A285401 Positions of 0 in A285177; complement of A285402.

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

Author

Clark Kimberling, Apr 26 2017

Keywords

• nonn
• easy

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)

Examples

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

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A285177, A285402, A285403.

Programs

Mathematica

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