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A285519
Positions of 0 in A285518; complement of A285520.
3
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
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
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]]
w1 = StringReplace[w, {"11" -> "1", "00" -> "0"}]
s1 = ToCharacterCode[w1] - 48 (* A285518 *)
Flatten[Position[s1, 0]] (* A285519 *)
Flatten[Position[s1, 1]] (* A285520 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 30 2017
STATUS
approved