A285960 - OEIS

A285960

{01->1}-transform of the Thue-Morse word A010060.

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

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OFFSET

COMMENTS

From Michel Dekking, Apr 19 2022: (Start)

This sequence is a morphic sequence, i.e., the letter-to-letter image of the fixed point of a morphism mu. Here is a proof.

The Thue-Morse sequence x:=A010060 is a concatenation of the four return words A=011, B=010, C=0110, and D=01, of the word 01 in x. (These are the words occurring in x starting with 01, and having no other occurrences of 01 in them). By applying the Thue-Morse morphism 0->01, 1->10 to the return words one induces the derived morphism

tau: A -> AB, B -> CD, C -> ABD, D -> C.

It is clear that to obtain (a(n)) from x one has to apply the morphism delta to x written as ABCDABD... given by

A-> 11, B-> 10, C-> 110, D-> 1.

In the paper "Morphic words, Beatty sequences and integer images of the Fibonacci language" delta is called a decoration map. It is well-known that decorated fixed points are morphic sequences, and the 'natural' algorithm to achieve this yields a morphism on an alphabet of 2+2+3+1 = 8 symbols. In this particular case one can reduce the number of symbols to 4, say {a,b,c,d}, and obtain the morphism mu

mu: a->ab, b->cd, c->abd, d->c.

Let y be the fixed point of mu starting with the letter a. Then (a(n)) = lambda(y), where the letter-to-letter map lambda is defined by

lambda: a->1, b->1, c->1, d->0.

Proof of the bisection formula below: the derived morphisms for the return words of 01, respectively 011 in A285960 and A286046 both are the same morphism tau. Moreover, the decoration morphism A-> 11, B-> 10, C-> 110, D-> 1 to obtain (a(n)) corresponds exactly to the letters in the decoration morphism A-> 1010, B-> 1001, C-> 101001, D->10 occurring at the odd positions. This implies a(n) = A286046(2n-1), n = 1,2,...

(End)

LINKS

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

M. Dekking, Morphic words, Beatty sequences and integer images of the Fibonacci language, Theoretical Computer Science 809, 407-417 (2020).

FORMULA

a(n) = A286046(2n-1), n = 1,2,...

EXAMPLE

As a word, A010060 = 0110100110010110100101100..., and replacing each 01 by 1 gives 1110110111101110111011011101110...

MATHEMATICA

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 8] (* Thue-Morse, A010060 *)

w = StringJoin[Map[ToString, s]]

w1 = StringReplace[w, {"01" -> "1"}]