Timeline for On the number of divisors in a given range
Current License: CC BY-SA 3.0
11 events
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| Nov 25, 2015 at 7:26 | vote | accept | CommunityBot | ||
| Nov 25, 2015 at 7:04 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 25, 2015 at 6:44 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 25, 2015 at 6:43 | comment | added | GH from MO | @Arul: See my added section, which answers your second question up to the precise value of the best constant $c$. | |
| Nov 25, 2015 at 6:38 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 25, 2015 at 4:33 | comment | added | user76479 | It will be useful for my purposes if I have a good bound | |
| Nov 25, 2015 at 3:04 | comment | added | GH from MO | @Arul: There is no such bound, i.e. you can get $(\log N)^{1000}$ divisors if you wish. In fact the number of divisors in the given range can be much larger than polylogarithmic, but I have not attempted to determine the maximum order of this quantity. | |
| Nov 25, 2015 at 2:04 | vote | accept | CommunityBot | ||
| Nov 25, 2015 at 5:48 | |||||
| Nov 24, 2015 at 23:15 | comment | added | GH from MO | Yes. If $n\geq t$ and $n\in\mathbb{Z}$, then $n\geq\lceil t\rceil$. Similarly, if $n\leq t$ and $n\in\mathbb{Z}$, then $n\leq\lfloor t\rfloor$. Hence my last display is clear with $\leq$ around $Mp_1p_2p_3p_4p_5$. However, we talk about a product of different primes, so strict inequality also follows (namely for $p_2$, $p_3$, $p_4$ the intermediate inequalities are strict). Note also that $\leq$ is sufficient in the last display. | |
| Nov 24, 2015 at 22:22 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 24, 2015 at 21:34 | history | answered | GH from MO | CC BY-SA 3.0 |