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

%I A284387 #15 May 01 2019 05:52:13
%S A284387 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 A284387 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 A284387 0,2,2,1,0,2,1,0,2,2,1,0,2,2,1,0,2,1
%N A284387 {010->2}-transform of the infinite Fibonacci word A003849.
%C A284387 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 A284387 From _Michel Dekking_, Apr 29 2019: (Start)
%C A284387 This sequence is the unique fixed point of the morphism
%C A284387     0->10, 1->2, 2->2210.
%C A284387 To prove this, let phi2 be the square of the Fibonacci morphism given by
%C A284387     phi2(0)=010,  phi2(1)=01.
%C A284387 Then xF := A003849 = 0100101001... is the unique fixed point of phi2.
%C A284387 We introduce the morphism beta with fixed point xB := A188432 = 00100101... given by
%C A284387     beta(0) = 001,  beta(1) = 01,
%C A284387 and also the morphism psi given by
%C A284387     psi(0) = 010,  psi(1) = 10.
%C A284387 CLAIM:  psi(xB) = xF.
%C A284387 This claim can be proved by showing with induction that for n>0
%C A284387     psi(beta^n(0)) = phi2^{n+1}(0),
%C A284387     psi(beta^n(01)) = phi2^{n+1}(10).
%C A284387 Why is this claim useful? Well, it implies directly that
%C A284387     (a(n)) = delta(xB),
%C A284387 where delta is the 'decoration' morphism given by
%C A284387     delta(0) = 2, delta(1) = 10.
%C A284387 Now double the 1's in xB: 1->11'. Then beta induces a 'substitution' S
%C A284387     0 -> 0011', 11' -> 011'.
%C A284387 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 A284387     sigma(0) = 0011', sigma(1) = 0, sigma(1') = 11'.
%C A284387 The decoration morphism delta gives rise to a letter-to-letter map gamma given by
%C A284387     gamma(0) = 2, gamma(1) = 1, gamma(1') = 0.
%C A284387 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 A284387 (End)
%C A284387 This sequence is the {0->2, 1->10}-transform of A188432. - _Michel Dekking_, Apr 29 2019
%H A284387 Clark Kimberling, <a href="/A284387/b284387.txt">Table of n, a(n) for n = 1..10000</a>
%e A284387 As a word, A003849 = 01001010010010100..., and replacing consecutively (not simultaneously!) each 010 by 2 gives 2210221021...
%t A284387 s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 13]  (* A003849 *)
%t A284387 w = StringJoin[Map[ToString, s]]
%t A284387 w1 = StringReplace[w, {"010" -> "2"}]
%t A284387 st = ToCharacterCode[w1] - 48 (* A284387 *)
%t A284387 Flatten[Position[st, 0]]  (* A214971 *)
%t A284387 Flatten[Position[st, 1]]  (* A003231 *)
%t A284387 Flatten[Position[st, 2]]  (* A276886 *)
%Y A284387 Cf. A003231, A003849, A214971, A276886.
%K A284387 nonn,easy
%O A284387 1,1
%A A284387 _Clark Kimberling_, May 02 2017
%E A284387 Comment edited by _Clark Kimberling_, Oct 14 2017