Here is a bad heuristic argument for the prime number theorem. Let $n$ be a positive integer and assume that PNT holds up to $n$. Then $n$ itself is prime if and only if for each prime $p<n$ the event $p\mid n$ fails to hold. Assuming that all these events are independent, the probability that n is prime should be around $\prod_{p<n}(1-1/p)$. Minus the logarithm of this is approximately $\sum_{p<n}1/p$. Since the $r$th prime is about $r\log r$ (by PNT), this is about $\sum_{r<n/\log n}1/r\log r$, which is about $\log\log n$. Therefore, the probability that $n$ is prime is about $\exp(-\log\log n)$, which is $1/\log n$.

Here are a few criticisms of the above argument. The most glaring problem is that for the approximations to be valid, we need our estimates to be correct up to $o(1)$, and they are not. For instance, it is known that $\sum_{p<n}1/p$ is not $\log\log n+o(1)$, but rather $\log\log n+M$, where M is the Meissel–Mertens constant. We can break this failure down into two subfailures. The first is that minus the logarithm of $\prod_{p<n}(1-1/p)$ differs from $\sum_{p<n}1/p$ by a non-zero constant (plus $o(1)$). The second is that minus the logarithm of $\prod_{p<n}(1-1/p)$ differs from $\log\log n$ by $\gamma+o(1)$, where $\gamma$ is the Euler–Mascheroni constant.

The second problem is more devastating, since it shows that the independence assumption is seriously flawed. (Everything I have said, by the way, is a well known and often pointed out observation.) My question is whether, despite all these problems, some kind of heuristic argument like this can be salvaged. For instance, it's clear that if $p$ and $q$ are two primes that are fairly large and fairly close, then there will be a repulsion between the events $p\mid n$ and $q\mid n$. Can we say heuristically what the effects of these repulsions should be, and thereby understand where the $\gamma$ comes in?

Just to be clear, I am looking for a simple and non-rigorous argument that does not use the zeta function (except perhaps making use of the product formula in a very elementary way, but I'd rather avoid it altogether) that predicts that if PNT holds up to $n$ then the probability that $n$ is prime should be around $1/\log n$. I'm asking the question because I'm pretty sure the answer will be known, and pretty standard, to many people. It just isn't to me.

  • 3
    $\begingroup$ Possibly related, T. Tao discussed this on his blog a while ago. terrytao.wordpress.com/2009/09/24/… $\endgroup$ Feb 26, 2010 at 11:18
  • $\begingroup$ That may indeed be exactly what I am looking for. I'll have to digest it carefully to see. $\endgroup$
    – gowers
    Feb 26, 2010 at 12:35

3 Answers 3


Here is a more precise heuristic argument for the probability that $n$ is prime. Let $f(k, n)$ represent our subjective probability that $n$ is $k$-rough, i.e. has no divisors smaller than $k$. We are interested in $f(n, n)$.

We have: $$f(k+1, n) = f(k, n) (1 - \mathbb{P}(\text{k is prime}) \mathbb{P}(\text{$k|n$ given that $k$ is prime and $n$ is $k$-rough}))).$$

From here on out we’ll assume $k$ is prime and write $p$ instead. Your “bad heuristic argument” treats $p|n$ and “$n$ is $p$-rough” as independent. But as you point out they aren’t independent and we can get a better estimate by taking the correlation into account.

To do this we apply Bayes’ rule: $$\mathbb{P}(\text{$p|n$ given $n$ is $p$-rough}) = \mathbb{P}(p|n)\frac{\mathbb{P}(\text{$n$ is $p$-rough given $p|n$})}{\mathbb{P}(\text{$n$ is $p$-rough})}$$

Note that if $p|n$, then $n$ is $p$-rough if and only if $n/p$ is $p$-rough. So $\mathbb{P}(\text{$n$ is $p$-rough given $p|n$})$ is just $f(p, n/p)$. The denominator $\mathbb{P}(\text{$n$ is $p$-rough})$ is $f(p, n)$. Thus: $$\mathbb{P}(\text{$p|n$ given $n$ is $p$-rough}) = \frac 1p \frac{f(p, n/p)}{f(p, n)}.$$ This differs from your naive estimate by a factor of $f(p, n/p) / f(p, n)$, reflecting the correlation between different divisibility events. The correction factor is is 1 if $p$ is much smaller than $n$, indicating approximate independence. It’s 0 if $p > \sqrt{n}$, reflecting the common crude approximation where we stop looking for prime factors at $\sqrt{n}$. But you get a softer interpolation between those regimes.

