Timeline for What is known about the largest prime divisor of the product of $k$ consecutive integers?
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Nov 3, 2015 at 17:47 | comment | added | René Gy | @Gerhard That is very interesting, thanks. Probably above my current level though. | |
| Nov 2, 2015 at 18:18 | comment | added | Gerhard Paseman | You might be interested in a paper of Filip Najman at arxiv.org/abs/1108.3710 . The largest prime divisor of the product n+1 to n +f(k) is looked at and shown to be larger than k when n is larger than k. f(k) is pretty small and conjectured to be O(log(k)^2). So you don't need as many as k consecutive integers. Gerhard "Sees This As Smooth Intervals" Paseman, 2015.11.02 | |
| Oct 31, 2015 at 16:08 | vote | accept | René Gy | ||
| Oct 31, 2015 at 15:30 | answer | added | so-called friend Don | timeline score: 8 | |
| Oct 31, 2015 at 14:58 | history | edited | René Gy | CC BY-SA 3.0 |
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| Oct 31, 2015 at 13:22 | comment | added | Wojowu | I was asking because when you said "...I could not find an elementary (my level) proof.", I thought you might have found a hard proof and you were just looking for a simpler one. | |
| Oct 31, 2015 at 13:21 | comment | added | René Gy | @Wojowu. No, but I am not professional mathematician, so my knowledge of (and access to) math literature is very limited. If you know of relevant papers, I 'd be interested. | |
| Oct 31, 2015 at 13:09 | comment | added | Wojowu | Do you have any reference to a "nonelementary" proof? | |
| Oct 31, 2015 at 10:42 | history | asked | René Gy | CC BY-SA 3.0 |