Preliminaries

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Section 48.2 Preliminaries

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The Fibonacci sequence is defined as such;

$f_1=f_2=1$ and

$f_n=f_{n-1}+f_{n-2}$ for all

$n\geq2\text{.}$

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The Fibonacci set is

$F=\{f_1,f_2,f_3...\}\text{.}$ Let

$F_E$ be the set of even Fibonacci numbers, then

$F_E=\{f_{3n}\}_{n=1}^\infty\text{.}$

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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

$i\leq k\text{.}$

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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

$\text{doa }(S)\text{.}$

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Bine's formula is an explicit equation for finding Fibonacci numbers. It requires the golden ratio,

$\varphi={1\over2}(1+\sqrt{5})\text{:}$

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

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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,

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

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which completes the induction.