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-a3)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}$ . All numbers in the sequence can be written in the $ $ x_i=x_1+d\cdot 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 $ k_i\geq 1$ such that each $ x_i$ can be written as $ $ x_i = x_{i-1}+d\cdot \prod_{j_3}^{i}k_j$ $

**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 $ n\geq 7$ , for every colouring $ $ \chi:[n]\rightarrow 2$ $ There exist a monochromatric sequence from $ R_3=\{(x,x+d,x+rd)\mid x\in[n],\ d\geq1, \ r\geq2\}$ . 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’=\frac{7(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 $ =\frac{7(k+1)!}{24}$ . Therefore in each term, I know that there exist a monochromatic sequence from $ 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.