Gamma[1/3]/(2^(1/3) Gamma[2/3]^2)
So it's only natural that a formula of Bruce Berndt should give
Hypergeometric2F1[1/3, 2/3, 1, -1] ==
-(((-1)^(11/12) 3^(
3/4) (2 (674484539 - 581632 Sqrt[2])^(2/3) -
1734 (-674484539 + 581632 Sqrt[2])^(1/3) +
769097 I (I + Sqrt[3]))^(1/4)
EllipticK[1/32 (16 - \[Sqrt](6 (-14867 +
17 (-674484539 - 581632 Sqrt[2])^(1/3) -
17 (-1)^(2/3) (674484539 - 581632 Sqrt[2])^(
1/3))))])/(2 Sqrt[2] (674484539 - 581632 Sqrt[2])^(
1/12) \[Pi]))
--rwg
(-: Explanation under construction)
