A160498 Number of cubic primitive Dirichlet characters modulo n.
1, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 4, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0
Offset: 1
Examples
From _Jianing Song_, Mar 02 2019: (Start) Let w = (-1 + sqrt(3)*i)/2 be one of the primitive 3rd root of unity. For n = 7, the 2 cubic primitive Dirichlet characters modulo n are [0, 1, w, w^2, w^2, w, 1] and [0, 1, w^2, w, w, w^2, 1], so a(7) = 2. For n = 9, the 2 cubic primitive Dirichlet characters modulo n are [0, 1, w, 0, w^2, w^2, 0, w, 1] and [0, 1, w^2, 0, w, w, 0, w^2, 1], so a(9) = 2. (End)
Links
- Jianing Song, Table of n, a(n) for n = 1..10000
- C. David, J. Fearnley and H. Kisilevsky,On the vanishing of twisted L-functions of elliptic curves, Experim. Math. 13 (2004) 185-198.
- Steven R. Finch, Cubic and quartic characters.
- Steven R. Finch, Cubic and quartic characters.
- Steven R. Finch, Quartic and Octic Characters Modulo n, arXiv:0907.4894 [math.NT], 2016.
- Vaclav Kotesovec, Plot of Sum_{k=1..n} a(k) / n for n = 1..10000000
Crossrefs
Programs
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Mathematica
A060839[n_] := Sum[If[Mod[k^3 - 1, n] == 0, 1, 0], {k, 1, n}]; a[n_] := Sum[ MoebiusMu[n/d]*A060839[d], {d, Divisors[n]}]; Table[a[n], {n, 2, 81}] (* Jean-François Alcover, Jun 19 2013 *) f[3, 2] = 2; f[p_, e_] := If[Mod[p, 3] == 1 && e == 1, 2, 0]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 16 2020 *)
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PARI
a(n)=sum(d=1, n, if(n%d==0, moebius(n/d)*sum(i=1, d, if((i^3-1)%d, 0, 1)), 0)) \\ Steven Finch, Jun 09 2009
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PARI
A005088(n)=my(f=factor(n)[,1]); sum(i=1,#f,f[i]%3==1) A060839(n)=3^((n%9==0)+A005088(n)) a(n)=sumdiv(n,d,moebius(n/d)*A060839(d)) \\ Charles R Greathouse IV, Aug 26 2009
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PARI
a(n) = my(L=factor(n), w=omega(n)); for(i=1, w, if(!((L[i, 1]%3==1 && L[i, 2]==1) || L[i, 1]^L[i, 2] == 9), return(0))); 2^w \\ Jianing Song, Apr 03 2021
Formula
Multiplicative with a(p^e) = 2 if p^e = 9 or p == 1 (mod 3) and e = 1, otherwise 0. - Jianing Song, Mar 02 2019
a(n) = 2*A343023(n) for n >= 2. - Jianing Song, Apr 03 2021
Extensions
a(1) = 1 prepended by Jianing Song, Feb 27 2019
Comments