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 )
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 sqrt(-n) is a UFD! Apparently you can UFD-test n by solving the eta identity for a and seeing if it satisfies a cubic (with integer coefficients).
Actually, this works for n down to 5. 7 gives a = 14, which satisfies a cubic!-)
There are similar identities for eta(%e^-(%pi*sqrt(n))), eta(%e^-(4*%pi*sqrt(n))), and those log-derivative Lambert type sums.
Joerg> RWG: congrats on that 163 thing! Thanks! Oddity: DedekindEta[I*Sqrt[13]]==(2^(5/104)*Gamma[7/52]*Gamma[15/52]* (Cos[Pi/26]*Cos[(5*Pi)/26]*Gamma[11/52]*Gamma[19/52])^(3/4)* Sqrt[(Gamma[23/52]*Gamma[31/52]*Gamma[35/52])/ (13*Pi*Gamma[5/52]*Gamma[17/52]*Gamma[25/52]*Gamma[29/52])]* ((Gamma[27/52]*Sin[(2*Pi)/13])/(Gamma[37/52]*Gamma[10/13]* Gamma[45/52]*Gamma[53/52]))^(1/4))/(Gamma[2/13]^(3/4)*Gamma[21/52]) I.e., no algebraic factor other than the trigs, which are all intances of Gamma reflection. Pretty clearly not in simplest form. By contrast, DedekindEta[I*Sqrt[17]] == (Gamma[9/34]^(3/8)*(Gamma[15/34]/ ((5 + Sqrt[17] + Sqrt[26 + 10*Sqrt[17]])*Gamma[2/17]* Gamma[4/17]*Gamma[8/17]))^(1/8)*(17*Gamma[3/68]*Gamma[7/68]* Gamma[11/68]*Gamma[23/68]*Gamma[13/34]*Gamma[27/68]*Gamma[31/68]* Gamma[39/68]*Gamma[63/68])^(1/4)*Sqrt[Gamma[35/34]])/ (2*2^(15/68)*Pi^(11/8)*Gamma[1/17]^(5/8)), and the algebraic factor appears not to be expressible in Gammas. I.e., I couldn't eliminate it with LatticeReduce. ---rwg