OFFSET

1

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

CROSSREFS

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, May 05 2017

STATUS

approved