Timeline for Is it true that $\sum_{k=m}^n\frac{\sigma(k)}k\not\in\mathbb Z$ for all derangements $\sigma\in S_n$ and $1\le m\le n$?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2021 at 12:19 | vote | accept | Zhi-Wei Sun | ||
| Feb 28, 2021 at 12:21 | |||||
| Nov 27, 2018 at 14:13 | answer | added | WhatsUp | timeline score: 13 | |
| Nov 27, 2018 at 13:15 | comment | added | Nathaniel Johnston | I think that this approach based on Bertrand's postulate shows that your conjecture is true as long as the interval contains a prime (then $\sigma$ does not fix the largest prime $p$ in the interval so you can split the sum into one with $p$ in the denominator and one without). | |
| Nov 27, 2018 at 13:09 | comment | added | WhatsUp | If $m$ is smaller than the last prime before $n$, then this prime appears in the denominator. | |
| Nov 27, 2018 at 12:35 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |