A287726 - OEIS

%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