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Positions of 0 in A285969; complement of A285970.
3

%I #4 May 09 2017 00:13:51

%S 2,6,8,11,14,18,21,24,26,30,32,35,38,41,43,47,50,54,56,59,62,66,69,72,

%T 74,78,81,85,87,90,93,96,98,102,104,107,110,114,117,120,122,126,128,

%U 131,134,137,139,143,146,150,152,155,158,161,163,167,169,172,175

%N Positions of 0 in A285969; complement of A285970.

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

%H Clark Kimberling, <a href="/A285971/b285971.txt">Table of n, a(n) for n = 1..10000</a>

%e As a word, A285969 = 0100010100100......, in which 10 is in positions 2,6,8,11,14,...

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

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

%t w1 = StringReplace[w, {"0110" -> "0"}]

%t st = ToCharacterCode[w1] - 48 (* A285969 *)

%t Flatten[Position[st, 0]] (* A285970 *)

%t Flatten[Position[st, 1]] (* A285971 *)

%Y Cf. A010060, A285969, A285970.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, May 08 2017