%I #4 May 09 2017 00:14:30

%S 3,5,9,10,15,17,22,23,27,29,33,34,39,40,44,46,51,53,57,58,63,65,70,71,

%T 75,77,82,84,88,89,94,95,99,101,105,106,111,113,118,119,123,125,129,

%U 130,135,136,140,142,147,149,153,154,159,160,164,166,170,171,176

%N Positions of 0 in A286064; complement of A286066.

%C Conjecture: a(n)/n -> 3.

%H Clark Kimberling, <a href="/A286065/b286065.txt">Table of n, a(n) for n = 1..10000</a>

%e As a word, A286064 = 110101110011110101111001110......, in which 0 is in positions 3,4,9,10,15,...

%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 9] (* A010060 *)

%t w = StringJoin[Map[ToString, s]]

%t w1 = StringReplace[w, {"0110" -> "1"}]

%t st = ToCharacterCode[w1] - 48 (* A286064 *)

%t Flatten[Position[st, 0]] (* A286065 *)

%t Flatten[Position[st, 1]] (* A286066 *)

%Y Cf. A010060, A286064, A286066.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, May 08 2017