A079343 - OEIS (original) (raw)

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

COMMENTS

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

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

EXAMPLE

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.

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 *)

PROG

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

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