A286665 - OEIS

%I #29 Aug 26 2018 04:46:16

%S 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,

%T 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,

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

%N {0->01}-transform of the Pell word, A171588.

%C From _Michel Dekking_, Mar 11 2018: (Start)

%C 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).

%C 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).

%C 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.

%C 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.

%C (End)

%H Clark Kimberling, <a href="/A286665/b286665.txt">Table of n, a(n) for n = 1..10000</a>

%H Michel Dekking, <a href="http://arxiv.org/abs/1705.08607">Substitution invariant Sturmian words and binary trees</a>, arXiv:1705.08607 [math.CO], 2017.

%H Michel Dekking, <a href="http://math.colgate.edu/~integers/sjs7/sjs7.Abstract.html">Substitution invariant Sturmian words and binary trees</a>, Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A17.

%H M. Lothaire, <a href="http://tomlr.free.fr/Math%E9matiques/Fichiers%20Claude/Auteurs/aaaDivers/Lothaire%20-%20Algebraic%20Combinatorics%20On%20Words.pdf">Algebraic combinatorics on words</a>, Cambridge University Press. Online publication date: April 2013; Print publication year: 2002.

%F 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

%e As a word, A171588 = 001001000100100010010010001001000..., and replacing each 0 by 01 gives 010110101101010110101101010110...

%e From _Michel Dekking_, Mar 11 2018: (Start)

%e To see that (a(n)) is fixed by sigma, iterate sigma starting with 01:

%e   sigma(01) = 01010^{-1}01 = 01011,

%e sigma^2(01) = 01010^{-1}0101010^{-1}0101 = 010110101101. (End)

%t s = Nest[Flatten[# /. {0 -> {0, 0, 1}, 1 -> {0}}] &, {0}, 6] (* A171588 *)

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

%t w1 = StringReplace[w, {"0" -> "01"}]

%t st = ToCharacterCode[w1] - 48 ; (* A286665 *)

%t p0 = Flatten[Position[st, 0]];  (* A286666 *)

%t p1 = Flatten[Position[st, 1]];  (* A286667 *)

%Y Cf. A171588, A286666, A286667.

%K nonn,easy

%O 1

%A _Clark Kimberling_, May 13 2017