TITLE: Is there a conjectured dependence on $n$ in van der Waerden's conjecture?
QUESTION [6 upvotes]: Bhargava 2021 proves van der Waerden's conjecture about Galois groups of random integer polynomials: over all $x^n + a_{n-1} x^{n-1} + \cdots + a_0 = 0$ with $a_k \in \{-H, \ldots, H\}$, the number of polynomials $E_n(H)$ with Galois group not equal to the full symmetric group $S_n$ is $O(H^{n-1})$.

First (and this is a very basic question), am I correct that the $O$ here is in terms of $H$ only, so that in detail we have $E_n(H) < c_n H^{n-1}$ for some constants $c_n$ that depend on $n$?

Second, is there a conjectured growth rate for $c_n$?


From Bhargava 2021 it seems like in the worst case $c_n$ might have a factor related to the number of nonisomorphic groups of order $n$ (possibly with other factors), but presumably most of the count concentrates on fewer of the groups.

REPLY [10 votes]: There are two different questions you could ask here.
One is the dependence on $n$ in the conjectured asymptotic best constant, i.e. the $c_n$ such that we have $E_n (H) < c_n H^{n-1} + o(H^{n-1})$ with the little $o$ depending on $n$. This one has a precise answer, because we do not need to worry about all the groups that are proven or expected to occur with frequency $o(H^{n-1})$. So we just need to deal with the reducible polynomials, where it was computed by Chela that we can take
$$ c_n = 2^n (\zeta(n-1) - 1/2) + 2 k_n $$
where $k_n$ is the volume of the region in $\mathbb R^{n-1}$ consisting of tuples $x_1,\dots, x_{n-1}$ with $$-1 \leq x_1,\dots, x_{n-1}, \sum_{i=1}^{n-1} x_i \leq 1 .$$
It is straightforward to check that this has growth rate proportional to $2^n$ in $n$.
The other is the dependence on $n$ of the conjectured best constant valid for all $H$. This is a somewhat annoying question because it will depend heavily on small values of $H$. For example, we must take $c_n \geq 3^{n-1}$ to deal with the case where $H$ is $1$ (and the constant coefficient vanishes). It's possible that we can take $c_n \leq 3^n$ - I don't see a reason against it.