A288708 Positions of 0 in A288707; complement of A288709.

1, 2, 4, 5, 6, 8, 9, 10, 11, 12, 14, 15, 16, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 30, 31, 32, 34, 35, 36, 37, 38, 40, 41, 42, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 56, 57, 58, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 72, 73, 74, 76, 77, 78, 79, 80, 82

Offset

1,2

Comments

Conjecture: a(n)/n -> -1 + sqrt(5), and if m denotes this number, then -1 < m - a(n)/n < 1 for n >= 1.

Proof of this conjecture. One follows the same strategy as in the proof of the conjecture in A288709. We have a(n)/n -> -1 + sqrt(5) if and only if n/a(n) -> (1+sqrt(5))/4, which is the frequency of 0's in A288707. If alpha = (3-sqrt(5))/2 is the frequency of 1's in y = A189661 = 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, ... then (2*(1-alpha)+alpha)/2 is the frequency of 0's in A288707, since every 0 in y produces 2 zeros in A288707, and every 1 just 1 zero (and one 1). One computes (2*(1-alpha)+alpha)/2 = (1+sqrt(5))/4. The bounds follow from the exponential convergence of the frequencies of 0 and 1 in y. - Michel Dekking, Oct 19 2018

Links

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

Mathematica

s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n - 1], {"00" -> "1000", "10" -> "00"}]
Table[w[n], {n, 0, 8}]
st = ToCharacterCode[w[10]] - 48   (* A288707 *)
Flatten[Position[st, 0]]  (* A288708 *)
Flatten[Position[st, 1]]  (* A288709 *)

Crossrefs

Cf. A288707, A288709.

Sequence in context: A135668 A276216 A226946 * A039138 A285670 A030124

Adjacent sequences: A288705 A288706 A288707 * A288709 A288710 A288711

Keyword

nonn,easy

Author

Clark Kimberling, Jun 16 2017

Status

approved