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 A286665 {0->01}-transform of the Pell word, A171588. 4
 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1 COMMENTS From Michel Dekking, Mar 11 2018: (Start) Let psi_8 be the elementary Sturmian morphism given by psi_8(0)=01, psi_(8)1=1, and let x = A171588 be the Pell word. Then, by definition, (a(n)) = psi_8(x). Now x is a Sturmian sequence x = s(alpha, rho) with slope alpha = 1-sqrt(2)/2 and intercept rho = alpha. This implies that (a(n)) is a Sturmian sequence with slope alpha' = 1/(2-alpha) = 2-sqrt(2), and intercept rho' = rho/(2-alpha) = 3-2*sqrt(2) (cf. Lothaire Lemma 2.2.18). Since the algebraic conjugate of rho' is equal to 3+2*sqrt(2), which is larger than the algebraic conjugate of alpha', (a(n)) is NOT a fixed point of a morphism, by Yasutomi's criterion. However, (a(n)) IS a fixed point of an automorphism sigma of the free group generated by 0 and 1. In fact sigma: 0-> 01010^{-1},  1-> 01. To see this, let pi be the Pell morphism given by pi(0)=001, pi(1)=0. Then pi(x) = x, and so psi_8(pi(x)) = psi_8(x) = a, implying that sigma := psi_8 pi psi_8^{-1} fixes a = (a(n)). One easily computes psi_8^{-1}: 0->01^{-1}, 1->1, which gives sigma. (End) LINKS Clark Kimberling, Table of n, a(n) for n = 1..10000 Michel Dekking, Substitution invariant Sturmian words and binary trees, arXiv:1705.08607 [math.CO], 2017. Michel Dekking, Substitution invariant Sturmian words and binary trees, Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A17. M. Lothaire, Algebraic combinatorics on words, Cambridge University Press. Online publication date: April 2013; Print publication year: 2002. FORMULA a(n) = [n*alpha+rho]-[(n-1)*alpha+rho], where alpha = 2-sqrt(2), and rho = 3-2*sqrt(2). - Michel Dekking, Mar 11 2018 EXAMPLE As a word, A171588 = 001001000100100010010010001001000..., and replacing each 0 by 01 gives 010110101101010110101101010110... From Michel Dekking, Mar 11 2018: (Start) To see that (a(n)) is fixed by sigma, iterate sigma starting with 01:   sigma(01) = 01010^{-1}01 = 01011, sigma^2(01) = 01010^{-1}0101010^{-1}0101 = 010110101101. (End) MATHEMATICA s = Nest[Flatten[# /. {0 -> {0, 0, 1}, 1 -> {0}}] &, {0}, 6] (* A171588 *) w = StringJoin[Map[ToString, s]] w1 = StringReplace[w, {"0" -> "01"}] st = ToCharacterCode[w1] - 48 ; (* A286665 *) p0 = Flatten[Position[st, 0]];  (* A286666 *) p1 = Flatten[Position[st, 1]];  (* A286667 *) CROSSREFS Cf. A171588, A286666, A286667. Sequence in context: A188027 A193496 A284533 * A096270 A308185 A159689 Adjacent sequences:  A286662 A286663 A286664 * A286666 A286667 A286668 KEYWORD nonn,easy AUTHOR Clark Kimberling, May 13 2017 STATUS approved

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Last modified December 12 15:11 EST 2019. Contains 329960 sequences. (Running on oeis4.)