Minimum Number of Colours to Avoid k-Term Monochromatic Arithmetic Progressions

Abstract

By recalling van der Waerden theorem, there exists a least a positive integer $w=w(k;r)$ such that for any $n\ge w$, every r-colouring of $[1,n]$ admits a monochromatic k-term arithmetic progression. Let $k\ge 2$ and $r_k\left(n\right)$ denote the minimum number of colour required so that there exists a $r_k\left(n\right)$-colouring of $[1,n]$ that avoids any monochromatic k-term arithmetic progression. In this paper, we give necessary and sufficient conditions for $r_k(n+1)=r_k\left(n\right)$. We also show that $r_k\left(n\right)=2$ for all $k\le n\le 2{(k-1)}^2$ and give an upper bound for $r_p\left(p^m\right)$ for any prime $p\ge 3$ and integer $m\ge 2$.

1. Introduction

Let $\mathbb{N}$ be the set of positive integer, i.e., $\mathbb{N}=\{1,2,3,\dots \}$. A subset $A\subseteq \mathbb{N}$ is said to have an arithmetic progression (a.p.) of length k, if there exist $a,d\in \mathbb{N}$ such that

$$\begin{array}{c}\hfill a,a+d,a+2d,\dots ,a+(k-1)d\in A.\end{array}$$

The arithmetic progression of length k is also called a k-term a.p. with difference d, or we write ${\{a+ld\}}_{0\le l\le k-1}$.

There are many problems related to arithmetic progressions. One of them is to find an arrangement of $\left[n\right]=\{1,2,\dots ,n\}$ that avoids k-term arithmetic progressions. An arrangement of $\left[n\right]$ is a sequence $a_1,a_2,\dots ,a_n$ such that $\{a_1,a_2,\dots ,a_n\}=\left[n\right]$. Davis et al. [1] discovered an arrangement of $\left[n\right]=\{1,2,\dots ,n\}$ that avoids three-term monotone arithmetic progressions. By using the construction of certain magic square, Sim and Wong [2] proposed an arrangement of $\left[n\right]$ that avoids a k-term monotone arithmetic progression, but contains a $(k-1)$-term monotone arithmetic progression. For the infinite case, Davis et al. [1] and Sidorenko [3] showed that there is arrangement of $\mathbb{N}$ that avoids three-term monotone arithmetic progressions. Here, an arrangement of $\mathbb{N}$ is a sequence $a_1,a_2,\dots ,a_n,\dots ,$ such that $\{a_1,a_2,\dots ,a_n,\dots \}=\mathbb{N}$. They [1] also showed that there exists an arrangement of $\mathbb{N}$ that avoid five-term monotone arithmetic progressions. Recently, Geneson [4] constructed an arrangement of the set of integers that avoids six-term monotone arithmetic progressions. Up to now, it is still unknown whether there exists an arrangement of positive integers that avoids four-term monotone arithmetic progressions [5].

Another interesting problem is the colouring problem, which we will describe shortly. For a positive integer t, we denote the set $\{1,2,3,\dots ,t\}$ by $[1,t]$. A r-colouring of a set S is a function $\chi :S\to [1,r]$. A monochromatic k-term a.p. refers to a k-term a.p. such that all of its elements are of the same colour.

We now state the van der Waerden’s theorem, which was proved in 1927, see [6].

Theorem 1.

(Van der Waerden’s Theorem) Let $k,r\ge 2$ be integers. There exists a least positive integer $w=w(k;r)$ such that for any $n\ge w$, every r-colouring of $[1,n]$ admits a monochromatic k-term arithmetic progression.

In an effort to investigate the van der Waerden theorem, Erdős and Turán [7] approached the problem in the opposite direction: Given a positive integer n, what is the maximum size of a subset of $[1,n]$ that does not contain an arithmetic progression of length k? They define the following.

Definition 1.

For $k\ge 2$ and $n\ge 3$, let $\mathcal{S}$ be the collection of sets $S\subseteq [1,n]$ such that $\mathcal{S}$ does not contain an arithmetic progression of length k. Then ${\nu }_k\left(n\right)=\max \left(\right\{\left|S\right|:S\in \mathcal{S}\left\}\right)$.

Note that ${\nu }_k\left(n\right)$ is monotone increasing in $k\ge 3$. Suppose ${\nu }_k\left(n\right)=w$, then there is a subset of $w+1$ integers in $[1,n]$ which may contain only k-term a.p. but definitely does not contain a $k+1$ a.p. Hence ${\nu }_{k+1}\left(n\right)>{\nu }_k\left(n\right)$. They [7] showed that ${\nu }_3\left(2n\right)\le n$ for $n\ge 8$ and conjectured ${\nu }_3\left(n\right)=o\left(n\right)$. The conjectured was solved by Roth [8] and has been strengthened (see [9,10]). Studying ${\nu }_k\left(n\right)$ and having an upper bound on these numbers would lead to an upper bound on $w(k;r)$. In 2018, Blankenship, Cummings, and Taranchuk [11] showed that for $p\ge 5$ and q be primes, then $w(p+1;q)\ge p(q^p-1)+1$. The current best known general lower bound of this type is due to Kozik and Shabanov [12], who in 2016 proved that, for all $r\ge 2$

$$w(k;r)\ge c{r}^{k-1}$$

(1)

for some constant $c>0$.

