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Section 48.4 More About Even Fibonacci Numbers

First we extend the Fibonacci numbers by adding \(f_0=f^E_0=0\text{,}\) and producing a recurrence sequence for even Fibonacci numbers which is \(f^E_n=4f^E_{n-1}+f^E_{n-2}\text{.}\)

\begin{equation*} f^E_n=f_{3n}=2f_{3n-2}+f_{3n-3}\\ =3f_{3n-3}+f_{3n-4}+f_{3n-5}+f_{3n-6}\\ =4f_{3n-3}+f_{3n-6}=4f^E_{n-1}+f^E_{n-2}\\ \end{equation*}

Indeed, if we arrange the first \(p^2+2\) Fibonacci numbers into \(p^2+1\) consecutive doubles, \((f^E_{i-1},f^E_i)\) for all \(0\leq i\leq p^2+1\text{.}\) Then by Pigeonhole Principle there must exist two pairs with \((f^E_{i-1},f^E_i)\equiv(f^E_{j-1},f^E_j)~~(\bmod{p},\bmod{p})\text{.}\)

But this is true for the pairs of any consecutive \(p^2+2\) Fibonacci numbers, so there exists infinitely many \(f\in F^E\) such that \(p|f\text{.}\)

There is a nice proof in [48.7.2] (See Remark 2) that the modular colouring have infinitely long \(F^E\)–diffsequence. I independently discovered this fact while looking for a \(2\)–colouring that is not \(F^E\) accessible.

Define a modular coloring as any colouring that repeats itself after a period \(p\text{,}\) so if the colouring function is \(\chi\text{,}\) then \(\chi(a)=\chi(a+p)\text{.}\) Then by Lemma 1 there exists a \(f\in F^E\) such that \(p|f\) so the set \(\{a+nf\}_{n=0}^\infty\) is monochromatic under \(\chi\text{.}\) \qed

Remark 48.4.2.

A \(2\)–colouring of \(\{1,2,\ldots,9\}\) produces two monochromatic values that are a distance of either \(2,8\in F^E\text{.}\) Consecutive odd numbers have different colours otherwise there exists monochromatic digits a distance of 2 apart. So 1 and 9 which are a distance of 8 apart share the same colour by \(1\equiv9\pmod{4}\text{.}\) So if \(F^E\) is not 2–accessible then there exists infinitely many \(F^E\)–diffsequences of length at least two. To complicate matters for all \(i\not\equiv j\pmod{2}\text{,}\) \(f^E_i\) and \(f^E_j\) will have this same issue.

This complicates creating a colouring that avoids arbitrarily long \(F^E\)–diffsequences. Because if the proof follows the same theme as Theorem 3 in [48.7.2]. Then it likely requires creating a finite partition of the set of natural numbers, \(\{C_1,C_2,\ldots,C_k\}\) with \(C_i\cap C_j=\emptyset\text{,}\) \(i\not= j\) and \(\bigcup\limits_{i=1}^kC_k=\mathbb{Z}^+\) such that it is possible to find a monochromatic \(F^E\)–diffsequences but the next terms fall in a strictly higher hierarchy with digits in \(C_k\) being unable to make an even Fibonacci jump to a monochromatic digit.

Written in mathematical notation: if \(\{x_i\}_{i=0}^n\) is a monochromatic \(F^E\)–diffsequence in our \(2\)–colouring \(\chi\) then:

  • \(\forall w\in[1,n~~]\) \(x_{w-1}\in C_i\text{,}\) \(x_{w}\in C_j\rightarrow j\gt i\text{;}\) and

  • \(\forall x\in C_k, \forall f\in F^E~~\)\(\chi(x)\neq \chi(x+f)\text{.}\)

Thus all \(F^E\)–diffsequences have a maximum length of \(k\text{,}\) which would prove \(F^E\) is not \(2\)–accessible. This to me is very difficult, so I suspect a proof that has a different style then Theorem 3 would be better situated to prove \(F^E\) is not \(2\)–accessible (if true).