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Positions of 0 in A284622; complement of A284626.
3

%I #7 Mar 31 2022 03:12:53

%S 1,4,6,7,8,10,13,15,16,18,21,22,24,25,26,28,31,32,34,35,37,40,42,43,

%T 44,46,49,51,52,54,57,58,60,61,63,66,68,69,70,72,75,76,78,79,80,82,85,

%U 87,88,90,93,94,96,97,98,100,103,104,106,107,109,112,114,115

%N Positions of 0 in A284622; complement of A284626.

%C a(n) - a(n-1) is in {1,2,3} for n>=2. Conjecture: a(n)/n -> 9/5.

%C This conjecture follows directly from the corresponding result for the positions of 1, proved in A284626. - _Michel Dekking_, Mar 31 2022

%H Clark Kimberling, <a href="/A284623/b284623.txt">Table of n, a(n) for n = 1..10000</a>

%e As a word, A284622 = 011010001011010010..., in which 0 is in positions 1,4,6,7,8,10,...

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

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

%t w1 = StringReplace[w, {"0011" -> "0"}]

%t st = ToCharacterCode[w1] - 48 (* A284622 *)

%t Flatten[Position[st, 0]] (* A284623 *)

%t Flatten[Position[st, 1]] (* A284626 *)

%Y Cf. A010060, A284622, A284626.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, May 10 2017