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A287664 Positions of 0's in A287663; complement of A287665. 3

%I #33 Jan 22 2022 08:45:32

%S 2,3,4,7,8,9,11,12,13,16,17,18,21,22,23,25,26,27,30,31,32,34,35,36,39,

%T 40,41,44,45,46,48,49,50,53,54,55,58,59,60,62,63,64,67,68,69,71,72,73,

%U 76,77,78,81,82,83,85,86,87,90,91,92,94,95,96,99,100,101

%N Positions of 0's in A287663; complement of A287665.

%C Conjecture: -2 < n*r - a(n) < 1 for n >= 1, where r = (7 + sqrt(5))/6. [Corrected by _Clark Kimberling_, Aug 19 2019]

%C From _Michel Dekking_, Aug 22 2019: (Start)

%C Computation of a formula for a(n): we will show that (a(n)) is a union of three generalized Beatty sequences.

%C Let T denote the morphism {0->1, 1->000}.

%C The Fibonacci word xF:=A003849 is a concatenation of the two words v = 100 and w = 10 (ignoring xF(0)). It is known that the order of these words is given by the Fibonacci word itself (cf. A003622). Now note that

%C T(v) = T(100) = 00011, T(w) = T(10) = 0001.

%C We see from this that the sequence of first differences of a = A287664 for the first 0 in each triple 000 is given by 5,4,5,5,4,5,4,5,5,4,..., and that in fact these two letters occur as the Fibonacci word on the alphabet {5,4}.

%C Let {b^(j)} for j = 1,2,3 be the sequence of positions of the j-th 0 in the triples 000. From Lemma 8 in the paper by Allouche and Dekking it follows that these sequences are given by

%C b^(j)(n) = floor(n*phi) + 3*n + r^(j),

%C where r^(1) = -2, r^(2) = -1, and r^(3) = 0.

%C Note that a(3*n+j-3) = b^(j)(n) for n=1,2,3,... and j=1,2,3. (End)

%C From _Michel Dekking_, Aug 22 2019: (Start)

%C Proof of Kimberling's conjecture. {} denotes fractional part.

%C Let r = (7+sqrt(5))/6 = phi/3+1. Using the formula above we have

%C (3*n+j-3)*r - a(3*n+j-3) =

%C (n+j/3-1)*(phi+3) - (n*phi - {n*phi} + 3n + r^(j)) =

%C (j/3-1)*phi + j - 3 - r^(j) + {n*phi}.

%C Since ({n*phi}) is equidistributed on (0,1), this implies that a(3*n+j-3) takes values in the interval (-(2/3)*phi, -(2/3)*phi+1) for j=1, in the interval (-(1/3)*phi, -(1/3)*phi+1) for j=2, and in the interval (0,1) for j=3. Combining these, we see that a(n) takes values in (-(2/3)*phi, 1), and that this is best possible. Since -(2/3)*phi = -1.078689..., this improves on the lower bound conjectured by Kimberling.

%C (End)

%H Clark Kimberling, <a href="/A287664/b287664.txt">Table of n, a(n) for n = 1..10000</a>

%H J.-P. Allouche, F. M. Dekking, <a href="https://arxiv.org/abs/1809.03424">Generalized Beatty sequences and complementary triples</a>, arXiv:1809.03424 [math.NT], 2018.

%F a(3*n+j-3) = b^(j)(n) for j=1,2,3, with b^(1)(n) = floor(n*phi) + 3*n - 2, b^(2)(n) = floor(n*phi) + 3*n - 1, b^(3)(n) = floor(n*phi) + 3*n. - _Michel Dekking_, Aug 22 2019

%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" -> "000"}]

%t st = ToCharacterCode[w1] - 48 (* A287663 *)

%t Flatten[Position[st, 0]] (* A287664 *)

%t Flatten[Position[st, 1]] (* A287665 *)

%Y Cf. A287663, A287665.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jun 02 2017

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Last modified March 29 11:14 EDT 2024. Contains 371278 sequences. (Running on oeis4.)