Dgdrxrt

Here is the problem, as given

Let $R_k$ be the set of increasing sequences $x_1<x_2<ldots<x_k$ of length $k$ such that there are integers $a_3, a_4,ldots,a_k$ (depending on the sequence) such that
$$x_3=a_3x_2+(1-a_3)x_1, x_4=a_4x_3+(1-a_4)x_2,ldots, x_k=a_kx_k-1+(1-a_k)x_k-2$$
Prove that every $2$ colouring of $[n]$ with $n geq 7(k + 1)!/24$ contains a monochromatic member of $R_k$.

Notations

We denote $[n]=1,2,ldots,n$.

For simplicity, to compare with Van der Wearden numbers $W(2,k)$, I'll call $P(k)$ the minimum $n$ such that every $2$ colouring of $[n]$ contains a monochromatic sequence of $R_k$.

Comments

We can first notice that the case $a_i=2$ for all $i$ induces all artithmetic progressions. Therefore $P(k)<W(2,k)$.

We can also rewrite $R_k$ (I found it easier to read as follow, it might not be the case for everyone). In such a sequence $x_i$'s, the difference between two consecutive term $x_i-x_i-1$ is a divisor of the difference $x_i-x_i-2$. Each number in the sequence can be written in the form$$x_i=x_1+dcdot r_i$$ with some constraints on the $r_i$. More precisely, $R_k$ is a set of sequence of the form $x_1,x_2,ldots,x_k$ with

• 
  $x_1$ is the initialisation,


• 
  $x_2$ defines the difference $d=x_2-x_1$,


• and then there exist a set of constants $a_igeq 2$ such that each $x_i$ can be written as
  $$x_i = x_i-2+dcdot a_icdotprod_j=3^i-1(a_j-1)$$
  Or equivalently
  $$x_i = x_i-1+dcdotprod_j=3^i(a_j-1)$$
  

Visualization : Therefore the sequence of $R_k$ are sequences of integers, where the difference of two consecutive number is a multiple of the difference between the two previous numbers:

$$ldots x_i overbraceqquad^alpha x_i+1 overbraceqquad^alphabetax_i+2ldots$$

Proof: Now working on the actual problem, I think that induction might be a good way to start. Looking then at the base case, $k=3$. I need to show that for $ngeq 7$, for every colouring
$$chi:[n]rightarrow 2$$
there exists a monochromatric sequence from $R_3=(x,x+d,x+rd)mid xin[n], dgeq1, rgeq2$.
This is easy enough by elimination (I don't write the details here, but noting that having two consecutive numbers with the same colouring is "bad", we know that 1,2,3,4 have alternating colors, then 5 must be colored as 2 and 4, 6 must be colored as 1,3, and whatever color for 7 will lead to a monochromatic sequence from $R_3$) .

Induction

This is where I fail. Suppose we have a colouring of $n'=frac7(k+2)!24$. I think a good way to start is to split $[n']$ as follow
$$underbrace[n], [n]+n, [n]+2n,ldots, [n]+(k+1)n_(k+2)text terms$$
With $n=frac7(k+1)!24$. Therefore in each term, I know that there exist a monochromatic sequence from $R_k$. Even more, for every constant $c$, the set $c+1,c+2,ldots,c+n$ contains a monochromatic sequence of $R_k$.

I should be able to construct the sequence from $R_k+1$ from theses sequences, but I'm not sure to see how.

Let $n_k=frac7(k+1)!24$. Suppose that for a given $k$, any 2-colouring of $[n_k]$ contains a monochromatic sequence from $R_k$. Suppose we have a colouring of $[n_k+1]$.

By induction hypothesis, in $[n_k]$ there exist a monochromatic sequence $x_1,ldots,x_k$ from $R_k$, w.l.g make it red.

Let $d=x_k-x_k-1$, the last difference in our original sequence. Because the $x_i$'s sequence is in $[n_k]$, we have $d<n(k)$. Let define
$$left{ beginarrayll
y_1=x_k+d\
y_i=x_k+id=y_i-1+d&text for i=1,ldots,k+1\
endarrayright.$$

Note that $n_k+1=frac7(k+2)!24=n_kcdot(k+2)$, therefore
$$y_k+1=x_k+(k+1)d < (k+2)n_k=n_k+1$$
and
$$forall iin1,ldots,k+1, y_i in [n_k+1]$$

If for any $i$, $y_i$ is red, then we are done as the sequence $x_1,ldots,x_k,y_i$ is a sequence of $R_k+1$. If they are all blue, then the sequence $y_i_i=1^k+1$ is a monochromatic $(k+1)$-arithmetic progression (with common difference $d$) in $[n_k+1]$, proving the induction step.

Therefore every $2$-colouring of $[n]$ with $n geq frac7(k + 1)!24$ contains a monochromatic member of $R_k$.