%I #32 Aug 10 2023 02:21:20
%S 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,
%T 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,
%U 8,4,5,11,8,11,2,35,1,12,6,10,3,5,2,13,4,9,20
%N The number of 6th powers in the multiplicative group modulo n.
%C The size of the set of numbers j^6 mod n, gcd(j,n)=1, 1 <= j <= n.
%C A000010(n) / a(n) is another multiplicative integer sequence.
%H R. J. Mathar, <a href="/A293483/b293483.txt">Table of n, a(n) for n = 1..10132</a>
%H Richard J. Mathar, <a href="/A293482/a293482.pdf">Size of the Set of Residues of Integer Powers of Fixed Exponent</a>, 2017.
%F 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
%F a(n) = A000010(n)/A319100(n). This implies that the conjecture above is true. - _Jianing Song_, Nov 10 2019
%p A293483 := proc(n)
%p local r,j;
%p r := {} ;
%p for j from 1 to n do
%p if igcd(j,n)= 1 then
%p r := r union { modp(j &^ 6,n) } ;
%p end if;
%p end do:
%p nops(r) ;
%p end proc:
%p seq(A293483(n),n=1..120) ;
%t a[n_] := EulerPhi[n]/Count[Range[0, n - 1]^6 - 1, k_ /; Divisible[k, n]];
%t Array[a, 100] (* _Jean-François Alcover_, May 24 2023 *)
%t 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 *)
%Y 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).
%Y Cf. A052275, A319100, A000010.
%K nonn,mult
%O 1,5
%A _R. J. Mathar_, Oct 10 2017