Are there cube-free numbers $n$, for which the number of groups of order $n$ is unknown?

For squarefree $n$, there is a formula allowing to compute the number of groups of order $n$.

I do not think that such a formula exists for cubefree numbers. If a cubefree number $n$ has the property, that there is no prime power $p^k|n$, such that $p^k\equiv 1\ (\ mod\ q\ )$ for some prime $q|n$, then every group of order $n$ is abelian and $n$ is called an abelian number.

In OEIS, I found a slightly different definition of abelian numbers. Is the criterion I mentioned correct ?

The number of abelian groups of order $n$ can be easily calculated (assuming the prime factorization of $n$ is known). But what is the situation for general cubefree numbers $n$ ?

Is the cubefree case easy enough that the number of groups can be efficiently calculated, or are there cubefree numbers $n$ (of course with known factorization), for which the number of groups of order $n$ is unknown ?

• Yes the condition you wrote down for an abelian number is correct. It is both necessary and sufficient. But perhaps you should say that $k=1$ or $2$ (you don't want to allow $k=0$). – Derek Holt Dec 2 '15 at 13:35
• Is the number of groups of order $p^2q^2\ ,\ p<q$ bounded from above by a constant or depending on p ? – Peter Dec 2 '15 at 16:45
• It must depend on $p$. The number of groups of order $pq^2$ where $p|q-1$ increases with $p$. (It is just over $(p-1)/2$. – Derek Holt Dec 2 '15 at 18:54
• @Derek Holt how can we conclude that the number of groups of order $p^2q^2$ is at least the number of groups of order $pq^2$ ? – Peter Dec 2 '15 at 18:59
• If $m|n$ then the number of groups of order $n$ is at least as large as the number of order $m$. As you say, you can use just take direct products with $C_{n/m}$. The Krull-Schmidt Theorem ensures that $G \times C_k \cong H \times C_k \Rightarrow G \cong H$. – Derek Holt Dec 2 '15 at 23:05