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Positions of 0 in A289112; complement of A289114.
3

%I #4 Jun 28 2017 13:24:11

%S 1,2,4,5,6,8,9,11,12,13,15,18,19,21,22,25,28,29,31,34,35,37,39,40,43,

%T 44,46,49,52,53,55,58,61,62,64,66,67,69,72,75,76,79,80,82,85,88,89,92,

%U 93,95,98,101,102,104,105,106,109,110,113,114,116,119,120

%N Positions of 0 in A289112; complement of A289114.

%C Conjecture: a(n)/n -> 2.14..., and if m denotes this number, then 0 < a(n)/n < m - 1 for n >= 1.

%H Clark Kimberling, <a href="/A289113/b289113.txt">Table of n, a(n) for n = 1..10000</a>

%t z = 11; (* number of iterates *)

%t s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];

%t w[n_] := StringReplace[w[n - 1], {"00" -> "0010", "01" -> "011", "10" -> "001"}]

%t TableForm[Table[w[n], {n, 0, 10}]]

%t st = ToCharacterCode[w[z]] - 48 (* A289112 *)

%t Flatten[Position[st, 0]] (* A289113 *)

%t Flatten[Position[st, 1]] (* A289114 *)

%Y Cf. A289112, A289114.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jun 28 2017