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Positions of 1 in A285568; complement of A285569.
3

%I #9 Feb 22 2021 05:05:33

%S 1,2,4,5,8,9,11,12,13,14,16,17,20,21,23,24,26,27,30,31,33,34,36,37,40,

%T 41,43,44,45,46,48,49,52,53,55,56,58,59,62,63,66,67,70,71,73,74,76,77,

%U 80,81,83,84,85,86,88,89,92,93,95,96,98,99,102,103,105

%N Positions of 1 in A285568; complement of A285569.

%C Conjecture: a(n)/n -> (1+sqrt(5))/2 = golden ratio (A001622).

%C Proof of this conjecture: the limit is 1/f0 where f0 is the frequency of 0 in (a(n)). Since the generating morphism of (a(n)) has the same incidence matrix as the generating morphism 0->11, 1->0101 of the sequence A285518, it follows that 1/f0 = phi = golden ratio. - _Michel Dekking_, Feb 22 2021

%H Clark Kimberling, <a href="/A286031/b286031.txt">Table of n, a(n) for n = 1..10000</a>

%e As a word, A285568 = 110110011011..., in which 1 is in positions 1,2,4,5,8,...

%t s = Nest[Flatten[# /. {0 -> {1, 1}, 1 -> {0, 1, 1, 0}}] &, {0}, 9] (* A285568 *)

%t Flatten[Position[s, 0]] (* A285569 *)

%t Flatten[Position[s, 1]] (* A286031 *)

%Y Cf. A001622, A285566, A285568, A285569.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Apr 30 2017