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A091338
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a(n) = (3/n), where (k/n) is the Kronecker symbol.
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8
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1, -1, 0, 1, -1, 0, -1, -1, 0, 1, 1, 0, 1, 1, 0, 1, -1, 0, -1, -1, 0, -1, 1, 0, 1, -1, 0, -1, -1, 0, -1, -1, 0, 1, 1, 0, 1, 1, 0, 1, -1, 0, -1, 1, 0, -1, 1, 0, 1, -1, 0, 1, -1, 0, -1, 1, 0, 1, 1, 0, 1, 1, 0, 1, -1, 0, -1, -1, 0, -1, 1, 0, 1, -1, 0, -1, -1, 0, -1, -1, 0, 1, 1, 0, 1, 1, 0, -1, -1, 0, -1, 1, 0, -1, 1, 0, 1, -1, 0, 1, -1, 0
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OFFSET
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1,1
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COMMENTS
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a(2n+1) has period 6, i.e., if n == 1 (mod 2) then a(n+12) = a(n). A.H.M. Smeets, Jan 23 2018
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LINKS
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FORMULA
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If n==0 (mod 3) a(n)=0; for p ==1 or 11 (mod 12) (i.e., p>3 in A038874), a(p)=+1; for p==2, 5 or 7 (mod 12) (i.e., p in A038875), a(p)=-1. - Benoit Cloitre, Jan 03 2004
Conjecture:
a(n) = 0 if and only if (n mod 3 = 0),
a(n) = 1 if (n mod 12 = 1 or n mod 12 = 11 or n mod 48 = 4 or n mod 48 = 44),
a(n) = -1 if (n mod 12 = 5 or n mod 12 = 7 or n mod 48 = 20 or n mod 48 = 28),
a(2) = -1, a(12*n+10) = -a(12*n+2) and a(12*n+14) = a(12*n+10) for n >= 0,
a(24*n+8) = -a(12*n+4) and a(24*n+16) = -a(12*n+4) for n >= 0. (End)
a(2*n+1) = 1 if and only if (n mod 6 = 0 or n mod 6 = 5),
a(2*n+1) = -1 if and only if (n mod 6 = 2 or n mod 6 = 3),
a(2*n+1) = 0 if and only if n mod 3 = 1,
a(2*n) = -a(n). (End)
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MAPLE
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numtheory[jacobi](3, n) ;
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MATHEMATICA
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PROG
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(PARI) a(n)=kronecker(3, n)
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CROSSREFS
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KEYWORD
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sign,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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