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:


$N_8(x) > N_k(x)$ 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?

  • $\begingroup$ what is $N_k$ exactly? $\endgroup$ – SK19 May 10 '18 at 10:51
  • $\begingroup$ @SK19 $N_k$ is the number of integers $\le x$ which have $k$ divisors. $\endgroup$ – Nilotpal Sinha May 10 '18 at 10:52
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    $\begingroup$ If I understand the notation correctly, then $N_3(x)=\pi(\sqrt x)$. $\endgroup$ – Barry Cipra May 10 '18 at 11:07
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    $\begingroup$ This mathoverflow question gives an estimate for the number of numbers less than $x$ with exactly $k$ distinct prime factors $$ \pi_k(x)\sim\frac{(\log\log x)^{k-1}}{(k-1)!} \cdot \frac{x}{\log x} $$ I imagine $N_{2^k}$ should be closely related to $\pi_k$. I'm not sure if that relationship is asymptotically just a constant multiple or not. $\endgroup$ – user14972 May 10 '18 at 11:14
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    $\begingroup$ There is nothing special about 8. Go higher, and eventually some other number will take over. $\endgroup$ – Ivan Neretin May 10 '18 at 15:16

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.


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