login
Positions of 0 in A289235; complement of A289151.
3

%I #8 Jun 29 2017 12:07:57

%S 1,2,4,5,6,7,8,9,11,12,13,15,17,18,19,20,22,24,25,27,28,29,30,32,33,

%T 34,36,37,38,39,40,41,42,44,45,46,47,49,50,51,53,54,55,56,57,58,60,61,

%U 62,63,64,65,67,68,69,71,72,73,75,76,77,78,79,80,82,84,85

%N Positions of 0 in A289235; complement of A289151.

%C Conjecture: a(n)/n -> 1.28..., and if m denotes this number, then -1 < m - a(n)/n < m - 1 for n >= 1.

%H Clark Kimberling, <a href="/A289002/b289002.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" -> "100", "10" -> "000"}]

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

%t st = ToCharacterCode[w[z]] - 48 (* A289235 *)

%t Flatten[Position[st, 0]] (* A289002 *)

%t Flatten[Position[st, 1]] (* A289151 *)

%Y Cf. A289235, A289151.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jun 28 2017