) = ( 2 k k ) ∑ j = 0 k ( k j ) ∑ m = 0 j ( j m ) 3 = ( 2 k k ) ∑ j = 0 k ( k j ) 2 ( 2 j j ) = 1 , 6 , 90 , 1860 , 44730 , … {\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 ) α 3 ( 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 \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 c ( k ) = ( 2 k k ) {\displaystyle c(k)={\tbinom {2k}{k}}} with: first, the Franel numbers ∑ j = 0 k ( k j ) 3 {\displaystyle \textstyle \sum _{j=0}^{k}{\tbinom {k}{j}}^{3}} ; second, OEIS : A002893 , and third, ( − 1 ) k {\displaystyle (-1)^{k}} OEIS : A093388 . Note that the second sequence, α 2 (k ) is also the number of 2n -step polygons on a cubic lattice . Their complements,
α 2 ′ ( k ) = ( 2 k k ) ∑ j = 0 k ( k j ) ( − 1 ) k − j ∑ m = 0 j ( j m ) 3 = 1 , 2 , 42 , 620 , 12250 , … {\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 } α 3 ′ ( k ) = ( 2 k k ) ∑ j = 0 k ( k j ) ( 8 ) k − j ∑ m = 0 j ( j m ) 3 = 1 , 20 , 636 , 23840 , 991900 , … {\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,
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}{\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 ,
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}{\binom {k}{j}}^{2}{\binom {2(k-j)}{k-j}}{\binom {2j}{j}}=1,-4,28,-256,2716,\ldots } (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}\,{\frac {(3j)!}{j!^{3}}}{\binom {k}{3j}}{\binom {k+j}{j}}=1,-3,9,-3,-279,2997,\ldots } (OEIS : A125143 )where
( 3 j ) ! j ! 3 = ( 2 j j ) ( 3 j j ) {\displaystyle {\frac {(3j)!}{j!^{3}}}={\binom {2j}{j}}{\binom {3j}{j}}}
Identities The modular functions can be related as,
P = ∑ k = 0 ∞ α 1 ( k ) 1 ( j 6 A ( τ ) ) k + 1 2 = ∑ k = 0 ∞ α 2 ( k ) 1 ( j 6 A ( τ ) + 4 ) k + 1 2 = ∑ k = 0 ∞ α 3 ( k ) 1 ( j 6 A ( τ ) − 32 ) k + 1 2 {\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}}}}}} Q = ∑ k = 0 ∞ s 6 B ( k ) 1 ( j 6 B ( τ ) ) k + 1 2 = ∑ k = 0 ∞ s 6 C ( k ) 1 ( j 6 C ( τ ) ) k + 1 2 = ∑ k = 0 ∞ s 6 D ( k ) 1 ( j 6 D ( τ ) ) k + 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,
P = Q = ∑ k = 0 ∞ α 2 ′ ( k ) 1 ( j 6 A ( τ ) − 4 ) k + 1 2 = ∑ k = 0 ∞ α 3 ′ ( k ) 1 ( j 6 A ( τ ) + 32 ) k + 1 2 {\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}}}}}} which implies,
∑ k = 0 ∞ α 2 ( k ) 1 ( j 6 A ( τ ) + 4 ) k + 1 2 = ∑ k = 0 ∞ α 2 ′ ( k ) 1 ( j 6 A ( τ ) − 4 ) k + 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}}}}}} and similarly using α3 and α'3 .
Examples One can use a value for j 6A in three ways. For example, starting with,
Δ = j 6 A ( − 17 6 ) = 198 2 − 4 = ( 140 2 ) 2 = 39200 {\displaystyle \Delta =j_{6A}\left({\sqrt {\frac {-17}{6}}}\right)=198^{2}-4=\left(140{\sqrt {2}}\right)^{2}=39200} and noting that 3 ⋅ 17 = 51 {\displaystyle 3\cdot 17=51} then,
1 π = 24 3 35 ∑ k = 0 ∞ α 1 ( k ) 51 ⋅ 11 k + 53 ( Δ ) k + 1 2 1 π = 4 3 99 ∑ k = 0 ∞ α 2 ( k ) 17 ⋅ 560 k + 899 ( Δ + 4 ) k + 1 2 1 π = 3 2 ∑ k = 0 ∞ α 3 ( k ) 770 k + 73 ( Δ − 32 ) k + 1 2 {\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}}} as well as,
1 π = 12 3 9799 ∑ k = 0 ∞ α 2 ′ ( k ) 11 ⋅ 51 ⋅ 560 k + 29693 ( Δ − 4 ) k + 1 2 1 π = 6 3 613 ∑ k = 0 ∞ α 3 ′ ( k ) 51 ⋅ 770 k + 3697 ( Δ + 32 ) k + 1 2 {\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,
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)\,\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}} 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}\left({\sqrt {\frac {-1}{3}}}\right)=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+{\frac {1}{2}}}}},\quad j_{6D}\left({\sqrt {\frac {-1}{2}}}\right)=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}{\binom {k}{j}}^{2}{\binom {2j}{k}}{\binom {k+j}{j}}=1,4,48,760,13840,\ldots } (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 )&=\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}}} Example:
1 π = 7 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 = − 10647 {\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} No pi formula has yet been found using j 7B .
