* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Aug 28. 2008 13:29]:
In case anybody besides me missed it,
Gamma(1/3) = 2^(11/18)*(sqrt(3)+1)^(1/3)*%pi^(2/3)/(3^(1/4)*agm(sqrt(2)*(sqrt(3)+1),1)^(1/3))
11/18 1/3 2/3 1 2 (sqrt(3) + 1) %pi gamma(-) = ------------------------------------- 3 1/4 1/3 3 AGM (sqrt(2) (sqrt(3) + 1), 1)
follows from Borwein&Borwein, pp 15(a) and 28(d), so Gamma(1/3) is nearly as easy as pi. Similarly,
Gamma(1/4) = 2^(3/4)*%pi^(3/4)/sqrt(agm(sqrt(2),1))
3/4 3/4 1 2 %pi Gamma(-) = --------------------- 4 sqrt(AGM(sqrt(2), 1))
finishes off Gamma(n/12). Interesting that, unlike pi, there are no hypergeometric series for these, although the sum(n/(%e^(2*sqrt(3)*%pi*n)-1),n,1,inf) is a degenerate bibasic series.
cf. http://arxiv.org/abs/math/0403510 (section 5) gives similar expressions also for gamma(1/8) and gamma(1/24)
[... AGM and eta ...]
lovely!