<s>[INST] Determine the sequence generated by a generating function $A(z)=2z-1+\frac{1}{2z-2z^2}$
I have no clue how to solve this, I tried looking at other examples but I am just stuck, could anyone be so kind and explain how to solve this step by step?
 [/INST] If you want to see the generating function:
$\begin{array}\\
A(z)
&=2z-1+\frac{1}{2z-2z^2}\\
&=2z-1+\frac1{2z}\frac{1}{1-z}\\
&=2z-1+\frac1{2z}\sum_{n=0}^{\infty} z^n\\
&=2z-1+\frac1{2z}+\sum_{n=1}^{\infty} \frac{z^{n-1}}{2}\\
&=2z-1+\frac1{2z}+\sum_{n=0}^{\infty} \frac{z^{n}}{2}\\
&=2z-1+\frac1{2z}+\frac12+\frac{z}{2}+\sum_{n=2}^{\infty} \frac{z^{n}}{2}\\
&=\frac53 z-\frac12+\frac1{2z}+\sum_{n=2}^{\infty} \frac{z^{n}}{2}\\
\end{array}
$
If you want to invert the function,
that is,
find $z$ in terms of $A$:
$\begin{array}\\
A(z)
&=2z-1+\frac{1}{2z-2z^2}\\
&=\frac{(2z-1)(2z-2z^2)+1}{2z-2z^2}\\
&=-\frac{4 z^3 - 6 z^2 + 2 z - 1
}{2z-2z^2}\\
\end{array}
$
so
$-A(2z-2z^2)
=4 z^3 - 6 z^2 + 2 z - 1
$
or
$4 z^3 - (2A+6) z^2 + (2a+2) z - 1
=0
$.
Wolfy gives some very complicated expressions
for $z$ in terms of $A$.
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