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

Remark48.4.2.

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).