Timeline for Largest number of k-arithmetic progressions without a (k+1)-arithmetic progression
Current License: CC BY-SA 3.0
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| Aug 26, 2013 at 1:38 | comment | added | Jacob Fox | I should add that the upper bound can be improved further to $o(p^2N^2)$, where the $o$ term can be explicitly computed. However, the proof is a little more involved and requires using the following lemma: every set $B$ of integers with no $(k+1)$-term arithmetic progression (again $k$ is fixed) has $o(|B|^2)$ $3$-term arithmetic progressions. The proof uses the Balog-Szemerédi-Gowers theorem, Freiman's theorem, and Szemerédi's theorem. | |
| Aug 25, 2013 at 19:23 | history | edited | Jacob Fox | CC BY-SA 3.0 |
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| Aug 25, 2013 at 17:40 | history | edited | Jacob Fox | CC BY-SA 3.0 |
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| Aug 25, 2013 at 17:32 | comment | added | Andrés E. Caicedo | Nice seeing you here! | |
| Aug 25, 2013 at 17:18 | comment | added | Will Sawin | If $f(k,N)$ is the maximum number of $k$-term progressions in subsets of $\{1,\dots,N\}$ with no $k+1$-term progressions, we have: $$ N^2 c^k e^{-k\left(\log_C N\right)^{1/k}} \leq f(k,N) \leq N^2 \frac{2}{k-1} (\log_2 \log_2 N) ^{ - 2^{-2^{k+10}}} $$ This lower bound improves on the lower bound which I noted follows from Joel Moreira's construction. | |
| S Aug 25, 2013 at 17:07 | review | Late answers | |||
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| S Aug 25, 2013 at 17:07 | review | First posts | |||
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| Aug 25, 2013 at 16:49 | history | answered | Jacob Fox | CC BY-SA 3.0 |