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Oct 16, 2017 at 15:31 answer added Peter Heinig timeline score: 3
Nov 19, 2010 at 1:04 history edited user3409 CC BY-SA 2.5
deleted 68 characters in body; edited title
Nov 14, 2010 at 0:40 history edited user3409 CC BY-SA 2.5
title; improved formatting
Oct 24, 2010 at 16:22 history edited Yaroslav Bulatov
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Oct 24, 2010 at 13:10 answer added user3409 timeline score: 3
Oct 22, 2010 at 10:16 vote accept CommunityBot
Oct 22, 2010 at 1:39 answer added Gerry Myerson timeline score: 5
Oct 22, 2010 at 1:28 comment added Gwyn Whieldon As a rough count, if you've got n vertices, you've got less than $n^{n-2}$ trees. :)
Oct 22, 2010 at 0:03 comment added Gerry Myerson To expand on Lukasz' comment: the upper bound is infinity, and the lower bound is zero. If the number of vertices is $n$, the upper bound is whatever you get for the complete graph, the lower bound is still zero. Maybe a good question would be, if you fix $c$, $0\lt c\lt1/2$, and ask about graphs with $n$ vertices and roughly $cn^2$ edges, what upper and lower bounds do you get as functions of $c$ and $n$. But we shouldn't have to write your question for you.
Oct 21, 2010 at 18:32 comment added Łukasz Grabowski You should tell in what terms do you want to get the bounds.
Oct 21, 2010 at 18:01 answer added ohai timeline score: 0
Oct 21, 2010 at 17:57 history asked user3409 CC BY-SA 2.5