Given two positive integers n and k, the anti-van der Waerden number aw$\left(\right[n],k)$ is the smallest r such that every exact r-colouring of $\left[n\right]$ contains a rainbow k-a.p., where an exact r-colouring is a colouring using all the r colours. For more results related to rainbow arithmetic progressions, see [13,14,15,16]. Recently, Li et al. [17] investigated the integer colourings with no rainbow 3-term arithmetic progression. They obtained the asymptotic number of r-colourings of $\left[n\right]$ without rainbow 3-term arithmetic progressions, and showed that the typical colourings with this property are 2-colourings.

Theorem 2

([5]). Let $k\ge 2,r\ge 2$ and assume ${\nu }_k\left(m\right)\le f(k,m)$, with $f(k,m)\le \frac{m-1}{r}$. Then $w(k;r)\le rf(k,m)+1$.

Let $k\ge 2$ and $r_k\left(n\right)$ denote the minimum number of colour required so that there exists a $r_k\left(n\right)$-colouring of $[1,n]$ that avoids any monochromatic k-term a.p. Since $[1,n]$ is coloured by $r_k\left(n\right)$ colours, clearly

$${\nu }_k\left(n\right)\ge ⌈{\displaystyle \frac{n}{r_k\left(n\right)}}⌉.$$

From [18], $r_k\left(n\right)<\frac{2nlogn}{{\nu }_k\left(n\right)}\left(1+o\left(1\right)\right)$ is obtained from the consequence of Symmetric Hypergraph Theorem. For sufficiently large n, $r_k\left(n\right)<\frac{3nlogn}{{\nu }_k\left(n\right)}$, see [19]. In this paper, we are focusing on the the minimum number of colour in $[1,n]$ to avoid a monochromatic k-term a.p. From the known results of $w(3;2)=9$, $w(3;3)=27, w(3;4)=76$ and $w(3;5)\ge 171$, see [20,21,22], we have $r_3\left(9\right)=3, r_3\left(27\right)=4, r_3\left(76\right)\ge 5$. Additionally, if $1<n_1<n_2$, then $r_k\left(n_1\right)\le r_k\left(n_2\right)$.

In this paper, we give necessary and sufficient conditions for $r_k(n+1)=r_k\left(n\right)$ (Theorem 3). We also show that $r_k\left(n\right)=2$ for all $k\le n\le 2{(k-1)}^2$ (Theorem 4). Finally, we give an upper bound for $r_p\left(p^m\right)$ for any prime $p\ge 3$ and integer $m\ge 2$ (Theorem 5).

2. Main Results

Note that $r_2\left(n\right)=n$. Therefore, we shall assume that $k\ge 3$.

Lemma 1.

$$r_k(n_1+n_2)\le r_k\left(n_1\right)+r_k\left(n_2\right).$$

Proof. 

Colour $[1,n_1]$ with $r_k\left(n_1\right)$ colours that avoids any monochromatic k-term a.p. and colour $[n_1+1,n_1+n_2]$ with different $r_k\left(n_2\right)$ colours that avoids any monochromatic k-term a.p.. Note that this colouring avoids any monochromatic k-term a.p. in $[1,n_1+n_2]$. Hence, the lemma follows. □

The following corollary is a consequence of Lemma 1 by noting that $r_k\left(i\right)=1$ for $1\le i\le k-1$.

Corollary 1.

For $1\le i\le k-1$,

$$r_k(n+i)\le r_k\left(n\right)+1.$$

Since $r_k\left(n\right)\le r_k(n+1)$, we have the following corollary.

Corollary 2.

$r_k(n+1)=r_k\left(n\right)$ or $r_k\left(n\right)+1$.

Corollary 3.

$r_k(w(k;l)-1)=l$ and $r_k\left(n\right)=l+1$ for $w(k;l)\le n\le w(k;l+1)-1$.

Proof. 

There is a l-colouring of $[1,w(k;l)-1]$ that avoids any k-term a.p.. Therefore, $r_k(w(k;l)-1)\le l$. If $r_k(w(k;l)-1)\le l-1$, then by Corollary 2, $r_k\left(w(k;l)\right)\le l$. This is not possible as any l-colouring of $[1,w(k;l\left)\right]$ has a monochromatic k-term a.p. Hence, $r_k(w(k;l)-1)=l$.

