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 A287675 Positions of 0 in A287674; complement of A287676. 4
 2, 3, 7, 8, 11, 12, 16, 17, 21, 22, 25, 26, 30, 31, 34, 35, 39, 40, 44, 45, 48, 49, 53, 54, 58, 59, 62, 63, 67, 68, 71, 72, 76, 77, 81, 82, 85, 86, 90, 91, 94, 95, 99, 100, 104, 105, 108, 109, 113, 114, 118, 119, 122, 123, 127, 128, 131, 132, 136, 137, 141 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Conjecture: -1 < n*r - a(n) < 2 for n >= 1, where r = (7 + sqrt(5))/4. From Michel Dekking, Feb 12 2021: (Start) Let T be the transform given by T(0) = 1, T(1) = 001 that defines the underlying sequence A287674 by A287674(n) = T(A003849(n)). The Fibonacci word A003849 is fixed point of the morphism sigma: 0->01, 1->0, and therefore also of the morphism sigma^2: 0->010, 1->01. Now note that      T(sigma^2(0)) = T(10011) = 10111,  T(sigma^2(1)) = T(01) = 1001. We see from this that the sequence (a(2n+1)-a(2n-1)) of first differences 5,4,5,5,4,4,5,4,5,5,4,..., of every second occurrence of a 0 is a sequence on the letters 4 and 5, and that in fact these two letters occur as the Fibonacci word on the alphabet {5,4}. From Lemma 8 in the paper by Allouche and Dekking it follows that (a(2n-1)) is the generalized Beatty sequence given for n = 1,2,... by       a(2n-1) = floor((n*phi) + 3n - 2 = A287726(n), and (a(2n)) is the generalized Beatty sequence given for n = 1,2,... by         a(2n) = floor(n*phi) + 3n - 1. From this we derive directly Kimberling's conjecture. Note that         r = (7 + sqrt(5))/4 = phi/2 +3/2, where phi is the golden mean. So       (2n-1)*r - a(2n-1) =  (n-1/2)(phi+3) - floor(n*phi) - 3n + 2 =       n*phi - floor(n*phi) - 1/2*phi + 1/2, which lies in the interval (1/2-phi/2, 3/2-phi/2). Also we have       2n*r - a(2n) = n*(phi+3) - floor(n*phi) - 3n + 1 =       n*phi - floor(n*phi) + 1, which lies in the interval (1,2). The two intervals  (1/2-phi/2, 3/2-phi/2) = (-0.3090...,0.6090...) and (1,2) are subintervals of the interval (-1,2) in Kimberling's conjecture, and there are no better bounds, since the fractional part of n*phi is equidistributed modulo 1. (End) LINKS Clark Kimberling, Table of n, a(n) for n = 1..10000 J.-P. Allouche and F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424 [math.NT], 2018. J.-P. Allouche and F. M. Dekking, Generalized Beatty sequences and complementary triples, Moscow J. Comb. Number Th. 8, 325-341, 2019. FORMULA a(2n-1) = floor(n*phi) + 3n - 2 , a(2n) = floor(n*phi) + 3n - 1 . - Michel Dekking, Feb 12 2021 a(2n-1) = A287726(n),  a(2n) =  A287726(n) + 1. - Michel Dekking, Feb 12 2021 MATHEMATICA s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* A003849 *) w = StringJoin[Map[ToString, s]] w1 = StringReplace[w, {"0" -> "1", "1" -> "001"}] st = ToCharacterCode[w1] - 48    (* A287674 *) Flatten[Position[st, 0]]  (* A287675 *) Flatten[Position[st, 1]]  (* A287676 *) CROSSREFS Cf. A287675, A287676, A287726. Sequence in context: A051468 A002274 A329783 * A273944 A214324 A102664 Adjacent sequences:  A287672 A287673 A287674 * A287676 A287677 A287678 KEYWORD nonn,easy AUTHOR Clark Kimberling, Jun 02 2017 STATUS approved

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Last modified July 26 01:49 EDT 2021. Contains 346294 sequences. (Running on oeis4.)