Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers

Theorem
Let $f \left({n}\right)$ and $g \left({n}\right)$ be arbitrary arithmetic functions.

Let $\left\langle{a_n}\right\rangle$ be the sequence defined as:
 * $a_n = \begin{cases}

0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 1} + a_{n - 2} + f \left({n - 2}\right) & : n > 1 \end{cases}$

Let $\left\langle{b_n}\right\rangle$ be the sequence defined as:
 * $b_n = \begin{cases}

0 & : n = 0 \\ 1 & : n = 1 \\ b_{n - 1} + b_{n - 2} + g \left({n - 2}\right) & : n > 1 \end{cases}$

Let $\left\langle{c_n}\right\rangle$ be the sequence defined as:
 * $c_n = \begin{cases}

0 & : n = 0 \\ 1 & : n = 1 \\ c_{n - 1} + c_{n - 2} + x f \left({n - 2}\right) + y g \left({n - 2}\right) & : n > 1 \end{cases}$

where $x$ and $y$ are arbitrary.

Then $\left\langle{c_n}\right\rangle$ can be expressed in Fibonacci numbers as:
 * $c_n = x a_n + y b_n + \left({1 - x - y}\right) F_n$

Lemma
Hence also:


 * $b_n = F_n + \displaystyle \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} g \left({k}\right)$

Thus: