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Section 48.2 Preliminaries

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.