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A131743
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Period 4: repeat [0, 1, 0, 2].
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5
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0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0
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OFFSET
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0,4
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COMMENTS
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Least positive integer k such that n^k == 1 (mod 4), or 0 if GCD(n,4) > 1. - Bruno Berselli, Mar 22 2016
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LINKS
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Table of n, a(n) for n=0..104.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).
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FORMULA
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a(n) = (1/8)*(5*(n mod 4)-3*((n+1) mod 4)+3*((n+2) mod 4)-((n+3) mod 4)). - Paolo P. Lava, Oct 02 2007
G.f.: x*(1+2*x^2)/ ((1-x)*(x+1)*(x^2+1)). - R. J. Mathar, Nov 15 2007
a(n) = 3/4-1/2*sin(1/2*Pi*n)+3/4*(-1)^(1+n). - R. J. Mathar, Nov 15 2007
a(n) = (1-(-1)^n)*(3+I^(n+1))/4. - Paolo P. Lava, Jul 17 2008, Bruno Berselli, Mar 14 2011
a(n) = Fibonacci(2*n) mod 3. - Gary Detlefs Feb 13 2011
a(n) == A006368(n) (mod 3). - Philippe Deléham, Oct 24 2011
a(n) = a(n-4) for n>3. - Wesley Ivan Hurt, Jul 09 2016
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MAPLE
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seq(op([0, 1, 0, 2]), n=0..50); # Wesley Ivan Hurt, Jul 09 2016
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MATHEMATICA
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PadRight[{}, 106, {0, 1, 0, 2}] (* Harvey P. Dale, Apr 06 2012 *)
CoefficientList[Series[x (1 + 2 x^2)/((1 - x) (x + 1) (x^2 + 1)), {x, 0, 120}], x] (* Michael De Vlieger, Jul 09 2016 *)
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PROG
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(MAGMA) &cat [[0, 1, 0, 2]^^30]; // Bruno Berselli, Mar 22 2016
(PARI) x='x+O('x^200); concat(0, Vec(-x*(1+2*x^2)/((x-1)*(x+1)*(x^2+1)))) \\ Altug Alkan, Mar 22 2016
(PARI) a(n)=n%2*(n%4+1)/2 \\ Charles R Greathouse IV, Mar 22 2016
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CROSSREFS
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Cf. A006368.
Sequence in context: A194591 A070105 A111397 * A147648 A113686 A193403
Adjacent sequences: A131740 A131741 A131742 * A131744 A131745 A131746
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KEYWORD
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nonn,easy
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AUTHOR
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Paul Curtz, Sep 20 2007
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STATUS
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approved
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