eta(%e^-(2*%pi*sqrt(n))) = (prod(gamma(k/n)^((jacobi(k,n)+1)/4),k,1,n-1))/(sqrt(2)*(a*2^n*%pi^(n+1))^(1/8))
n - 1 jacobi(k, n) + 1 /===\ ---------------- | | 4 k | | gamma (-) | | n - 2 %pi sqrt(n) k = 1 eta(%e ) = ------------------------------ n n + 1 1/8 sqrt(2) (a 2 %pi )
[...]
This can be simplified to
eta(%e^-(2*%pi*sqrt(n)))^8 = prod(gamma(k/n)^(2*jacobi(k,n)),k,1,n-1)/(32*%pi^2*a*n)
n - 1 /===\ | | 2 jacobi(k, n) k | | Gamma (-) | | n 8 - 2 pi sqrt(n) k = 1 eta (e ) = ---------------------------- . 2 32 pi a n
where a is the real root of
3*a^2*\j((%i*sqrt(n)+1)/2)^(1/3)*n+2*a^3 = n^3
2 1/3 %i sqrt(n) + 1 3 3 3 a J (--------------) n + 2 a = n 2
and J is Klein's absolute invariant,
IFF n in {7, 11, 19, 43, 67, 163} (= -Heegners > 3).
But this prod(gamma(k/n)^jacobi(k,n),k,1,n-1) construct appears in the 60 year old(!) Chowla-Selberg formula for certain products of etas, and the n = 11 case of the above is (hairily!) *derived* (as opposed to guessed from numeric evidence) in http://www.warwick.ac.uk/~masfaw/BillRobin2.pdf ,
I should have cautioned that there is a bogus reciprocation at eqn (17) near the end of this derivation.
wherein it is suggested that Robin Chapman and Alf van der Poorten have been doing these for >= 9 years. What I haven't seen them claim is the the cubicsurd(J^(1/3)) business, but I wouldn't be surprised.
At this point I think I could automate the guessing process for eta(rational+sqrt(-rational)). Also for J(such),
Actually, this remark is redundant because (\j(-%i*log(q)/%pi)^(1/3) = (16*eta(q^2)^24+eta(q)^24)/(12*eta(q)^8*eta(q^2)^16)) = (256*eta(q^4)^24+eta(q^2)^24)/(12*eta(q^2)^16*eta(q^4)^8) 24 2 24 24 4 24 2 1/3 i log(q) 16 eta (q ) + eta (q) 256 eta (q ) + eta (q ) J (- --------) = ----------------------- = ------------------------- pi 8 16 2 16 2 8 4 12 eta (q) eta (q ) 12 eta (q ) eta (q )
e.g., In[158]:= KleinInvariantJ[((Sqrt[13]*I + 1)/2)]== -((125*(9*Sqrt[13] - 31)^3)/8)
Out[158]= 1 125 3 KleinInvariantJ[- (1 + I Sqrt[13])] == -(---) (-31 + 9 Sqrt[13]) 2 8 --rwg
Apparently you can UFD-test n by solving the eta identity for a and seeing if it satisfies a cubic (with integer coefficients).
There are similar identities for eta(%e^-(%pi*sqrt(n))), eta(%e^-(4*%pi*sqrt(n))), and those log-derivative Lambert type sums.
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