A285519 - OEIS

A285519

Positions of 0 in A285518; complement of A285520.

3, 5, 9, 11, 15, 17, 21, 23, 25, 27, 29, 31, 35, 37, 41, 43, 45, 47, 49, 51, 55, 57, 61, 63, 67, 69, 73, 75, 79, 81, 85, 87, 89, 91, 93, 95, 99, 101, 105, 107, 109, 111, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 153, 155, 157, 159

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OFFSET

1,1

COMMENTS

Conjecture: a(n)/n -> (3+sqrt(5))/2.

From Michel Dekking, Feb 22 2021: (Start)

Here is a proof of this conjecture. The limit of a(n)/n as n tends to infinity is equal to the inverse 1/f0 of the frequency f0 of the letter 0 in the sequence (a(n)) (see also the argument in A285401).

From the Perron-Frobenius theorem we know that the frequency vector (f0,f1) is the normalized eigenvector associated to the Perron-Frobenius eigenvalue lambda of the incidence matrix M of the generating morphism 0->11, 1->0101.

Here M = | 0 2|

| 2 2|.

The Perron-Frobenius eigenvalue of M is lambda = 1+sqrt(5), and the vector [u,v]:= [2, 1+sqrt(5)] is an eigenvector with lambda. So

1/f0 = (u+v)/u = (3+sqrt(5))/2 = phi^2, where phi = (1+sqrt(5))/2.

(End)

LINKS

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

EXAMPLE

As a word, A285518 = 11010111010111010111..., in which 0 is in positions 3,5,9,11,15,...

MATHEMATICA

s = Nest[Flatten[# /. {0 -> {1, 1}, 1 -> {0, 0, 1, 1}}] &, {0}, 7] (* A285504 *)

w = StringJoin[Map[ToString, s]]