A284753 Positions of 1 in A284751; complement of A284752.

2, 6, 8, 10, 12, 16, 18, 22, 24, 28, 30, 34, 36, 38, 40, 44, 46, 50, 52, 54, 56, 60, 62, 66, 68, 70, 72, 76, 78, 82, 84, 86, 88, 92, 94, 98, 100, 104, 106, 110, 112, 114, 116, 120, 122, 126, 128, 130, 132, 136, 138, 142, 144, 148, 150, 154, 156, 158, 160

Offset

1,1

Comments

Conjecture: -2 < n*r - a(n) < 3 for n >= 1, where r = 1 + sqrt(3).

From Michel Dekking, Feb 27 2021: (Start)

The formula a(n) = 2*A026363(n) follows from the two facts that (a(n)) gives the positions of 1 in A284751, fixed point of the morphism mu given by

mu: 0 -> 01, 1 -> 0001,

and A026363 gives the positions of 1 in the fixed point of the morphism tau given by

tau: 0->11, 1 -> 101.

The non-overlapping words of length 2 occurring in (a(n)) at odd positions are a:=00 and b:=01. Their occurrences can be read from the fixed point of the morphism nu induced by mu, and are given by

nu: a =00 -> 0101 = bb, b =01 -> 010001 = bab.

The fact that tau is obtained under the alphabet change {a,b} -> {0,1} proves that (a(n)) is twice the sequence of occurrences of 1 in the fixed point of tau.

Interestingly, the conjecture: -2 < n*(1+sqrt(3))- a(n) < 3 , implies a conjecture: -1 < n*(1 + sqrt(3))/2 - A026363(n) < 3/2, which is stronger than the conjecture -1 < n*(1 + sqrt(3))/2 - a(n) < 2 given in the comments of A026363.

The best bounds are not simple to obtain, but the fact that (n*(1+sqrt(3))- a(n)) is a bounded sequence follows by a general symbolic discrepancy result, Theorem 1 in Adamscewski's 2004 paper. To apply that theorem, one writes a(n) as a sum of differences 1 and 2, corresponding to the words 1 and 10 in the fixed point 10111101... of tau. The induced morphism is given by 1->21, 2->2111, with eigenvalues of its incidence matrix 1+sqrt(3) and 1-sqrt(3). The boundedness result is then implied by the fact that the second eigenvalue is smaller than 1 in absolute value.

(End)

In the Fokkink-Joshi paper, this sequence is the Cloitre (0,2,4,2)-hiccup sequence, i.e., a(1) = 2; for m < n, a(n) = a(n-1)+4 if a(m) = n, else a(n) = a(n-1)+2. - Michael De Vlieger, Jul 28 2025

See Fokkink-Joshi p. 16, Theorem 19, for proof of Kimberling's conjecture, listed first in these comments. - Michael De Vlieger, Feb 02 2026

Links

Clark Kimberling, Table of n, a(n) for n = 1..10000

Boris Adamczewski, Symbolic discrepancy and self-similar dynamics, Annales de l'Institut Fourier 54 (2004), 2201-2234.

Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, Ramanujan J. 69 (2026), 40. See pp. 8-9, 13-16, 25, esp. p. 14, Table 2. See also arXiv:2507.16956 [math.CO], 2025. See pp. 7, 9, 12-13, 17.

Formula

a(n) = 2*A026363(n). - Michel Dekking, Feb 27 2021

a(n) = A086398(n) + 1. [Fokkink-Joshi] - Michael De Vlieger, Feb 02 2026

Example

As a word, A284751 = 010001..., in which 1 is in positions 2,6,...

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0, 0, 0, 1}}] &, {0}, 6] (* A284751 *)
Flatten[Position[s, 0]]  (* A284752 *)
Flatten[Position[s, 1]]  (* A284753 *)

Crossrefs

Cf. A026363, A086398, A284751, A284752.

Sequence in context: A216032 A076300 A049637 * A258663 A166447 A075332

Adjacent sequences: A284750 A284751 A284752 * A284754 A284755 A284756

Keyword

nonn,easy

Author

Clark Kimberling, Apr 13 2017

Status

approved