A284387 - OEIS

%I #15 May 01 2019 05:52:13

%S 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,

%T 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,

%U 0,2,2,1,0,2,1,0,2,2,1,0,2,2,1,0,2,1

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

%C 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.

%C From _Michel Dekking_, Apr 29 2019: (Start)

%C This sequence is the unique fixed point of the morphism

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

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

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

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

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

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

%C and also the morphism psi given by

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

%C CLAIM:  psi(xB) = xF.

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

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

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

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

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

%C where delta is the 'decoration' morphism given by

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

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

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

%C 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

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

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

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

%C 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.

%C (End)

%C This sequence is the {0->2, 1->10}-transform of A188432. - _Michel Dekking_, Apr 29 2019

%H Clark Kimberling, <a href="/A284387/b284387.txt">Table of n, a(n) for n = 1..10000</a>

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

%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 13]  (* A003849 *)

%t w = StringJoin[Map[ToString, s]]

%t w1 = StringReplace[w, {"010" -> "2"}]

%t st = ToCharacterCode[w1] - 48 (* A284387 *)

%t Flatten[Position[st, 0]]  (* A214971 *)

%t Flatten[Position[st, 1]]  (* A003231 *)

%t Flatten[Position[st, 2]]  (* A276886 *)

%Y Cf. A003231, A003849, A214971, A276886.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, May 02 2017

%E Comment edited by _Clark Kimberling_, Oct 14 2017