On Wed, Sep 30, 2015 at 10:46 PM, Bill Gosper <billgosper@gmail.com> wrote:

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)

Meanwhile, using http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/05... Hypergeometric2F1[1/3, 2/3, 1, 1 - GoldenRatio] == (2 Pi Sqrt[GoldenRatio])/ (5^(7/12) Gamma[1/3] Gamma[11/15] Gamma[14/15]) == 3^(3/4) EllipticK[1/64 (47 - 17 I Sqrt[3] - 21 Sqrt[5] + 7 I Sqrt[15])]* E^(I ArcCot[1/8 (1+Sqrt[3])^4 (Sqrt[3]+Sqrt[5])^2])/(Pi Sqrt[GoldenRatio]) which, being two identities, is rather nice in TraditionalForm: gosper.org/K=phi.png especially when you see its mother. --rwg