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

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

Cf. A171588, A286665, A286666, A317204.

Sequence in context: A303589 A106829 A190228 * A083120 A001614 A284535

Adjacent sequences: A286664 A286665 A286666 * A286668 A286669 A286670

Keyword

nonn,easy

Author

Clark Kimberling, May 13 2017

Status

approved