

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
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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 = 1sqrt(2)/2 and intercept rho = alpha. This implies that (a(n)) is a Sturmian sequence with slope alpha' = 1/(2alpha) = 2sqrt(2), and intercept rho' = rho/(2alpha) = 32*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][(n1)*alpha+rho], where alpha = 2sqrt(2), and rho = 32*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



