OFFSET
1
COMMENTS
This is a 3-automatic sequence. See Allouche et al. link. - Michel Dekking, Oct 05 2020
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
J.-P. Allouche, F. M. Dekking, and M. Queffélec, Hidden automatic sequences, arXiv:2010.00920 [math.NT], 2020.
FORMULA
Conjecture: a(n) = A284905(n+1). - R. J. Mathar, May 08 2017
From Michel Dekking, Jan 26 2024: (Start)
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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 25 2017
STATUS
approved