A328900 Decimal expansion of s = 1.507126591638653..., solution to 2^s + 3^s = 4^s.

1, 5, 0, 7, 1, 2, 6, 5, 9, 1, 6, 3, 8, 6, 5, 3, 1, 3, 3, 9, 8, 6, 8, 8, 3, 3, 6, 0, 8, 3, 8, 6, 3, 1, 1, 6, 4, 3, 7, 3, 9, 9, 4, 0, 9, 4, 4, 8, 5, 6, 5, 6, 8, 9, 6, 6, 7, 5, 3, 6, 4, 3, 5, 9, 4, 4, 3, 8, 1, 4, 7, 3, 3, 8, 0, 4, 8, 5, 1, 5, 7, 2, 5, 9, 2, 2, 8

Offset: 1

Author

M. F. Hasler, on suggestion from Artur Jasinski, Oct 30 2019

Keywords

• nonn
• cons

Comments

Equivalently, solution to 1/(1 - 2^-s) = 1 + 2^-s + 3^-s, related to partial sums and Euler product approximating zeta(s).

When a + b = c, then the only solution to a^x + b^x = c^x is trivially x = 1. The solution to 1 + 2^x = 4^x is log_2(Phi) = A242208.

See A328904, A328905 for (a, b, c) = (1, 3, 5) and (1, 2, 5).

Examples

			1.5071265916386531339868833608386311643739940944856568966753643594438147338...
		

Links

• M. F. Hasler, Table of n, a(n) for n = 1..1000
• M. F. Hasler, Solutions to a^x + b^x = c^x, OEIS wiki, Nov. 2019

Crossrefs

Cf. A328913 (continued fraction).

Cf. A242208 (1 + 2^x = 4^x), A328912 (continued fraction thereof).

Cf. A328904 (1 + 3^x = 5^x), A328905 (1 + 2^x = 5^x), A328906 (1 + 2^x = 6^x), A328907 (1 + 3^x = 6^x).

Programs

Mathematica

RealDigits[ x /. FindRoot[2^x + 3^x == 4^x, {x, 1.5}, WorkingPrecision -> 100]][[1]] (* Artur Jasinski, Oct 30 2019 *)

PARI

solve(s=1,2,2^s+3^s-4^s) \\ use e.g. \p200 to get more digits