Ramanujan–Sato series

Level 1

Examples for levels 1–4 were given by Ramanujan in his 1917 paper. Given q = e 2 π i τ 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 + …

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

s 1 A ( k ) = ( 2 k k ) ( 3 k k ) ( 6 k 3 k ) = 1 , 120 , 83160 , 81681600 , … ( 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 , …

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

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 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

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

Level 2

Using Zagier’s notation 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 − ⋯

Note that the coefficient of the linear term of j_2A(τ) is one more than 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 , … ( 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 , …

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

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 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

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

Level 3

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 + …

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

s 3 A ( k ) = ( 2 k k ) ( 2 k k ) ( 3 k k ) = 1 , 12 , 540 , 33600 , 2425500 , … ( 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 , …

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 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

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 + …

where the first is the 24th power of the Weber modular function f ( τ ) . And,

s 4 A ( k ) = ( 2 k k ) 3 = 1 , 8 , 216 , 8000 , 343000 , … ( 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 , … ( 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 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

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 + …

and,

s 5 A ( k ) = ( 2 k k ) ∑ j = 0 k ( k j ) 2 ( k + j j ) = 1 , 6 , 114 , 2940 , 87570 , … 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 , … ( A229111)

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

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 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

Modular functions

In 2002, Sato established the first results for level > 4. Interestingly, it involved Apéry numbers which were first used to establish the irrationality of ζ ( 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 + … 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 + … 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 + … 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 + … 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 + …

J. Conway and S. Norton showed 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

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

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 ) and  A002893 as σ₁(k),

σ 1 ( k ) = ( 2 k k ) ∑ j = 0 k ( k j ) 2 ( 2 j j ) = 1 , 6 , 90 , 1860 , 44730 , … ( A002896)

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

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

σ 3 ( k ) = ( 2 k k ) ∑ j = 0 k ( k j ) 3 = 1 , 4 , 60 , 1120 , 24220 , … ( 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 , … ( 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 , … ( 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 , … ( A125143)

Identities

The modular functions can be related as 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

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

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

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 1 π = 1 2 ∑ k = 0 ∞ s 6 C ( k ) 3 k + 1 32 k , j 6 C ( − 1 3 ) = 32 1 π = 2 3 ∑ k = 0 ∞ s 6 D ( k ) 4 k + 1 81 k + 1 / 2 , j 6 D ( − 1 2 ) = 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 , … ( 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 + …

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

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 + … .

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 , … s 8 A ′ ( k ) = ∑ j = 0 k ( − 1 ) k ( k j ) 2 ( 2 j k ) 2 = 1 , − 4 , 40 , − 544 , 8536 , …

where the first is the product of the central binomial coefficient and a sequence related to an arithmetic-geometric mean ( 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 1 π = − 2 70 ∑ k = 0 ∞ s 4 B ( k ) 58 ⋅ 13 ⋅ 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

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 + …

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 , … 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 , …

where the first is the product of the central binomial coefficients and  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 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

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 + … 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 + … 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 + … 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 + … 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 + …

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

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

Sequences

Let,

v 1 ( k ) = ∑ j = 0 k ( k j ) 4 = 1 , 2 , 18 , 164 , 1810 , … ( 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 , … 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 , …

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 , … 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 , …

and,

s 10 B ( k ) = 1 , − 2 , 10 , − 68 , 514 , − 4100 , 33940 , … s 10 C ( k ) = 1 , − 1 , 1 , − 1 , 1 , 23 , − 263 , 1343 , − 2303 , … s 10 D ( k ) = 1 , 3 , 25 , 267 , 3249 , 42795 , 594145 , …

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

Identities

The modular functions can be related as,

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 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

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

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

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

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

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 + …

where,

F = η ( 3 τ ) η ( 33 τ ) η ( τ ) η ( 11 τ )

and,

s 11 A ( k ) = 1 , 4 , 28 , 268 , 3004 , 36784 , 476476 , …

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

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

Example:

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

Higher levels

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

Similar series

R. Steiner found an example using Catalan numbers C k ,

1 π = 1 4 ∑ k = 0 ∞ C k 2 k + 1 2 4 k

(sequence A013709 in the OEIS) by using a linear combination of higher parts of Wallis-Lambert series for 4/Pi and Euler series for the circumference of an ellipse. Equivalently,

1 π = 1 4 ∑ k = 0 ∞ ( 2 k k ) 2 k + 1 1 2 4 k

and is related to the fact that,

π = lim k → ∞ 2 4 k k ( 2 k k ) 2

which is a consequence of Stirling's approximation.