A334413 - OEIS

%I #35 May 30 2026 16:40:49

%S 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,

%T 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,

%U 1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1

%N First differences of A101803.

%C 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."

%C 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.

%C From _Michel Dekking_, Aug 31 2020: (Start)

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

%C 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,....

%C 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.

%C 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.

%C Claim: (a(n)) and y generate the same language, i.e., any subword of (a(n)) occurs in y, and conversely.

%C For a proof, consider the cube of the morphism nu: lambda:=nu^3: 0->101, 1->10110.

%C 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.

%C (End)

%H T. C. Brown, <a href="https://doi.org/10.4153/CMB-1991-006-4">A characterization of the quadratic irrationals</a>, Canad. Math. Bull, 1991, 34(1), 36-41.

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

%F a(n) = A006340(n) - 1. - _Michel Dekking_, Aug 31 2020

%e a(0) = A101803(1) - A101803(0) = 1 - 0 = 1.

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

%t Differences[Round[Range[0,100]/GoldenRatio]] (* _Harvey P. Dale_, Jan 30 2025 *)

%o (PARI) a(n) = my(x=(sqrt(5)-1)/2); round((n+1)*x) - round(n*x); \\ _Michel Marcus_, May 23 2020

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

%Y Binary complement of A334414.

%K nonn

%O 0

%A _Aresh Pourkavoos_, Apr 28 2020