# Simple generating functions without recurrence

I'm still getting to grips with generating functions. I think I can grasp how to form them when you have a recurrence relation, but what if you don't?

Consider the sequence $\{a_n\}_{n\geq0}$ with generating function $a(x)$. What would the generating functions be for sequences like $$1,2,3,a_3,a_4,a_5,...\\ a_0,-a_1,a_2,-a_3,...\\ \{a_n+2a_{n+2}\}_{n\geq0}$$ where they are not given any equality to help manipulate the expressions?

For the first, I feel like some sort of division is necessary from simple forms, for instance, since $\frac{1}{(1-x)^2}=\sum_{n\geq0}nx^n$, then the terms after 3 could somehow be divided out. I also am not sure whether something as simple as $a(x)+(1-a_0)+(2-a_1)x+(3-a_2)x^2$ would count as a generating function?

For the second, I thought perhaps $\sum_{n\geq0}a_nx^n(-1)^n$, but I'm not sure if this counts as a generating function or if it would need to be in a simpler form.

For the third, I treated it as a recurrcence and arrived at $a(x)+2\left(\frac{a(x)-a_0-a_1x}{x^2}\right)$, but without any other side to factor out and manipulate, I am not sure whether that is any use.

Your first idea for the first problem doesn’t really go anywhere, but your second is fine:

$$a(x)+(1-a_0)+(2-a_1)x+(3-a_2)x^2$$

is a fine generating function, and there really isn’t anything else to do without further information about the numbers $a_n$.

For the second problem you want $\sum_{n\ge 0}(-1)^na_nx^n$. To relate it to $a(x)$, notice that

$$\sum_{n\ge 0}(-1)^na_nx^n=\sum_{n\ge 0}a_n(-x)^n=a(-x)\;:$$

the desired generating function is simply $a(-x)$, the composition $a\circ m$, where $m(x)=-x$ for all $x$.

What you’ve done in the third problem is fine. You could manipulate your answer algebraically in various ways, e.g., combining the $a(x)$ terms, but there’s really no point until you know what you want to do with the generating function.