Timeline for Largest number of k-arithmetic progressions without a (k+1)-arithmetic progression
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2013 at 16:55 | comment | added | Will Sawin | $A$ has at least $2^{n+1}-1$ arithmetic progressions and occupies an interval of length $(2k)^n$, so if $N$ is the length of the interval, this gives a lower bound of about $N^{\log 2/\log 2k}$. | |
| Mar 4, 2013 at 18:48 | comment | added | Joel Moreira | I think one can in principle use a Varnavides-type argument, but I am not sure how that would work out exactly (I am using the term "Varnavides-type" from a section on this blog post: terrytao.wordpress.com/2008/02/10/…) | |
| Mar 4, 2013 at 16:23 | comment | added | Marcin Kotowski | I upvote the answer, but the conclusion "any subset B⊂A with positive relative density has a k-long arithmetic progression." seems too weak to say anything about the number of such k-progressions in the whole set. | |
| Mar 4, 2013 at 15:56 | history | answered | Joel Moreira | CC BY-SA 3.0 |