by
Mewto55555, Jun 15, 2012, 4:08 PM
by
hyperbolictangent, Jun 16, 2012, 12:54 PM
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be the 
t
be the 
and ![\[ x_n = \sum_{k=1}^{n} \left( (2k-1)! \binom{2n-1}{2k-1} x_{n-k} \right)\]](http://latex.artofproblemsolving.com/8/7/b/87babe3e9e5d131dc178ce3030027d88b7d3a6d7.png)
.
.
satisfies the recursion.
gives the number of ways to pair each number from 
to 
with another, so 
gives the number of ways to pair each row with another row and column with another column in a 
such that row number 
is paired with row number 
,
,
,
is paired with column number 
.
ways to choose the remaining 
elements, 
ways to permute them, and 
rows and columns.![\[ (2n - 1)!!^2 = \sum_{k = 1}^{n} (2k - 1)!\binom{2n - 1}{2k - 1}(2n - 2k - 1)!!^2\]](http://latex.artofproblemsolving.com/6/1/e/61e931562f6d7f092157108c2d2302ab7026de41.png)
or ![\[x_n =\sum_{k = 1}^{n} (2k - 1)!\binom{2n - 1}{2k - 1}x_{n - k}\]](http://latex.artofproblemsolving.com/d/9/d/d9dbe8c552fab3f7fe2b1be19d3ab27c0c502ff5.png)
which is the original recursion. Since it satisfies the initial constraint 
,