

A187950


[nr+kr]  [nr]  [kr], where r = (1+sqrt(5))/2, k = 4, [.]=floor.


64



1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1
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OFFSET

1


COMMENTS

Suppose r is a positive irrational number and k is a positive integer, so that the sequence given by a(n) = [nr+kr]  [nr]  [kr] consists of zeros and ones (the fractional part of nr and kr being < 1). Let b(n) = (position of the nth 0) and c(n) = (position of the nth 1), so that b and c are a complementary pair of sequences.
Examples of a, b, c using r = (1+sqrt(5))/2:
k = 1: a = A005614 (infinite Fibonacci word),
b = A001950 (upper Wythoff sequence),
c = A000201 (lower Wythoff sequence);
Returning to arbitrary positive irrational r, let s(n) = [nr+r]  [nr]  [r], this being a(n) when k = 1. For k >= 2, the sequence a(n) = [nr+kr]  [nr]  [kr] is a shifted sum of shifted copies of s: a(n) = s(n) + s(n+1) + ... + s(n+k1)  (constant).
It would be more natural to start the sequence with offset n = 0.  A periodic pattern of length 21 seems to appear at n = 35 but it remains only up to n = 105.  M. F. Hasler, Oct 12 2017
This sequence is a morphic sequence, i.e., the lettertoletter projection lambda of the fixed point of a morphism.
The fixed point is A276757=1,2,3,4,5,1,2,3,1,2,3..., the fixed point of the 4block Fibonacci substitution on the alphabet {1,2,3,4,5} given by
1>12, 2>3, 3> 45, 4>12, 5>3.
The lettertoletter projection lambda is given by
1>1, 2>0, 3>1, 4>0, 5>0.
The reason that this works, is that the words of length 4 in the infinite Fibonacci word A003849 = 0100101001..., and their codings in the alphabet {1,2,3,4,5} are given by
0100 <> 1, 1001 <> 2, 0010 <> 3, 0101 <> 4, 1010 <> 5.
The difference sequence A014755 of the lower Wythoff sequence w = A000201, given by w(n) = [n*phi] is equal to the Fibonacci word on the alphabet {2,1} (modulo a minor offset problem). This gives that the difference between [(n+4)*phi] and [n*phi] is equal to the sum w(n)+w(n+1)+w(n+2)+w(n+3), which is 6 or 7. After subtracting [4r] = floor(4*phi) = 6, these sums are equal to a(n) = lambda(A276757(n)). (End)


LINKS



FORMULA

a(n) = floor(nr+4r)floor(nr)6, where r = (1+sqrt(5))/2.
a(n) = lambda(A276757(n)), where lambda(2) = lambda(4) = lambda(5) = 0, lambda(1) = lambda(3) = 1.  Michel Dekking, Apr 02 2020


EXAMPLE

a......1..0..1..1..0..1..0..1..1..0..1..1..0..1...
b......2..4..5..7..10..12.. (positions of 0 in a)
c......1..3..6..8..9..11... (positions of 1 in a).
As noted in Comments, a(n) = [nr+4r]  [nr]  [4r] is also obtained in another way: by adding left shifts of the infinite Fibonacci word s = A005614 and then down shifting:
s(n)......1..0..1..1..0..1..0..1..1..0..1..1..0..1...
s(n+1)....0..1..1..0..1..0..1..1..0..1..1..0..1..0...
s(n+2)....1..1..0..1..0..1..1..0..1..1..0..1..0..1...
s(n+3)....1..0..1..0..1..1..0..1..1..0..1..0..1..1...
sum.......3..2..3..2..2..3..2..3..3..2..3..2..2..3...
sum2.....1..0..1..0..0..1..0..1..1..0..1..0..0..1... [Corrected by M. F. Hasler, Oct 12 2017]


MATHEMATICA

r = (1+5^(1/2))/2;
A187950 = Table[Floor[(n+4)r]Floor[n*r]6, {n, 1, 220}]


PROG

(Python)
from __future__ import division
from gmpy2 import isqrt
return int((isqrt(5*(n+4)**2)+n)//2 (isqrt(5*n**2)+n)//2  4) # Chai Wah Wu, Oct 07 2016


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