Since any l-colouring of $[1,w(k;l\left)\right]$ has a monochromatic k-term a.p., by Corollary 2, $r_k\left(w(k;l)\right)=l+1$. By first part of the corollary, $r_k(w(k;l+1)-1)=l+1$. Again, by Corollary 2, $r_k\left(n\right)=l+1$ for $w(k;l)\le n\le w(k;l+1)-1$. □

For $l\ge 1$, a family of sets $\{A_1,A_2,\dots ,A_l\}$ is called a partition of $[1,n]$ if

(a)

$A_i\cap A_j=⌀$ for $i\ne j$;

(b)

${\bigcup }_{i=1}^l{A}_i=[1,n]$.

Here, we allow $A_j=⌀$ for some j. If each $A_j$ avoids any k-term a.p., the partition $\{A_1,A_2,\dots ,A_l\}$ is called a free k-term a.p. partition.

Let $r_k\left(n\right)=l$. Then, there exists a l-colouring of $[1,n]$ that avoids any monochromatic k-term a.p. For $1\le i\le l$, let

$$\begin{array}{c}\hfill A_i=\{a\in [1,n]:a\mathrm{is}\mathrm{coloured}\mathrm{with}\mathrm{colour}i\}.\end{array}$$

Note that $\{A_1,A_2,\dots ,A_l\}$ is a free k-term a.p. partition of $[1,n]$. So, $r_k\left(n\right)$ can be considered as the minimum size of a free k-term a.p. partition of $[1,n]$.

A subset $A\subseteq \mathbb{N}$ is said to have a pure arithmetic progression of length k (or pure k-term a.p.), if there exists a $d\in \mathbb{N}$ such that

$$\begin{array}{c}\hfill d,2d,3d,\dots ,kd\in A.\end{array}$$

Given a subset $A\subseteq \mathbb{N}$ and $c\in \mathbb{N}$, we write

$$\begin{array}{c}\hfill c-A=\{c-a:a\in A\}\mathrm{and}c+A=\{c+a:a\in A\}.\end{array}$$

Lemma 2.

Let $\{A_1,A_2,\dots ,A_l\}$ be a partition of $[1,n]$ such that each $A_i$ does not contain any k-term a.p. Suppose there is a $A_{i_0}$ that does not contain any pure $(k-1)$-term a.p. Let $B_{i_0}=((n+1)-A_{i_0})\cup \{n+1\}$ and $B_j=(n+1)-A_j$ for all $j\ne i_0$. Then, $\{B_1,B_2,\dots ,B_l\}$ is a free k-term a.p. partition of $[1,n+1]$.

Proof. 

Note that $(n+1)-[1,n]=[1,n]$. So, it is not hard to see that $\{B_1,B_2,\dots ,B_l\}$ is a partition of $[1,n+1]$. Now, if $B_j$ contains a k-term a.p. for some $j\ne i_0$, say ${\{a+sd\}}_{0\le s\le k-1}$, then $A_j$ contains the k-term a.p. ${\{((n+1)-a-(k-1)d)+ld\}}_{0\le l\le k-1}$, a contradiction. So, we may assume that $B_j$ does not contain any k-term a.p. for all $j\ne i_0$.

Suppose $B_{i_0}$ contains a k-term a.p. for some $j\ne i_0$, say ${\{a^{\prime }+l{d}^{\prime }\}}_{0\le l\le k-1}$. If $a^{\prime }+(k-1)d^{\prime }\ne n+1$, then $A_{i_0}$ contains the k-term a.p. ${\{((n+1)-a^{\prime }-(k-1)d^{\prime })+s{d}^{\prime }\}}_{0\le s\le k-1}$, a contradiction. So, we may assume that $a^{\prime }+(k-1)d^{\prime }=n+1$. This implies that $A_{i_0}$ contains the pure $(k-1)$-term a.p. ${\left\{s{d}^{\prime }\right\}}_{1\le s\le k-1}$, a contradiction. □

Theorem 3.

Suppose $r_k\left(n\right)=l$. Then, $r_k(n+1)=l$ if and only if there exists a free k-term a.p. partition $\{A_1,A_2,\dots ,A_l\}$ of $[1,n]$ such that $A_{i_0}$ does not contain any pure $(k-1)$-term a.p. for some $i_0\in [1,l]$.

Proof. 

If there exists a free k-term a.p. partition $\{A_1,A_2,\dots ,A_l\}$ of $[1,n]$ and $A_{i_0}$ does not contain any pure $(k-1)$-term a.p., then by Lemma 2, $\{B_1,B_2,\dots ,B_l\}$ is a free k-term a.p. partition of $[1,n+1]$ where $B_{i_0}=((n+1)-A_{i_0})\cup \{n+1\}$ and $B_j=(n+1)-A_j$ for all $j\ne i_0$. It follows from Corollary 2 that $r_k(n+1)=l$.

