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Positions of 0 in A286665; complement of A286667.
5

%I #16 Mar 16 2022 05:50:10

%S 1,3,6,8,11,13,15,18,20,23,25,27,30,32,35,37,40,42,44,47,49,52,54,56,

%T 59,61,64,66,69,71,73,76,78,81,83,85,88,90,93,95,97,100,102,105,107,

%U 110,112,114,117,119,122,124,126,129,131,134,136,139,141,143,146

%N Positions of 0 in A286665; complement of A286667.

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

%C This conjecture follows easily from the fact that (a(n)) is a Beatty sequence, see my comments in A286665. - _Michel Dekking_, Mar 11 2018

%C Numbers with no trailing 0's in their minimal representation in terms of the positive Pell numbers (A317204). - _Amiram Eldar_, Mar 16 2022

%H Clark Kimberling, <a href="/A286666/b286666.txt">Table of n, a(n) for n = 1..10000</a>

%e As a word, A286665 = 010110101101010110101101..., in which 0 is in positions 1,3,6,8,11,...

%t s = Nest[Flatten[# /. {0 -> {0, 0, 1}, 1 -> {0}}] &, {0}, 6] (* A171588 *)

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

%t w1 = StringReplace[w, {"0" -> "01"}]

%t st = ToCharacterCode[w1] - 48 ; (* A286665 *)

%t p0 = Flatten[Position[st, 0]]; (* A286666 *)

%t p1 = Flatten[Position[st, 1]]; (* A286667 *)

%Y Cf. A171588, A286665, A286667, A317204.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, May 13 2017