Timeline for Sets whose elements are mutually "weakly" coprime?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 26, 2013 at 3:38 | vote | accept | Dustin G. Mixon | ||
| Dec 25, 2013 at 21:13 | answer | added | Lucia | timeline score: 6 | |
| Dec 25, 2013 at 20:39 | comment | added | Dustin G. Mixon | Not sure how rigorous this can be made, but it seems like when $n$ is large, $\nu_p$ of a random member of $\{1,\ldots,n\}$ nearly behaves like a geometric random variable with success probability $1-1/p$. As such, I would think that each member of $S$ will tend to have a large (i.e., uncommon) prime divisor, and the size of $S$ would be dictated by the collision probability of these large primes. | |
| Dec 25, 2013 at 10:50 | answer | added | Konstantinos Gaitanas | timeline score: 3 | |
| Dec 25, 2013 at 3:32 | comment | added | Greg Martin | The following is a sufficient condition for $S$ to have that property - one that could probably be probabilistically analyzed without too much difficulty. Let $\nu_p(s)$ denote the power of the prime $p$ dividing $s$. Then if $\sum_{s\in S} \nu_p(s) > 2\max_{s\in S} \nu_p(s)$ for every prime $p$, the set $S$ has the required property. | |
| Dec 25, 2013 at 3:16 | comment | added | Greg Martin | Probabilistic number theory seems like the right kind of math. I also observe that this seems to be an interesting question even for $k=1$. | |
| Dec 25, 2013 at 0:29 | history | edited | Dustin G. Mixon | CC BY-SA 3.0 |
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| Dec 24, 2013 at 23:21 | comment | added | The Masked Avenger | If S is the set of primes greater than n/2, you can get your property surely, if the y's chosen are at most n. | |
| Dec 24, 2013 at 22:58 | comment | added | Boris Bukh | Is S same as A? | |
| Dec 24, 2013 at 22:47 | history | asked | Dustin G. Mixon | CC BY-SA 3.0 |