Timeline for Products of relative prime numbers with least sum
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 16, 2015 at 9:30 | vote | accept | J Fabian Meier | ||
| Jun 22, 2015 at 16:21 | comment | added | The Masked Avenger | Ben's suggestion gives a good starting approximation. You will need to use prime powers at some point; the answer I give tells you what to look for for a more precise answer. | |
| Jun 22, 2015 at 16:14 | answer | added | The Masked Avenger | timeline score: 2 | |
| Jun 22, 2015 at 12:28 | comment | added | Ben Barber | Wikipedia asserts that the primorial grows like $e^{(1+o(1))k\log k}$, and the sum of the first $k$ primes looks like $k^2 \log k / 2$, so $f(n) =O( (\log n)^2/(\log \log n))$. | |
| Jun 22, 2015 at 11:27 | comment | added | J Fabian Meier | So if we really take 2*3*5*7*... until we exceed n, can we get we get an estimate of the sum in terms of n (log(n)?) from some prime number distribution estimate? | |
| Jun 22, 2015 at 10:05 | comment | added | Ben Barber | I think taking the first few primes is the asymptotically cheapest way to get large products. I'd be interested to see precise bounds, as this also comes up in a problem I have on the back burner. | |
| Jun 22, 2015 at 9:00 | comment | added | user35593 | Obviously a minimizer $P$ must have the property that all its elements are primes or prime powers. | |
| Jun 22, 2015 at 8:31 | history | asked | J Fabian Meier | CC BY-SA 3.0 |