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Jul 27, 2013 at 16:41 comment added The Masked Avenger I don't know if n + O(log(n)) is best possible. In view of (k! - 1)!k! |(k!)!, I would say it is pretty darn good though.
Jul 27, 2013 at 16:28 history edited Seva CC BY-SA 3.0
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Jul 27, 2013 at 16:03 history edited Seva CC BY-SA 3.0
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Jul 27, 2013 at 15:47 comment added GH from MO I was in the process of writing up a bound $a+b\leq n+O(n/\log n)$ when I saw your update. At any rate, I would formulate your result in the sharper form $a+b\leq n+O(\log n)$.
Jul 27, 2013 at 15:43 history edited Seva CC BY-SA 3.0
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Jul 27, 2013 at 15:11 history answered Seva CC BY-SA 3.0