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May 7, 2017 at 23:16 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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Apr 7, 2017 at 23:08 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Mar 8, 2017 at 22:52 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Feb 7, 2017 at 13:46 comment added Fedor Petrov @GerryMyerson Yes, I am not completely satisfied by this double logarithm gap between lower and upper bound.
Feb 6, 2017 at 22:51 comment added Gerhard Paseman I think he wants a proof that it is O(log p/ log log p). I don't have a proof, but a darned good idea (augmentation) in that direction. Gerhard "It's The Idea That Counts" Paseman, 2017.02.06.
Feb 6, 2017 at 22:10 comment added Gerry Myerson So, you need something better than just saying it's between $\log p$ and $\log p/\log\log p$?
Feb 6, 2017 at 20:25 answer added Gerhard Paseman timeline score: 1
Feb 6, 2017 at 19:12 comment added Fedor Petrov Yes, but such constructions can not give better bounds, since after an elementary transform the rows of our matrix (almost all of them) have bounded norms.
Feb 6, 2017 at 14:32 comment added js21 Alternatively, the bound $k = O(\log p)$ can be achieved by writing $p=Q(2)$ where $Q$ is a polynomial with coefficients in $\{ 0 ,1 \}$ (base $2$ expansion), and then by noting that $p = \det(2 I - M)$ where $M$ is the companion matrix of $Q$. The matrix $ 2I - M$ has coefficients in $\{-1,0,1,2,3 \}$.
Feb 6, 2017 at 8:01 history asked Fedor Petrov CC BY-SA 3.0