A293483 The number of 6th powers in the multiplicative group modulo n.

1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 5, 1, 2, 1, 2, 2, 8, 1, 3, 2, 1, 5, 11, 1, 10, 2, 3, 1, 14, 2, 5, 4, 5, 8, 2, 1, 6, 3, 2, 2, 20, 1, 7, 5, 2, 11, 23, 2, 7, 10, 8, 2, 26, 3, 10, 1, 3, 14, 29, 2, 10, 5, 1, 8, 4, 5, 11, 8, 11, 2, 35, 1, 12, 6, 10, 3, 5, 2, 13, 4, 9, 20

Offset: 1

Author

R. J. Mathar, Oct 10 2017

Keywords

• nonn
• mult

Comments

The size of the set of numbers j^6 mod n, gcd(j,n)=1, 1 <= j <= n.

A000010(n) / a(n) is another multiplicative integer sequence.

Links

• R. J. Mathar, Table of n, a(n) for n = 1..10132
• Richard J. Mathar, Size of the Set of Residues of Integer Powers of Fixed Exponent, 2017.

Crossrefs

The number of k-th powers in the multiplicative group modulo n: A046073 (k=2), A087692 (k=3), A250207 (k=4), A293482 (k=5), this sequence (k=6), A293484 (k=7), A293485 (k=8).

Cf. A052275, A319100, A000010.

Programs

Maple

A293483 := proc(n)
    local r,j;
    r := {} ;
    for j from 1 to n do
        if igcd(j,n)= 1 then
            r := r union { modp(j &^ 6,n) } ;
        end if;
    end do:
    nops(r) ;
end proc:
seq(A293483(n),n=1..120) ;

Mathematica

a[n_] := EulerPhi[n]/Count[Range[0, n - 1]^6 - 1, k_ /; Divisible[k, n]];
Array[a, 100] (* Jean-François Alcover, May 24 2023 *)
f[p_, e_] := (p - 1)*p^(e - 1)/If[Mod[p, 6] == 1, 6, 2]; f[2, e_] := If[e <= 3, 1, 2^(e - 3)]; f[3, e_] := If[e <= 2, 1, 3^(e - 2)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 10 2023 *)

Formula

Conjecture: a(2^e) = 1 for e <= 3; a(2^e) = 2^(e-3) for e >= 3; a(3^e) = 1 for e <= 2; a(3^e) = 3^(e-2) for e >= 2; a(p^e) = (p-1)*p^(e-1)/2 for p == 5 (mod 6); a(p^e) = (p-1)*p^(e-1)/6 for p == 1 (mod 6). - R. J. Mathar, Oct 13 2017

a(n) = A000010(n)/A319100(n). This implies that the conjecture above is true. - Jianing Song, Nov 10 2019