Timeline for Probability a polynomial $v(t)$ is divisible either by $1-t$ or by $1+t^{2^{j-1}}$, for some $j$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S May 23, 2014 at 13:23 | history | bounty ended | Simd | ||
| S May 23, 2014 at 13:23 | history | notice removed | Simd | ||
| May 16, 2014 at 16:02 | answer | added | Marco Golla | timeline score: 4 | |
| S May 16, 2014 at 12:37 | history | bounty started | Simd | ||
| S May 16, 2014 at 12:37 | history | notice added | Simd | Draw attention | |
| May 13, 2014 at 21:45 | comment | added | Igor Rivin | @MarcoGolla I think the divisibility by $1+t^k$ is actually equivalent to the sum of the $k$ sums of $k$-space coefficients being equal. I agree that this should not be hard. | |
| May 13, 2014 at 20:57 | comment | added | Marco Golla | This shouldn't be too hard: divisibility by $1-t$ is equivalent to the sum of the coefficients being 0, while divisibility by $1+t^k$ for some $k$ has the same probability of being divisible by $1-t^k$, and this is equivalent to the vanishing of the sum of $k$-spaced coefficients. One should get a good asymptotic control on these two conditions holding both separately and simultaneously. | |
| May 13, 2014 at 20:30 | review | First posts | |||
| May 13, 2014 at 20:30 | |||||
| May 13, 2014 at 20:28 | comment | added | Simd | @IgorRivin Upper and lower bounds that apply for large $n$ would be great. If I had to choose, I would pick the upper band though. | |
| May 13, 2014 at 20:24 | comment | added | Igor Rivin | What sort of estimate are you looking for (that is: upper bound, lower bound, both?) | |
| May 13, 2014 at 20:12 | history | asked | Simd | CC BY-SA 3.0 |