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Let ${Z}$ denote an integer triangle (like Pascal's), ${Z}_{n,k}$ for $0\leq k \leq n$ and let $f$ be an exponential generating function for polynomials $p_{n}(x)$ with $[x^k]p_{n}(x)= Z_{n,k}$.

$$\sum_{n \ge 0}{{p}_{n}(x) \frac{z^n}{n!}} = EG({Z}(n,k);x,z) = f$$

Assume that $Z$, as a lower triangular matrix, is invertible and is also an integer matrix.

What is the exponential generating function of the inverse triangle, $f$ given?

Edit: By request of Gerry Myerson and to clarify with examples found on OEIS:

Example 1: Let $f1 = \exp(z(1/\sqrt{1-2x}-1))$. f1 is the bivariate exponential generating function of the triangle A035342 (apart from the first column). The inverse of this triangle is A046089 (apart from the first column and the signs). How can I derive from f1 the bivariate egf of (the modified) A046089?

Example 2: Let $f2 = \exp(z\log(1+x))$. f2 is the bivariate exponential generating function of the triangle A048994. The inverse of this triangle is A048993. How can I derive from f2 the bivariate egf of A048993?

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What is $(x)$ in $Z(n,k)(x)$ and does the sum run over $n$? –  draks ... Jul 24 '12 at 17:53
    
Have you computed any examples to get a feeling for what's going on? –  Gerry Myerson Jul 24 '12 at 23:04
    
One note is that for your lower triangular matrix to be unimodular (matrix and inverse both have integer entries), your diagonal elements can only be either $1$ or $-1$. –  J. M. Jul 25 '12 at 12:08
    
Have you, BTW, looked into Riordan arrays? –  J. M. Jul 25 '12 at 12:09

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