Here is the relation- $$jB(n,j)=(j+2)(j+1)B(n,j+2) + \frac{1}{3}B(n-1,j-3) -\sum_{p=1}^{n-1} B(n-p,j)B(p,1)$$
$B(n,j) = 0$ for $n=0$ or $j= 0$ or $(n<0)$ or $(j<0)$
$B(n,j)=0$, for $j>3n$
$B(0,0) =1$
$B(n,1) =0$ for $n-odd$
I got this recurrence relation while trying to figure out a series expansion of energy eigen value for the anharmonic Oscillator having a cubic term in potential using the method of Bender and Wu, in their paper-(PhysRev.184.1231), they did for quartic term ($x^4$) in potential, I am trying to do the same for $x^3$.( A self-study project work) B(n,1) are related to the coefficient of a perturbation series for Energy Eigen value. I have tried the following-
Explicit formulas like this can be derived- $$B(n,3n) = \frac{1}{3^{n+1} n!}$$
but this does not work for n=1 as $B(n,1) =0$ for all odd n, so I should get $B(1,3) =0$ based on the recurrence relation.
Also, it can be checked that $B(n, 3+2m) = 0$ for all odd $n$ and for $m=1, 2, 3....$ I get the feeling that $B(n,j) = 0$ for all odd $n$ as $B(1,j)= 0$ for all $j$ which can be checked easily -
$B(1, 1)= B(1,3)= B(1,5) =...= 0.$
$B(1,4)= (something) B(1,2) = 0$ and so on for $B(1,2n).$
But if $B(n,j) = 0$ for all $n$- odd then How do I solve for $B(n,1)$?
The basic idea would be to obtain an explicit form as $B(n, 3n-1)$ or something and then proceeding using a computer program to solve this matrix $B(n,j)$ from right to left, but this vanishing of $B(n,j)$ for n- odd is the problem. How do I obtain such a closed form for even n? Any help here?