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:

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

@@ Line 3: / Line 3: @@
 <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>
+<math>W(n):=\sum_{i=0}^{n-1}({n-1 \choose i}W(i)(x-i-1)!(2^{x-i-1}))</math>
 Then:
 <math>W(n)=(2n-1)!!</math>