Suppose I have the following definition: $$\frac{x^2/2!}{e^x-1-x}=\sum_{k=0}^{\infty}A_k\frac{x^k}{k!}$$
I want to find a convolution identity for these coefficients $A_k$, but I've never studied convolutions before. Essentially, what I would like to find is a way to write $A_n$ in terms of a finite convolution, where $$A_n=C\sum_{k=0}^{n-1}c_{nk}A_kA_{n-k}$$ or$$f(n)A_n=\sum{A_kA_{n-k}}$$or something along those lines...
I don't even know if it is possible, but what I am looking for is guidance to approach these problems. What are the best methods for developing convolution identities and what approaches work well?
EDIT: I was able to find a convolution identity using Cauchy products and an identity to come up with $$\sum_{j=2}^{k-2}\binom{k}{j}A_jA_{k-j}=-k\left(A_k+\frac1{3}A_{k-1}\right)$$ which is kind of bittersweet since (1) I found one (2) it is not an identity in only $A_k$. Any thoughts on how I can find a similar formula that kills the $A_{k-1}$ term OR that removes the binomial coefficients in the summation and leaves me with $\sum{A_jA_{k-j}}=f(k)A_k$?
EDIT 2:
I feel like my desire is to have a convolution identity of the form
$$A_n=g(n)\sum{A_kA_{n-k}}$$ or something of this form. My trouble with this lies in the fact that the numbers $A_k$ are defined by an exponential power series which places $k!$ into the denominator of the coefficients of the power series. can anyone think of a way around this?