# A Comment on the Prime Number Theorem

(Note: I’m describing here a topic where I can’t personally follow the proofs. Think of this as being a second-hand account)

The prime number theorem states that asymptotically the number of primes less than $N$ is approximately $N / \log(N)$. It is fairly easy to prove that this is true up to a constant, that is, letting $\pi(x)$ denote the number of primes less than $x$, that $\frac{1}{C} \frac {N} {\log(N)} < \pi(N) < C \frac {N} {\log(N)}$

for some constant $C$. So the interesting thing is the constant factor.

When you think about it, it’s rather strange that the constant factor is exactly one. Consider a similar case: counting the number of twin primes less than a given $N$. The value is conjectured to be approximately $C N / \log(N)^2$, where $C = 2 \prod_{p\text{ prime}, p \geq 3} 1 - \frac{1} {(p-1)^2}$

This constant $C$ can be generated as follows: We want to determine the probability that $q, q+2$ are both prime for $q < N-2$. The probability for $q$ to be prime and for $q+2$ to be prime are both $\frac{1} {\log(N}$, so assuming they are independent the probability for both being prime is $\frac{1} {\log(N)^2}$. However, these probabilities are not independent. For example, assuming $q$ and $q+2$ are independent would lead to a probability of $\frac{1}{4}$ of them both
being odd, whereas the actual probability is $\frac{1}{2}$. Taking this into account, the probability estimate should be twice what it was before, up to $\frac{2} {\log{N}^2}$. Similarly, it is possible to calculate that the probability that a certain prime $p \geq 3$ divides neither $q$ nor $q+2$ is different by a factor of $1-1/q^2$ than it would under the indepence assumption. Multiplying all these factors out you get the constant $C$.

You would expect something similar to happen when counting primes: That there should be some constant factor that is some product over all primes, and that by default it would be a fairly random-looking number rather than exactly 1. The naive estimate $\pi (N) \approx N \prod_p (1-1/p) = 0$ fails. Still, it seems strange that the fact that each prime number is odd, which reduces the frequency of prime by a half, and that they are all 1 or 2 mod 3, which reduces by $\frac{2}{3}$, and so, all reduce to a constant factor of exactly 1.

The explanation is this: The frequency of the primes is self-adjusting. For instance, in the interval $[1, N]$ there are less primes than there are supposed to be, that would make any number in $[N, 2 N]$ more likely to be prime. Thus facts like how the specific number 2 is prime in the long run make no difference in the density of prime.