A287675 Positions of 0 in A287674; complement of A287676.

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

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]]  (* this sequence *)
Flatten[Position[st, 1]]  (* A287676 *)

Crossrefs

Cf. A003849, A287674, 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