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A284387 {010->2}-transform of the infinite Fibonacci word A003849. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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

LINKS

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

EXAMPLE

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

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 *)

CROSSREFS

Cf. A003231, A003849, A214971, A276886.

Sequence in context: A342707 A282947 A287200 * A143667 A299485 A246785

Adjacent sequences:  A284384 A284385 A284386 * A284388 A284389 A284390

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, May 02 2017

EXTENSIONS

Comment edited by Clark Kimberling, Oct 14 2017

STATUS

approved

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Last modified July 25 02:17 EDT 2021. Contains 346273 sequences. (Running on oeis4.)