cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A293484 The number of 7th powers in the multiplicative group modulo n.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, 6, 18, 8, 12, 10, 22, 8, 20, 12, 18, 12, 4, 8, 30, 16, 20, 16, 24, 12, 36, 18, 24, 16, 40, 12, 6, 20, 24, 22, 46, 16, 6, 20, 32, 24, 52, 18, 40, 24, 36, 4, 58, 16, 60, 30, 36, 32, 48, 20, 66, 32, 44, 24, 10, 24, 72, 36, 40, 36
Offset: 1

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Author

R. J. Mathar, Oct 10 2017

Keywords

Comments

The size of the set of numbers j^7 mod n, gcd(j,n)=1, 1 <= j <= n.
A000010(n) / a(n) is another multiplicative integer sequence (size of the kernel of the isomorphism of the multiplicative group modulo n to the multiplicative group of 7th powers modulo n).

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), A293483 (k=6), this sequence (k=7), A293485 (k=8).

Programs

  • Maple
    A293484 := proc(n)
        local r,j;
        r := {} ;
        for j from 1 to n do
            if igcd(j,n)= 1 then
                r := r union { modp(j &^ 7,n) } ;
            end if;
        end do:
        nops(r) ;
    end proc:
    seq(A293484(n),n=1..120) ;
  • Mathematica
    a[n_] := EulerPhi[n]/Count[Range[0, n - 1]^7 - 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, 7] == 1, 7, 1]; f[2, e_] := 2^(e-1); f[7, 1] = 6; f[7, e_] := 6*7^(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 <= 1; a(2^e) = 2^(e-1) for e >= 1; a(7^e) = 6 for e=1; a(7^e) = 6*7^(e-2) for e >= 2; a(p^e) = (p-1)*p^(e-1) for p == {2,3,4,5,6} (mod 7); a(p^e) = (p-1)*p^(e-1)/7 for p == 1 (mod 7). - R. J. Mathar, Oct 13 2017
a(n) = A000010(n)/A319101(n). This implies that the conjecture above is true. - Jianing Song, Nov 10 2019