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Positions of 0 in A286493; complement of A286495.
3

%I #4 May 11 2017 20:11:27

%S 1,4,6,7,10,11,12,13,15,16,17,18,19,22,24,25,28,29,30,31,32,35,37,38,

%T 39,40,42,43,46,47,48,49,51,52,53,54,55,58,60,61,62,63,65,66,69,70,71,

%U 72,73,76,78,79,82,83,84,85,87,88,89,90,91,94,96,97,100

%N Positions of 0 in A286493; complement of A286495.

%C a(n) - a(n-1) is in {1,2,3} for n>=2. Conjecture: a(n)/n -> 3/2.

%H Clark Kimberling, <a href="/A286494/b286494.txt">Table of n, a(n) for n = 1..10000</a>

%e As a word, A286493 = 01101001100001000001..., in which 0 is in positions 1,4,6,7,10,...

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

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

%t w1 = StringReplace[w, {"1011" -> "0"}]

%t st = ToCharacterCode[w1] - 48 (* A286493 *)

%t Flatten[Position[st, 0]] (* A286494 *)

%t Flatten[Position[st, 1]] (* A286495 *)

%Y Cf. A010060, A286493, A286495.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, May 11 2017