A293484 The number of 7th powers in the multiplicative group modulo n.
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
Links
- R. J. Mathar, Table of n, a(n) for n = 1..10116
- Richard J. Mathar, Size of the Set of Residues of Integer Powers of Fixed Exponent, 2017.
Crossrefs
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
Comments