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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.)