We say that an positive integer $n$ is a
- simple number if there exist a non abelian simple group of order $n$. Denote by $\mathfrak{s}$ this set.
- prime-power number if it is of the form $n=p^a$, with $p$ prime. Denote by $\mathfrak{pp}$ this set.
- Sylow congruent number if it is a non prime-power number for which the congruence conditions of Sylow's theorem do not force at least one Sylow subgroup to be normal. In other words we can have $n_p\neq 1$ for all primes $p$ dividing $n$. Denote by $\mathfrak{sc}$ this set.
Also denote by $\mathfrak{p}$ the set of prime numbers.
Even though we know of the finite simple groups (I don't but some do), it seems unrealistic to expect an clean characterization of $\mathfrak{s}$.
For a set of positive integers $\mathfrak{t}$ we define $\pi_{\mathfrak{t}}$ the counting function of that set, i.e. $\pi_{\mathfrak{t}}(x)$ gives the number of elements in $\mathfrak{t}$ less than or equal to $x$, for any real number $x$.
It is well-known that $\pi_{\mathfrak{p}}(x)\sim\frac{x}{\ln x}$ (Prime number theorem).
Do we know the asymptotic laws of distribution for the other sets defined above, i.e. asymptotic behaviors of $\pi_{\mathfrak{pp}}(x)$, $\pi_{\mathfrak{s}}(x)$ and $\pi_{\mathfrak{sc}}(x)$, or $\pi_{\mathbb{N}\setminus \mathfrak{pp}}(x)$, $\pi_{\mathbb{N}\setminus \mathfrak{s}}(x)$ and $\pi_{\mathbb{N}\setminus \mathfrak{sc}}(x)$?
I suspect that we have $\pi_{\mathbb{N}\setminus \mathfrak{sc}}(x)\sim x$, or $\pi_{\mathbb{N}\setminus \mathfrak{sc}}(x)\sim k x$ for some $0