Including this correction factor, we get the recurrence $$f(k+1, n) - f(k, n) = - \mathbb{P}(\text{$k$ is prime}) \frac {f(k, n/k)}{k}.$$

If we don’t assume the prime number theorem is true up to $n$, and just substitute $\mathbb{P}(\text{$k$ is prime}) = f(k, k)$, then it turns out that any solution of this equation satisfies $f(k, k) = (1+o(1))/\log(k)$.

But that doesn’t really answer your question or show that we’ve fixed the bogus independence assumption. The reason it works out is that the resulting differential equation is self-correcting---if the density of primes is more than $1/\log(k)$ for a while then it exerts downwards pressure on $\mathbb{P}(\text{$n$ is prime})$ and vice versa. So even if we make some bogus independence assumptions leading to $O(1)$ multiplicative error, we will still get the correct $\frac {1 + o(1)}{\log n}$ asymptotics.

To answer your question, and show that we’ve actually fixed the problem with the independence assumption in the naive heuristic estimate, let’s instead use the correct distribution for the primes up to $n$ and show that our recurrence gives the correct estimate for $\mathbb{P}(\text{$n$ is prime})$. That is, let’s assume that for $k < n$ we have:

  • $\mathbb{P}(\text{$k$ is prime}) = 1/\log(k).$
  • $\prod_{p < k}(1 - 1/p) = \exp(-\gamma)/\log(k).$

We can exactly work out $f(k, n)$ for small values of $k$. In this regime, the events $p|n$ are exactly independent, and so $f(k, n) = \prod_{p < k}\left(1 - 1/p\right) = \exp(-\gamma)/\log(k)$.

For large values of $k$ we can approximate $f(k+1, n) - f(k, n) \approx \frac {d}{dk} f(k, n)$.

So now we have a differential equation: $\frac {d}{dk} f(k, n) = -\frac{f(k, n/k)}{k \log k}$, with boundary condition $f(k, n) = \frac{\exp(-\gamma)}{\log k}$ for $ k \ll n$.

I claim the solution is exactly $$f(k, n) = \frac {\omega\left(\frac {\log n}{\log k}\right)}{\log k}$$for $ k \leq n$, and $0$ for $k > n$, where $\omega$ is Buchstab's function. Plugging in $k = n$, and using $\omega(1) = 1$, we get the desired $\mathbb{P}(\text{$n$ is prime}) = \frac 1{\log n}$.

To see that this is a solution:

  • If $k \ll n$, then we have $f(k, n) = \frac {\exp(–\gamma)}{\log k}$, since $\omega(u) \rightarrow \exp(-\gamma)$ for large $u$.
  • If $k > \sqrt{n}$, then $f(k, n) = \frac {\omega\left(\frac {\log n}{\log k}\right)}{\log k} = \frac {1}{\log n}$, because $\omega(u) = \frac 1u$ for $u < 2$. Thus $\frac{d}{dk} f(k, n) = 0$. This is correct, because $k > k/n$ and hence $f(k, n/k) = 0$.
  • If $k$ is large but $k < \sqrt{n}$, then we have $\frac {d}{dk} f(k, n) = \frac {d}{dk} \frac{\omega\left(\frac {\log n}{\log k}\right) \frac {\log n}{\log k}}{\log n}$. Now we can use the fact that the Buchstab function $\omega$ is defined so that $\frac {d}{du} \omega(u) u = \omega(u - 1)$. Applying this with $u = \frac {\log n}{\log k}$, and using the fact that $\frac d{dk} \frac {\log n}{\log k} = -\frac {\log n}{k \log^2 k}$, we have: $$\frac {d}{dk} f(k, n) = -\frac {\omega\left(\frac{\log n}{\log k}-1\right)\frac {\log n}{k \log^2 k}}{\log n} = -\frac {\omega\left(\frac {\log n/k}{\log k}\right)}{k \log^2 k} = - \frac{f(k, n/k)}{k \log k},$$ as desired.

