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OFFSET

1,2


COMMENTS

Conjecture: a(n)/n > (61sqrt(3))/26 = 2.279...
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 PerronFrobenius 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



EXAMPLE

As a word, A285177 = 001001..., in which 0 is in positions 1,2,4,5,7,...


MATHEMATICA

s = Nest[Flatten[# /. {0 > {1, 1}, 1 > {0, 0, 1}}] &, {0}, 10] (* A285177 *)
Flatten[Position[s, 0]] (* A285401 *)
Flatten[Position[s, 1]] (* A285402 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



