Number of groups of squarefree order

This post is a sort of footnote to the previous post, Estimating the number of groups of a given order.

The following is taken from an answer to a question on Stack Exchange.

In general there is no formula f(n) for the number of groups of order up to isomorphism. However, if n is squarefree (no prime to the power 2 divides n), then Otto Hölder in 1893 proved the following amazing formula

f(n) = \sum_{m \mid n} \prod_p \frac{p^{c(p)} - 1}{p-1}

where p is a prime divisor of n/m and c(p) is the number of prime divisors q of m that satisfy q = 1 (mod p).

In a sense that can be made precise, about 60% of integers are square free [1]. So Hölder’s formula is often useful, but there are a lot of values it can’t compute.

In this post I develop Python code to implement Hölder’s formula. I’ve tested the code below by comparing it to a table of f(n) for n = 1, 2, 3, …, 100.

from sympy import mobius, divisors, factorint

def squarefree(n):
    return n > 1 and mobius(n) != 0

def prime_divisors(n):
    return list(factorint(n).keys())

def c(p, m):
    return len( [q for q in prime_divisors(m) if q % p == 1] )

def holder(n):
    if not squarefree(n):
        print("Sorry. I can't help you.")
        return None

    s = 0
    for m in divisors(n):
        prod = 1
        for p in prime_divisors(n // m):
            prod *= (p**c(p, m) - 1) // (p - 1)
        s += prod
    return s

[1] The number of square-free integers between 1 and x is asymptotically equal to 6x/π² as x → ∞.