A189572 - OEIS

%I #19 Nov 03 2018 02:10:00

%S 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,

%T 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,

%U 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

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

%C Is this a shifted version of A004641? - _R. J. Mathar_, May 16 2011

%C 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

%C (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

%C 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

%H Wieb Bosma, Michel Dekking, Wolfgang Steiner, <a href="http://math.colgate.edu/~integers/sjs4/sjs4.Abstract.html">A remarkable sequence related to Pi and sqrt(2)</a>, Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A4.

%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 Takao Komatsu, <a href="https://projecteuclid.org/euclid.tjm/1270041624">Substitution invariant inhomogeneous Beatty sequences</a>, Tokyo J. Math. 22 (1999), 235-243.

%F 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

%e 0->01->01001->010010101001->

%t t = Nest[Flatten[# /. {0->{0,1}, 1->{0, 0, 1}}] &, {0}, 6] (*A189572*)

%t f[n_] := t[[n]]

%t Flatten[Position[t, 0]] (*A189573*)

%t Flatten[Position[t, 1]] (* A080652 conjectured *)

%Y Cf. A189573, A080762.

%Y See A188068 and the mirror image of A275855 for a similar pair of fixed points of morphisms. - _Michel Dekking_, Feb 07 2017

%K nonn

%O 1

%A _Clark Kimberling_, Apr 23 2011