|
%I #21 Sep 30 2023 09:18:03
%S 2,3,7,8,11,12,16,17,21,22,25,26,30,31,34,35,39,40,44,45,48,49,53,54,
%T 58,59,62,63,67,68,71,72,76,77,81,82,85,86,90,91,94,95,99,100,104,105,
%U 108,109,113,114,118,119,122,123,127,128,131,132,136,137,141
%N Positions of 0 in A287674; complement of A287676.
%C Conjecture: -1 < n*r - a(n) < 2 for n >= 1, where r = (7 + sqrt(5))/4.
%C From _Michel Dekking_, Feb 12 2021: (Start)
%C 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)).
%C 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.
%C Now note that
%C T(sigma^2(0)) = T(10011) = 10111, T(sigma^2(1)) = T(01) = 1001.
%C 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}.
%C 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
%C a(2n-1) = floor((n*phi) + 3n - 2 = A287726(n),
%C and (a(2n)) is the generalized Beatty sequence given for n = 1,2,... by
%C a(2n) = floor(n*phi) + 3n - 1.
%C From this we derive directly Kimberling's conjecture. Note that
%C r = (7 + sqrt(5))/4 = phi/2 + 3/2,
%C where phi is the golden mean. So
%C (2n-1)*r - a(2n-1) = (n-1/2)*(phi+3) - floor(n*phi) - 3n + 2 =
%C n*phi - floor(n*phi) - (1/2)*phi + 1/2,
%C which lies in the interval (1/2 - phi/2, 3/2 - phi/2). Also we have
%C 2n*r - a(2n) = n*(phi+3) - floor(n*phi) - 3n + 1 =
%C n*phi - floor(n*phi) + 1,
%C which lies in the interval (1,2).
%C 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.
%C (End)
%H Clark Kimberling, <a href="/A287675/b287675.txt">Table of n, a(n) for n = 1..10000</a>
%H J.-P. Allouche and F. M. Dekking, <a href="https://arxiv.org/abs/1809.03424">Generalized Beatty sequences and complementary triples</a>, arXiv:1809.03424 [math.NT], 2018.
%H J.-P. Allouche and F. M. Dekking, <a href="https://msp.org/moscow/2019/8-4/p02.xhtml">Generalized Beatty sequences and complementary triples</a>, Moscow J. Comb. Number Th. 8, 325-341, 2019.
%F a(2n-1) = floor(n*phi) + 3n - 2, a(2n) = floor(n*phi) + 3n - 1. - _Michel Dekking_, Feb 12 2021
%F a(2n-1) = A287726(n), a(2n) = A287726(n) + 1. - _Michel Dekking_, Feb 12 2021
%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* A003849 *)
%t w = StringJoin[Map[ToString, s]]
%t w1 = StringReplace[w, {"0" -> "1", "1" -> "001"}]
%t st = ToCharacterCode[w1] - 48 (* A287674 *)
%t Flatten[Position[st, 0]] (* this sequence *)
%t Flatten[Position[st, 1]] (* A287676 *)
%Y Cf. A003849, A287674, A287676, A287726.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Jun 02 2017
|
