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An old paper by Ernest Cesàro provides a suggested approximation of the n'th prime. The expression and the reference currently appears in the Wikipedia article on the Prime Number Theorem.

It is inspired by an even earlier and difficult to locate article by one Monsieur Pervouchine aka Ivan Mikheevich Pervushin in a Russian Journal.

Cesàro points out that his own original approximation that appeared in a paper in Actes de l'Académie des Sciences de Naples in 1893: $$\frac{p_n}{n} = \log p_n - 1 - \frac{1}{\log p_n} - \frac{3}{(\log p_n)^2} - \cdots$$ (presumably with the implicit understanding that all subsequent terms add up to $o(1/(\log p_n)^2)$) implies Pervushin's approximation.

What I would like to know is whether there exists a modern or at least better accessible account of this type of approximation. And I am curious why it seems much easier to get the attempted asymptotics right numerically when I change the "3" to a "2"?

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    $\begingroup$ Please double-check the attributions. As I read it, the first, unnumbered, formula in Cesàro's paper, on p. 848, is due to Mr. Pervouchine, who worked three years on it and had recently published it in the Memoirs of the Kazan Physico-mathematical Society. Cesàro goes on to say that Pervouchine's formula can be deduced from a known formula of unspecified authorship, numbered (1) and reproduced above, which had been published in Actes de l'Académie des Sciences de Naples in 1893. $\endgroup$
    – njuffa
    Dec 3, 2023 at 2:34
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    $\begingroup$ The "recent paper" referenced by Cesàro is by Ivan Mikheevich Pervushin (1827-1900): "Les formules pour la détermination approximative des nombres premiers, de leur somme et de leur différence d’apres le numéro de ces nombres." Bulletin de la Société physico-mathématique de Kasan, 2nd Series, Vol. 4, 1894, pp. 94-96. This publication used Russian and French at the time. I have been unable to find a scan of volume 4 so far. $\endgroup$
    – njuffa
    Dec 3, 2023 at 3:20
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    $\begingroup$ The "known formula" is by Cesàro himself. It is equation (40) in Ernesto Cesàro, "Nuova contribuzione ai principii fondamentali dell'Aritmetica assintotica." Atti della Reale Accademia delle scienze fisiche e matematiche di Napoli, 2nd Series, Vol. 6, No. 11, 1894. $\endgroup$
    – njuffa
    Dec 3, 2023 at 4:23
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    $\begingroup$ A followup publication by (EN/FR/DE) Pervushin / Pervouchine / Perwuschin: J. Pervouchine (tr. A. Vassilief), "Formules pour la détermination approximative des nombres premiers, de leur somme et de leur différence d’après le numéro de ces nombres." In Ferdinand Rudio (ed.), Verhandlungen des ersten internationalen Mathematiker-Kongresses, Zurich, Aug. 9-11, 1897, p. 166-167 $\endgroup$
    – njuffa
    Dec 3, 2023 at 4:44
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    $\begingroup$ It seems strange to me to call that formula an approximation of $p_n$, when you have to know $p_n$ to evaluate the right side. $\endgroup$ Dec 3, 2023 at 22:34

2 Answers 2

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In this paper, it is shown that the prime counting function $\pi(x)$ satisfies $$ \pi(x)=\frac{x}{\log x-1-\frac{k_1}{\log x} - \frac{k_2}{\log^2 x} -\ldots-\frac{k_m}{\log^m x}+o\!\left(\frac{1}{\log^m x}\right) } $$ for any fixed $m\ge 1$ as $x\to+\infty$. Here $$ k_m+1!\cdot k_{m-1}+2!\cdot k_{m-2}+\ldots+(m-1)!\cdot k_1=m\cdot m! $$ for $m\ge 1$. In particular, $k_1=1$, $k_2=3$, $k_3=13$, $k_4=71$. Substituting $x=p_n$ and re-arranging the resulting equality yields $$ \frac{p_n}{n}=\log p_n-1-\frac{k_1}{\log p_n} - \frac{k_2}{\log^2 p_n} -\ldots-\frac{k_m}{\log^m p_n}+o\!\left(\frac{1}{\log^m p_n}\right) $$ for any fixed $m\ge 1$ as $n\to+\infty$. This is a rigorous interpretation of Cesàro's result.

Addendum. An explicit asymptotic expansion comes from the asymptotic inversion of the (offset) logarithmic integral $\operatorname{Li}(x)$: $$ \frac{p_n}{n} \sim \log n + \sum\limits_{k = 0}^\infty {\frac{{P_k (\log \log n)}}{{\log ^k n}}} $$ as $n\to+\infty$, where $P_0 (x) = x - 1$, $P_1 (x) = x - 2$, and \begin{align*} P_k (x) = kP_{k - 1} (x) & - P'_{k - 1} (x)\\ & + \frac{1}{k}\sum\limits_{m = 1}^{k - 1} {m((m - 1)P_{m - 1} (x) - P_m (x) - P'_{m - 1} (x))P_{k - m - 1} (x)} \end{align*} for $k\ge 2$. This result is originally due to Cipolla. The recurrence for the polynomials $P_k(x)$ was derived in this paper.

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    $\begingroup$ For added info, the sequence $1,3,13,71,461,3447,$ etc is A233824 or Panaitopol's formula for pi(x). And for $n>0$, is also sequence A003319 which is about connected permutations. Interesting that, except for the initial terms, they are the same. $\endgroup$ Dec 4, 2023 at 7:29
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    $\begingroup$ @Gary Thanks a lot! This appears to be what I am looking for. $\endgroup$ Dec 6, 2023 at 13:25
  • $\begingroup$ One should be careful, in that solving the last display for $p_n$ gives an asymptotic $p_n = n\log n + n\log\log n - n + O(\frac{n\log\log n}{\log n})$, with a perhaps unexpected $n\log\log n$ term. See Montgomery and Vaughan (Multiplicative Number Theory I), section 6.2, exercise 5. $\endgroup$ Dec 6, 2023 at 19:56
  • $\begingroup$ @TommyR.Jensen If this answers your question, consider accepting it. $\endgroup$
    – Gary
    Dec 6, 2023 at 22:28
  • $\begingroup$ @Gary I think that I still need some help to understand it. Let me describe what I am seeing: Ingham Thm 23 in " The Distribution..." states $\pi(x)=\mbox{li}(x) + O(xe^{-a\sqrt{\log x}})$ ($a > 0$). Vinogradov in "A new estimate..." states the improvement $\pi(x) = \int_2^x \frac{dx}{x} + O(xe^{-\alpha (\ln x)^{0.6}})$ ($\alpha > 0$). While Panaitopol cites Vinogradov for $\pi(x) = \mbox{Li}(x) + O(x \exp (-a \log x)^\alpha)$ ($a > 0$ and $0 < \alpha < 0.6$) and suggests to apply integration by parts to this last expression. The argument to Panaitopol's big-O is almost never a real number. $\endgroup$ Dec 7, 2023 at 12:48
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A more modern account (from 1962, though still widely cited) is Rosser and Schoenfeld's classic paper Approximate formulas for some functions of prime numbers. See Theorem 3 for example.

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  • $\begingroup$ Rosser and Schoenfeld, and include Dusart as well, seem to approach the question more from the point of view of obtaining deterministic bounds. Naturally such bounds are too weak to derive an asymptotic behavior. So I have not looked into this too much. $\endgroup$ Dec 6, 2023 at 13:22

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