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Sep 3, 2015 at 17:00 history edited Igor Rivin CC BY-SA 3.0
fixed buggy formula
Sep 3, 2015 at 17:00 comment added Igor Rivin @EricNaslund That's what happens when you don't sleep :(
Sep 3, 2015 at 16:53 comment added Eric Naslund "Is on the order of $\log N$" I believe you mean $N/\log N$, and the resulting sum of squares has order $N^3/\log N$.
Sep 3, 2015 at 16:11 history edited Igor Rivin CC BY-SA 3.0
typo
Sep 3, 2015 at 15:58 history answered Igor Rivin CC BY-SA 3.0