TITLE: A special type of generating function for Fibonacci
QUESTION [14 upvotes]: Notation. Let $[x^n]G(x)$ be the coefficient of $x^n$ in the Taylor series of $G(x)$.
Consider the sequence of central binomial coefficients $\binom{2n}n$. Then there two ways to recover them:
$$\binom{2n}n=[x^n]\left(\frac1{\sqrt{1-4x}}\right) \tag{1-1}$$
and
$$\binom{2n}n=[x^n]\left((1+x)^2\right)^n. \tag{1-2}$$
This time, take the Fibonacci numbers $F_n$ then, similar to (1-1), we have
$$F_n=[x^n]\left(\frac1{1-x-x^2}\right). \tag{2-1}$$
I would like to ask:

QUESTION. Does there exist a function $F(x)$, similar to (1-2), such that
  $$F_n=[x^n]\left(F(x)\right)^n? \tag{2-2}$$

REPLY [5 votes]: The power series $F(x)$ is closely related to the series of the "exponential reversion of Fibonacci numbers" $$R(x)=\sum_{n\ge1}r_n\frac{x^n}{n!}$$  (the $r_n$ are A258943, quoted in a comment). In fact it appears that, again in the notation of Henri Cohen, $$a_{n+1}=nr_n,$$ equivalently $$F'(x)=xR'(x).$$
So if the Fibonacci numbers are encapsulated by $$x=\sum_{n\ge1}F_n\frac{y^n}{n!}=y+\frac{y^2}{2!}+2\frac{y^3}{3!}+3\frac{y^4}{4!}+5\frac{y^5}{5!}+\cdots,$$ the reverse series of this is 
$$ 
y=R(x)=\sum_{n\ge1}r_n\frac{x^n}{n!}=x-\frac{x^2}{2!}+\frac{x^3}{3!}+\color{red}{2}\frac{x^4}{4!}-\color{red}{25}\frac{x^5}{5!}+-\cdots,$$
while $$\begin{align}F(x)=1+\sum_{n\ge1}a_n {x^n} &=1 + x+\frac{x^2}{ 2!}-2\frac{x^3}{ 3!}+3\frac{x^4}{4!}+8\frac{x^5}{5!}-125\frac{x^6}{6!}+-\cdots\\
&=1 + x+\frac{x^2}{ 2!}-2\frac{x^3}{ 3!}+3\frac{x^4}{4!}+4\cdot\color{red}{2}\frac{x^5}{5!}-5\cdot\color{red}{25}\frac{x^6}{6!}+-\cdots\end{align}.$$
Possibly this relationship is not even specific to the Fibonacci numbers.  
EDIT: It looks like the sequence $\{a_n\}$ has finally yielded its secrets. Given the conjectured "smoothness" of these coefficients (i.e. all prime factors are relatively small) as mentioned in Henri Cohen's answer, I have looked again into the factors and those quadratic sequences mentioned in the comments, and fortunately there are enough primes in them, such that finally I was able to find the pattern! We have for the sequence $\{r_n\}$ $${ r_n=\begin{cases} {(-1)^k} \prod\limits_{j=1}^k(n^2-5nj+5j^2) \quad\text{for 
 }\ n=2k-1, \\ \\  {(-1)^kk\cdot} \prod\limits_{j=1}^{k-1}(n^2-5nj+5j^2) \quad\text{for  }\ n=2k. \end{cases}}$$ Once found, it should not be hard to prove that rigorously.
As pointed out by Agno in a comment, we can reduce to linear factors and write the product in terms of the Gamma function simply as $$r_n= {\sqrt5^{ \,n-1 }\frac { \Gamma \left( \frac{5-\sqrt {5}}{10}n \right)}{  \Gamma \left( 1-\frac{5+\sqrt {5}}{10}n \right) }}.$$More generally, if we start with a Lucas sequence $$f_0=0,\ f_1=1,\ f_n=pf_{n-2}+qf_{n-1}\quad(n\ge2),$$ the reversed series has  $$\boxed{r_n= {\sqrt{4p+q^2}^{ \,n-1 }\frac { \Gamma \left[\dfrac n2 \Bigl(1-\dfrac{q}{\sqrt{4p+q^2}}  \Bigr)\right]}{ \Gamma \left[1-\dfrac n2 \Bigl(1+\dfrac{q}{\sqrt{4p+q^2}}  \Bigr)\right]}}}.$$ Note that whenever the argument in the denominator is a negative integer, the coefficient $r_n$ vanishes, e.g. this happens when $p=3,q=2$ for all $n\equiv0\pmod4$.  
As far as the sequence of the signs, it is after all quite regular and is in fact self-similar (that is, unless $\sqrt{4p+q^2}$ is rational). This self-similar behaviour of the signs can be seen by virtue of the (negative) argument of the Gamma function in the denominator, knowing that $\Gamma$ changes signs at each negative integer and the multiples of $\sqrt{4p+q^2}$ occurring in the argument do the rest. (Think e.g. of the self similarity features of the Wythoff sequence.)
For the reversion of the original Fibonacci sequence, I have displayed here the signs of the first $1500$ even and then the first $1500$ odd coefficients and found that their quasi periodicity comes out nicely when putting exactly $76$ in each row (writing "o" instead of "$-$" for better visibility). The "longest pairings" are colored:
all patterns "$++--++--++$" in yellow and
all patterns "$--++--++--$" in blue.