|
|
A085002
|
|
a(n) = floor(phi*n) - 2*floor(phi*n/2) where phi is the golden ratio.
|
|
7
|
|
|
1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
A fractal sequence.
Usually an integer sequence is called 'fractal' if it has the self-generating properties of a morphic sequence, i.e., the letter to letter projection of a fixed point of a morphism. Indeed, take the alphabet {1,2,...,8} and the morphism eta defined by
eta: 1->5, 2->7, 3->8, 4->6, 5->53, 6->71, 7->82, 8->64.
Then eta has fixed point
x = (5,3,8,6,4,7,1,6,8,2,5,71,6,4,7,5,3,...).
Let pi be the projection morphism
pi(1)=1, pi(2)=0, pi(3)=1, pi(4)=0, pi(5)=1, pi(6)=0, pi(7)=1, pi(8)=0.
Then pi(x) = (a(n)).
To prove this, one may use my paper "Iteration of maps by an automaton".
The two maps are phi_a and phi_b defined by
phi_a(0) = 1, phi_a(1) = 0, phi_b(0) = 0, phi_b(1) = 1.
The substitution is the Fibonacci substitution sigma given by
sigma(a) = b, sigma(b) = ba.
Since the first differences of the lower Wythoff sequence are given by the Fibonacci substitution 1->2, 2-> 21, it follows from replacing 1 with a and 2 with b that (a(n)) is generated by iterating the two maps phi_a and phi_b according to the fixed point babba...of sigma. The two maps phi_a and phi_b are commuting bijections on {a,b}, exactly as in the Example on page 85 of "Iteration of maps by an automaton". It follows as in that example that (a(n)) is generated by the projection of a fixed point of a substitution on an alphabet of 8 letters, and a simple computation as on that page yields the morphism eta. (End)
|
|
LINKS
|
|
|
FORMULA
|
Proof that this sequence is the parity sequence of the lower Wythoff sequence:
if n*phi/2 = M + e, with 0 < e < 1, then 2*floor(phi*n/2) = 2M, and
floor(phi*n) = floor(2M+2e) = 2M or 2M+1.
So floor(phi*n) - 2*floor(phi*n/2) = 0 if floor(phi*n) is even, and equals 1 if floor(phi*n) is odd. (End)
|
|
MATHEMATICA
|
Table[Floor[GoldenRatio n] - 2 Floor[GoldenRatio n/2], {n, 110}] (* Harvey P. Dale, Dec 11 2012 *)
|
|
PROG
|
(Python)
from math import isqrt
|
|
CROSSREFS
|
Characteristic function of A283766.
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|