OFFSET
1,2
COMMENTS
Conjecture: a(n)/n -> (61-sqrt(3))/26 = 2.279...
From Michel Dekking, Feb 10 2021: (Start)
This conjecture is false. In fact,
a(n)/n --> (5+sqrt(17))/4 = 2.28077...
Let mu be the defining morphism for A285177, i.e,
mu(0) = 11, mu(1) = 001.
The sequence A285177 is the fixed point x = 0010010010011111... starting with 0 of mu^2:
mu^2(0) = 001001, mu^2(1) = 1111001.
The 0's in x are at positions a(1)=1, a(2)=2, a(3)=4, etc.
Now suppose that N_0(K) = n is the number of 0's in a prefix x[1,K] of length K of x. Then obviously a(n) = K +/- 6.
Also N_0(K) + N_1(K) = K, where N_1(K) is the number of 1's in x[1,K].
So
K/N_0(K) = a(n)/n +/- 6/n.
Letting n tend to infinity, we find that
a(n)/n --> 1/f0,
where f0 is the frequency of 0's in x.
It is well known that these exist and are equal to the normalized eigenvector of the Perron-Frobenius eigenvalue of the incidence matrix of the morphism mu.
A simple computation yields that f0 = 4/(5+sqrt(17)).
It follows that a(n)/n --> (5+sqrt(17))/4.
(End)
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
EXAMPLE
As a word, A285177 = 001001..., in which 0 is in positions 1,2,4,5,7,...
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 26 2017
STATUS
approved