- #1

- 107

- 1

[tex]b_{n,j+1} (j+1)(2n-1)(2n-2) = (2n-2+j)(2n-1+j)b_{n-1,j}[/tex]

which has the solution

[tex]b_{n,j} = \frac{(2n-1+j)!}{(2n-1)!j!}[/tex]

Is there any algorithm that can be used to derive this result? I have a recurrence relation which is a bit more complex than this one.