- Let $$ s(n,z)=\sum\limits_{j=0}^{n}L(n,j,z) $$ where $$ L(n,j,z)=\sum\limits_{p=0}^{n-j-1}f(p,z)L(n-j-1,p,z), \\ L(n,n,z)=1 $$
- Now let $s(n,z)$ be an arbitrary function such that $s(0, z)=1$. It means that we can compute $f(n, z)$ recursively: $$ f(n,z)=s(n+1,z)-\sum\limits_{j=1}^{n+1}L(n+1,j,z)-\sum\limits_{p=0}^{n-1}f(p,z)L(n,p,z) $$ But how can we use this function? Here are a few examples:
- If $s(n, z) = 1 + nz$, then unsigned coefficients of $f(n, z)$ are given in A007318.
- If $s(n, z) = nz$, then unsigned coefficients of $f(n, z)$ are given in A085478.
- If $s(n, z) = (n+1)z$, then unsigned coefficients of $f(n, z)$ are given in A056242.
- If $s(n, z) = 1 + n^2z$, then unsigned coefficients of $f(n, z)$ are given in A208904.
- If $s(n, z) = -1 + nz$, then unsigned coefficients of $f(n, z)$ are given in A124237.
- If $s(n, z) = 1 + (n+1)z$, then unsigned coefficients of $f(n, z)$ are given in A121462.
Here is the PARI/GP prog to check it numerically:
s(n,z)=1 + n*z
f_upto(n)=my(v1, v2, z='z); v1=vector(n+1, i, vector(i, j, i==j)); v2=vector(n, i, s(i,z)); v3=vector(n, i, 0); for(i=1, n, for(j=1, i-1, v1[i+1][j+1]=sum(p=0, i-j-1, v3[p+1]*v1[i-j][p+1])); my(A=sum(p=0, i-2, v3[p+1]*v1[i][p+1])); v3[i]=v2[i] - vecsum(v1[i+1]) - A; v1[i+1][1]=A + v3[i]); v3=vector(n, i, abs(Vecrev(v3[i])))
Note that some of the sequences given above are related with Riordan arrays.
Is there a way to compute $f(n, z)$ from $s(n, z)$ without using $L(n, j, z)$? In other words, is it possible to simplify computing of $f(n,z)$ from $s(n,z)$?
