Timeline for The lower bound for prime gaps
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| May 12, 2023 at 18:19 | comment | added | GH from MO | @AndrejLeško As I said, $\frac{g_n}{\log g_n}\leq (2+o(1))\log n$ follows from $g_n>\left( \frac{p_{n+1}}{p_n}\right) ^{\frac{n}{2}}$. If you are not familiar with the $o(1)$ notation, it denotes an unspecified function tending to zero as $n\to\infty$. I don't think that $2+o(1)$ can be improved to $2$ here. | |
| May 12, 2023 at 15:24 | comment | added | Andrej Leško | ADITIONAL QUESTION: Can $\frac{g_n}{\log{g_n}}<2\log{n}$,$ n\geq 5$ be deduced from $g_n>\left( \frac{p_{n+1}}{p_n}\right) ^{\frac{n}{2}} , n\geq 2$? The oposite is obvious. | |
| May 12, 2023 at 13:37 | vote | accept | Andrej Leško | ||
| May 11, 2023 at 20:43 | history | edited | GH from MO | CC BY-SA 4.0 |
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| May 11, 2023 at 20:34 | history | edited | GH from MO | CC BY-SA 4.0 |
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| May 11, 2023 at 20:28 | history | answered | GH from MO | CC BY-SA 4.0 |