Level 8
Modular functions Levels 2 , 4 , 8 {\displaystyle 2,4,8} are related since they are just powers of the same prime. Define,
j 4 B ( τ ) = j 2 A ( 2 τ ) = ( j 4 D ( τ ) + 8 j 4 D ( τ ) ) 2 − 16 = ( j 8 A ( τ ) − 4 j 8 A ( τ ) ) 2 = ( j 8 A ′ ( τ ) + 4 j 8 A ′ ( τ ) ) 2 = ( η ( 2 τ ) η ( 4 τ ) ) 12 + 2 6 ( η ( 4 τ ) η ( 2 τ ) ) 12 = 1 q + 52 q + 834 q 3 + 4760 q 5 + 24703 q 7 + ⋯ j 4 D ( τ ) = ( η ( 2 τ ) η ( 4 τ ) ) 12 = 1 q − 12 q + 66 q 3 − 232 q 5 + 639 q 7 − 1596 q 9 + ⋯ 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 + ⋯ 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 B ( τ ) = ( η 2 ( 4 τ ) η ( 2 τ ) η ( 8 τ ) ) 12 = j 4 A ( 2 τ ) = 1 q + 12 q + 66 q 3 + 232 q 5 + 639 q 7 + ⋯ j 8 E ( τ ) = ( η 3 ( 4 τ ) η ( 2 τ ) η 2 ( 8 τ ) ) 4 = 1 q + 4 q + 2 q 3 − 8 q 5 − q 7 + 20 q 9 − 2 q 11 − 40 q 13 + ⋯ {\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,
j 4 B − j 4 D − j 8 A − j 8 A ′ + 2 j 8 E = 0 {\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 j 8 A ′ {\displaystyle j_{8A'}} is not a moonshine function. However, it is related to one as,
j 8 A ′ ( τ ) = − j 8 A ( τ + 1 2 ) {\displaystyle j_{8A'}(\tau )=-j_{8A}{\Big (}\tau +{\tfrac {1}{2}}{\Big )}}
Sequences 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)={\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 } s 4 D ( k ) = ( 2 k k ) 3 = 1 , 8 , 216 , 8000 , 343000 , … {\displaystyle s_{4D}(k)={\binom {2k}{k}}^{3}=1,8,216,8000,343000,\ldots } s 8 A ( k ) = ∑ j = 0 k ( k j ) 2 ( 2 j k ) 2 = 1 , 4 , 40 , 544 , 8536 , … {\displaystyle s_{8A}(k)=\sum _{j=0}^{k}{\binom {k}{j}}^{2}{\binom {2j}{k}}^{2}=1,4,40,544,8536,\ldots } (OEIS : A290575 ) s 8 B ( k ) = ∑ j = 0 k ( 2 j j ) 3 ( 2 k − 4 j k − 2 j ) = 1 , 2 , 14 , 36 , 334 , … {\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[2] of the central binomial coefficient and a sequence related to an arithmetic-geometric mean (OEIS : A081085 ).
Identities The modular functions can be related as,
± ∑ k = 0 ∞ s 4 B ( k ) 1 ( j 4 B ( τ ) + 16 ) k + 1 2 = ∑ k = 0 ∞ s 4 D ( k ) 1 ( j 4 D ( τ ) ) 2 k + 1 2 = ∑ k = 0 ∞ s 8 A ( k ) 1 ( j 8 A ( τ ) ) k + 1 2 = ∑ k = 0 ∞ ( − 1 ) k s 8 A ( k ) 1 ( j 8 A ′ ( τ ) ) k + 1 2 {\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 ( j 4 D ( τ ) ) 2 k + 1 2 {\displaystyle \left(j_{4D}(\tau )\right)^{2k+{\frac {1}{2}}}} from the others.
