Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers

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Theorem

Let $\map f n$ and $\map g n$ be arbitrary arithmetic functions.

Let $\sequence {a_n}$ be the sequence defined as:

$a_n = \begin{cases}

0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 1} + a_{n - 2} + \map f {n - 2} & : n > 1 \end{cases}$


Let $\sequence {b_n}$ be the sequence defined as:

$b_n = \begin{cases}

0 & : n = 0 \\ 1 & : n = 1 \\ b_{n - 1} + b_{n - 2} + \map g {n - 2} & : n > 1 \end{cases}$


Let $\sequence {c_n}$ be the sequence defined as:

$c_n = \begin{cases}

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

where $x$ and $y$ are arbitrary.


Then $\sequence {c_n}$ can be expressed in Fibonacci numbers as:

$c_n = x a_n + y b_n + \paren {1 - x - y} F_n$


Proof

Lemma

Let $\map f n$ be an arbitrary arithmetic function.

Let $\sequence {a_n}$ be the sequence defined as:

$a_n = \begin{cases}

0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 1} + a_{n - 2} + \map f {n - 2} & : n > 1 \end{cases}$

Then:

$a_n = F_n + \ds \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \map f k$

$\Box$


Hence also:

$b_n = F_n + \ds \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \map g k$


Thus:

\(\ds c_n\) \(=\) \(\ds F_n + \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \paren {x \map f k + y \map g k}\)
\(\ds \) \(=\) \(\ds F_n + x \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \map f k + y \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \map g k\)
\(\ds \) \(=\) \(\ds F_n + x \paren {a_n - F_n} + y \paren {b_n - F_n}\)
\(\ds \) \(=\) \(\ds x a_n + y b_n + \paren {1 - x - y} F_n\)

$\blacksquare$


Sources

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