Is the expected number of primes in a specific interval $[p_n^2,p_{n+1}^2]$ approximately $p_n$?

Question

The expected (average?) number of primes in the interval $[p_n^2,p_{n+1}^2]$ is approximately $p_n$.

While thinking about a completely different problem, I noticed the above relationship, which I regard as very pretty. I suspect that the relationship is well known, although I have found no mention of it in a cursory search of references.

My question: I want to confirm that the reasoning that led me to this observation is valid.

My reasoning: For sufficiently large $p_n$, the expected gap $p_{n+1}-p_n \approx \ln{p_n}$. Hence, $p_{n+1} \approx p_n+\ln{p_n} \Rightarrow p_{n+1}^2 \approx p_n^2+(2\ln{p_n})\cdot p_n+(\ln{p_n})^2$. The interval between the two squares would have the size $p_{n+1}^2-p_n^2 \approx (2\ln{p_n})\cdot p_n$. The average difference between primes at numbers of the magnitude $p_n^2$ is just $\ln{p_n^2}=2\ln{p_n}$. Ergo, the expected number of primes in the interval is $\approx p_n$.

Of course, this is a very general statement, not a hard and fast rule. It would not apply, for example, when $p_n,p_{n+1}$ are twin primes, or otherwise unusually close together. By the same token, there would be occasional gaps between consecutive primes of unusually large size that would engender intervals between their primes containing a comparative excess of primes. I just want to know if the general relationship is a valid inference.

Answer 1

Saying that the average number of primes in the intervall $\left[ p_{n}^2, p_{n+ 1}^2 \right]$ is $p_n$ should mean that $$ \frac{1}{n}\sum\limits_{k = 1}^n {(\pi (p_{k + 1}^2 ) - \pi (p_k^2 ))} \approx p_n . $$ However, by the prime number theorem $$ \pi (x) = \frac{x}{{\log x}}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\log x}}} \right)} \right), $$ $$ p_{n + 1} ,p_n = n\log n\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right), $$ $$ \log p_{n + 1} ,\log p_n = \log n\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right), $$ whence \begin{align*} \frac{1}{n}\sum\limits_{k = 1}^n {(\pi (p_{k + 1}^2 ) - \pi (p_k^2 ))} & = \frac{1}{n}(\pi (p_{n + 1}^2 ) - \pi (4)) = \frac{{p_{n + 1}^2 }}{{2n\log p_{n + 1} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\log n}}} \right)} \right) \\ & = \frac{{p_n }}{2}\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right). \end{align*} Thus, the average is half of what you expected.

Addendum: For the different average, as a lower bound, we have \begin{align*} & \sum\limits_{k = 1}^n {\frac{{\pi (p_{k + 1}^2 ) - \pi (p_k^2 )}}{{p_k }}} = \frac{{\pi (p_{n + 1}^2 )}}{{p_n }} - \frac{{\pi (4)}}{2} - \sum\limits_{k = 1}^{n - 1} {\pi (p_{k + 1}^2 )\left( {\frac{1}{{p_{k + 1} }} - \frac{1}{{p_k }}} \right)} \\ & = \frac{{\pi (p_{n + 1}^2 )}}{{p_n }} - \frac{{\pi (4)}}{2} + \sum\limits_{k = 1}^{n - 1} {\frac{{\pi (p_{k + 1}^2 )}}{{p_{k + 1} p_k }}(p_{k + 1} - p_k )} \\ & = \frac{n}{2}\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right) + \sum\limits_{k = 1}^{n - 1} {\frac{{\pi (p_{k + 1}^2 )}}{{p_{k + 1} p_k }}(p_{k + 1} - p_k )} \\ & \ge \frac{n}{2}\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right) + \sum\limits_{k = 1}^{n - 1} {\frac{1}{{p_{k + 1} p_k}}\frac{{p^2_{k + 1} }}{{\log p_{k + 1}^2 }}(p_{k + 1} - p_k )} \\ & \ge \frac{n}{2}\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right) + \sum\limits_{k = 1}^{n - 1} {\frac{1}{{p_{k + 1} p_k }}\frac{{p_{k + 1} p_k }}{{\log p_{k + 1}^2 }}(p_{k + 1} - p_k )} \\ & = \frac{n}{2}\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right) + \frac{1}{2}\sum\limits_{k = 1}^{n - 1} {\frac{1}{{\log p_{k + 1} }}(p_{k + 1} - p_k )} \\ & \ge \frac{n}{2}\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right) + \frac{1}{2}\frac{1}{{\log p_n }}\sum\limits_{k = 1}^{n - 1} {(p_{k + 1} - p_k )} \\ & = n\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right). \end{align*} Here, I used summation by parts, the above asymptotics, and the lower bound $\pi(x)\geq x/\log x$ ($x\geq 1$). Thus $$ \mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\pi (p_{k + 1}^2 ) - \pi (p_k^2 )}}{{p_k }}} \ge 1. $$ We still need the corresponding upper bound to prove your claim.

Answer 2

We apply the Prime Number Theorem of the form $$ \pi(x)=\sum_{p\leq x} 1 \sim \frac x{\log x}, and $$ the $n$-th prime $p_n$ satisfies $p_n\sim n \log n$.

The average of prime counting function on the interval $(p_n, p_{n+1}]$ can be written as $$ \frac1{\pi(x)}\sum_{p\leq x} (\pi(p^2)-\pi((p')^2)) $$ where $p'$ is the previous prime to $p$.

This sum is a telescoping sum and hence the sum is in fact, $$ \frac1{\pi(x)}\pi(p_{\pi(x)}^2). $$ Applying Prime Number Theorem, the above is asymptotic to $$ \frac{\log x}x \frac{p_{\pi(x)}^2}{\log(p_{\pi(x)}^2)}\sim \frac{p_{\pi(x)}}{2}. $$ Hence, your average with $n=\pi(x)$ is $\sim p_n/2$.