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%I #14 Nov 27 2019 05:39:18
%S 1,1,1,34,1,1,2,1,1,1,2,3,28,2,1,1,2,4,3,2,7,2,35,3,1,1,2,1,2,53,1,33,
%T 1,1,1,2,2,2,35,10,52,1,1,1,2,3,1,1,2,2,1,2,1,1,3,1,1,1,18,1,1,7,2,14,
%U 2,84,1,4,5,3,2,3,1,2,2,1,2,40,1,3,5
%N Continued fraction expansion of A328900 = 1.50712659... solution to 2^x + 3^x = 4^x.
%C This number is also the solution to 1 + 1.5^x = 2^x or 1/(1 - 2^-x) = 1 + 2^-x + 3^-x, see A328900.
%e A328900 = 1.50712659... = 1 + 1/(1 + 1/(1 + 1/(34 + 1/(1 + 1/(1 + 1/(2 + ...))))))
%t ContinuedFraction[ x /. FindRoot[2^x + 3^x == 4^x, {x, 1.5}, WorkingPrecision -> 100]] (* _Robert G. Wilson v_, Nov 12 2019 *)
%o (PARI) contfrac(solve(s=1,2,1+1.5^s-2^s)) \\ Use e.g. \p999 to get more terms.
%Y Cf. A328900, A328912 (if 3 is replaced by 1).
%K nonn,cofr
%O 0,4
%A _M. F. Hasler_, Oct 31 2019
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