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Aug 25, 2015 at 6:46 vote accept David Sun
Aug 25, 2015 at 5:07 vote accept David Sun
Aug 25, 2015 at 6:46
Aug 25, 2015 at 5:07 vote accept David Sun
Aug 25, 2015 at 5:07
Aug 9, 2015 at 14:32 answer added Max Alekseyev timeline score: 2
Aug 9, 2015 at 13:34 answer added Fedor Petrov timeline score: 6
Aug 8, 2015 at 17:17 comment added Fedor Petrov determinant equals $x^{1+\dots+(2n-1)}=x^{n^2}=(-1)^n$, or what do I miss?
Aug 8, 2015 at 15:35 history edited David Sun
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Aug 8, 2015 at 10:41 answer added Peter Mueller timeline score: 10
Aug 7, 2015 at 19:34 comment added benblumsmith By a computer calculation, the determinants also repeat mod 6, and in fact for $n=0\mod 6$ the product is always the identity.
Aug 7, 2015 at 15:04 review First posts
Aug 7, 2015 at 15:09
Aug 7, 2015 at 15:01 history asked David Sun CC BY-SA 3.0