A359421 a(n) = number of abelian groups of order p^2 - 1, where p = prime(n).

1, 3, 3, 5, 3, 3, 14, 6, 5, 3, 11, 6, 5, 3, 7, 9, 3, 3, 3, 10, 10, 7, 3, 10, 22, 6, 5, 9, 9, 7, 44, 3, 5, 3, 6, 10, 3, 15, 5, 3, 6, 6, 15, 15, 12, 20, 3, 11, 3, 3, 10, 7, 14, 18, 30, 5, 9, 21, 3, 5, 3, 6, 6, 5, 5, 3, 3, 14, 3, 6, 11, 10, 7, 3, 9, 22, 3, 6, 14

Offset: 1

Author

Ali Ramsey, Dec 31 2022

Keywords

• nonn

Examples

			For p = 5, p^2 - 1 = 24 = 2^3 * 3^1. The number of abelian groups of order 24 = (the number of partitions of 3)*(the number of partitions of 1) = 3*1 = 3.
		

Crossrefs

Subsequence of A000688.

Cf. A000040, A000041, A084920.

Programs

Maple

a:= n-> mul(combinat[numbpart](i[2]), i=ifactors(ithprime(n)^2-1)[2]):
seq(a(n), n=1..79);  # Alois P. Heinz, Dec 31 2022

Mathematica

A000688[n_] := Times @@ (PartitionsP /@ FactorInteger[n][[All, 2]]);
a[n_] := A000688[Prime[n]^2 - 1];
Table[a[n], {n, 1, 79}] (* Jean-François Alcover, Feb 03 2025 *)

Python

from sympy import factorint, npartitions
from math import prod
def A359421(n): return prod(npartitions(d) for d in factorint(prime(n)**2-1).values()) # Chai Wah Wu, Jan 12 2023

Formula

a(n) = A000688(prime(n)^2-1) = A000688(A000040(n)^2-1) = A000688(A084920(n)).