Suppose $r_k(n+1)=l$. Then, there exists a free k-term a.p. partition $\{C_1,C_2,\dots ,C_l\}$ of $[1,n+1]$. By relabelling if necessary, we may assume that $(n+1)\in C_1$. Let $D_1=(n+1)-(C_1\setminus \{n+1\})$ and $D_j=(n+1)-C_j$ for $2\le j\le l$. Note that $\{D_1,D_2,\dots ,D_l\}$ is a free k-term a.p. partition of $[1,n]$. If $D_1$ contains a pure $(k-1)$-term a.p., say ${\left\{sd\right\}}_{1\le s\le k-1}$, then ${\{(n+1)-(k-1)d+sd\}}_{0\le s\le k-1}$ is a k-term a.p. in $C_1$, a contradiction. Hence, $D_1$ does not contain any pure $(k-1)$-term a.p. □

Corollary 4.

Suppose $r_k\left(n\right)=l$ and $r_k(n+1)=l+1$. Then, $r_k(n+j)=l+1$ for all $1\le j\le 2k-2$.

Proof. 

Let $\{A_1,A_2,\dots ,A_l\}$ be a free k-term a.p. partition of $[1,n]$. Since $r_k(n+1)=l+1$, by Theorem 3, each $A_i$ contains a pure $(k-1)$-term a.p. Let

$$\begin{array}{cc}\hfill B_i& =(k-1)+A_i\mathrm{for}\mathrm{all}1\le i\le l;\hfill \\ \hfill B_{l+1}& =\{1,2,\dots ,k-1,n+k,n+k+1,\dots ,n+2k-2\}.\hfill \end{array}$$

Note that $\{B_1,B_2,\dots ,B_{l+1}\}$ is a free k-term a.p. partition of $[1,n+2k-2]$. Therefore, $r_k(n+j)\le l+1$ for all $1\le j\le 2k-2$. The equality follows from Corollary 2. □

Corollary 5.

$$w(k;l+1)\ge k+2l(k-1).$$

Proof. 

By Corollary 3, $r_k(w(k;l)-1)=l$ and $r_k\left(n\right)=l+1$ for $w(k;l)\le n\le w(k;l+1)-1$. It follows from Corollary 4 that

$$\begin{array}{c}\hfill w(k;l+1)-1\ge \left(w\right(k;l)-1)+2k-2,\end{array}$$

which is equivalent to $w(k;l+1)\ge w(k;l)+2k-2$. Since $w(k;1)=k$, we have

$$\begin{array}{cc}\hfill w(k;l+1)& \ge w(k;l)+2(k-1)\hfill \\ \hfill & \ge w(k;l-1)+4(k-1)\hfill \\ \hfill & \ge w(k;l-2)+6(k-1)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \ge w(k;1)+2l(k-1)=k+2l(k-1).\hfill \end{array}$$

 □

Let $a,b\in \mathbb{N}$ with $a<b$. We shall write $[a,b]=\{a,a+1,\dots ,b\}$ and $[a,a]=\left\{a\right\}$.

Theorem 4.

$r_k\left(n\right)=2$ for all $k\le n\le 2{(k-1)}^2$.

Proof. 

For $1\le i\le k-1$ and $0\le j\le k-2$, let

$$\begin{array}{cc}\hfill A(i,j)& =[1,i]\cup \bigcup _{s=1}^j[i+2s(k-1)-(k-2),i+2s(k-1)]\hfill \\ \hfill B(i,j)& =[i+2j(k-1)+1,2i+2j(k-1)]\cup \bigcup _{s=1}^j[i+(2s-1)(k-1)-(k-2),i+(2s-1)(k-1)].\hfill \end{array}$$

In particular, when $j=0$

$$\begin{array}{cc}\hfill A(i,0)& =[1,i]\hfill \\ \hfill B(i,0)& =[i+1,2i].\hfill \end{array}$$

Note that $\left\{A\right(i,j),B(i,j\left)\right\}$ is a partition of $[1,2i+2j(k-1\left)\right]$. Now,

$$\begin{array}{c}\hfill (2i+2j(k-1)+1)-[i+2j(k-1)+1,2i+2j(k-1\left)\right]=[1,i]\end{array}$$

and

$$\begin{array}{cc}\hfill (2i+2j(k-1)+1)-& [i+(2s-1\left)\right(k-1)-(k-2),i+(2s-1\left)\right(k-1\left)\right]\hfill \\ \hfill & =[i+2(j-s+1\left)\right(k-1)-(k-2),i+2(j-s+1\left)\right(k-1\left)\right].\hfill \end{array}$$

So, $(2i+2j(k-1)+1)-B(i,j)=A(i,j)$.

