%I #15 Nov 30 2019 17:20:49
%S 1,2,4,5,6,8,9,10,11,12,14,15,16,18,19,20,21,22,24,25,26,27,28,30,31,
%T 32,34,35,36,37,38,40,41,42,44,45,46,47,48,50,51,52,53,54,56,57,58,60,
%U 61,62,63,64,66,67,68,69,70,72,73,74,76,77,78,79,80,82
%N Positions of 0 in A288707; complement of A288709.
%C Conjecture: a(n)/n -> -1 + sqrt(5), and if m denotes this number, then -1 < m - a(n)/n < 1 for n >= 1.
%C Proof of this conjecture. One follows the same strategy as in the proof of the conjecture in A288709. We have a(n)/n -> -1 + sqrt(5) if and only if n/a(n) -> (1+sqrt(5))/4, which is the frequency of 0's in A288707. If alpha = (3-sqrt(5))/2 is the frequency of 1's in y = A189661 = 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, ... then (2*(1-alpha)+alpha)/2 is the frequency of 0's in A288707, since every 0 in y produces 2 zeros in A288707, and every 1 just 1 zero (and one 1). One computes (2*(1-alpha)+alpha)/2 = (1+sqrt(5))/4. The bounds follow from the exponential convergence of the frequencies of 0 and 1 in y. - _Michel Dekking_, Oct 19 2018
%H Clark Kimberling, <a href="/A288708/b288708.txt">Table of n, a(n) for n = 1..10000</a>
%t s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
%t w[n_] := StringReplace[w[n - 1], {"00" -> "1000", "10" -> "00"}]
%t Table[w[n], {n, 0, 8}]
%t st = ToCharacterCode[w[10]] - 48 (* A288707 *)
%t Flatten[Position[st, 0]] (* A288708 *)
%t Flatten[Position[st, 1]] (* A288709 *)
%Y Cf. A288707, A288709.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Jun 16 2017