A288382 Positions of 0 in A288381; complement of A288383.

1, 2, 3, 5, 8, 13, 22, 39, 72, 137, 266, 523, 1036, 2061, 4110, 8207, 16400, 32785, 65554, 131091, 262164, 524309, 1048598, 2097175, 4194328, 8388633, 16777242

Offset: 1

Author

Clark Kimberling, Jun 10 2017

Keywords

• nonn
• easy

Comments

a(n+1)/a(n)-> 2.

Appears to be the same as A052968 (apart from the offset). - R. J. Mathar, Jun 14 2017

This conjecture is proved in A288381. - Michel Dekking, Feb 18 2021

Crossrefs

Cf. A288381, A288383.

Programs

Mathematica

s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n - 1], {"00" -> "0001", "1" -> "11"}]
Table[w[n], {n, 0, 8}]
st = ToCharacterCode[w[11]] - 48   (* A288381 *)
Flatten[Position[st, 0]]  (* A288382 *)
Flatten[Position[st, 1]]  (* A288383 *)

Formula

a(n) = -1 + A288133(n-1) for n >= 2.

Conjectures from Colin Barker, Jun 10 2017: (Start)

G.f.: x*(1 - 2*x + x^3 - x^4) / ((1 - x)^2*(1 - 2*x)).

a(n) = -1 + 2^(n-3) + n for n>2.

a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3) for n>5.

(End)

Barker's conjectures are implied by Mathar's conjecture. - Michel Dekking, Feb 18 2021