Clearly, $\left\{A\right(i,0),B(i,0\left)\right\}$ is a free k-term a.p. partition for $1\le i\le k-1$. Suppose $\{A(i_0,j_0),B(i_0,j_0)\}$ is a free k-term a.p. partition for some $1\le i_0\le k-2$ and $0\le j_0\le k-2$. We claim that $\{A(i_0+1,j_0),B(i_0+1,j_0)\}$ is also a free k-term a.p. partition.

Since $i_0\le k-2$, $A(i_0,j_0)$ does not contain ${\left\{m\right\}}_{1\le m\le k-1}$. Suppose $A(i_0,j_0)$ contains a pure $(k-1)$-term a.p. ${\left\{md\right\}}_{1\le m\le k-1}$ where $d\ge 2$. Note that

$$d(k-1)\in [i_0+2s^{\prime }(k-1)-(k-2),i_0+2s^{\prime }(k-1)],$$

for some $1\le s^{\prime }\le j_0$. If $2\le d\le k-1$, there exists a $m^{\prime }\in [1,k-2]$ such that

$$m^{\prime }d\in [i_0+(2s^{\prime }-1)(k-1)-(k-2),i_0+(2s^{\prime }-1)(k-1)]\subseteq B(i_0,j_0),$$

a contradiction. So, we may assume that $d\ge k$. If both $m_{1}d$ and $(m_1+1)d$ are in

$$[i+2s(k-1)-(k-2),i+2s(k-1\left)\right],$$

for some $1\le s\le j_0$, then $d\le k-2$, a contradiction. So, we may assume that for each s, $[i+2s(k-1)-(k-2),i+2s(k-1\left)\right]$ contains at most one element of the form $md$ where $1\le m\le k-1$. This implies that $d\in [1,i_0]$ and $d\le i_0\le k-2$, a contradiction. Hence, $A(i_0,j_0)$ does not contain any pure $(k-1)$-term a.p. By Lemma 2,

$$\begin{array}{cc}\hfill \{((2i_0+2j_0(k-1)+1)-A(i_0,j_0))& \cup \{2i_0+2j_0(k-1)+1\},(2i_0+2j_0(k-1)+1)-B(i_0,j_0)\}\hfill \\ \hfill & =\{B(i_0,j_0)\cup \{2i_0+2j_0(k-1)+1\},A(i_0,j_0)\},\hfill \end{array}$$

is a free k-term a.p. partition. Note that

$$\begin{array}{c}\hfill ((2i_0+2j_0(k-1)+2)-A(i_0,j_0))\cup \{2i_0+2j_0(k-1)+2\}=B(i_0+1,j_0),\end{array}$$

and

$$\begin{array}{c}\hfill (2i_0+2j_0(k-1)+2)-(B(i_0,j_0)\cup \{2i_0+2j_0(k-1)+1\})=A(i_0+1,j_0).\end{array}$$

By Lemma 2, $\{A(i_0+1,j_0),B(i_0+1,j_0)\}$ is a free k-term a.p. partition.

Suppose $\{A(k-1,j_0),B(k-1,j_0)\}$ is a free k-term a.p. partition for some $0\le j_0\le k-3$. We claim that $\{A(1,j_0+1),B(1,j_0+1)\}$ is also a free k-term a.p. partition. Note that

$$\begin{array}{cc}\hfill A(k-1,j_0)=[1,k-1]\cup & \bigcup _{s=1}^{j_0}[1+2s(k-1),(2s+1)(k-1)]\hfill \\ \hfill B(k-1,j_0)=[k+2j_0(k-1),& 2(j_0+1)(k-1)]\cup \hfill \\ \hfill & \bigcup _{s=1}^{j_0}[1+(2s-1)(k-1),2s(k-1)].\hfill \end{array}$$

Clearly, $B(k-1,j_0)$ does not contain ${\left\{m\right\}}_{1\le m\le k-1}$. Suppose $B(k-1,j_0)$ contains a pure $(k-1)$-term a.p., say ${\left\{md\right\}}_{1\le m\le k-1}$ where $d\ge 2$. If $2\le d\le k-1$, then $d\in [1,k-1]\subseteq A(k-1,j_0)$, a contradiction. So, we may assume that $d\ge k$. If both $m_{1}d$ and $(m_1+1)d$ are in

$$[1+(2s-1\left)\right(k-1),2s(k-1\left)\right],$$

for some $1\le s\le j_0$, then $d\le k-2$, a contradiction. So, we may assume that for each s, $[1+(2s-1\left)\right(k-1),2s(k-1\left)\right]$ contains at most one element of the form $md$ where $1\le m\le k-1$. Since $j_0\le k-3$, we must have both $(k-2)d$ and $(k-1)d$ in $[k+2j_0(k-1),2(j_0+1)(k-1)]$. This implies that $d\le k-2$, a contradiction. Hence, $B(k-1,j_0)$ does not contain any pure $(k-1)$-term a.p. Note that

