|
%I #31 Dec 01 2022 11:00:31
%S 1,8,3,2,5,24,7,8,3,40,11,6,13,56,15,2,17,24,19,10,21,88,23,24,5,104,
%T 3,14,29,120,31,8,33,136,35,6,37,152,39,40,41,168,43,22,15,184,47,6,7,
%U 40,51,26,53,24,55,56,57,232,59,30,61,248,21,2,65,264,67,34,69,280,71,24,73,296,15,38,77,312,79,10,3,328,83,42,85,344,87,88,89,120,91,46,93,376,95,24,97,56,33,10
%N Period of Kronecker symbol modulo n.
%C From _Jianing Song_, Nov 24 2018: (Start)
%C The sequence {Kronecker(k,n)} forms a Dirichlet character modulo n if and only if n !== 2 (mod 4).
%C Let n = 2^t*s, s odd, then a(n) = A117888(n) if and only if t is odd or s == 1 (mod 4) (or both); a(n) = A117889(n) if and only if t is odd or s == 3 (mod 4) (or both). (End)
%H Jianing Song, <a href="/A302138/b302138.txt">Table of n, a(n) for n = 1..10000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Kronecker_symbol">Kronecker symbol</a>.
%F Multiplicative with a(p^e) = p, p > 2; a(2^e) = 2 for even e and 8 for odd e.
%F a(n) = A007947(n) if A007814(n) is even, else 4*A007947(n).
%F Sum_{k=1..n} a(k) ~ c * n^2, where c = (49/50) * Product_{p prime} (1 - 1/(p*(p+1))) = (49/50) * A065463 = 0.690353... . - _Amiram Eldar_, Dec 01 2022
%e The Kronecker symbol modulo 2 is 1, 0, -1, 0, -1, 0, 1, 0 with period 8, so a(2) = 8.
%e The Kronecker symbol modulo 9 is 1, 1, 0 with period 3, so a(9) = 3.
%t Array[Apply[Times, FactorInteger[#] /. {p_, e_} /; p > 0 :> If[p == 2, 2 + 6 Boole[OddQ@ e], p]] &, 100] (* _Michael De Vlieger_, Nov 25 2018 *)
%o (PARI) a(n)={my(f=factor(n)); prod(i=1, #f~, if(f[i,1]==2 && f[i,2]%2, 8, f[i,1]))} \\ _Andrew Howroyd_, Apr 29 2018
%Y Cf. A007947, A065463.
%Y Cf. A117888 (period of Kronecker(n,k)), A117889 (period of Kronecker(-n,k)).
%K nonn,easy,mult
%O 1,2
%A _Jianing Song_, Apr 02 2018
|
