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So I am using exponential generating functions and have a question about taking the product of more than 2 exponential generating functions.

I know that the product of 2 exponential generating functions is

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What I'm having a hard time with is that taking the product of more than 2 gets messy with the sum notation. So my question; is there a more efficient way to determine the coefficient of \$$a_n$ given that \$$a_n$ is a product of more than 2 exponential generating functions?

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  • $\begingroup$ Yes this seems to work out perfectly. I guess my problem was that for some arbitrary n of the form n=3k, I would be able to find an explicit formula for the coefficient. $\endgroup$ – tamefoxes Jun 24 '14 at 20:55
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$$\left(\sum_{k=0}^\infty a_{k}^{(1)}\frac{x^k}{k!}\right)\cdots\left(\sum_{k=0}^\infty a_k^{(r)}\frac{x^k}{k!}\right)=\sum_{n=0}^\infty\left[\sum_{k_1+\cdots+k_r=n}\binom{n}{k_1,\cdots,k_r}a_{k_1}^{(1)}\cdots a_{k_r}^{(r)}\right]\frac{x^n}{n!}$$

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