Difference between revisions of "User talk:Baijiangchen"

Revision as of 23:21, 21 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: $W(n)=(2n-1)!!$

Revision as of 23:20, 21 July 2012 (view source) Baijiangchen (talk \| contribs) (Created page with "If: <math>W(0):=1</math> <math>W(n):=\sum_{i=0}^{n-1}({n-1 \choose i}W(i)(x-i-1)!(2^{x-i-1})</math> Then: W(n)=(2n-1)!!")		Revision as of 23:21, 21 July 2012 (view source) Baijiangchen (talk \| contribs) Newer edit →
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	Then:		Then:
−	W(n)=(2n-1)!!	+	<math>W(n)=(2n-1)!!</math>