Denote : $gnu(n)$ = number of groups of order $n$
It is much easier to decide whether a natural number $n$ is group-deficient ($gnu(n)<n$) , group-perfect ($gnu(n)=n$) or group-abundant ($gnu(n)>n$) than to actually calculate $gnu(n)$.
For example $gnu(2048$) is unknown, but it is known that $gnu(2048)>2048$, so $2048$ is group-abundant.
Squarefree numbers are always group-deficient ($gnu(n)<n$) and it seems that all numbers $n$ , for which there is no prime $p$ with $p^4|n$ are group-deficient.
I am not sure whether it is known for all prime powers $p^k$ , whether $p^k$ is group-deficient, group-perfect or group-abundant.
The cubefree numbers upto $50,000$ are group-deficient and the odd numbers upto $9,000$ except $3^7$ and $3^8$ are also group-deficient. (Perhaps, someone can doublecheck it ?)
There are no group-perfect numbers (except $1$) upto $2499$. (Again, can someone please doublecheck it ?)
Is there an efficient method for the decision problem (lower or upper bounds would do the job in most cases) ?
Is it true, that $m|n$ implies $gnu(m)\le gnu(n)$ ?
Is it true that $p<q$ implies $gnu(p^m)<gnu(q^m)$ for $m>4$?