To summarize: if we take into account the correlation of divisibility events using Bayes’ rule in the most naive possible way, we end up with a correction term like $f(k, n/k) / f(k, n)$. Solving the resulting equation gives you the the correct heuristic estimate for the probability that $n$ is $k$-rough, and in particular the probability that $n$ is prime. The missing factor of $\exp(\gamma)$ you are looking for comes from the limit of the Buchstab function, which appears as the solution of the differential equation corresponding to this correction factor.

At first this felt to me like an unexplained coincidence or a bit of magical reverse causality. I've effectively just argued heuristically that for small $k$, $\mathbb{P}(\text{$n$ is prime}) = \frac {\exp(\gamma) \log k}{\log n}\prod_{p < k} (1 - 1/p)$. But why does $\prod_{p < k} (1 - 1/p)$ happen to have an exactly compensating factor of $\exp(-\gamma)$?

This comes directly from the self-correcting nature of the differential equation we mentioned before. As long as $\prod_{p < k} (1 - 1/p) > \exp(-\gamma)/\log(k)$, then $\mathbb{P}(\text{$n$ is prime}) > 1/\log(n)$ and so $\prod_{p < k}(1 - 1/p)$ will shrink faster than $1/\log(k)$. And similarly as long as $\prod_{p < k}(1 - 1/p)< \exp(-\gamma)/\log(k)$ then it will shrink slower than $1/\log(k)$. So I intuitively view the $\exp(\gamma)$ correction from the correlation as being the "logically prior" fact, with the $\exp(-\gamma)$ factor in $\prod_{p < k}(1 - 1/p)$ arising from the control system compensating for that correction.

  • 2
    $\begingroup$ I've been thinking about how to formalize this kind of probabilistic heuristic argument, see here: arxiv.org/abs/2211.06738. I ended up digging into this question because it appears to be one of the most famous examples of a heuristic argument giving a wrong answer without a clear explanation for why. I'm not currently aware of any unexplained examples, but I'm very interested in candidates. $\endgroup$ Nov 15, 2022 at 20:46

I'm not an expert in this area, but this may be a start.

Rather than $\prod_{p\lt n}$, you can use $\prod_{p\le \sqrt n}$.

$\log\log \sqrt n + \gamma \lt \log\log \sqrt n +\log 2 = \log\log n $

That gets you a little closer, since now you are off by $\log 2 - \gamma \approx 0.116$.

The heuristic probability that $n$ is prime is not

$$\prod_{p\lt n} (1-Pr(p|n))$$

It is the product of probabilities

$$\prod_{p\lt n} (1-Pr(p|n \text{ given no smaller prime divides } n))$$

For $p$ small, the term you get may be close to $(1-1/p)$, but I that's not the case for $p$ large.

For $\sqrt n \lt p$, the term corresponding to $p$ in the product is just $1$.

For $\sqrt[3]n \lt p \le \sqrt n$, if $p$ is the smallest prime dividing $n$, then $n/p$ must be prime, too. Perhaps that means that by strong induction, we should discount these terms by the probability $n/p$ is prime, about $1/\log \frac np$, so that those terms in the product are $(1-1/(p \log \frac np))$.

It looks like you get some sums/integrals if you try to extend this to more terms, and I don't know whether you can expect to get the desired accuracy at the end.

  • $\begingroup$ Gjergji, I think you dropped a factor of x here. $\endgroup$ Feb 26, 2010 at 14:33
  • $\begingroup$ Yes! And it's already mentioned in the answer, I don't know why I wrote it.. $\endgroup$ Feb 26, 2010 at 14:45

You might look at Courant and Robbins in the section at the end, "The Prime Number Theorem Obtained by Statistical Methods".

Also, there is an article by Montgomery and Wagon, in which they mention the "Mertens Paradox" and extend the argument of Courant and Robbins.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.