Given a rational function, the coefficient of its Taylor series is a sequence. Often the sequence can be recursively defined. For a rational function $f(x)/g(x)$, we can consider its Taylor expansion at $x=0$, i.e. $$ \frac{f(x)}{g(x)}=a_{0}+a_{1}x+a_{2}x^{2}+\cdots $$ How can we go from the right to the left?
That is, if we are given a sequence $(a_{n})$, and we know how $(a_{n})$ is recursively defined, then how can we find a rational function whose Taylor expansion is $(a_{n})$.
More specifically, the $(a_{n})$ that I really care about is defined as the following: \begin{cases} a_{0}=1,\; a_{1}=0,\; a_{2}=1 \\[2pt] a_{n}=a_{n-1}+2a_{n-2}-a_{n-3}, & n \geq 3 \end{cases}
Let $F(x)=\frac{f(x)}{g(x)}$. Consider $G(x)=xF(x)+2x^2F(x)-x^3F(x)$ and assume $G(x)=b_0+b_1x+b_2x^2+...\;$ . Then for $n\geq 3$ we have $a_n=b_n$. Note that $b_0=0$, $b_1=a_0=1$ and $b_2=a_1+2a_0=2$, so $$G(x)=x+2x^2+a_2x^2+a_3x^3+\ldots=x+2x^2+(F(x)-1-x^2).$$ This means $xF(x)+2x^2F(x)-x^3F(x)=-1+x+x^2+F(x),$ and solving for $F(x)$ gives $F(x)=\frac{-1+x+x^2}{-1+x+2x^2-x^3}$.