Explicit partition of $\mathbb{N}$ to demonstrate that $\{x,y,3x-y\}$ is not a Ramsey family

Question

Let $k, s \in \mathbb{N}$, and let $f_{1}, \ldots, f_{k}: \mathbb{N}^{s} \rightarrow \mathbb{Z}$. We call $\left\{f_{1}, \ldots, f_{k}\right\}$ a Ramsey family if for any finite coloring $\mathbb{N}=C_{1} \cup \cdots \cup C_{r}$, there exist $\mathrm{x} \in \mathbb{N}^{s}$ and $i \in\{1, \ldots, r\}$ such that $\left\{f_{1}(\mathrm{x}), \ldots, f_{k}(\mathrm{x})\right\} \subset C_{i}.$ For example $\{x,x+1\}$ is not a Ramsey family because the partition to odd and even naturals would be a counter example. By Rado's theorem we know that $\{x,y,3x-y\}$ is not a Ramsey family but I have been unable to construct an explicit partition of $\mathbb{N}$ to demonstrate this. Is there a way to do it?

Answer 1

Suppose you have a coloring of $\mathbb N$ with the property that whenever $x,y \in \mathbb N$ and $1.5x \le y \le 3x$, $x$ and $y$ are a different color. In such a coloring, a triple $\{x, y, 3x-y\}$ can never be monochromatic: both $y$ and $3x-y$ are at most $3x$, and the larger of them must be at least $1.5x$.

Can we find such a coloring? Yes! Define a sequence $(a_n)$ by the recurrence relation $a_n = \lceil \frac32 a_{n-1}\rceil$, with $a_1=1$. This sequence begins $1, 2, 3, 5, 8, \dots$ and it's not the Fibonacci numbers because the next term is $12$: it is sequence A061419 in the OEIS.

Now color the intervals $[a_n, a_{n+1})$ by cycling through four different colors. If $x \in [a_n, a_{n+1})$ and $1.5x \le y \le 3x$, then either $y \in [a_{n+1}, a_{n+2})$ or $y \in [a_{n+2}, a_{n+3})$ or $y \in [a_{n+3}, a_{n+4})$, because $a_{n+4} > 3a_{n+1}$. So $y$ is in an interval of a different color from $x$.