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 A328912 Continued fraction expansion of log_2(sqrt(5)+1)/2) = 0.6942419... = A242208. 7

%I

%S 0,1,2,3,1,2,3,2,4,2,1,2,11,2,1,11,1,1,134,2,2,2,1,4,1,1,3,1,7,1,13,1,

%T 3,5,1,1,1,8,1,3,4,1,1,1,3,4,1,3,1,4,1,4,1,3,40,1,1,5,4,3,3,1,3,1,2,6,

%U 1,1,2,28,11,1,69,1,1,6,5,1,1,5,2,1,2

%N Continued fraction expansion of log_2(sqrt(5)+1)/2) = 0.6942419... = A242208.

%C This number is also the solution to 1 + 2^x = 4^x, or 1 + 1/2^x = 2^x, which clarifies the relation to Phi = (sqrt(5)+1)/2, solution to 1 + 1/x = x.

%e log_2(sqrt(5)+1)/2) = 0.6942419... = 0 + 1/(1 + 1/(2 + 1/(3 + 1/(1 + ...))))

%o (PARI) contfrac(log(sqrt(5)+1)/log(2)) \\ Use e.g. \p999 to get more terms.

%Y Cf. A242208, A001622 (decimals of Phi), A000012 (cont. frac. of Phi).

%K nonn,cofr

%O 0,3

%A _M. F. Hasler_, Oct 31 2019

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Last modified November 28 13:47 EST 2021. Contains 349413 sequences. (Running on oeis4.)