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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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9
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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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 (paoloplava(AT)gmail.com), 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). [From Gary Detlefs (gdetlefs(AT)aol.com), 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
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EXAMPLE
| a(6) = F(6) mod 4 = 8 mod 4 = 0.
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MATHEMATICA
| f[n_]:=Mod[Fibonacci[n], 4]; lst={}; Do[AppendTo[lst, f[n]], {n, 0, 6!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Feb 18 2010]
PadLeft[{}, 108, {0, 1, 1, 2, 3, 1}] (* From Harvey P. Dale, Aug 10 2011 *)
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PROG
| (PARI) for (n=0, 100, print1(fibonacci(n)%4", "))
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CROSSREFS
| 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
| nonn
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AUTHOR
| Jon Perry (perry(AT)globalnet.co.uk), Jan 04 2003
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