Difference between revisions of "User talk:Baijiangchen"

Revision as of 00:21, 22 July 2012

If:

$W(0):=1$

$W(n):=\sum_{i=0}^{n-1}({n-1 \choose i}W(i)(x-i-1)!(2^{x-i-1})$

Then:

$W(n)=(2n-1)!!$

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	<math>W(0):=1</math>		<math>W(0):=1</math>