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Thursday, October 31, 2019

Level 10

I. Moonshine functions: Define, a(τ)=(d22d5d1d210)2,b(τ)=(d2d55d1d510),c(τ)=(d31d5d2d310) then (d22d5d1d210)21=(d2d55d1d510)(d22d5d1d210)25=(d31d5d2d310) so a1=b,a5=c and we get the moonshine functions j10A(τ)=j10B(τ)+42j10B(τ)+8j10B(τ)=(d1d5d2d10)4=bcaj10C(τ)=(d1d2d5d10)2=acbj10D(τ)=(d2d5d1d10)6=abcj10E(τ)=(d22d5d1d210)2=a Example: j10A(19/10)=762

II. Relations: Let a,b,c as defined above and m=1,n=5, then the system S1 in the introduction explains all four trinomial identities of level 10 (denoted t10) found by Somos. For S2, let a=j10E then, j10B=a+5a6j10C=a14a13j10D=a5+20a5+9 as well as S3, j10A=j10B+42j10B+8=j10C+52j10C+6=j10D+1j10D2 which implies the linear relation between 5 moonshine functions,
j10A+2j10E=j10B+j10C+j10D+8
III. Special property. As mentioned in the Intro, there are special triples {a,b,c} of level N=6,10,12,18,30 that, using a common formula, can generate another triple {a,b,c} of level 2N. For N=10,
a(τ)=a(12+τ)=(d1d4d10d2d5d20)2=j20C(τ)b(τ)=b(12+τ)=(d1d4d1010d22d55d520)c(τ)=c(12+τ)=(d82d31d34d5d20) such that a+1=b,a+5=c.

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