It's well-known that for each prime number $p$ there are exactly two groups of order $p^2$, five of order $p^3$, and fifteen of order $p^4$ (at least when $p>3$).
I know that the classification of $p$-groups gets much harder for higher exponents. But I wonder if $p$-groups with higher dimensions are still structured enough that they would at least theoretically allow a classification.
In particular: Is there, for every natural $n$, a prime $p_0(n)$ and a number $m(n)$ such that for every $p \ge p_0(n)$ the number of groups of order $p^n$ equals $m(n)$?
(For example, $m(2)=2,m(3)=5,m(4)=15$ for $p_0(2)=p_0(3)=2$ $p_0(4)=5$.)
The answer is no. For $n \ge 5$ the number of isomorphism classes of groups of order $p^n$ increases with $p$.
The Higman PORC conjecture (polynomial on residue classes) is that, for large enough $p$, this number is a polynomial function of $p$, and of the value of $p$ modulo $n_p$ for some finite number of constants $n_p$. For example, for $n=5$ and $p \ge 5$, the number of groups is
$2p + 61 + 2 \gcd(p-1, 3) + \gcd(p-1, 4)$.
The PORC conjecture has been proved for $n=5,6,7$, but is open for $n=8$.
Recent work of Vaughan-Lee and du Sautoy suggests that the conjecture is likely to be false for large $n$. See users.ox.ac.uk/~vlee/PORC/porcsurvey.pdf