I thought about one relation while doing some other problem -
$ \displaystyle f_{n}(x) = x(f_{n-1}(x) + f_{n-2}(x)); \ f_0(x) = 1, f_1(x) = x $
I don't know the general procedure to solve such relations. What I tried is to compare the coefficients on both sides and get a recurrence for each of them.
$ [x^{n}]{f_{n}(x)} = [x^{n-1}]{f_{n-1}(x)} + [x^{n-1}]{f_{n-2}(x)} $
$ [x^n]{f_{n}(x)} = 1 $
Similarly,
$ [x^{n-1}]{f_{n}(x)} = [x^{n-2}]{f_{n-1}(x)} + [x^{n-2}]{f_{n-2}(x)} $
$ [x^{n-1}]{f_n(x)} = n-1 $
Similarly,
$ [x^{n-2}]{f_n(x)} = \dfrac{(n-2)(n-3)}{2} $
$ [x^{n-3}]{f_n(x)} = \dfrac{(n-4)(n-5)(n-6)}{6} $
Now, a general formula can be found relating coefficients.
$ \displaystyle [x^k]{f_n(x)} = \sum_{m}^{n-k}{[x^{k+1}]{f_m(x)}}$
Thus,
$ \displaystyle f_n(x) = x^n + \frac{(n-1)}{1!} x^{n-1} + \frac{(n-2)(n-3)}{2!} x^{n-2} + \frac{(n-3)(n-4)(n-5)}{3!} x^{n-3} + \cdots $
You can write it in rising/falling factorial if you like that notation.
This function looks interesting. And I would like to investigate more of its properties but I don't know what/how to search for.
Consider a second variable $y$, and rewrite the sequence as $f_0 = 1$, $f_1 = x$, and $f_n = x f_{n-1} + xyf_{n-2}$. Then each $f_n$ is homogeneous of degree $n$ in $x, y$. Then consider the formal series $f_0 + f_1 + f_2 + \cdots$. By rearranging the terms, you will find that this is a geometric series, equal to $1 + x(1+y) + \big(x(1+y)\big)^2 + \cdots$. If you expand this, you take the homogeneous parts of degree $0, 1, ...$ and specialize to $y=1$ you will find your original sequence. This is also why you get the binomial coefficients which you already found on your own.