I've seen many special cases of this, so I thought it would be good to provide a more general version.
If this has been done before, that's OK.
If $f(x) =\sum_{n=0}^{\infty} a_nx^n $ where, for $n \ge m$, $a_n =\sum_{k=1}^m c_ka_{n-k} $, what is a closed form for $f(x)$?
My answer is $f(x) =\dfrac{\sum_{n=0}^{m-1} x^n(a_n-\sum_{k=1}^n c_ka_{n-k})}{1-\sum_{k=1}^m c_kx^k} $.
Here is my derivation.
$\begin{array}\\ f(x) &=\sum_{n=0}^{\infty} a_nx^n\\ &=\sum_{n=0}^{m-1} a_nx^n+\sum_{n=m}^{\infty} a_nx^n\\ &=\sum_{n=0}^{m-1} a_nx^n+\sum_{n=m}^{\infty} x^n\sum_{k=1}^m c_ka_{n-k}\\ &=\sum_{n=0}^{m-1} a_nx^n+\sum_{k=1}^m c_k\sum_{n=m}^{\infty} x^na_{n-k}\\ &=\sum_{n=0}^{m-1} a_nx^n+\sum_{k=1}^m c_kx^k\sum_{n=m}^{\infty} x^{n-k}a_{n-k}\\ &=\sum_{n=0}^{m-1} a_nx^n+\sum_{k=1}^m c_kx^k\sum_{n=m-k}^{\infty} x^{n}a_{n}\\ &=\sum_{n=0}^{m-1} a_nx^n+\sum_{k=1}^m c_kx^k\left(\sum_{n=0}^{\infty} x^{n}a_{n}-\sum_{n=0}^{m-k-1} x^{n}a_{n}\right)\\ &=\sum_{n=0}^{m-1} a_nx^n+\sum_{k=1}^m c_kx^k\left(f(x)-\sum_{n=0}^{m-k-1} x^{n}a_{n}\right)\\ &=\sum_{n=0}^{m-1} a_nx^n+f(x)\sum_{k=1}^m c_kx^k-\sum_{k=1}^m c_kx^k\sum_{n=0}^{m-k-1} x^{n}a_{n}\\ &=\sum_{n=0}^{m-1} a_nx^n+f(x)\sum_{k=1}^m c_kx^k-\sum_{k=1}^m c_k\sum_{n=0}^{m-k-1} x^{n+k}a_{n}\\ &=\sum_{n=0}^{m-1} a_nx^n+f(x)\sum_{k=1}^m c_kx^k-\sum_{k=1}^m c_k\sum_{n=k}^{m-1} x^{n}a_{n-k}\\ &=\sum_{n=0}^{m-1} a_nx^n+f(x)\sum_{k=1}^m c_kx^k-\sum_{n=1}^{m-1}\sum_{k=1}^n c_k x^{n}a_{n-k}\\ &=\sum_{n=0}^{m-1} a_nx^n+f(x)\sum_{k=1}^m c_kx^k-\sum_{n=1}^{m-1}x^n\sum_{k=1}^n c_ka_{n-k}\\ &=\sum_{n=0}^{m-1} x^n(a_n-\sum_{k=1}^n c_ka_{n-k})+f(x)\sum_{k=1}^m c_kx^k\\ \text{so}\\ f(x) &=\dfrac{\sum_{n=0}^{m-1} x^n(a_n-\sum_{k=1}^n c_ka_{n-k})}{1-\sum_{k=1}^m c_kx^k}\\ \end{array} $
If $m = 2$ then
$\begin{array}\\ f(x) &=\dfrac{\sum_{n=0}^{m-1} x^n(a_n-\sum_{k=1}^n c_ka_{n-k})}{1-\sum_{k=1}^m c_kx^k}\\ &=\dfrac{ a_0+x(a_1-c_1a_0)}{1-c_1x-c_2x^2}\\ \end{array} $
If $c_1 = c_2 = 1, a_0 = a_1 = 1$ (Fibonacci sequence),
$\begin{array}\\ f(x) &=\dfrac{ a_0+x(a_1-c_1a_0)}{1-c_1x-c_2x^2}\\ &=\dfrac{ 1}{1-x-x^2}\\ \end{array} $