With Neil's more recent microhint "56", I *finally* found DedekindEta[I Sqrt[14]] == ( 14 2^(5/16) 7^(5/32) (1 + Sqrt[2] - Sqrt[-1 + 2 Sqrt[2]])^(1/4) Sqrt[((1/56)! (9/56)! (25/56)!)/(1/8)!])/( 5 3^(5/8) (1 + Sqrt[7])^( 3/8) ((2 Sqrt[2] + Sqrt[7]) (1 + 2 Sqrt[7]))^(1/8) \[Pi]^(3/4)) DedekindEta[I Sqrt[7/2]] == ( 14 2^(9/16) 7^(5/32) (1 + Sqrt[2] + Sqrt[-1 + 2 Sqrt[2]])^(1/4) Sqrt[((1/56)! (9/56)! (25/56)!)/(1/8)!])/( 5 3^(5/8) (1 + Sqrt[7])^( 3/8) ((2 Sqrt[2] + Sqrt[7]) (1 + 2 Sqrt[7]))^(1/8) \[Pi]^(3/4)) which distilled out of a terrifying surdberg. The final step was 7 + 4 Sqrt[2] + Sqrt[77 + 56 Sqrt[2]] == 32/(1 + Sqrt[2] - Sqrt[-1 + 2 Sqrt[2]])^4 It's been so long that I forget why I wanted this! Probably some atypical ellipse circumference. Merry Christmas from the mathgods! —rwg On Mon, Dec 10, 2018 at 7:53 PM Bill Gosper <billgosper@gmail.com> wrote:
DedekindEta[I Sqrt[5/2]] == 2 2^(7/8) 5^(3/8) (1 + Sqrt[10])^(1/4) (7/40)! (3/8)! Sqrt[(√2 + √5) (1/20)! (1/5)! (1/4)! (2/5)! (9/20)!/(3 (1/10)!)]/(3 π^(5/4) (3/40)! (7/20)!)
I had not imagined needing (7/40)! —rwg But I still can't find DedekindEta[I Sqrt[7/2]]