$$\begin{array}{cc}\hfill ((2(j_0+1)(k-1)+1)-B(k-1,j_0))& \cup \{2(j_0+1)(k-1)+1\}\hfill \\ \hfill & =A(k-1,j_0)\cup \{2(j_0+1)(k-1)+1\},\hfill \end{array}$$

and

$$\begin{array}{c}\hfill (2(j_0+1)(k-1)+1)-A(k-1,j_0)=B(k-1,j_0).\end{array}$$

By Lemma 2, $\{A(k-1,j_0)\cup \{2(j_0+1)(k-1)+1\},B(k-1,j_0)\}$ is a free k-term a.p. partition. Note that

$$\begin{array}{c}\hfill (2(j_0+1)(k-1)+2)-B(k-1,j_0))\cup \{2(j_0+1)(k-1)+2\}=B(1,j_0+1),\end{array}$$

and

$$\begin{array}{c}\hfill (2i_0+2j_0(k-1)+2)-(A(k-1,j_0)\cup \{2(j_0+1)(k-1)+1\})=A(1,j_0+1).\end{array}$$

By Lemma 2, $\{A(i_0+1,j_0),B(i_0+1,j_0)\}$ is a free k-term a.p. partition. By double induction, $\left\{A\right(i,j),B(i,j\left)\right\}$ is a free k-term a.p. partition of $[1,2i+2j(k-1\left)\right]$. In particular, $\left\{A\right(k-1,k-2),B(k-1,k-2\left)\right\}$ is a free k-term a.p. partition of $[1,2{(k-1)}^2]$. Hence, $r_k\left(n\right)=2$ for all $k\le n\le 2{(k-1)}^2$. □

Note that Theorem 4 is best possible because when $k=3$, $r_3\left(n\right)=2$ for all $3\le n\le 8$ and $r_3\left(9\right)=3$. Note also that for $k\ge 4$, $2{(k-1)}^2\ge k^2$, therefore $r_k\left(k^2\right)=2$.

The following corollary follows from Corollary 3.

Corollary 6.

$w(k;2)\ge 2{(k-1)}^2+1$.

Lemma 3.

Let $p\ge 3$ be a prime and $A\subseteq \mathbb{N}$. For each $i\in [1,p]$, let

$$\begin{array}{c}\hfill A_i=\{b\in A:b\equiv imodp\}.\end{array}$$

Suppose $A_{i_0}=⌀$ for some $i_0$. If A contains a p-term a.p., ${\{a+sd\}}_{0\le s\le p-1}$, then $gcd(p,d)=p$ and the p-term a.p is contained entirely in $A_{a_0}$ for some $a_0\in [1,p]$ with $a_0\equiv amodp$.

Proof. 

Suppose $gcd(p,d)=1$. Then,

$$\begin{array}{c}\hfill \{a,a+d,\dots ,a+(p-1\left)d\right\}\equiv \{1,2,\dots ,p\}modp.\end{array}$$

This implies that $A_i\ne ⌀$ for all i, a contradiction. Hence, $gcd(p,d)=p$ and the p-term a.p is contained entirely in $A_{a_0}$. □

For each real number x, let $⌈x⌉$ be the smallest integer such that $⌈x⌉-1<x\le ⌈x⌉$.

Lemma 4.

Let $p\ge 3$ be a prime. If $p\le k$, then

$$r_k\left(np\right)\le r_k\left(n\right)+⌈\frac{r_k\left(n\right)}{p-1}⌉$$

Proof. 

For each $j\in [1,p]$, let

$$\begin{array}{c}\hfill A_j=\{a\in [1,np]:a\equiv jmodp\}.\end{array}$$

and

$$\begin{array}{c}\hfill f_j:[1,n]\to A_j\end{array}$$

be defined by $f_j\left(u\right)=j+(u-1)p$ for all $u\in [1,n]$. Note that $f_j$ is a bijection.

Let $r_k\left(n\right)=l$ and $\{B_1,B_2,\dots ,B_l\}$ be a free k-term a.p. partition of $[1,n]$. Then,

$$\{f_j\left(B_1\right),f_j\left(B_2\right),\dots ,f_j\left(B_l\right)\}$$

is a free k-term a.p. partition of $A_j$. Let

$$\begin{array}{c}\hfill A_{i,j}=f_j\left(B_i\right),\end{array}$$

for all $1\le i\le l$ and $1\le j\le p$. Then, $A_j={\bigcup }_{i=1}^l{A}_{i,j}$.

