A120614 a(n) = g(n+1) - g(n) where g(k) = floor(phi*floor(k/phi)) and phi = (1+sqrt(5))/2.

1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0

Offset: 1

Author

Benoit Cloitre, Jun 17 2006

Keywords

• nonn

Comments

From Michel Dekking, Oct 29 2018: (Start)

Here is a proof that (a(n)) is fixed point of the morphism 0->102, 1->102, 2->02.

Let alpha:=phi-1. Then alpha*phi = 1. So

g(k) = floor(phi*floor(k*alpha)).

Write k*alpha = floor(k*alpha) + {k*alpha}, i.e., {k*alpha} is the fractional part of k*alpha. Then

g(k) = floor(phi*(k*alpha-{k*alpha})) = k + floor(-phi*{k*alpha}).

Thus

a(n) = n+1 +floor(-phi*{(n+1)*alpha})-n -floor(-phi*{n*alpha}).

It follows that

a(n) = 1 - floor(phi*{(n+1)*alpha}) + floor(phi*{n*alpha}).

The difference -floor(phi*{(n+1)*alpha}) + floor(phi*{n*alpha}) is equal to -1, 0 or 1, since floor(phi*{n*alpha}) is equal to 0 or 1.

In fact, phi*{n*alpha} can only take values between 0 and 1.619, and floor(phi*{n*alpha}) = 0 if and only if

{n*alpha} < 1/phi = alpha.

This is the same (putting rho:=1-alpha) as requiring

{n*alpha+rho} < 1-alpha.

Via the rotation description of Sturmian sequences (see, e.g., Lothaire), one sees that this sequence is the inhomogeneous Sturmian sequence s(alpha, rho), but with offset 1, and with 0 and 1 exchanged. Since rho+alpha=1, it follows that s(alpha, rho) with offset 2 equals s(1-alpha, 1-alpha), the classical Fibonacci sequence xF:=A003849, fixed point of 0->01, 1->0. We have found that

a(n+1)=0 iff xF(n)=0, xF(n+1)=1,

a(n+1)=1 iff xF(n)=0, xF(n+1)=0,

a(n+1)=2 iff xF(n)=1, xF(n+1)=0.

This means that (a(n+1)) equals the 3-symbol Fibonacci sequence A270788 on the alphabet {0,2,1}. Then Proposition 5 in "Morphisms, Symbolic Sequences, and Their Standard Forms" yields that (a(n)) is fixed point of the morphism 0->102, 1->102, 2->02. (End)

Links

• Muniru A Asiru, Table of n, a(n) for n = 1..1000
• F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
• M. Lothaire, Algebraic Combinatorics on Words, Cambridge, 2002.

Crossrefs

Cf. A003849, A120613, A120615, A270788.

Programs

Magma

[Floor((1+Sqrt(5))*Floor(2*(k+1)/(1+Sqrt(5)))/2) -
Floor((1+Sqrt(5))*Floor(2*k/(1+Sqrt(5)))/2): k in [1..100]]; // G. C. Greubel, Oct 23 2018

Maple

g:=k->floor((1+sqrt(5))/2*floor(k/((1+sqrt(5))/2))): seq(g(n+1)-g(n),n=1..110); # Muniru A Asiru, Oct 21 2018

Mathematica

#[[2]]-#[[1]]&/@Partition[Table[Floor[GoldenRatio*Floor[n/GoldenRatio]],{n,0,110}],2,1] (* Harvey P. Dale, Dec 14 2012 *)

PARI

{phi=(1+sqrt(5))/2; g(k)=floor(phi*floor(k/phi))};
vector(100, n, g(n+1)-g(n)) \\ G. C. Greubel, Oct 23 2018

Python

from math import isqrt
def A120614(n): return ((m:=(n+1+isqrt(5*(n+1)**2)>>1)-n-1)+isqrt(5*m**2)>>1)-((k:=(n+isqrt(5*n**2)>>1)-n)+isqrt(5*k**2)>>1) # Chai Wah Wu, May 22 2025

Formula

a(floor(k*phi)+k+1)=0; a(floor(k*phi)+k+2)=2, if n is not in {floor(k*phi)+k+1}U{floor(k*phi)+k+2}_{k>=1} a(n)=1.

(a(n)) is a fixed point of the morphism 02-->10202 and 102-->10210202. [Corrected by Michel Dekking, Oct 29 2018]

Fixed point of the morphism 0->102, 1->102, 2->02. - Michel Dekking, Oct 21 2018

Extensions

Initial 0 removed from data by Michel Dekking, Oct 22 2018