A079343 Period 6: repeat [0, 1, 1, 2, 3, 1]; also F(n) mod 4, where F(n) = A000045(n).

0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1

Offset: 0

Author

Jon Perry, Jan 04 2003

Keywords

• nonn
• easy

Comments

This sequence shows that every sixth Fibonacci number (A134492) is divisible by 4. - Alonso del Arte, Jul 27 2013

Examples

			a(5) = F(5) mod 4 = 5 mod 4 = 1.
a(6) = F(6) mod 4 = 8 mod 4 = 0.
a(7) = F(7) mod 4 = 13 mod 4 = 1.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Jon Maiga, Upper bound of Fibonacci entry points, (2019).
• Index entries for linear recurrences with constant coefficients, signature (1,-1,1,-1,1).

Crossrefs

Cf. A000045, A079344, A079345, A134492.

Programs

Magma

[Fibonacci(n) mod 4: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

Maple

A079343:=n->[0, 1, 1, 2, 3, 1][(n mod 6)+1]: seq(A079343(n), n=0..100); # Wesley Ivan Hurt, Jun 20 2016

Mathematica

PadLeft[{}, 108, {0, 1, 1, 2, 3, 1}] (* Harvey P. Dale, Aug 10 2011 *)
Table[Mod[Fibonacci[n], 4], {n, 0, 127}] (* Alonso del Arte, Jul 27 2013 *)
LinearRecurrence[{1, -1, 1, -1, 1},{0, 1, 1, 2, 3},105] (* Ray Chandler, Aug 27 2015 *)

PARI

for (n=0,100,print1(fibonacci(n)%4","))

Formula

a(n) = 2^(1 - P(3, n) + P(6, n+2))*3^P(6, n+3) - 1, where P(k, n) = floor(1/2*cos(2*n*Pi/k) + 1/2). [Gary Detlefs, May 16 2011]

a(n) = 4/3 - cos(Pi*n/3) - sin(Pi*n/3)/sqrt(3) - cos(2*Pi*n/3)/3 + sin(2*Pi*n/3)/sqrt(3). - R. J. Mathar, Oct 08 2011

G.f.: x*(1+2*x^2+x^3) / ( (1-x)*(1-x+x^2)*(1+x+x^2) ). - R. J. Mathar, Jul 14 2012

a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) for n>4. - Wesley Ivan Hurt, Jun 20 2016

E.g.f.: 2*(2*exp(x) - sqrt(3)*sin(sqrt(3)*x/2)*sinh(x/2) - cos(sqrt(3)*x/2)*(sinh(x/2) + 2*cosh(x/2)))/3. - Ilya Gutkovskiy, Jun 20 2016