A140081 Period 4: repeat [0, 1, 1, 2].

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

Offset: 0

Author

Nadia Heninger and N. J. A. Sloane, Jun 03 2008

Keywords

• nonn
• easy

Comments

Also fix e = 4; then a(n) = minimal Hamming distance between the binary representation of n and the binary representation of any multiple k*e (0 <= k <= n/e) which is a child of n.

A number m is a child of n if the binary representation of n has a 1 in every position where the binary representation of m has a 1.

Links

• Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Crossrefs

Cf. A140201. - Reinhard Zumkeller, Feb 21 2010

Programs

Haskell

a140081 n = div (mod n 4 + mod n 2) 2
a140081_list = cycle [0, 1, 1, 2]  -- Reinhard Zumkeller, Aug 15 2015

Maple

A140081:=n->floor((3*(n mod 4)+1)/4); seq(A140081(n), n=0..100); # Wesley Ivan Hurt, Mar 27 2014

Mathematica

PadLeft[{}, 100, {0,1,1,2}] (* Harvey P. Dale, Aug 19 2011 *)
Table[Floor[(3 Mod[n, 4] + 1)/4], {n, 0, 100}] (* Wesley Ivan Hurt, Mar 27 2014 *)

PARI

a(n)=n%4-n%4\2 \\ Jaume Oliver Lafont, Aug 28 2009

Formula

a(n) = 1 + a(n - 1 - a(n-1)) + 2*a(a(n-1)) - 2*a(n-1), a(0)=0. - Ramasamy Chandramouli, Jan 31 2010

a(n) = A047624(n+2) - A047624(n+1) - 1. - Reinhard Zumkeller, Feb 21 2010

a(n) = 1 - cos(Pi*n/2)/2 - sin(Pi*n/2)/2 - (-1)^n/2. - R. J. Mathar, Oct 08 2011

a(n) = ((n mod 4) + (n mod 2))/2. - Gary Detlefs, Apr 21 2012

From Colin Barker, Jan 13 2013: (Start)

a(n) = a(n-4).

G.f.: -x*(2*x^2+x+1) / ((x-1)*(x+1)*(x^2+1)). (End)

a(n) = floor((3*(n mod 4) + 1)/4). - Wesley Ivan Hurt, Mar 27 2014

From Wesley Ivan Hurt, Apr 22 2015: (Start)

a(n) = floor(1/2 + (n mod 4)/2).

a(n) = 1 - (-1)^n/2 - (-1)^(n/2 - 1/4 + (-1)^n/4)/2. (End)

a(n) = n - floor(n/2) - 2*floor(n/4). - Ridouane Oudra, Oct 30 2019