Counting equivalent expressions
David Wilson
dwilson at gambitcomm.com
Thu Aug 21 18:33:17 CEST 2008
Mitch Harris wrote:
> [My quote elided]
>
> Unordered binary trees? Double factorial...
>
> http://www.research.att.com/~njas/sequences/A001147
>
> Not obvious (but if memory serves, not a terrible combinatorial
> explanation, see Stanley reference)
>
>
The light went on when I read your note. There would be (2n-2)!/(n-1)!
trees each with with n leaves and n-1 interior nodes. Each of these
interior nodes could be flipped in one of two orders by commutativity,
so that's 2^(n-1) trees per equivalence class, or (2n-2)!/(2^n (n-1))!
equivalence classes. I'm not sure I've got it exactly right, but this is
close enough to the double factorial expression that I'm willing to take
your word.
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