Let $l=q(p-1)+e$ where $0\le e<p-1$. Let

$$\begin{array}{cc}\hfill C_1& =\{1,2,\dots ,p-1\}\hfill \\ \hfill C_2& =\{1+(p-1),2+(p-1),\dots ,p-1+(p-1\left)\right\}\hfill \\ \hfill & ⋮\hfill \\ \hfill C_q& =\{1+(q-1\left)\right(p-1),2+(q-1\left)\right(p-1),\dots ,p-1+(q-1\left)\right(p-1\left)\right\}\hfill \\ \hfill C_{q+1}& =\{1+q(p-1),2+q(p-1),\dots ,e+q(p-1\left)\right\}.\hfill \end{array}$$

Then, $\{C_1,C_2,\dots ,C_{q+1}\}$ is a partition of $[1,l]$ if $e\ge 1$ and $\{C_1,C_2,\dots ,C_q\}$ is a partition of $[1,l]$ if $e=0$. Let $q_0=⌈\frac{l}{p-1}⌉$. Then, $q_0=q+1$ if $e\ge 1$ and $q_0=q$ if $e=0$.

For each $b\in \mathbb{N}$, let $\overline{b}\in [1,p]$ be such that $\overline{b}\equiv bmodp$. Now, for each $t\in [1,q_0]$, let

$$\begin{array}{c}\hfill D_t=\bigcup _{b\in C_t}{A}_{b,\overline{b}}\end{array}$$

Since $|C_t|=p-1$, there is a $i_0$ such that $D_t\left(i_0\right)=\{d\in D_t:d\equiv i_{0}modp\}=⌀$. If $D_t$ contains a k-term a.p., say ${\{a+sd\}}_{0\le s\le k-1}$, then it contains the p-term a.p. ${\{a+sd\}}_{0\le s\le p-1}$. By Lemma 3, $gcd(p,d)=p$. This implies that $A_{b_1,{\overline{b}}_1}$ contains the k-term a.p. ${\{a+sd\}}_{0\le s\le k-1}$ for some $b_1\in C_t$. Since $A_{b_1,{\overline{b}}_1}=f_{{\overline{b}}_1}\left(B_{b_1}\right)$, $B_{b_1}$ contains the k-term a.p. ${\{a^{\prime }+s{d}^{\prime }\}}_{0\le s\le k-1}$ where $a^{\prime }=\frac{a-{\overline{b}}_1}{p}+1$ and $d^{\prime }=\frac{d}{p}$. This is not possible as $\{B_1,B_2,\dots ,B_l\}$ is a free k-term a.p. partition of $[1,n]$. Thus, $D_t$ does not contain any k-term a.p.

For each $i\in [1,l]$, let

$$\begin{array}{c}\hfill E_i=\left(\bigcup _{j=1}^p{A}_{i,j}\right)\setminus A_{i,\overline{i}}\end{array}$$

Clearly, $E_i\left(\overline{i}\right)=\{d\in E_i:d\equiv \overline{i}modp\}=⌀$. If $E_i$ contains a k-term a.p., say ${\{a+sd\}}_{0\le s\le k-1}$, then it contains the p-term a.p. ${\{a+sd\}}_{0\le s\le p-1}$. By Lemma 3, $gcd(p,d)=p$. This implies that $A_{i,j_1}$ contains the k-term a.p. ${\{a+sd\}}_{0\le s\le k-1}$ for some $j_1\in [1,p]\setminus \left\{\overline{i}\right\}$. Since $A_{i,j_1}=f_{j_1}\left(B_i\right)$, $B_i$ contains the k-term a.p. ${\{a^{\prime }+s{d}^{\prime }\}}_{0\le s\le k-1}$ where $a^{\prime }=\frac{a-j_1}{p}+1$ and $d^{\prime }=\frac{d}{p}$. This is not possible as $\{B_1,B_2,\dots ,B_l\}$ is a free k-term a.p. partition of $[1,n]$. Thus, $E_i$ does not contain any k-term a.p.

Now,

$$\{E_1,E_2,\dots ,E_l,D_1,D_2,\dots ,D_{q_0}\}$$

is a free k-term a.p. partition of $[1,pn]$. Hence, $r_k\left(np\right)\le l+q_0$ and the lemma follows. □

Theorem 5.

Let $p\ge 3$ be a prime. Then, $r_p\left(p\right)=2$,

$$\begin{array}{c}\hfill r_p\left(p^2\right)=\left\{\begin{array}{cc}2,\hfill & \mathrm{for}p\ge 5;\hfill \\ \hfill & \\ 3,\hfill & \mathrm{for}p=3,\hfill \end{array}\right.\end{array}$$

and for$m\ge 3$,

$$\begin{array}{c}\hfill r_p\left(p^m\right)<\left\{\begin{array}{cc}(p+1){\left(\frac{p}{p-1}\right)}^{m-2}-p+1,\hfill & \mathrm{for}p\ge 5;\hfill \\ \hfill & \\ 5{\left(\frac{3}{2}\right)}^{m-2}-2,\hfill & \mathrm{for}p=3.\hfill \end{array}\right.\end{array}$$

Proof. 

