## Section48.2Preliminaries

The Fibonacci sequence is defined as such; $f_1=f_2=1$ and $f_n=f_{n-1}+f_{n-2}$ for all $n\geq2\text{.}$

The Fibonacci set is $F=\{f_1,f_2,f_3...\}\text{.}$ Let $F_E$ be the set of even Fibonacci numbers, then $F_E=\{f_{3n}\}_{n=1}^\infty\text{.}$

A sequence is called a diffsequence of $S$ and written as $S$–diffsequence, if the sequence $\{x_i\}_{i=0}^k$ is strictly increasing and $x_i-x_{i-1}\in S$for all $i\leq k\text{.}$

A set, $S\text{,}$ is called $r$–accessible if for all colouring's of $\mathbb{Z}^+$ there exists arbitrarily long monochromatic $S$–diffsequence. A set is called accessible if a set is $r$–accessible for all $r\in\mathbb{Z}^+\text{.}$ The degree of accessibility of a set is the largest $r$ for which the set is $r$–accessible. and is written as $\text{doa }(S)\text{.}$

Bine's formula is an explicit equation for finding Fibonacci numbers. It requires the golden ratio, $\varphi={1\over2}(1+\sqrt{5})\text{:}$

\begin{equation*} f_n={1\over \sqrt{5}}\left(\varphi^n-(-\varphi)^{-n}\right)\text{.} \end{equation*}

There are multiple proofs including: induction, linear algebra and discrete methods. I will prove by induction, the base case is proving Binet's formula is true for $n=1$ and $n=2$ the algebra is left as an exercise for the reader. For induction, suppose that Binet's formula holds for $f_{n-1}$ and $f_{n-2}\text{.}$ It follows,

\begin{equation*} f_n=f_{n-1}+f_{n-2}= {1\over\sqrt{5}}[\varphi^{n-1}-(-\varphi)^{n-1}]+{1\over\sqrt{5}}[\varphi^{n-2}-(-\varphi)^{n-2}]\\ = {1\over\sqrt{5}}[\varphi^{n-2}(1+\varphi)-(-\varphi)^{n-2}(1-(-\varphi)^{-1})]\\ = {1\over\sqrt{5}}[\varphi^n-(-\varphi)^{-n}]\text{,}\\ \end{equation*}

which completes the induction.