# Lucas Number as Element of Recursive Sequence

## Contents

## Theorem

Let $L_k$ be the $k$th Lucas number, defined as the sum of two Fibonacci numbers:

- $L_n = F_{n - 1} + F_{n + 1}$

Then $L_n$ can be defined as the $n$th element of the recursive sequence:

- $L_n = \begin{cases} 2 & : n = 0 \\ 1 & : n = 1 \\ L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$

## Proof

Proof by induction:

Let $L_n$ be the Lucas number defined as the sum of two Fibonacci numbers:

- $L_n = F_{n - 1} + F_{n + 1}$

For all $n \in \N$, let $\map P n$ be the proposition:

- $L_n = \begin{cases} 2 & : n = 0 \\ 1 & : n = 1 \\ L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$

### Basis for the Induction

We have that:

- $L_0 = F_{-1} + F_1 = 1 + 1 = 2$
- $L_1 = F_0 + F_2 = 0 + 1 = 1$
- $L_2 = F_1 + F_3 = 1 + 2 = 3$

Thus $\map P 0$, $\map P 1$ and $\map P 2$ hold.

$\map P 3$ is the case:

- $L_3 = F_2 + F_4 = 1 + 3 = 4$

So $\map P 3$, as $L_3 = L_1 + L_2$.

This is our basis for the induction.

### Induction Hypothesis

Let us make the supposition that, for some $k \in \N: k \ge 1$, the proposition $\map P j$ holds for all $j \in \N: 1 \le j \le k$.

We shall show that it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

- $\forall 1 \le j \le k: L_j = L_{j - 1} + L_{j - 2}$

Then we need to show:

- $L_{k + 1} = L_k + L_{k - 1}$

### Induction Step

This is our induction step:

\(\displaystyle L_{k + 1}\) | \(=\) | \(\displaystyle F_k + F_{k + 2}\) | by hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle F_{k - 2} + F_{k - 1} + F_k + F_{k + 1}\) | Definition of Fibonacci Number | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {F_{k - 2} + F_k} + \paren {F_{k - 1} + F_{k + 1} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle L_{k - 1} + L_k\) | by hypothesis |

Hence $L_n = L_{n - 2} + L_{n - 1}$ follows by the Second Principle of Mathematical Induction.

That is: $\sequence {L_n}$ is the sequence defined as:

- $L_n = \begin{cases} 2 & : n = 0 \\ 1 & : n = 1 \\ L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$

$\blacksquare$

## Sources

- 1971: George E. Andrews:
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