If the prime factorisation of a number is $n = p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$ then the number of divisors of $n$ is $\tau_n = (a_1+1)(a_2+1)\ldots(a_k+1)$. Thus number of divisors of a number is 2 if an only if the number is a prime. In other words, the number of integers $\le x$ that have exactly two divisors is equal to $\pi(x)$ the number of primes $\le x$.
Let $N_k(x)$ be the number of integers $\le x$ which have $k$ divisors. Thus for $x = 2000000$, $N_8(x) = 448777$ and $N_4(x) = 407091$. Is there an asymptotic formula for the number integers $\le x$ which have exactly $k$ divisors?
I observed that 8 is the most common number of divisors. More specifically, we have:
Conjecture:
$N_8(x) > N_k(x)$ for all $x > 248770$ and $k \ne 8$.
or more elegantly:
Eight is the most common number of divisors of integers.
I have verified this conjecture for $x = 2*10^{10}$.
Question: Can this conjecture be proved/disproved or is there any heuristic arguments against or in support of it?
For a number to have exactly 8 divisors, it must be of the form $p^7$, or $p^3q$, or $pqr$, where $p,q,r$ are distinct primes. The first two types are much rarer in the long run, so the count is dominated by the numbers of the form $pqr$.
It's probably not hard to prove that numbers of the form $pqrs$, with $p,q,r,s$ distinct primes, eventually outnumber numbers of the form $pqr$, and when that happens (or soon thereafter), numbers with 16 divisors will take over from numbers with 8 divisors.