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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 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 = StringJoin[Map[ToString, s]]; w[n_] := StringReplace[w[n - 1], {"00" -> "1000", "10" -> "00"}] Table[w[n], {n, 0, 8}] st = ToCharacterCode[w] - 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 October 27 09:36 EDT 2021. Contains 348272 sequences. (Running on oeis4.)