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Oct 4, 2013 at 2:09 history edited Brendan McKay CC BY-SA 3.0
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Oct 4, 2013 at 2:02 comment added Brendan McKay @Jernej: Agreed if you count oriented cycles. I'm posting counts for unoriented cycles.
Oct 3, 2013 at 23:21 comment added Jernej The related integer sequence appears to be $4,12,24,40,40,96,96$
Oct 3, 2013 at 18:20 comment added Boris Bukh Interesting question! I communicated it to Michael Krivelevich. He does not know, but suggests that the methods from ams.org/mathscinet-getitem?mr=2520275 might be useful
Oct 3, 2013 at 18:16 comment added Gerhard Paseman How about (3n/2) choose (n/2)? Another silly bound, but it might lead somewhere. You could also divide the graph into connected groups of 5 edges and note that a cycle can intersect each group of edges in only 8 ways. (I assume each group is acyclic, perhaps a bad move.) Gerhard "Spins Me Round Right Round" Paseman, 2013.10.03
Oct 3, 2013 at 13:44 history asked Brendan McKay CC BY-SA 3.0