Timeline for answer to "half arithmetic progressions" in dense sets by Boris Bukh
Current License: CC BY-SA 2.5
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| Nov 6, 2009 at 22:52 | comment | added | Mark Lewko | Boris, thanks for your comments! I like your proof of the k=6 case. It seems, however, that the k=6 argument can't (at least naively) be extended to the case of k=8. To see this, find the 4-AP with a maximal largest element and (of these) a minimal smallest element. Now pass to the sub-progression containing this 4-AP (and relabel the progression as {0,1,2,3}). The intersection of our set with this progression could look like {-3,0,1,2,3,7}, which doesn't contain a 8-half-AP. Establishing the k=8 case would be very interesting. | |
| Nov 6, 2009 at 20:31 | history | answered | Boris Bukh | CC BY-SA 2.5 |