A319597 Number of conjugacy classes for a non-abelian group of order p^3, where p is prime: a(n) = p^2 + p - 1 where p = prime(n).

5, 11, 29, 55, 131, 181, 305, 379, 551, 869, 991, 1405, 1721, 1891, 2255, 2861, 3539, 3781, 4555, 5111, 5401, 6319, 6971, 8009, 9505, 10301, 10711, 11555, 11989, 12881, 16255, 17291, 18905, 19459, 22349, 22951, 24805, 26731, 28055, 30101, 32219, 32941, 36671

Offset: 1

Author

Juan Lanfranco, Sep 23 2018

Keywords

• nonn

Comments

For a non-abelian group of order p^3, we can use the class equation, p-group has nontrivial center result, group modulo center is cyclic implies group is abelian result, and the orbit-stabilizer theorem to give the number of conjugacy classes and number of elements in each conjugacy class.

The elements of A028387 with prime index.

Examples

			For p^3=2^3=8, the conjugacy classes of the Dihedral group = <r, s | r^4=1, s^2=1, srs=r^{-1}> are {1}, {r^2}, {r, r^3}, {s, sr^2}, {sr, sr^3}.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A028387, A065480, A065489.

Programs

Maple

A028387:= n -> n^2+n-1:
seq(A028387(ithprime(i)),i=1..50); # Robert Israel, Dec 23 2018

Mathematica

f[n_]:=n^2 + n - 1 ; f[Prime[Range[43]]] (* Amiram Eldar, Nov 21 2018 *)

PARI

a(n) = {my(p = prime(n)); p^2 + p - 1; } \\ Amiram Eldar, Nov 07 2022

Formula

From Amiram Eldar, Nov 07 2022: (Start)

a(n) = A028387(A000040(n)-1).

Product_{n>=1} (1 + 1/a(n)) = A065489.

Product_{n>=1} (1 - 1/a(n)) = A065480. (End)