Timeline for In which orders can the numbers of prime factors of consecutive integers be?
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6 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2014 at 16:50 | comment | added | Joni Teräväinen | Yes, $k=2$ is quite trivial. In the case $k=3, \sigma=id$, the choise $n+1=2^m$ leads to an elementary proof. For $k>3$ and $\sigma=id$ I haven't found an elementary proof. | |
| Jun 11, 2014 at 9:35 | comment | added | joro | Isn't $k=2$ easy with primes? Take $n=p-1$, $\sigma(1)=1$. | |
| Jun 11, 2014 at 3:41 | comment | added | Joni Teräväinen | This is very helpful. Thanks to both of you. | |
| Jun 11, 2014 at 2:34 | comment | added | Lucia | To add a tiny bit to Terry Tao's comment: Look for example at this paper by Rizwan Khan (cms.math.ca/cmb/v53/khanB9034.pdf ) where a problem on the simultaneous distribution of $\omega(n+i)$ is considered. Khan wants all of these to be very close to $\log \log (n+i)$ (and with some uniformity) but the method will work in your problem too. | |
| Jun 11, 2014 at 2:25 | comment | added | Terry Tao | I think just about any of the standard proofs of the Erdos-Kac theorem would extend to give the joint gaussian distribution of the $\frac{\omega(n+i)-\log\log n}{\sqrt{\log\log n}}$, which would give a positive anwswer to your questions. | |
| Jun 11, 2014 at 1:49 | history | asked | Joni Teräväinen | CC BY-SA 3.0 |