A284387 {010->2}-transform of the infinite Fibonacci word A003849.

2, 2, 1, 0, 2, 2, 1, 0, 2, 1, 0, 2, 2, 1, 0, 2, 2, 1, 0, 2, 1, 0, 2, 2, 1, 0, 2, 1, 0, 2, 2, 1, 0, 2, 2, 1, 0, 2, 1, 0, 2, 2, 1, 0, 2, 2, 1, 0, 2, 1, 0, 2, 2, 1, 0, 2, 1, 0, 2, 2, 1, 0, 2, 2, 1, 0, 2, 1, 0, 2, 2, 1, 0, 2, 1, 0, 2, 2, 1, 0, 2, 2, 1, 0, 2, 1

Offset: 1

Author

Clark Kimberling, May 02 2017

Keywords

• nonn
• easy

Comments

It appears that the sequences p = A214971, q = A003231, r = A276886 give the positions of 0, 1, 2, respectively. Let t,u,v be the slopes of p, q, r, respectively. Then t = (5+sqrt(5))/2, u = (5+sqrt(5))/2, v = sqrt(5), and 1/t + 1/u + 1/v = 1. If 1 is removed from p (or from r), the resulting three sequences partition the set of positive integers.

From Michel Dekking, Apr 29 2019: (Start)

This sequence is the unique fixed point of the morphism

0->10, 1->2, 2->2210.

To prove this, let phi2 be the square of the Fibonacci morphism given by

phi2(0)=010, phi2(1)=01.

Then xF := A003849 = 0100101001... is the unique fixed point of phi2.

We introduce the morphism beta with fixed point xB := A188432 = 00100101... given by

beta(0) = 001, beta(1) = 01,

and also the morphism psi given by

psi(0) = 010, psi(1) = 10.

CLAIM: psi(xB) = xF.

This claim can be proved by showing with induction that for n>0

psi(beta^n(0)) = phi2^{n+1}(0),

psi(beta^n(01)) = phi2^{n+1}(10).

Why is this claim useful? Well, it implies directly that

(a(n)) = delta(xB),

where delta is the 'decoration' morphism given by

delta(0) = 2, delta(1) = 10.

Now double the 1's in xB: 1->11'. Then beta induces a 'substitution' S

0 -> 0011', 11' -> 011'.

Since 1 is always followed by 1', and 1' always preceded by 1, the action of S is equivalent to the action of the morphism sigma defined by

sigma(0) = 0011', sigma(1) = 0, sigma(1') = 11'.

The decoration morphism delta gives rise to a letter-to-letter map gamma given by

gamma(0) = 2, gamma(1) = 1, gamma(1') = 0.

Now the change of alphabet gamma gives the morphism we have been looking for, since delta(xB) = gamma(xS), where xS is the unique fixed point of sigma.

(End)

This sequence is the {0->2, 1->10}-transform of A188432. - Michel Dekking, Apr 29 2019

Examples

			As a word, A003849 = 01001010010010100..., and replacing consecutively (not simultaneously!) each 010 by 2 gives 2210221021...
		

Links

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

Crossrefs

Cf. A003231, A003849, A214971, A276886.

Programs

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 13]  (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"010" -> "2"}]
st = ToCharacterCode[w1] - 48 (* A284387 *)
Flatten[Position[st, 0]]  (* A214971 *)
Flatten[Position[st, 1]]  (* A003231 *)
Flatten[Position[st, 2]]  (* A276886 *)

Extensions

Comment edited by Clark Kimberling, Oct 14 2017