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does anyone knows if there are studies on the number of primes between prime $p_n$ and $p_{n}^2$, where $p_n$ is the $n$-th prime?

I am studying it through the following formula:

\begin{align} \pi(p^2_n)-\pi(p_n) \simeq (p_{n}^2-1) \prod_{i=1}^{n} \frac{(p_i-1)}{p_i}, \end{align}

where $p_n$ is the $n$-th prime number, $p_i$ the $i$-th primer number as well, and $\pi(n)$, for positive integer $n$, is the prime counting function.

I have seen a related question here:

The values of $n$ which satisfy an inequality about prime numbers

EDIT:

Since there are questions about the interest of this, let me explain why it is interesting for me:

All the integers up to $p_n^2$ have only prime factors in $\{2,3,\ldots,p_n\}$. So if we exclude the even numbers (i.e. $1/2$), then the multiples of $3$ (i.e. $2/3$), then the multiples of $5$ (i.e. $4/5$) and so on ... up to $p_n$ we should obtain exactly the prime numbers in the interval $(p_n,p_n^2)$. However it seems not to be as easy as it appears.

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  • $\begingroup$ The formula you wrote down does not make sense. What are $N$ and $n$ on the right hand side? $\endgroup$ Jul 14, 2023 at 9:59
  • $\begingroup$ Ups, yes, sorry my fault. $\endgroup$ Jul 14, 2023 at 10:06
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    $\begingroup$ I find it hard to really grasp the interest of this question. $\endgroup$
    – YCor
    Jul 14, 2023 at 13:37
  • $\begingroup$ I find it interesting since the integers in the interval $p_i$ up to $p_n$ are not divisible by any prime in this interval, so we know all its factors and I find quite intriguing why we can't then count them exactly. $\endgroup$ Jul 20, 2023 at 15:54

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There is nothing special about the difference $\pi(p_n^2) - \pi(p_n)$; we can take the input to be any large positive number. That is, it makes sense to consider

$$\displaystyle \pi(x^2) - \pi(x), x \gg 1.$$

By the prime number theorem we have

$$\displaystyle \pi(y) \sim \frac{y}{\log y},$$

and if we use the logarithmic integral we have the more precise form

$$\displaystyle \pi(y) = \int_2^y \frac{dt}{\log t} + O \left(y \exp \left( - A (\log y)^{\frac{3}{5}} \right) \right)$$

for some $A > 0$. Therefore, we see that

$$\displaystyle \pi(x^2) \sim \pi(x^2) - \pi(x)$$

for all $x$ sufficiently large. Assuming the Riemann hypothesis, one can expect that the fluctuations of

$$\displaystyle \pi(x^2) = \int_2^{x^2} \frac{dt}{\log t} + O_\varepsilon \left(x^{1 + \varepsilon} \right)$$

will overwhelm the main term of $\pi(x) \sim x (\log x)^{-1}$.

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  • $\begingroup$ Thanks @stanley, perhaps I missed the context. I find it interesting since the integers in the interval $p_i$ up to $p_n$ are not divisible by any prime in this interval, so we know all its divisors and I find quite intriguing why we can't then count them exactly. $\endgroup$ Jul 20, 2023 at 15:56
  • $\begingroup$ We can readily count them exactly! If you want you can even turn that into a 'closed form' using inclusion-exclusion, in the usual way of such things. But that formula isn't particularly useful. It sounds like what you're expecting is a 'constant length' formula ($O(1)$ or at worst $O(\log^k n)$ to evaluate, for some small $k$) for $\pi(p_n^2)-\pi(p_n)$ and the simple observations about size of divisors just turn out to not be much help with that. $\endgroup$ Jul 30, 2023 at 2:25

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