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Wednesday, June 11, 2025

Entry 159

Given the McKay-Thompson series of Class 4A for the Monster (A097340)

j4A(τ)=(η2(2τ)η(τ)η(4τ))24=1q+24+276q+2048q2+

We take the 6th root, scale ττ2, and use the version, (xd)4=(η2(τ)η(τ2)η(2τ))4=2F1(12,12,1,λ(τ))η2(τ)=2π×K(kd)η2(τ)
where K(kd) is a complete elliptic integral of the first kind. Compared also to Ramanujan's G-function, 21/4Gn=η2(τ)η(τ2)η(2τ)=q124n>0(1+q2n1)
where Ramanujan used odd n for Gn and calculated a lot of n. Mathworld also has a partial list of closed-forms for K(kd). Knowing xd or Gn immediately leads to a closed-form for η2(τ) as well. Examples, 

2π×K(k7)=x27Γ(17)Γ(27)Γ(47)71/4(2π)2η2(7)=1x27Γ(17)Γ(27)Γ(47)71/4(2π)2

where x7=2. And

2π×K(k11)=x211Γ(111)Γ(311)Γ(411)Γ(511)Γ(911)111/4(2π)3η2(11)=1x211Γ(111)Γ(311)Γ(411)Γ(511)Γ(911)111/4(2π)3

where x11=1T+1 and T is the tribonacci constant, the real root of x3x2x1=0. For d=11,19,43,67,163, then xd is just the real root of a cubic.

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