A189572 Fixed point of the morphism 0->01, 1->001.

0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1

Offset

1

Comments

Is this a shifted version of A004641? - R. J. Mathar, May 16 2011

The answer is: yes. We have (a(n+1)) = A004641(n), for all n. Let S be the morphism 0->01, 1->001, and let T be the morphism 0->10, 1->100. By definition A004641 is the unique fixed point of T. The equality above is obviously implied by the following claim: for all n it holds that S^n(0)0 = 0T^n(0), and S^n(01)0 = 0T^n(10). Proof: S(0)0 = 010 = 0T(0), and S(01)0 = 010010 = 0T(10). Proceed by induction: S^(n+1)(0)0 = S^n(01)0 = 0T^n(10) = 0T^(n+1)(0) and S^(n+1)(01)0 = S^n(01)S^n(0)S^n(01)0 = 0T^n(10)T^n(0)T^n(10) = 0T^(n+1)(10). - Michel Dekking, Feb 07 2017

(a(n)) is the inhomogeneous Sturmian sequence with slope sqrt(2) - 1 and intercept 1 - sqrt(2)/2. Let psi be the morphism 0->01, 1->001. Then the parameters of the Sturmian sequence can be computed from psi = psi_1 psi_4, where psi_1 is the Fibonacci morphism, and psi_4 is the morphism 0->0, 1->10. See also Example 2 in Komatsu's paper, which considers the square 0->01001, 1->0101001 of psi. - Michel Dekking, Nov 03 2018

Proof of Kimberling's conjecture: one can use Lemma 1 in the Bosma, Dekking, Steiner paper, which gives the positions of 1 in a Sturmian sequence as a Beatty sequence. - Michel Dekking, Nov 03 2018

Links

Table of n, a(n) for n=1..116.

Wieb Bosma, Michel Dekking, Wolfgang Steiner, A remarkable sequence related to Pi and sqrt(2), Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A4.

Michel Dekking, Substitution invariant Sturmian words and binary trees, Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A17.

Takao Komatsu, Substitution invariant inhomogeneous Beatty sequences, Tokyo J. Math. 22 (1999), 235-243.

Formula

a(n) = floor(alpha*n+beta) - floor(alpha*(n-1)+beta), with alpha = sqrt(2)-1, beta = 1-sqrt(2)/2 - Michel Dekking, Nov 03 2018

Example

0->01->01001->010010101001->

Mathematica

t = Nest[Flatten[# /. {0->{0, 1}, 1->{0, 0, 1}}] &, {0}, 6] (*A189572*)
f[n_] := t[[n]]
Flatten[Position[t, 0]] (*A189573*)
Flatten[Position[t, 1]] (* A080652 conjectured *)

Crossrefs

Cf. A189573, A080762.

See A188068 and the mirror image of A275855 for a similar pair of fixed points of morphisms. - Michel Dekking, Feb 07 2017

Sequence in context: A288600 A284468 A260455 * A287028 A327202 A357448

Adjacent sequences: A189569 A189570 A189571 * A189573 A189574 A189575

Keyword

nonn

Author

Clark Kimberling, Apr 23 2011

Status

approved