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Given a generating function $G(z)$ for $\sum_{n=0}^{\infty} a_n z^n$, what could be said about the generating function of $\sum_{n=0}^{\infty} a_n^2 z^n$, what algebraic form should it have?

For example, given the generating function for $G(z) = \sum_{n=0}^{\infty} a_n z^n$, then the generating function for its cumulative sum, i.e. $\sum_{n=0}^{\infty} \left(\sum_{k}^n a_k\right) z^n$ is $\frac{1}{1-z} G(z)$, which could be derived by the formulae for the convolution product of two generating functions $$ F(z) G(z) = \sum_n \left( \sum_{k=0}^n f_k g_{n-k} \right) z^n $$ applied on the well know generating function of $(1,1,1,1,\ldots)$ $$ \sum_{n=0}^{\infty} z^n = \frac{1}{1-z}. $$ But I am looking for a formulae for the pointwise product to solve my problem?

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    $\begingroup$ There seems to be something off in your title. $\endgroup$ – user14082 Mar 4 '13 at 13:39
  • $\begingroup$ sorry, my fault... $\endgroup$ – StefanH Mar 5 '13 at 2:06
  • $\begingroup$ The "Hadamard product" of $\sum a_nz^n$ and $\sum b_nz^n$ is $\sum a_nb_nz^n$, so you are asking about the square of a power series (with respect to the Hadamard product). Searching for Hadamard product (and power series) may turn up something useful. See also math.stackexchange.com/questions/4744/… $\endgroup$ – Gerry Myerson Mar 5 '13 at 2:37
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This is the Hadamard product, in terms of generating functions:

$\begin{align*} A(z) &= \sum_{n \ge 0} a_n z^n \\ B(z) &= \sum_{n \ge 0} b_n z^n \\ C(z) &= \sum_{n \ge 0} a_n b_n z^n \end{align*}$

Sorry, but except for very special sequences there isn't any nice expression for $C(z)$ in terms of $A(z)$ and $B(z)$.

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