A328900 - OEIS

%I #22 Nov 11 2019 14:05:39

%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,

%T 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,

%U 1,4,7,3,3,8,0,4,8,5,1,5,7,2,5,9,2,2,8

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

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

%C 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.

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

%H M. F. Hasler, <a href="/A328900/b328900.txt">Table of n, a(n) for n = 1..1000</a>

%H M. F. Hasler, <a href="/wiki/Solutions to a^x + b^x = c^x">Solutions to a^x + b^x = c^x</a>, OEIS wiki, Nov. 2019

%e 1.5071265916386531339868833608386311643739940944856568966753643594438147338...

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

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

%Y Cf. A328913 (continued fraction).

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

%Y 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).

%K nonn,cons

%O 1,2

%A _M. F. Hasler_, on suggestion from _Artur Jasinski_, Oct 30 2019