OOPS, no! 3 () () () = () () () () () () is the product of ((2 - Power[3, (3)^-1]) (1 + Power[3, (3)^-1]))/((Power[3, (3)^-1] - 1) (1 + 3^(2/3))) == 1 and ((1+Power[2, (3)^-1])^3 (1+2^(2/3)) (Sqrt[3]-2^(2/3)) (Sqrt[3]+2^(2/3)))/(1+3 Power[2, (3)^-1])==3 . More: <http://gosper.org/surdburger.pdf> My objective is to use PSLQ to replace giant Chowla-Selberg surdbergs with smaller surdburgers in DedekindEta valuations. Besides, surdburgers are neat. —rwg On Tue, Jan 29, 2019 at 3:51 PM Bill Gosper <billgosper@gmail.com> wrote:
Another tasty morsel: (1 + 3^(2/3))^2 == ((1 + 3^(1/3)) (1 + 2 3^(1/3))) (!)—rwg
On Sun, Jan 27, 2019 at 9:13 AM Bill Gosper <billgosper@gmail.com> wrote:
(Empirical)
Bogus
claim: This identity is "primitive" or "irreducible" in the sense that it
holds for no other rational exponents of the 3 and the binomials if any exponent is 0, unless they all are. —rwg
On Sat, Jan 26, 2019 at 7:54 AM Bill Gosper <billgosper@gmail.com> wrote:
3 (1 + 3 2^(1/3)) (2 - 3^(1/3)) (1 + 3^(1/3)) = = (1 + 2^(1/3))^3 (1 + 2^(2/3)) (-1 + 3^(1/3)) (-2^(2/3) + Sqrt[3]) (2^(2/3) + Sqrt[3]) (1 + 3^(2/3)) —rwg