$$x = -\sum_{k=1}^p \binom{p}{k} p^{k-2}\bigl(-xA(x)\bigr)^k$$

For degree $n>1$, the left hand side of the equation is equal to $0$. Setting $0$ equal to the degree $n$ term of the right hand side of the equation gives this equation

$$0=-\sum_{k=1}^p \binom{p}{k} p^{k-2} \sum_{i_1+\ldots+i_k=n} \prod_{j=1}^k a(i_j-1) (-1)^kx^n.$$

This was taken from the article entitled "Super Patalan numbers" by Thomas M. Richardson...

I want to see the full details of the transformation and shifting


The power series $A(x)$ is the generating function of the Patalan numbers $a(n)$. Thus the coefficient of $x^{i_j}$ in the product $xA(x)$ is $a(i_j - 1)$; the factor $x$ results in the index shift. Then we are pulling out the factor $(-1)^k$ from the product, and writing the expression for the degree $n$ term of the $k^{\rm th}$ power of a series.


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