

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
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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 lettertoletter 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



