The meaning of Generating functions I had a look to this video on the field of series and sequences which I know not much about!
This guy looked for the generating function of $a_n = 2a_{n-1} + 4a_{n-2}$ with $a_0=1$ and $a_1=3$. The generating function is $$A(x)=\frac{1+x}{1-2x-4x^2}$$ (solution showed at 04:33 in the video)
I've been trying to understand what is the real meaning of this generating function he found with wikipedia but I got lost into thousands of pages and definitions.
Can we find with this formula the 18th number of the series? What does the x represents? How does this generating function "represents" the series?
Hope my question is not too broad and that you will easily find where is my gap!
 A: 
Can we find with this formula the 18th number of the series?

Yes we can. We first rewrite $A$. The denominator is $(1-x)^2-5x^2=(1-ux)(1+vx)$ with $u=\sqrt5+1$ and $v=\sqrt5-1$ hence, for some suitable $a$, using the fact that $A(0)=1$,
$$
A(x)=\frac{a}{1-ux}+\frac{1-a}{1+vx}=\sum_{n\geqslant0}(au^n+(-1)^n(1-a)v^n)x^n.
$$
In particular, for every $n\geqslant0$,
$$
a_n=au^n+(-1)^n(1-a)v^n.
$$
Since $a_1=3$, $au-(1-a)v=3$, hence
$$
a=\frac{3+v}{u+v}=\frac{\sqrt5+2}{2\sqrt5},
$$
in particular,
$$
a_{18}=\frac{(\sqrt5+2)(\sqrt5+1)^{18}+(\sqrt5-2)(\sqrt5-1)^{18}}{2\sqrt5}.
$$

What does the x represents?

Nothing.

How does this generating function "represents" the series?

As shown in this post, every $a_n$ can be recovered from $A(x)$, and, conversely, $(a_n)$ determines the function $A$. Hence both are equivalent.
A: This is probably the first example you see of what we call in mathematics a "transform".  There are many of these, and what they have in common is that they all take the information given by one function and express it in a new way.
So, what is $A(x)$, and how is it related to our series?
$A(x)$ is a rational function, which is also expressible as a polynomial of infinite degree.  When expressed as its polynomial sum, the coefficients of $A(x)$ give us the sequence in question.
Can we get the sequence by plugging in values for $x$?  Yes, in a way. Several ways in fact, but here's one in particular. Remember from calculus that $\frac{d}{dx}x^n=nx^{n-1}$.  Remember also that the constant term of any polynomial $p(x)$ is simply $p(0)$. Using these facts, you can show that the $n^{th}$ number in the sequence (starting from $n=0$) is given by
$$
a_n = \frac1{n!}\frac{d^nA}{dx^n}\mid_{x=0}=\frac1{n!}A^{(n)}(0)
$$
What does $x$ represent? It's hard to say satisfyingly. 
I encourage you to think of the opposite question: given a function $f(x)$ that is expressible as some infinite polynomial, we can extract the sequence of coefficients.  What does this sequence mean, in reference to the polynomial?  How does this sequence "represent" the polynomial?
