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If I have given the following generating function

$$B = \sum_{n > 0} x^n \sum_{k_1 + \cdots + k_c = n} a(k_1) \cdots a(k_c)$$

is it possible to obtain a nice convolution expression for $a(k_1) \cdots a(k_c)$ in terms of some generating function A?

What bothers me is the summation-constraint $k_1 + \cdots + k_c = n$ of the inner sum, thus it is not straight forward for me to come up with a convolution.

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This must be a homework, so I've tagged it as such. Feel free to roll back if it is not. – Sasha Mar 9 '12 at 1:45
@Sasha no its not, but I am glad if it looks easy, because I don't have an approach for it, but I am no expert in this field. – steven Mar 9 '12 at 1:47

1 Answer 1

up vote 2 down vote accepted

Here goes: $$ \begin{eqnarray} B(x) &=& \sum_{n > 0} x^n \sum_{k_1 + \cdots + k_c = n} a(k_1) \cdots a(k_c) = \sum_{n > 0} x^n \sum_{k_1,\ldots, k_c >0 } \delta_{n, k_1+\cdots+k_c} a(k_1) \cdots a(k_c) \\ &=& \sum_{k_1,\ldots, k_c >0 } \left( \sum_{n > 0} x^n \delta_{n, k_1+\cdots+k_c} \right) a(k_1) \cdots a(k_c) \\ &=& \sum_{k_1,\ldots, k_c >0 } x^{k_1 + \cdots+k_c} a(k_1) \cdots a(k_c) = \left( \sum_{k_1 >0 } x^{k_1} a(k_1)\right) \cdots \left( \sum_{k_c >0 } x^{k_c} a(k_c)\right) = (A(x))^c \end{eqnarray} $$

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