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A079343
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Period 6: repeat (0,1,1,2,3,1); also F(n) mod 4, where F(n) = A000045(n).
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12
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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
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OFFSET
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0,4
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COMMENTS
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This sequence shows that every sixth Fibonacci number (A134492) is divisible by 4. - Alonso del Arte, Jul 27 2013
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LINKS
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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).
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FORMULA
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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
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(PARI) for (n=0, 100, print1(fibonacci(n)%4", "))
(MAGMA) [Fibonacci(n) mod 4: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014
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CROSSREFS
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Cf. A079344, A079345.
Sequence in context: A124314 A059087 A030373 * A004566 A050074 A006705
Adjacent sequences: A079340 A079341 A079342 * A079344 A079345 A079346
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KEYWORD
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nonn,easy
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AUTHOR
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Jon Perry, Jan 04 2003
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STATUS
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approved
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