

A285345


Fixed point of the morphism 0 > 10, 1 > 1100.


6



1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1
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OFFSET

1


COMMENTS

This is a 3automatic sequence. See Allouche et al. link.  Michel Dekking, Oct 05 2020


LINKS



FORMULA

Proof of Mathar's conjecture: let alpha be the morphism 0 > 10, 1 > 1100, and let beta be the morphism 0 > 01, 1 > 1001, which has A284905 as its fixed point starting with 0. Note that alpha^n(0) tends to (a(n)) as n tends to infinity because alpha(0) starts with 1. It therefore suffices to prove the relation
(A) : 0 alpha^n(0) = beta^n(0) 0 for all n=1,2,3,...
To prove such a thing one uses the fact that alpha and beta are conjugate morphisms, i.e., there exists a word u such that
(B) beta(w) = u^{1} alpha(w) u.
Here u^{1} is the free group inverse of u.
We have in this case u:=1, and it suffices to prove (B) for the words w=0, and w=1. Indeed:
beta(0) = 01 = 1^{1} 10 1 = 1^{1} alpha(0) 1,
beta(1) = 1001 = 1^{1} 1100 1 = 1^{1} alpha(1) 1.
Next, we prove (A). For n=1, we do have 0 alpha(0) = 010 = beta(0) 0.
Suppose (A) has been proved till n. Then
0 alpha^{n+1}(0) = 1^{1} 10 alpha^{n+1}(0) (10)^{1} 10
= 1^{1} alpha(0) alpha^{n+1}(0) (alpha(0))^{1} 10
= 1^{1} alpha(0 alpha^n(0) 0^{1} ) 10
= 1^{1} alpha(beta^n(0) ) 1 0
= beta(beta^n(0)) 0 = beta^{n+1}(0) 0.
Here we used (B) with w = beta^n(0) in line three, and the induction hypothesis in line four. (End)


EXAMPLE

0 > 10> 1100 > 110011001010 > ...


MATHEMATICA

s = Nest[Flatten[# /. {0 > {1, 0}, 1 > {1, 1, 0, 0}}] &, {0}, 10]; (* this sequence *)
Flatten[Position[s, 0]]; (* A285346 *)
Flatten[Position[s, 1]]; (* A285347 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



