A307381 Number of sextic primitive Dirichlet characters modulo n.

1, 0, 1, 1, 1, 0, 5, 2, 4, 0, 1, 1, 5, 0, 1, 0, 1, 0, 5, 1, 5, 0, 1, 2, 0, 0, 0, 5, 1, 0, 5, 0, 1, 0, 5, 4, 5, 0, 5, 2, 1, 0, 5, 1, 4, 0, 1, 0, 0, 0, 1, 5, 1, 0, 1, 10, 5, 0, 1, 1, 5, 0, 20, 0, 5, 0, 5, 1, 1, 0, 1, 8, 5, 0, 0, 5, 5, 0, 5, 0, 0, 0, 1, 5, 1, 0, 1

Offset: 1

Author

Jianing Song, Apr 06 2019

Keywords

• nonn
• mult

Comments

a(n) is the number of primitive Dirichlet characters modulo n such that all entries are 0 or a six-power root of unity (1, (1 + sqrt(3)*i)/2, (-1 + sqrt(3)*i)/2, -1, (-1 - sqrt(3)*i)/2, (1 - sqrt(3)*i)/2).

Mobius transform of A319100.

Examples

			Let w = exp(2*Pi/6) = (1 + sqrt(3)*i)/2. For n = 19, the 5 sextic primitive Dirichlet characters modulo n are:
  Chi_1 = [0, 1, w, w, w - 1, -w, w - 1, 1, -1, w - 1, -w + 1, 1, -1, -w + 1, w, -w + 1, -w, -w, -1];
  Chi_2 = [0, 1, w - 1, w - 1, -w, w - 1, -w, 1, 1, -w, -w, 1, 1, -w, w - 1, -w, w - 1, w - 1, 1];
  Chi_3 = [0, 1, -1, -1, 1, 1, 1, 1, -1, 1, -1, 1, -1, -1, -1, -1, 1, 1, -1];
  Chi_4 = [0, 1, -w, -w, w - 1, -w, w - 1, 1, 1, w - 1, w - 1, 1, 1, w - 1, -w, w - 1, -w, -w, 1];
  Chi_5 = [0, 1, -w + 1, -w + 1, -w, w - 1, -w, 1, -1, -w, w, 1, -1, w, -w + 1, w, w - 1, w - 1, -1],
so a(19) = 5.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..65539

Crossrefs

Number of k-th power primitive Dirichlet characters modulo n: A114643 (k=2), A160498 (k=3), A160499 (k=4), A307380 (k=5), this sequence (k=6), A307382 (k=7), A329272 (k=8).

Cf. A319100 (number of solutions to x^6 == 1 (mod n)).

Programs

Mathematica

f[2, e_] := Which[e == 1, 0, e == 2, 1, e == 3, 2, e >= 4, 0]; f[3, e_] := Which[e == 1, 1, e == 2, 4, e >= 3, 0]; f[p_, 1] := If[Mod[p, 6] == 1, 5, 1]; f[p_, e_] := 0; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 16 2020 *)

PARI

a(n)={
    my(r=1, f=factor(n));
    for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]);
        if(p==2, if(e==3, r*=2, if(e!=2, r=0; return(r))));
        if(p==3, if(e==2, r*=4, if(e!=1, r=0; return(r))));
        if(p>3, if(p%6==1&&e==1, r*=5, if(e!=1, r=0; return(r))));
    );
    return(r);
} \\ Jianing Song, Nov 10 2019

Formula

Multiplicative with a(4) = 1, a(8) = 2, a(2^e) = 0 for e = 1 or e >= 4; a(3) = 1, a(9) = 4, a(3^e) = 0 for e >= 3; a(p) = 5 if p == 1 (mod 6) and 1 if p == 5 (mod 6), a(p^e) = 0 if p > 3 and e >= 2.