Number counting functions related to simple groups and asymptotic law of distribution 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 $0<k<1$ (based on some computation). At least if we restrict to odd numbers.
 A: Here is a simpler problem that should get you going: Instead of counting prime powers, let's just count prime squareds.
$$\pi_{\mathfrak{p}^2}(x) = \#\{ p^2 : p^2 < x, p \in \mathfrak{p} \} = \#\{ p^2 : p < \sqrt{x}, p \in \mathfrak{p} \} = \pi_{\mathfrak{p}}(\sqrt{x})$$
You can estimate this using the good ole PNT:
$$\pi_{\mathfrak{p}^2}(x) \sim \frac{2\sqrt{x}}{\ln(x)}$$

Now several of your sets are basically counting prime powers.  $\mathfrak{pp}$ for instance has a nice expression as:
$$\pi_{\mathfrak{pp}}(x) = \sum_{k=1}^{\log_2(x)} \pi_{\mathfrak{p}^k}(x) = \sum_{k=1}^{\log_2(x)} \pi_{\mathfrak{p}}(x^{(1/k)})$$
The asymptotics are the same:
$$\pi_{\mathfrak{pp}}(x) \sim \frac{x}{\ln(x)}$$
Similarly, I suspect there is no trouble with:
$$\pi_{\mathfrak{s}}(x) \sim \frac{3\sqrt[3]{x}}{\ln(x)}$$

Calculus fun:
$$\lim_{x\to\infty} \frac{ x/\ln(x) } { \sum_{k=1}^{\log_2(x)} kx^{(1/k)}/\ln(x) } = \lim_{x\to\infty} \frac{x}{\sum_{k=1}^{\log_2(x)} kx^{(1/k)}}$$
Clearly the operand of the limit is less than 1 for all $x > 1$, so we squeeze from below:
$$\begin{align} 1
&\geq \lim_{x\to\infty} \frac{x}{\sum_{k=1}^{\log_2(x)} k x^{(1/k)}} \\ \\
&\geq \lim_{x\to\infty} \frac{x}{x + \sqrt{x} \sum_{k=2}^{\log_2(x)} k} \\ \\
&\geq \lim_{x\to\infty} \frac{x}{x+\sqrt{x} (\log_2(x))^2} = 1
\end{align}$$
In the second line, I use that $\sqrt{x} \geq x^{(1/k)}$ for $k \geq 2$, and then in the third line I overestimate an arithmetic series.  In the last line, we divide top and bottom by x and the second term of the denominator goes to 0.
