OFFSET
1,2
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.
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
EXAMPLE
1.5071265916386531339868833608386311643739940944856568966753643594438147338...
MATHEMATICA
RealDigits[ x /. FindRoot[2^x + 3^x == 4^x, {x, 1.5}, WorkingPrecision -> 100]][[1]] (* Artur Jasinski, Oct 30 2019 *)
PROG
(PARI) solve(s=1, 2, 2^s+3^s-4^s) \\ use e.g. \p200 to get more digits
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
M. F. Hasler, on suggestion from Artur Jasinski, Oct 30 2019
STATUS
approved