How to find operator with Fibonacci eigenvalues? How can I find the operator that satisfies this equation?
$$F_nx^n=Dx^n$$
Summing over $n$ we can rewrite this as $$\frac1{1-x-x^2}=D\frac1{1-x}$$
I am unsure whether this can be solved.

I am trying to solve it in a similar way to something like this:
$$nx^n=Dx^n$$ has solution $$D=x\frac{\mathrm d}{\mathrm dx}$$
I am not sure really how else to describe it. Other than the fact that $D$ can't really depend on $n$ which might make this equation unsolvable.
 A: Using Binet's formula: 
$$F_n=\frac{1}{\sqrt{5}}\left(\varphi^n-\psi^n\right)$$
where $\varphi=\frac{1+\sqrt{5}}{2}$ and $\psi = \frac{1-\sqrt{5}}{2}$.
Then $$(Df)(x) = \frac{1}{\sqrt{5}}\left(f(\varphi x)-f(\psi x)\right)$$
If you write the scaling operators $(S_\alpha f)(x)=f(\alpha x)$, then you can write the above as:
$$D=\frac{1}{\sqrt{5}}\left(S_\phi-S_\psi\right)$$
I suspect there is no closed form in terms of just $\frac{d}{dx}$ and $x$.
Note that the $S_\alpha$ have the following properties:
$$\begin{align}\frac{d}{dx}S_\alpha &= \alpha S_\alpha \frac{d}{dx}\\
S_\alpha S_\beta &= S_{\alpha\beta}\\
xS_\alpha &= \frac{1}\alpha(S_\alpha x)\end{align}$$
A: Note: There are (at least) two ways of indexing the Fibonacci series.  Below I work with the $F_0 = F_1 = 1$ convention rather than the $F_0 = 0, F_1 = F_2 = 1$ convention.

I assume you want $D$ to be a differential operator, given the tags.  Then we have
$$D = \sum_{k = 0}^\infty f_k(x) \frac{d^k}{dx^k}$$
for some functions $f_k(x)$, and we want
$$F_n x^n = D x^n = \sum_{k = 0}^\infty f_k(x) \frac{d^k}{dx^k} x^n = \sum_{k = 0}^n (n \; P \; k) \; f_k(x) \; x^{n-k}$$
Let's suppose $f_k(x) = c_k x^k$ for some constant $c_k$, making this say
$$F_n x^n = \sum_{k = 0}^n (n \; P \; k) c_k x^n$$
or just
$$F_n = \sum_{k = 0}^n (n \; P \; k) c_k = \sum_{k=0}^n \frac{n!}{(n-k)!} c_k$$
This gives a recursive definition of the $c_k$.  Now if we find the first few $c_k$ we see that they're fractions with denominators that look like k!, so let's try defining $a_k := k! c_k$ so that our expression becomes
$$F_n = \sum_{k=0}^n \frac{n!}{(n-k)! k!} k! c_k = \sum_{k=0}^n \binom{n}{k} a_k$$
Calculating the first several $a_k$ we see that they are
$$1, 0, 1, -1, 2, -3, 5, -8, 13, -21, \ldots$$
To me, that looks like $a_0 = 1$ and $a_k = (-1)^k F_{k-2}$ for $k > 0$.  I imagine you can check this actually works using Pascal's identity or something, though I didn't actually check it.  This would then give the formula
$$D = 1 + \sum_{k=1}^\infty \frac{(-1)^k F_{k-2}}{k!} \; x^k \; \frac{d^k}{dx^k},$$
at least assuming I got all the indices correct.