Examples Recall that j 2 A ( − 58 2 ) = 396 4 , {\displaystyle j_{2A}\left({\tfrac {\sqrt {-58}}{2}}\right)=396^{4},} while j 4 B ( − 58 4 ) = 396 2 {\displaystyle j_{4B}\left({\tfrac {\sqrt {-58}}{4}}\right)=396^{2}} . Hence,
1 π = 2 2 13 ∑ k = 0 ∞ s 4 B ( k ) 70 ⋅ 99 k + 579 ( 396 2 + 16 ) k + 1 2 , j 4 B ( − 58 4 ) = 396 2 {\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}} 1 π = 2 2 ∑ k = 0 ∞ s 8 A ( k ) − 222 + 70 58 ( k + 1 2 ) ( 4 ( 99 + 13 58 ) 2 ) k + 1 2 , j 8 A ( − 58 4 ) = 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+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}} 1 π = 2 ∑ k = 0 ∞ ( − 1 ) k s 8 A ( k ) − 222 2 + 13 × 58 ( k + 1 2 ) ( 4 ( 1 + 2 ) 12 ) k + 1 2 , j 8 A ′ ( − 58 4 ) = 4 ( 1 + 2 ) 12 = 4 U 2 12 , {\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},} For another level 8 example,
1 π = 1 16 3 5 ∑ k = 0 ∞ s 8 B ( k ) 210 k + 43 ( 64 ) k + 1 2 , j 8 B ( − 7 4 ) = 2 6 = 64 {\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 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 )&=\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,
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)={\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 } 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}{\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).
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}{\left(-960-12\right)^{k+{\frac {1}{2}}}}},\quad j_{3C}\left({\frac {3+{\sqrt {-43}}}{6}}\right)=-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}}\,\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 Define,
j 10 A ( τ ) = ( j 10 D ( τ ) − 1 j 10 D ( τ ) ) 2 = ( j 6 B ( τ ) + 4 j 10 B ( τ ) ) 2 = ( j 10 C ( τ ) + 5 j 10 C ( τ ) ) 2 − 4 = 1 q + 4 + 22 q + 56 q 2 + ⋯ {\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}}} 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 )&=\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}}} 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 )&=\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}}} 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 )&=\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}}} 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 )&=\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 j 6 A {\displaystyle j_{6A}} , the function j 10 A {\displaystyle j_{10A}} is a square or a near-square of the others. Furthermore, 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}
β sequences Let,
β 1 ( k ) = ∑ j = 0 k ( k j ) 4 = 1 , 2 , 18 , 164 , 1810 , … {\displaystyle \beta _{1}(k)=\sum _{j=0}^{k}{\binom {k}{j}}^{4}=1,2,18,164,1810,\ldots } (OEIS : A005260 , labeled as s 10 in Cooper's paper) β 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 \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 } β 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 \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 } their complements,
β 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 \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 } β 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 \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 } and,
s 10 B ( k ) = 1 , − 2 , 10 , − 68 , 514 , − 4100 , 33940 , … {\displaystyle s_{10B}(k)=1,-2,10,-68,514,-4100,33940,\ldots } 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,\ldots } s 10 D ( k ) = 1 , 3 , 25 , 267 , 3249 , 42795 , 594145 , … {\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 The modular functions can be related as,[15]
U = ∑ k = 0 ∞ β 1 ( k ) 1 ( j 10 A ( τ ) ) k + 1 2 = ∑ k = 0 ∞ β 2 ( k ) 1 ( j 10 A ( τ ) + 4 ) k + 1 2 = ∑ k = 0 ∞ β 3 ( k ) 1 ( j 10 A ( τ ) − 16 ) k + 1 2 {\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}}}}}} 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}{\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,
U = V = ∑ k = 0 ∞ β 2 ′ ( k ) 1 ( j 10 A ( τ ) − 4 ) k + 1 2 = ∑ k = 0 ∞ β 3 ′ ( k ) 1 ( j 10 A ( τ ) + 16 ) k + 1 2 {\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 j n is positive.
Examples Just like level 6, the level 10 function j 10A can be used in three ways. Starting with,
j 10 A ( − 19 10 ) = 76 2 = 5776 {\displaystyle j_{10A}\left({\sqrt {\frac {-19}{10}}}\right)=76^{2}=5776} and noting that 5 ⋅ 19 = 95 {\displaystyle 5\cdot 19=95} then,
1 π = 5 95 ∑ k = 0 ∞ β 1 ( k ) 408 k + 47 ( 76 2 ) k + 1 2 1 π = 1 17 95 ∑ k = 0 ∞ β 2 ( k ) 19 ⋅ 1824 k + 3983 ( 76 2 + 4 ) k + 1 2 1 π = 1 6 95 ∑ k = 0 ∞ β 3 ( k ) 19 ⋅ 646 k + 1427 ( 76 2 − 16 ) k + 1 2 {\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}}} as well as,
1 π = 5 481 95 ∑ k = 0 ∞ β 2 ′ ( k ) 19 ⋅ 10336 k + 22675 ( 76 2 − 4 ) k + 1 2 1 π = 5 181 95 ∑ k = 0 ∞ β 3 ′ ( k ) 19 ⋅ 3876 k + 8405 ( 76 2 + 16 ) k + 1 2 {\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,
1 π = i 5 ∑ k = 0 ∞ s 10 C ( k ) 10 k + 3 ( − 5 2 ) k + 1 2 , j 10 C ( 1 + i 2 ) = − 5 2 {\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 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}+\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,
F = η ( 3 τ ) η ( 33 τ ) η ( τ ) η ( 11 τ ) {\displaystyle F={\frac {\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,\ldots } (OEIS : A284756 )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)\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}} 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+{\frac {1}{2}}}}},\quad j_{11A}\left({\frac {1+{\sqrt {\frac {-17}{11}}}}{2}}\right)=-44}
Higher levels As pointed out by Cooper,[16] there are analogous sequences for certain higher levels.
