A288133 Positions of 0 in A288132; complement of A288134.

1, 2, 4, 7, 12, 21, 38, 71, 136, 265, 522, 1035, 2060, 4109, 8206, 16399, 32784, 65553, 131090, 262163, 524308, 1048597, 2097174, 4194327, 8388632, 16777241, 33554458, 67108891, 134217756, 268435485

Offset: 1

Author

Clark Kimberling, Jun 07 2017

Keywords

• nonn
• easy

Comments

a(n+1)/a(n) -> 2. It appears that a(n) = A005126(n-2) for n >= 2.

This conjecture by Kimberling is proved in A288132. - Michel Dekking, Feb 18 2021

Crossrefs

Cf. A288132, A288134.

Programs

Mathematica

s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n - 1], {"00" -> "0010", "1" -> "11"}]
Table[w[n], {n, 0, 8}]
st = ToCharacterCode[w[11]] - 48   (* A288132 *)
Flatten[Position[st, 0]]  (* A288133 *)
Flatten[Position[st, 1]]  (* A288134 *)

Formula

Conjectures from Colin Barker, Jun 09 2017: (Start)

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

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

(End)

Colin Barker's conjectures are a consequence of

a(n) = 2^{n-2} + n - 1 = A005126(n-2) for n >= 2. - Michel Dekking, Feb 18 2021