Timeline for Divisibility of Stirling numbers
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2023 at 11:54 | history | edited | Martin Sleziak |
edited tags
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| Jan 4, 2023 at 21:48 | answer | added | Feldmann Denis | timeline score: 1 | |
| Jan 4, 2023 at 19:08 | history | became hot network question | |||
| Jan 4, 2023 at 18:56 | vote | accept | Feldmann Denis | ||
| Jan 4, 2023 at 18:56 | vote | accept | Feldmann Denis | ||
| Jan 4, 2023 at 18:56 | |||||
| Jan 4, 2023 at 17:04 | history | edited | Feldmann Denis | CC BY-SA 4.0 |
Sorry, stupid mistake in typing ; of course the result is trivially false for k=1
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| Jan 4, 2023 at 14:32 | answer | added | Peter Taylor | timeline score: 8 | |
| Jan 4, 2023 at 14:31 | comment | added | Richard Stanley | We have $x(x-1)\cdots (x-n+1)=\sum_{k=1}^n s_1(n,k)x^k$. If $s_1(n,k)$ is divisible by $n$ for all $1\leq k\leq n-1$, then $x(x-1)\cdots (x-n+1)=x^n+nf(x)$ for some polynomial $f(x)$ with integer coefficients. Putting $x=1$ gives a contradiction (namely, $1$ is divisible by $n$). | |
| Jan 4, 2023 at 11:06 | history | asked | Feldmann Denis | CC BY-SA 4.0 |