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A288708 Positions of 0 in A288707; complement of A288709. 5
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 (list; graph; refs; listen; history; text; internal format)
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 zero's 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

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Last modified June 15 16:13 EDT 2019. Contains 324142 sequences. (Running on oeis4.)