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A120614 a(n) = g(n+1) - g(n) where g(k) = floor(phi*floor(k/phi)) and phi = (1+sqrt(5))/2. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

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

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

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 *)

PROG

(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

(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

CROSSREFS

Cf. A003849, A120613, A120615, A270788.

Sequence in context: A194313 A127476 A140397 * A287516 A252055 A320836

Adjacent sequences:  A120611 A120612 A120613 * A120615 A120616 A120617

KEYWORD

nonn

AUTHOR

Benoit Cloitre, Jun 17 2006

EXTENSIONS

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

STATUS

approved

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Last modified November 15 13:07 EST 2018. Contains 317238 sequences. (Running on oeis4.)