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A112030
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a(n) = (2 + (-1)^n) * (-1)^floor(n/2).
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12
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3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1
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OFFSET
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0,1
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COMMENTS
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The fractions A112031(n)/A112032(n) give the partial sums of a(n)/floor((n+4)/2).
Sum of the two Cartesian coordinates from the image of the point (2,1) after n 90-degree counterclockwise rotations about the origin. - Wesley Ivan Hurt, Jul 06 2013
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LINKS
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Table of n, a(n) for n=0..77.
Index entries for linear recurrences with constant coefficients, signature (0,-1).
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FORMULA
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a(n) = A010684(n+1) * (-1)^floor(n/2).
O.g.f.: (3+x)/(1+x^2). - R. J. Mathar, Jan 09 2008
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MAPLE
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A112030 := proc(n)
(2 + (-1)^n) * (-1)^floor(n/2) ;
end proc: # R. J. Mathar, Jul 09 2013
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MATHEMATICA
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LinearRecurrence[{0, -1}, {3, 1}, 100] (* Jean-François Alcover, Nov 24 2020 *)
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PROG
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(PARI) a(n)=[3, 1, -3, -1][n%4+1] \\ Charles R Greathouse IV, Aug 21 2011
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CROSSREFS
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Cf. A010684, A016116, A112031, A112032, A112033.
Sequence in context: A063062 A066056 A153284 * A010684 A176040 A125768
Adjacent sequences: A112027 A112028 A112029 * A112031 A112032 A112033
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KEYWORD
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sign,easy
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AUTHOR
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Reinhard Zumkeller, Aug 27 2005
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STATUS
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approved
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