Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers/Lemma

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Lemma for Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers

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:

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


Proof

Trying out a few values:

\(\ds a_0\) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds F_0\)


\(\ds a_1\) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds F_1\)


\(\ds a_2\) \(=\) \(\ds a_1 + a_0 + \map f 0\)
\(\ds \) \(=\) \(\ds F_1 + F_0 + \map f 0\)
\(\ds \) \(=\) \(\ds F_2 + F_1 \map f 0\)


\(\ds a_3\) \(=\) \(\ds a_2 + a_1 + \map f 1\)
\(\ds \) \(=\) \(\ds F_2 + F_1 \map f 0 + F_1 + \map f 1\)
\(\ds \) \(=\) \(\ds F_3 + F_2 \map f 0 + F_1 \map f 1\)


\(\ds a_4\) \(=\) \(\ds a_3 + a_2 + \map f 2\)
\(\ds \) \(=\) \(\ds F_3 + F_2 \map f 0 + F_1 \map f 1 + F_2 + F_1 \map f 0 + \map f 2\)
\(\ds \) \(=\) \(\ds F_4 + F_3 \map f 0 + F_2 \map f 1 + F_1 \map f 2\)


\(\ds a_5\) \(=\) \(\ds a_4 + a_3 + \map f 3\)
\(\ds \) \(=\) \(\ds F_4 + F_3 \map f 0 + F_2 \map f 1 + F_1 \map f 2 + F_3 + F_2 \map f 0 + F_1 \map f 1 + \map f 3\)
\(\ds \) \(=\) \(\ds F_5 + F_4 \map f 0 + F_3 \map f 1 + F_2 \map f 2 + F_1 \map f 3\)


The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

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


$\map P 0$ is the case:

\(\ds F_0 + \sum_{k \mathop = 0}^{0 - 2} F_{0 - k - 1} \map f k\) \(=\) \(\ds F_0\) as the summation is vacuous
\(\ds \) \(=\) \(\ds 0\) Definition of Fibonacci Number: $F_0 = 0$
\(\ds \) \(=\) \(\ds a_0\) by definition

Thus $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

\(\ds F_1 + \sum_{k \mathop = 0}^{1 - 2} F_{1 - k - 1} \map f k\) \(=\) \(\ds F_1\) as the summation is vacuous
\(\ds \) \(=\) \(\ds 1\) Definition of Fibonacci Number: $F_1 = 1$
\(\ds \) \(=\) \(\ds a_1\) by definition


Thus $\map P 1$ is seen to hold.


$\map P 2$ is the case:

\(\ds F_2 + \sum_{k \mathop = 0}^{2 - 2} F_{2 - k - 1} \map f k\) \(=\) \(\ds F_2 + F_1 \map f 0\)
\(\ds \) \(=\) \(\ds F_2 + \map f 0\) Definition of Fibonacci Number: $F_1 = 1$
\(\ds \) \(=\) \(\ds F_1 + F_0 + \map f 0\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds 1 + 0 + \map f 0\) Definition of Fibonacci Number: $F_1 = 1, F_0 = 0$
\(\ds \) \(=\) \(\ds a_1 + a_0 + \map f 0\) by definition


Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.


So this is the induction hypothesis:

$a_r = F_r + \ds \sum_{k \mathop = 0}^{r - 2} F_{r - k - 1} \map f k$

and:

$a_{r - 1} = F_{r - 1} + \ds \sum_{k \mathop = 0}^{r - 3} F_{r - k - 2} \map f k$


from which it is to be shown that:

$a_{r + 1} = F_{r + 1} + \ds \sum_{k \mathop = 0}^{r - 1} F_{r - k} \map f k$


Induction Step

This is the induction step:


\(\ds a_{r + 1}\) \(=\) \(\ds a_{r - 1} + a_r + \map f {r - 1}\)
\(\ds \) \(=\) \(\ds \paren {F_{r - 1} + \sum_{k \mathop = 0}^{r - 3} F_{r - k - 2} \map f k} + \paren {F_r + \sum_{k \mathop = 0}^{r - 2} F_{r - k - 1} \map f k} + \map f {r - 1}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds F_{r + 1} + \sum_{k \mathop = 0}^{r - 3} F_{r - k - 2} \map f k + \sum_{k \mathop = 0}^{r - 2} F_{r - k - 1} \map f k + F_1 \map f {r - 1}\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds F_{r + 1} + \sum_{k \mathop = 0}^{r - 3} \paren {F_{r - k - 2} + F_{r - k - 1} } \map f k + F_{r - \paren {r - 2} - 1} \map f {r - 2} + F_1 \map f {r - 1}\) factoring out the summation
\(\ds \) \(=\) \(\ds F_{r + 1} + \sum_{k \mathop = 0}^{r - 3} F_{r - k} \map f k + F_1 \map f {r - 2} + F_1 \map f {r - 1}\) Definition of Fibonacci Number and simplifying
\(\ds \) \(=\) \(\ds F_{r + 1} + \sum_{k \mathop = 0}^{r - 3} F_{r - k} \map f k + F_2 \map f {r - 2} + F_1 \map f {r - 1}\) Definition of Fibonacci Number: $F_1 = F_2 = 1$
\(\ds \) \(=\) \(\ds F_{r + 1} + \sum_{k \mathop = 0}^{r - 1} F_{r - k} \map f k\)

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

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

$\blacksquare$

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