Difference between revisions of "User talk:Baijiangchen"

Revision as of 23:25, 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:

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

Let $W(n)=\sum_{i=1}^{n}(\binom{i-1}{n-1}W(i-1)(x-i)!(2^{x-i}))$

@@ Line 8: / Line 8: @@
 <math>W(n)=(2n-1)!!</math>
+==Sam's stuff==
+Let <math>W(n)=\sum_{i=1}^{n}(\binom{i-1}{n-1}W(i-1)(x-i)!(2^{x-i}))</math>