A286667 - OEIS

A286667

Positions of 1 in A286665; complement of A286666.

2, 4, 5, 7, 9, 10, 12, 14, 16, 17, 19, 21, 22, 24, 26, 28, 29, 31, 33, 34, 36, 38, 39, 41, 43, 45, 46, 48, 50, 51, 53, 55, 57, 58, 60, 62, 63, 65, 67, 68, 70, 72, 74, 75, 77, 79, 80, 82, 84, 86, 87, 89, 91, 92, 94, 96, 98, 99, 101, 103, 104, 106, 108, 109

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OFFSET

1,1

COMMENTS

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

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

Numbers with a positive number of trailing 0's in their minimal representation in terms of the positive Pell numbers (A317204). - Amiram Eldar, Mar 16 2022

LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000

EXAMPLE

As a word, A286665 = 010110101101010110101101..., in which 1 is in positions 2,4,5,7,9,...

MATHEMATICA

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

w = StringJoin[Map[ToString, s]]

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

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

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

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

CROSSREFS