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%I #14 Feb 22 2021 17:52:43
%S 1,2,4,6,7,8,10,12,13,14,16,18,19,20,22,24,26,28,30,32,33,34,36,38,39,
%T 40,42,44,46,48,50,52,53,54,56,58,59,60,62,64,65,66,68,70,71,72,74,76,
%U 77,78,80,82,83,84,86,88,90,92,94,96,97,98,100,102,103
%N Positions of 1 in A285518; complement of A285519.
%C Conjecture: a(n)/n -> (1+sqrt(5))/2 = golden ratio (A001622).
%C Proof of this conjecture: the limit equals 1/f1, where f1 is the frequency of 1 in (a(n)), and 1/f1 = phi = golden ratio. See A285519 for more details. - _Michel Dekking_, Feb 22 2021
%H Clark Kimberling, <a href="/A285520/b285520.txt">Table of n, a(n) for n = 1..10000</a>
%e As a word, A285518 = 11010111010111010111..., in which 1 is in positions 1,2,4,6,7,...
%t s = Nest[Flatten[# /. {0 -> {1, 1}, 1 -> {0, 0, 1, 1}}] &, {0}, 7] (* A285504 *)
%t w = StringJoin[Map[ToString, s]]
%t w1 = StringReplace[w, {"11" -> "1", "00" -> "0"}]
%t s1 = ToCharacterCode[w1] - 48 (* A285518 *)
%t Flatten[Position[s1, 0]] (* A285519 *)
%t Flatten[Position[s1, 1]] (* A285520 *)
%Y Cf. A285504, A285515, A285518, A285519.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Apr 30 2017
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