Counting equivalent expressions

[David Wilson]

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.