Density of nilpotent numbers A natural number $n$ is nilpotent if every group of order $n$ is nilpotent (equivalently, a direct product of Sylow subgroups). A natural number $n$ has nilpotent factorization if $\ell\not\equiv1$ mod $q$ for any two prime powers $\ell$ and $q$ dividing $n$. We know that a natural is nilpotent if and only if it has nilpotent factorization. Based on the folklore "almost all finite groups are $2$-groups" (which in particular means they are $p$-groups, which in particular means they are nilpotent), we might expect that nilpotent numbers have arithmetic density $1$, since otherwise nonnilpotent groups would have positive arithmetic density contrary to established belief. A little bit of data agrees with this idea.
Yet even if the "nilpotent number counting function" $N(x)$ satisfies $N(x)\sim x$, it wouldn't automatically mean "almost all finite groups are nilpotent," since we would also have to bound the number of ways that $p$-groups can be put together other than direct products for nonnilpotent $n$, and so we might reasonably expect $N(x)\sim x$ to be easier to prove than "almost all finite groups are nilpotent," and it seems like it would be amenable to pure number theory.
So, anybody have any ideas for establishing $N(x)\sim x$?
 A: The density of nilpotent numbers is at most $\frac{5}{6}$, since no number divisible by $6$ can be nilpotent. Actually I would be willing to bet that the density of nilpotent numbers is $0$. 
A: In a 1983 paper of M. Ram Murty and V. Kumar Murty it is shown that if 
$f(x)$ is any one of:
$\bullet$ $f_1(x)$: the number of $1 \leq n \leq x$ such that every group of order $n$ is cyclic
$\bullet$ $f_2(x)$: the number of $1 \leq n \leq x$ such that every group of order $n$ is abelian
$\bullet$ $f_3(x)$: the number of $1 \leq n \leq x$ such that every group of order $n$ is nilpotent,
then $\lim_{n \rightarrow \infty} \frac{f(x)}{x/\log \log \log x} = e^{-\gamma}$, where $\gamma$ is the Euler-Mascheroni constant.
The case of $f_1(x)$ is a 1948 result of Erdős.  The result for $f_3$ answers your question: the density of nilpotent numbers is zero...but not by much!
Added: These results were earlier derived by Michael E. Mays:

Mays, Michael E.
  Counting abelian, nilpotent, solvable, and supersolvable group orders.
  Arch. Math. (Basel) 31 (1978/79), no. 6, 536–538.

