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I'm looking for explicit upper bounds on the number of primes up to the square $m=p_n^2$ of the $n^\text{th}$ prime number.

Such estimates can rely on the knowledge of the exact number of primes up to $\sqrt{m}$, which is $n$. Since every composite number less than $m$ has a prime factor smaller than ${p_n}$, from the sieving perspective, the $n-1$ primes less than ${p_{n}}$ actually determine the primes between ${p_n}$ and $p_n^2$. Two very similar questions could be formulated as follows:

  1. knowing the exact number of primes less than or equal to a given natural number ${q}$, find an explicit upper bound on the number of primes up to ${q^2}$;

  2. knowing the exact number of primes less than a given natural number ${q}$, find an explicit upper bound on the number of primes in the interval $\mathopen]q,{q^2}\mathclose[$.

Edit: the goal would be, of course, to use the knowledge $\pi(q)=n$ to obtain sharper upper bound estimates on $\pi(q^2)$ than those obtainable without this information.

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    $\begingroup$ I don't think there is any better bound known for $\pi(p_n^2)$ than the usual approximation $\mathrm{li}(p_n^2)+O(\dots)$ coupled with a similar approximation for $p_n^2$ with the help of the prime number theorem. More generally, I don't think there is any simple relationship between $\pi(x)$ and $\pi(x^2)$, other than the usual approximations $\pi(x)=\mathrm{li}(x)+O(\dots)$ and $\pi(x^2)=\mathrm{li}(x^2)+O(\dots)$. The distribution of prime numbers is trickier than that. $\endgroup$
    – GH from MO
    Commented Jul 31 at 13:37
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    $\begingroup$ There are explicit versions of the prime number theorem, e.g. Trudgian (2016) proved that $ |\pi(x)-\mathrm{li}(x)|<x e^{-0.39\sqrt{\ln x}}$ for $x\geq 229$. You can use that to generate explicit bounds for $\pi(p_n^2)$ in terms of $n$. I don't think that it gets any better than that. $\endgroup$
    – GH from MO
    Commented Jul 31 at 14:33
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    $\begingroup$ To make myself clear, I don't think that the information $\pi(q)=n$ helps in any way to bound $\pi(q^2)$. $\endgroup$
    – GH from MO
    Commented Jul 31 at 15:04
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    $\begingroup$ Yes, I understood that is your opinion. And thank you for expressing it. $\endgroup$
    – Nautilus
    Commented Jul 31 at 15:08
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    $\begingroup$ Even if one can get some tiny improvement by comparing how $\pi(q)$ and $\pi(q^2)$ depend on the zeroes of the Riemann zeta function (the literal worst case configuration for one should not quite equal the worst case configuration for the other), this improvement will likely be smaller than the improvement you sacrifice by writing the bound in some readable fashion rather than some long and messy formula. and so wouldn't be useful. $\endgroup$
    – Will Sawin
    Commented Jul 31 at 15:20

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