A188014 - OEIS

A188014

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

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, 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

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OFFSET

COMMENTS

Suppose r is a positive irrational number and k is a positive integer, so that the sequence given by a(n)=[nr]-[nr-kr]-[kr] consists of zeros and ones. Let b(n)=(position of n-th 0) and c(n)=(position of n-th 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=A096270, b=A026352, c=A026351

k=2: a=A188009, b=A188010, c=A101866

k=3: a=A188011, b=A188012, c=A188013

k=4: a=A188014, b=A188015, c=A188016

Example using r=sqrt(2):

k=1: a=A188037, b=A083088, c=A080754.

Returning to arbitrary positive irrational r, let s(n)=[nr]-[nr-r]-[r], this being a(n) when k=1. For k>=2, the sequence a(n)=[nr]-[nr-kr]-[kr] is a shifted sum of shifted copies of s:

a(n)=s(n)+s(n-1)+...+s(n-k+1)-(constant). This sum matches the sum s(n)+s(n+1)+...+s(n+k-1)-(constant) at A187950.

Ratio of number of zeros to number of ones tends to sqrt(5)/2=1.118. Any (simple) proof? [Zak Seidov, Feb 14 2012] According to Charles R Greathouse IV (see formula), a(n)=1 if the fractional part of nr is less than w(1)=4r-6,and 0 otherwise, where r=(1+sqrt(5))/2. It means that probability of ones and zeros is, respectively, w(1) and w(0)=1-w(1). In the long run, number of ones and zeros is, respectively, w(1)*n and w(0)*n and, finally, the ratio of number of zeros to number of ones tends to w(0)/w(1)=sqrt(5)/2, qed. [Zak Seidov, Feb 17 2012]

Essentially the same as A187950. - Michel Dekking, Sep 17 2016

LINKS

Ivan Panchenko, Table of n, a(n) for n = 1..1000

FORMULA

a(n) = [nr]-[nr-4r]-[4r], where r=(1+sqrt(5))/2.

a(n) = 1 if the fractional part of nr is less than 4r - 6, and 0 otherwise. [Charles R Greathouse IV, Feb 14 2012]

a(n+4) = A187950(n) for n>=1. - Michel Dekking, Sep 17 2016

EXAMPLE

We get a=A188014, b=A188015, c=A188016 when

r=(1+sqrt(5))/2 and k=4:

a......0..1..0..0..1..0..1..0..0..1..0..1..1...

b......1..3..4..6..8..9..11..14.. (positions of 0 in a)

c......2..5..7..10..12..13..15... (positions of 1 in a).

As noted in Comments, a(n)=[nr]-[nr-4r]-[4r] is also obtained in another way: by adding right shifts of the infinite Fibonacci word s=A096270 (after left-extending it) and then down shifting:

s(n)......0..1..0..1..1..0..1..0..1..1..0..1..1..

s(n-1)....1..0..1..0..1..1..0..1..0..1..1..0..1..

s(n-2)....1..1..0..1..0..1..1..0..1..0..1..1..0..

s(n-3)....0..1..1..0..1..0..1..1..0..1..0..1..1..

sum.......2..3..2..2..3..2..3..2..2..3..2..3..3..

sum-2.....0..1..0..0..1..0..1..0..0..1..0..1..1..

(Right shifts s(n-k) of the infinite Fibonacci word are obtained from s(n)=[nr]-[nr-nk]-[nk] rather than the morphism 0->01, 1->011 at A096270. For an analogous example using left shifts, see A187950.)

MATHEMATICA

r=(1+5^(1/2))/2; k=4;

t=Table[Floor[n*r]-Floor[(n-k)*r]-Floor[k*r], {n, 1, 220}] (*A188014*)

Flatten[Position[t, 0]] (*A188015*)

Flatten[Position[t, 1]] (*A188016*)

PROG

(PARI) a(n)=my(r=(1+sqrt(5))/2); n*r-n*r\1<4*r-6 \\ Charles R Greathouse IV, Feb 14 2012

(Python)

from __future__ import division

from gmpy2 import isqrt

def A188014(n):

return int((isqrt(5*n**2)+n)//2 -(isqrt(5*(n-4)**2)+n)//2 - 4) if n > 3 else 1-(n % 2) # Chai Wah Wu, Oct 07 2016