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

Author

Steven Finch, May 15 2009

Keywords

• mult
• nonn

Comments

Also called primitive Dirichlet characters of order 3.

Mobius transform of A060839.

C. David, J. Fearnley & H. Kisilevsky prove that Sum_{k=1..n} a(k) ~ C*n, with C = (11*sqrt(3)/(18*Pi)) * Product_{primes p == 1 (mod 3)} (1 - 2/(p*(p+1))) = 0.3170565167922841205670156...; they credit Cohen, F. Diaz y Diaz, & M. Olivier 2002 (see Proposition 5.2. and Corollary 5.3.). - Charles R Greathouse IV, Aug 26 2009 [corrected by Vaclav Kotesovec, Sep 16 2020]

a(n) is the number of primitive Dirichlet characters modulo n such that all entries are 0 or a cubic root of unity: 1, w = (-1 + sqrt(3)*i)/2 or w^2 = (-1 - sqrt(3)*i)/2. - Jianing Song, Feb 27 2019

Every term is 0 or a power of 2. - Jianing Song, Mar 02 2019

From Jianing Song, Apr 03 2021: (Start)

For n >= 2, a(n) is the number of cyclic cubic fields with discriminant n^2. See A343023 for detailed information.

The first occurrence of 2^t is 9*A121940(t-1) for t >= 2. (End)

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

Cf. A114643 (number of quadratic primitive Dirichlet characters modulo n), A160499 (number of quartic primitive Dirichlet characters modulo n).

Cf. A060839 (number of solutions to x^3 == 1 (mod n)).

Cf. A121940, A343023.

Programs

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 *)

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

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

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