Difference between revisions of "User talk:Baijiangchen"

Line 1: Line 1:
 
If:
 
If:
 
<math>W(0):=1</math>
 
<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:
 
Then:
 +
 
<math>W(n)=(2n-1)!!</math>
 
<math>W(n)=(2n-1)!!</math>

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:

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