A284388 0-limiting word of the morphism 0 -> 1, 1 -> 001.

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

Offset

1

Comments

The morphism 0 -> 1, 1 -> 001 has two limiting words. If the number of iterations is even, the 0-word evolves from 0 -> 1 -> 001 -> 11001 -> 00100111001; if the number of iterations is odd, the 1-word evolves from 0 -> 1 -> 001 -> 11001 -> 00100111001 -> 110011100100100111001, as in A284391. The 0-limiting word results from the 1 limiting word by replacing the initial 00 by 1.

Conjecture: the limiting frequency of 0's in both limiting words is 1/2.

Conjecture: after the first term, same as one less than the run-lengths of the Thue-Morse sequence A010060. - George Beck, Mar 24 2021

From Adnan Baysal, Aug 06 2023: (Start)

Conjecture: Excluding the first two terms, if runs of 1s of length one are replaced by 0, and runs of 1s of length three are replaced by 1, we get the same sequence.

Conjecture: Index distance between two consecutive 1s is either one or three. Excluding the first two terms, if distances of one are replaced by 0, and distances of three are replaced by 1, we get the same sequence. (End)

Links

Clark Kimberling, Table of n, a(n) for n = 1..10000

Mathematica

s = Nest[Flatten[# /. {0 -> {1}, 1 -> {0, 0, 1}}] &, {0}, 8]; (* this sequence *)
Flatten[Position[s, 0]];  (* A284389 *)
Flatten[Position[s, 1]];  (* A284390 *)

Crossrefs

Cf. A010060, A284389, A284390, A284391, A003159, A026465.

Sequence in context: A273129 A288936 A064990 * A289174 A059125 A189820

Adjacent sequences: A284385 A284386 A284387 * A284389 A284390 A284391

Keyword

nonn,easy

Author

Clark Kimberling, Mar 30 2017

Status

approved