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%I #16 Feb 15 2021 01:59:39
%S 2,7,11,16,21,25,30,34,39,44,48,53,58,62,67,71,76,81,85,90,94,99,104,
%T 108,113,118,122,127,131,136,141,145,150,155,159,164,168,173,178,182,
%U 187,191,196,201,205,210,215,219,224,228,233,238,242,247,251
%N Positions of 0 in A287725; complement of A287727.
%C Conjecture: 2 < n*r - a(n) < 3 for n >= 1, where r = (7 + sqrt(5))/2.
%C From _Michel Dekking_, Feb 12 2021: (Start)
%C Let T be the transform given by T(0) = 1, T(1) = 011 that defines A287725.
%C The Fibonacci word A003849 is a 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 T(sigma^2(0)) = T(010) = 10111, T(sigma^2(1)) = T(01) = 1011.
%C We see from this that the sequence (a(n+1)-a(n)) of first differences 5,4,5,5,4,5,4,5,5,4,5,5,4..., of (a(n)) is a sequence on the two 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(n)) is the generalized Beatty sequence given by a(n) = floor(n*phi) + 3n - 2.
%C This immediately implies Kimberling's conjecture.
%C Note that r = (7 + sqrt(5))/2 = phi + 3, where phi is the golden mean. So
%C n*r - a(n) = n(phi+3) - floor(n*phi) + 3n - 2 = n*phi - floor(n*phi) - 2,
%C which lies in (2,3).
%C These bounds are best possible since the fractional part of n*phi is equidistributed modulo 1.
%C (End)
%H Clark Kimberling, <a href="/A287726/b287726.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(n) = floor(n*phi) + 3n - 2. - _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" -> "011"}]
%t st = ToCharacterCode[w1] - 48 (* A287725 *)
%t Flatten[Position[st, 0]] (* A287726 *)
%t Flatten[Position[st, 1]] (* A287727 *)
%Y Cf. A287725, A287727.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Jun 02 2017
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