%I #6 Oct 15 2022 15:20:33
%S 0,2,23,2,1,1,4,1,1,27,4,12,1,1,1,1,3,1,1,1,1,1,1,1,1,4,1,2,1,2,1,2,6,
%T 1,10,4,3,4,1,2,1,1,43,69,1,2,41,1,3,2,3,3,1,5,4,1,1,1,7,1,1,1,11,13,
%U 2,3,1,1,1,118,2,1,1,12,1,2,2,2,6,2,3,1,4,1,8,1,1,18,2,21,1,4,1,3,1,51,6,1,1,18,2,1,1,2,56,1,1,5,4,1,4,7,1,2,2,1,9,76,2,1,3,1,5,3,1,7,6
%N Continued fraction of A328906 = 0.4895363211996..., solution to 1 + 2^x = 6^x.
%e 0.4895363211996... = 0 + 1/(2 + 1/(23 + 1/(2 + 1/(1 + 1/(1 + 1/(4 + 1/...))))))
%t ContinuedFraction[x/.FindRoot[1+2^x==6^x,{x,.4},WorkingPrecision->1000],150] (* _Harvey P. Dale_, Oct 15 2022 *)
%o (PARI) contfrac(c=solve(x=0,1, 1+2^x-6^x))[^-1] \\ discarding possibly incorrect last term. Use e.g. \p999 to get more terms. - _M. F. Hasler_, Oct 31 2019
%Y Cf. A328912 (cont. frac. of A242208: 1 + 2^x = 4^x), A328913 (cont. frac. of A328900: 2^x + 3^x = 4^x), A329334 (cont. frac. of A328904: 1 + 3^x = 5^x).
%K nonn,cofr
%O 0,2
%A _M. F. Hasler_, Nov 11 2019