

A334413


First differences of A101803.


2



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

0


COMMENTS

This is a nonhomogenous Sturmian word, conjectured to be produced by the following recurrence relation: string S(0)=1; S(1)=10; S(n+2)=S(n+1)S(n)* for all natural n, where S(n)* denotes the reverse of S(n); a=S(infinity). Equivalently, a(n) is conjectured to be the fixed point of the morphism 0>101, 1>10101. See "A characterization of the quadratic irrationals."
Relatively long parts of this sequence appear in other OEIS sequences, but it does not appear that any of them match this sequence with only a different prefix. However, arbitrarily long subsequences appear to be contained within A005614. A rigorous proof is still needed, but the key steps are that, thanks to the Equidistribution Theorem, multiples of 1/phi come arbitrarily close to halfintegers, and round(r)=floor(r+1/2) for all real r.
Here is the rigorous proof and a precise version of the previous comment.
Since round(n*(phi1)) = round(n*phi)  n, this sequence essentially already occurs in the OEIS: a(n) = A006340(n)  1 for n = 0,1,....
The proof that (a(n)) is the fixed point of the morphism mu: 0>101, 1>10101 is given in my comments in A006340 from Mar 05 2018.
Let y = A005614 be the binary complement of the Fibonacci word. By definition, y is the fixed point of the morphism nu: 0>1, 1>10.
Claim: (a(n)) and y generate the same language, i.e., any subword of (a(n)) occurs in y, and conversely.
For a proof, consider the cube of the morphism nu: lambda:=nu^3: 0>101, 1>10110.
The crux is that mu and lambda are conjugate morphisms, i.e., there exists a word w such that lambda(v) w = w mu(v) for all words v. One can take w = 101, and it suffices to check this for v=0 and v=1.
(End)


LINKS



FORMULA

a(n) = round((n+1)/phi)  round(n/phi), where phi is the golden ratio.


EXAMPLE



PROG

(PARI) a(n) = my(x=(sqrt(5)1)/2); round((n+1)*x)  round(n*x); \\ Michel Marcus, May 23 2020


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



