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%I #47 Sep 08 2022 08:45:12
%S 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,
%T 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,
%U 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
%N a(n) = (3/n), where (k/n) is the Kronecker symbol.
%C 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
%H Vincenzo Librandi, <a href="/A091338/b091338.txt">Table of n, a(n) for n = 1..1000</a>
%H Jean-Paul Allouche, Leo Goldmakher, <a href="http://arxiv.org/abs/1608.03957">Mock characters and the Kronecker symbol</a>, arXiv:1608.03957 [math.NT], 2016.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/KroneckerSymbol.html">Kronecker Symbol</a>
%F 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
%F From _A.H.M. Smeets_, Aug 01 2018: (Start)
%F Conjecture:
%F a(n) = 0 if and only if (n mod 3 = 0),
%F a(n) = 1 if (n mod 12 = 1 or n mod 12 = 11 or n mod 48 = 4 or n mod 48 = 44),
%F a(n) = -1 if (n mod 12 = 5 or n mod 12 = 7 or n mod 48 = 20 or n mod 48 = 28),
%F a(2) = -1, a(12*n+10) = -a(12*n+2) and a(12*n+14) = a(12*n+10) for n >= 0,
%F a(24*n+8) = -a(12*n+4) and a(24*n+16) = -a(12*n+4) for n >= 0. (End)
%F From _A.H.M. Smeets_, Aug 01 2018: (Start)
%F a(2*n+1) = 1 if and only if (n mod 6 = 0 or n mod 6 = 5),
%F a(2*n+1) = -1 if and only if (n mod 6 = 2 or n mod 6 = 3),
%F a(2*n+1) = 0 if and only if n mod 3 = 1,
%F a(2*n) = -a(n). (End)
%p A091338 := proc(n)
%p numtheory[jacobi](3,n) ;
%p end proc: # _R. J. Mathar_, Nov 03 2011
%t Table[KroneckerSymbol[3, n], {n, 1, 100}] (* _Vincenzo Librandi_, Aug 16 2016 *)
%o (PARI) a(n)=kronecker(3,n)
%o (Magma) [KroneckerSymbol(3,n): n in [1..100]]; // _Vincenzo Librandi_, Aug 16 2016
%K sign,mult
%O 1,1
%A _Eric W. Weisstein_, Dec 30 2003
%E More terms from _Benoit Cloitre_, Jan 03 2004
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