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Jul 25, 2017 at 22:32 history edited GH from MO
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Apr 17, 2012 at 19:45 comment added Woett Then I said nothing :)
Apr 17, 2012 at 19:41 comment added Gerhard Paseman Woett, there likely was no 1/n in front when Brendan McKay made his comment. There certainly wasn't when I made my earlier comments. Gerhard "However, All Is Better Now" Paseman, 2012.04.17
Apr 17, 2012 at 17:59 vote accept unknown
Apr 17, 2012 at 15:24 comment added Woett @Brandon, look at the $\frac{1}{n}$-term in front
Apr 17, 2012 at 15:17 answer added Emil Jeřábek timeline score: 19
Apr 17, 2012 at 15:03 history edited unknown CC BY-SA 3.0
fixing important stuff
Apr 17, 2012 at 14:59 comment added Brendan McKay The $r=2$ term is $\frac12$ for every odd $n$, so how can can the limit be less than $\frac12$?
Apr 17, 2012 at 14:59 comment added unknown @Igor: thanks for spotting the typo. And no, I cannot prove this, this is purely experimental.
Apr 17, 2012 at 14:57 comment added Gerhard Paseman Further, I see no way of rescuing the problem without making the denominator substantially bigger, as in r^2 or n^2, or making the numerator correspondingly smaller. Gerhard "Ask Me About System Design" Paseman, 2012.04.17
Apr 17, 2012 at 14:57 history edited unknown CC BY-SA 3.0
edited body
Apr 17, 2012 at 14:56 comment added Igor Rivin Can the OP prove this, or is this an experimental fact?
Apr 17, 2012 at 14:56 comment added Igor Rivin The text is incorrect (should be the remainder of $n$ divided by $r$)
Apr 17, 2012 at 14:48 comment added Gerhard Paseman Without conditions on n, I have a hard time believing that lim inf of the sum is as large as you claim. I do not think the limit exists. Gerhard "Ask Me About System Design" Paseman, 2012.04.17
Apr 17, 2012 at 14:28 history asked unknown CC BY-SA 3.0