Clearly, $r_p\left(p\right)=2$. By Lemma 4,

$$\begin{array}{cc}\hfill r_p\left(p^m\right)& <r_p\left(p^{m-1}\right)+\frac{r_p\left(p^{m-1}\right)}{p-1}+1\hfill \\ \hfill & =\left(\frac{p}{p-1}\right)r_p\left(p^{m-1}\right)+1\hfill \\ \hfill & <\left(\frac{p}{p-1}\right)\left(\left(\frac{p}{p-1}\right)r_p\left(p^{m-2}\right)+1\right)+1\hfill \\ \hfill & ={\left(\frac{p}{p-1}\right)}^2{r}_p\left(p^{m-2}\right)+\left(\frac{p}{p-1}\right)+1\hfill \\ \hfill & ⋮\hfill \\ \hfill & <{\left(\frac{p}{p-1}\right)}^{m-2}{r}_p\left(p^2\right)+{\left(\frac{p}{p-1}\right)}^{m-3}+\dots +\left(\frac{p}{p-1}\right)+1.\hfill \end{array}$$

Suppose $p\ge 5$. Then, $2{(p-1)}^2\ge p^2$. By Theorem 4, $r_p\left(p^2\right)=2$. Thus,

$$\begin{array}{cc}\hfill r_p\left(p^m\right)& <2{\left(\frac{p}{p-1}\right)}^{m-2}+{\left(\frac{p}{p-1}\right)}^{m-3}+\dots +\left(\frac{p}{p-1}\right)+1\hfill \\ \hfill & =2{\left(\frac{p}{p-1}\right)}^{m-2}+\frac{{\left(\frac{p}{p-1}\right)}^{m-2}-1}{\left(\frac{p}{p-1}\right)-1}\hfill \\ \hfill & =2{\left(\frac{p}{p-1}\right)}^{m-2}+\left({\left(\frac{p}{p-1}\right)}^{m-2}-1\right)(p-1)\hfill \\ \hfill & =(p+1){\left(\frac{p}{p-1}\right)}^{m-2}-(p-1).\hfill \end{array}$$

Suppose $p=3$. Then, $r_3\left(3^2\right)=3$. So,

$$\begin{array}{cc}\hfill r_3\left(3^m\right)& <{\left(\frac{3}{2}\right)}^{m-2}{r}_3\left(3^2\right)+{\left(\frac{3}{2}\right)}^{m-3}+\dots +\left(\frac{3}{2}\right)+1\hfill \\ \hfill & =3{\left(\frac{3}{2}\right)}^{m-2}+{\left(\frac{3}{2}\right)}^{m-3}+\dots +\left(\frac{3}{2}\right)+1\hfill \\ \hfill & =3{\left(\frac{3}{2}\right)}^{m-2}+\left({\left(\frac{3}{2}\right)}^{m-2}-1\right)2\hfill \\ \hfill & =5{\left(\frac{3}{2}\right)}^{m-2}-2.\hfill \end{array}$$

 □

Corollary 7.

For a fixed $m\ge 2$, there exists an integer $n_0=n_0\left(m\right)$ such that for all prime $p\ge n_0$,

$$\begin{array}{c}\hfill r_p\left(p^m\right)\le m.\end{array}$$

Proof. 

Since $r_p\left(p^m\right)=2$ for $m=2$, we may assume that $m\ge 3$. Now, the corollary follows Theorem 5 and by noting that ${lim}_{p\to \infty }\left((p+1){\left(\frac{p}{p-1}\right)}^{m-2}-p+1\right)=m$. □

The following corollary follows from Corollaries 3 and 7.

Corollary 8.

For a fixed $m\ge 2$, there exists an integer $n_0=n_0\left(m\right)$ such that for all prime $p\ge n_0$,

$$\begin{array}{c}\hfill w(p;m)\ge p^m+1.\end{array}$$

3. Conclusions

In this paper, we gave the essential conditions for $r_k\left(n\right)=r_k(n+1)$ and gave an upper bound for $r_p\left(p^m\right)$ for any prime $p\ge 3$ and integer $m\ge 2$. Suppose $p\ge 3$ be a prime. We hope that exact values of $r_p\left(p^m\right)$ can be obtained for some small values of $m\ge 3$ in future. Here, we end this paper with the following conjecture. Note that the conjecture is true for $m=2$.

Conjecture 6.

For a fixed $m\ge 2$, there exists an integer $n_0=n_0\left(m\right)$ such that for all prime $p\ge n_0$,

$$\begin{array}{c}\hfill r_p\left(p^m\right)=m.\end{array}$$