A334413 First differences of A101803.

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

Offset

0

Comments

This is a non-homogenous 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 half-integers, and round(r)=floor(r+1/2) for all real r.

From Michel Dekking, Aug 31 2020: (Start)

Here is the rigorous proof and a precise version of the previous comment.

Since round(n*(phi-1)) = 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

Table of n, a(n) for n=0..88.

T. C. Brown, A characterization of the quadratic irrationals, Canad. Math. Bull, 1991, 34(1), 36-41.

Formula

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

a(n) = A006340(n) - 1. - Michel Dekking, Aug 31 2020

Example

a(0) = A101803(1) - A101803(0) = 1 - 0 = 1.
a(1) = A101803(2) - A101803(1) = 1 - 1 = 0.

Prog

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

Crossrefs

Cf. A001622 (golden ratio), A005614, A101803, A334415.

Binary complement of A334414.

Sequence in context: A288932 A155482 A174205 * A114483 A127822 A111967

Adjacent sequences: A334410 A334411 A334412 * A334414 A334415 A334416

Keyword

nonn

Author

Aresh Pourkavoos, Apr 28 2020

Status

approved