Timeline for Estimate for the $2n$-th consecutive prime number
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 25, 2023 at 18:50 | vote | accept | Andrej Leško | ||
| Apr 25, 2023 at 18:30 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 2 characters in body; edited title
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| Apr 25, 2023 at 17:14 | history | edited | Pace Nielsen | CC BY-SA 4.0 |
fixed math mode logs
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| Apr 25, 2023 at 17:09 | answer | added | Pace Nielsen | timeline score: 3 | |
| Apr 25, 2023 at 16:57 | comment | added | kodlu | please use $\log n$ etc, your equations look terrible | |
| Apr 25, 2023 at 13:30 | comment | added | Claude Chaunier | I think @DaveBenson means we already know $p_{2n} \sim 2n\log(n) \sim 2n\log(p_n) < c 2n\log(p_n)$ for any $c>1$ and large enough $n$ depending on $c$. Something like what you stated as an introduction to your question. I don't see either how it could answer your question ($c=1$). | |
| Apr 25, 2023 at 12:26 | comment | added | Andrej Leško | Sorry, could you be more explicit?,because i dont see the line of the proof betwen the asymptotic estimate and explicit inequality. Thank you. | |
| Apr 25, 2023 at 10:23 | comment | added | Dave Benson | Since $p_n \sim n\log(n)$, your inequality is certainly true for large enough $n$. | |
| Apr 25, 2023 at 9:52 | history | asked | Andrej Leško | CC BY-SA 4.0 |