# Given a generating function for $\sum a_n z^n$, what is the generating function of $\sum a_n^2 z^n$

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?

• There seems to be something off in your title. – user14082 Mar 4 '13 at 13:39
• sorry, my fault... – StefanH Mar 5 '13 at 2:06
• 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/… – Gerry Myerson Mar 5 '13 at 2:37

\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 $$B(z)$$.