cp's OEIS Frontend

A226162 a(n) = Kronecker Symbol (-5/n), n >= 0.

%I A226162 #36 Aug 07 2025 08:36:38
%S A226162 0,1,-1,1,1,0,-1,1,-1,1,0,-1,1,-1,-1,0,1,-1,-1,-1,0,1,1,1,-1,0,1,1,1,
%T A226162 1,0,-1,-1,-1,1,0,1,-1,1,-1,0,1,-1,1,-1,0,-1,1,1,1,0,-1,-1,-1,-1,0,-1,
%U A226162 -1,-1,-1,0,1,1,1,1,0,1,1,-1,1,0,-1,-1,-1,1,0,-1,-1,1,-1,0,1,-1,1,1,0,-1,1,1,1
%N A226162 a(n) = Kronecker Symbol (-5/n), n >= 0.
%C A226162 The number of -1's among the four terms following the 0 at a(5*k), for k >= 0, is 1, 2, 3, 3, 1, 0, 3, 2, 2, 1, 4, 4, 0, 1, 3, 3, 1, 1, 3, 4, ...
%C A226162 See the Weisstein link, where it is stated that the period length is 0.
%C A226162 In general, the sequence {(k/n)} is not periodic if and only if k == 3 (mod 4). - _Jianing Song_, Dec 30 2018
%H A226162 Vincenzo Librandi, <a href="/A226162/b226162.txt">Table of n, a(n) for n = 0..1000</a>
%H A226162 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 A226162 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/KroneckerSymbol.html">Kronecker Symbol</a> (contains this sequence).
%H A226162 Wikipedia, <a href="https://en.wikipedia.org/wiki/Kronecker_symbol">Kronecker Symbol</a>
%F A226162 Completely multiplicative with a(2) = -1, a(5) = 0, a(p) = 1 if p == 1, 3, 7, 9 (mod 20), a(p) = -1 if p == 11, 13, 17, 19 (mod 20). - _Jianing Song_, Dec 30 2018
%p A226162 0, seq(numtheory:-jacobi(-5, n), n=1..89); # _Peter Luschny_, Dec 30 2018
%t A226162 Table[KroneckerSymbol[-5, n],{n,0,89}]
%o A226162 (PARI) a(n)=kronecker(-5,n); \\ _Andrew Howroyd_, Jul 23 2018
%Y A226162 Cf. A035183 (inverse Moebius transform).
%Y A226162 Sequences related to Kronecker symbols that do not form a Dirichlet character: this sequence {(-5/n)}, A034947 {(-1/n)}, A091338 {(3/n)}, A089509 {(7/n)}.
%Y A226162 Cf. A080891 (5/n), A100047.
%K A226162 sign,mult
%O A226162 0,1
%A A226162 _Wolfdieter Lang_, May 29 2013
%E A226162 Keyword:mult added by _Andrew Howroyd_, Jul 23 2018