



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
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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 nonoverlapping 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 1sqrt(3). The boundedness result is then implied by the fact that the second eigenvalue is smaller than 1 in absolute value.
(End)


LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000
Boris Adamczewski, Symbolic discrepancy and selfsimilar dynamics, Annales de l'Institut Fourier 54 (2004), 22012234.


FORMULA

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


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. A284751, A284752, A026363.
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



