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%I #9 Nov 23 2013 04:10:32
%S 1,1,0,1,1,1,0,1,1,0,1,1,1,0,1,1,1,0,1,1,0,1,1,1,0,1,1,0,1,1,1,0,1,1,
%T 1,0,1,1,0,1,1,1,0,1,1,1,0,1,1,0,1,1,1,0,1,1,0,1,1,1,0,1,1,1,0,1,1,0,
%U 1,1,1,0,1,1,0,1,1,1,0,1,1
%N Generalized Fibonacci word. Binary complement of A221150.
%C Define strings S(0) = 1, S(1)= 110, thereafter S(n) = S(n-1)S(n-2); this sequence is the limit string S(infinity). See the examples below.
%F a(n) = floor((n + 2)/(3 - phi)) - floor((n + 1)/(3 - phi)), where phi = 1/2*(1 + sqrt(5)) is the golden ratio.
%F If we read the sequence as the decimal constant C = 0.11011 10110 11101 11011 01110 ... then C = sum {n >= 1} 1/10^floor(n*(3 - phi)).
%F 9*C has the simple continued fraction expansion [0; 1, 110, 10^1, 10^3, 10^4, 10^7, ..., 10^Lucas(n), ...].
%e S(0) = 1
%e S(1) = 110
%e S(2) = 110 1
%e S(3) = 1101 110
%e S(4) = 1101110 1101
%e S(5) = 11011101101 1101110
%e The sequence of word lengths [1, 2, 4, 7, 11, 18, ...] is A000204.
%p Digits := 50: u := evalf((5-sqrt(5))/2): A230603 := n->floor((n+2)/u)-floor((n+1)/u): seq(A230603(n), n = 0..80);
%Y Cf. A000204, A003849, A005614, A221150.
%K nonn,easy
%O 0
%A _Peter Bala_, Nov 22 2013
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