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A079343 Period 6: repeat (0,1,1,2,3,1); also F(n) mod 4, where F(n) = A000045(n). 12
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

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

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Index to sequences with linear recurrences with constant coefficients, signature (1,-1,1,-1,1).

FORMULA

a(n) = (1/90)*{23*[n mod 6]+38*[(n+1) mod 6] - 7*[(n+2) mod 6] - 7*[(n+3) mod 6] + 8*[(n+4) mod 6] - 7*[(n+5) mod 6]}, with n >= 0. - Paolo P. Lava, May 30 2007

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

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 *)

PROG

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

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

CROSSREFS

Cf. A079344, A079345.

Sequence in context: A124314 A059087 A030373 * A004566 A050074 A006705

Adjacent sequences:  A079340 A079341 A079342 * A079344 A079345 A079346

KEYWORD

nonn,easy

AUTHOR

Jon Perry, Jan 04 2003

STATUS

approved

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Last modified November 23 19:50 EST 2014. Contains 249865 sequences.