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A285401
Positions of 0 in A285177; complement of A285402.
9
1, 2, 4, 5, 7, 8, 10, 11, 17, 18, 20, 21, 23, 24, 26, 27, 29, 30, 36, 37, 39, 40, 42, 43, 45, 46, 48, 49, 55, 56, 58, 59, 61, 62, 64, 65, 67, 68, 74, 75, 81, 82, 88, 89, 95, 96, 102, 103, 105, 106, 108, 109, 111, 112, 114, 115, 121, 122, 124, 125, 127, 128
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
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
Clark Kimberling, Apr 26 2017
STATUS
approved