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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 30 2017
STATUS
approved