Similar series R. Steiner found examples using Catalan numbers C k {\displaystyle C_{k}} ,
1 π = ∑ k = 0 ∞ ( 2 C k − n ) 2 ( 4 z ) k + ( 4 2 n − 3 − ( 4 n − 3 ) z ) 16 k z ∈ Z , n ≥ 2 , n ∈ N {\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:
k = ( − 20 − 12 i ) + 16 n 16 , k = ( − 20 + 12 i ) + 16 n 16 {\displaystyle k={\frac {(-20-12{\boldsymbol {i}})+16n}{16}},\qquad k={\frac {(-20+12{\boldsymbol {i}})+16n}{16}}} Other similar series are
1 π = ∑ k = 0 ∞ ( 2 C k − 2 ) 2 3 k + 1 4 16 k {\displaystyle {\frac {1}{\pi }}=\sum _{k=0}^{\infty }\left(2C_{k-2}\right)^{2}{\frac {3k+{\frac {1}{4}}}{16^{k}}}} 1 π = ∑ k = 0 ∞ ( 2 C k − 1 ) 2 ( 4 z + 1 ) k − z 16 k z ∈ Z {\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} } 1 π = ∑ k = 0 ∞ ( 2 C k − 1 ) 2 − 1 k + 1 2 16 k {\displaystyle {\frac {1}{\pi }}=\sum _{k=0}^{\infty }\left(2C_{k-1}\right)^{2}{\frac {-1k+{\frac {1}{2}}}{16^{k}}}} 1 π = ∑ k = 0 ∞ ( 2 C k − 1 ) 2 0 k + 1 4 16 k {\displaystyle {\frac {1}{\pi }}=\sum _{k=0}^{\infty }\left(2C_{k-1}\right)^{2}{\frac {0k+{\frac {1}{4}}}{16^{k}}}} 1 π = ∑ k = 0 ∞ ( 2 C k − 1 ) 2 k 5 + 1 5 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}}}} 1 π = ∑ k = 0 ∞ ( 2 C k − 1 ) 2 k 3 + 1 6 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}}}} 1 π = ∑ k = 0 ∞ ( 2 C k − 1 ) 2 k 2 + 1 8 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}}}} 1 π = ∑ k = 0 ∞ ( 2 C k − 1 ) 2 2 k − 1 4 16 k {\displaystyle {\frac {1}{\pi }}=\sum _{k=0}^{\infty }\left(2C_{k-1}\right)^{2}{\frac {2k-{\frac {1}{4}}}{16^{k}}}} 1 π = ∑ k = 0 ∞ ( 2 C k − 1 ) 2 3 k − 1 2 16 k {\displaystyle {\frac {1}{\pi }}=\sum _{k=0}^{\infty }\left(2C_{k-1}\right)^{2}{\frac {3k-{\frac {1}{2}}}{16^{k}}}} 1 π = ∑ k = 0 ∞ ( 2 C k ) 2 k 16 + 1 16 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 4 π {\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
1 4 = ∑ k = 0 ∞ ( Γ ( 1 2 + k ) Γ ( 2 + k ) ) 2 ( 4 z k − ( 4 n − 3 ) z + 4 2 n − 3 ) z ∈ Z , n ≥ 2 , n ∈ N {\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} } ...
4 = ∑ k = 0 ∞ ( Γ ( 1 2 + k ) Γ ( 2 + k ) ) 2 ( k + 1 ) {\displaystyle 4=\sum _{k=0}^{\infty }{\left({\frac {\Gamma ({\frac {1}{2}}+k)}{\Gamma (2+k)}}\right)}^{2}(k+1)} .The last is also equivalent to,
1 π = 1 4 ∑ k = 0 ∞ ( 2 k k ) 2 k + 1 1 16 k {\displaystyle {\frac {1}{\pi }}={\frac {1}{4}}\sum _{k=0}^{\infty }{\frac {{\binom {2k}{k}}^{2}}{k+1}}\,{\frac {1}{16^{k}}}} and is related to the fact that,
lim k → ∞ 16 k k ( 2 k k ) 2 = π {\displaystyle \lim _{k\rightarrow \infty }{\frac {16^{k}}{k{\binom {2k}{k}}^{2}}}=\pi } which is a consequence of Stirling's approximation .
See also
References
External links