[Goaded by Mike Hirschhorn] A couple of years ago, a young friend, Corey Ziegler Hunts, wrote a radical denester which simplifies (1 + 3*Sqrt[5] - 2*Sqrt[10 + 2*Sqrt[5]])^(1/4), the fourth root of a trinomial in your expression for G(e^-(2π)), to √(Sqrt[2] 5^(1/4) - Sqrt[1 + Sqrt[5]]), the square root of a binomial. Similarly for H. Three lines later on p2 are two typos: "surprizingly" and "were were". Corey's younger brother Julian and I wrote a sextic solver which led to ∛(6 Cos[π/18] - 6 Cos[π/9] + 6 Cos[2π/9] - 6 Sin[2π/9] - 1) ~ 0.2170953869504416 as the value of Φ₁b(2,2;i/3) on p10. --Bill Gosper