In the family of finite groups of order less than $2000$, there are about 99% of order $1024$, so I have a question about $2$-groups:
Let $f(n)$ be the number of non-isomorphic finite groups of order $2^{\lfloor{\rm log}_2 n\rfloor}$, $g(n)$ be the number of non-isomorphic finite groups of order less than $n+1$.
Do we have $\lim\limits_{n\to\infty}f(n)/g(n)=1$ ?
The following is an empirical argument to show that the total number of (isomorphism classes of) groups of order less than $2^m$ is dwarfed by the number of order exactly $2^m$. The number of groups of order $2^m$ is $2^{\left(\frac{2}{27}+o(1)\right)m^3}$.
Pyber, "Enumerating finite groups of given order" (Annals 1993) has shown the following. If $n=\prod_{i=i}^k p_i^{g_i}$, let $\mu(n)$ be the maximum of the $g_i$. Then the number $f(n)$ of finite groups of order $n$ satisfies $f(n)\leqslant n^{\left(\frac{2}{27}+o(1)\right)\mu^2}$ as $\mu\to\infty$. This means that the only value of $n$ less than $2^m$ with anything approaching $f(2^m)$ is $n=3.2^{m-2}$. After that come $5.2^{n-3}$ and $7.2^{m-3}$, and so on.
Consider $f(n)$ for $2^{m-1}<n<2^m$. Then for just one value of $n$, namely $n=3.2^{m-2}$, we have $\mu(n)=m-2$. For two values of $n$ we have $\mu(n)=m-3$, ..., for $2^{i-2}$ values of $n$ we have $\mu(n)=m-i$. So for $m$ large, $\sum_{2^{m-1}<n<2^m}f(n)$ is bounded by $2^{\frac{2}{27}m(m-2)^2}+2.2^{\frac{2}{27}m(m-3)^2}+4.2^{\frac{2}{27}m(m-4)^2}+\cdots$ which is much less than $f(2^m)=2^{\left(\frac{2}{27}+o(1)\right)m^3}$. Inductively, if $\sum_{n<2^{m-1}}f(n)$ is much less than $f(2^{m-1})$ then adding in this and $f(2^{m-1})$ won't change this.
There are missing details in this argument, like taking care of the $o(1)$, but I think it should be capable of being made rigorous.
Edit: I suppose I should have looked at the sum of $f(n)$ up to $2^{m+1}-1$, but the argument is essentially the same.
