1. Introduction
Let
be the set of positive integer, i.e.,
. A subset
is said to have an arithmetic progression (a.p.) of length
k, if there exist
such that
The arithmetic progression of length k is also called a k-term a.p. with difference d, or we write .
There are many problems related to arithmetic progressions. One of them is to find an arrangement of
that avoids
k-term arithmetic progressions. An arrangement of
is a sequence
such that
. Davis et al. [
1] discovered an arrangement of
that avoids three-term monotone arithmetic progressions. By using the construction of certain magic square, Sim and Wong [
2] proposed an arrangement of
that avoids a
k-term monotone arithmetic progression, but contains a
-term monotone arithmetic progression. For the infinite case, Davis et al. [
1] and Sidorenko [
3] showed that there is arrangement of
that avoids three-term monotone arithmetic progressions. Here, an arrangement of
is a sequence
such that
. They [
1] also showed that there exists an arrangement of
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 by . A r-colouring of a set S is a function . 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 be integers. There exists a least positive integer such that for any , every r-colouring of 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
that does not contain an arithmetic progression of length
k? They define the following.
Definition 1. For and , let be the collection of sets such that does not contain an arithmetic progression of length k. Then .
Note that
is monotone increasing in
. Suppose
, then there is a subset of
integers in
which may contain only
k-term a.p. but definitely does not contain a
a.p. Hence
. They [
7] showed that
for
and conjectured
. The conjectured was solved by Roth [
8] and has been strengthened (see [
9,
10]). Studying
and having an upper bound on these numbers would lead to an upper bound on
. In 2018, Blankenship, Cummings, and Taranchuk [
11] showed that for
and
q be primes, then
. The current best known general lower bound of this type is due to Kozik and Shabanov [
12], who in 2016 proved that, for all
for some constant
.
Given two positive integers
n and
k, the
anti-van der Waerden number aw is the smallest
r such that every exact
r-colouring of
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
without rainbow 3-term arithmetic progressions, and showed that the typical colourings with this property are 2-colourings.
Theorem 2 ([
5])
. Let and assume , with . Then . Let
and
denote the minimum number of colour required so that there exists a
-colouring of
that avoids any monochromatic
k-term a.p. Since
is coloured by
colours, clearly
From [
18],
is obtained from the consequence of Symmetric Hypergraph Theorem. For sufficiently large
n,
, see [
19]. In this paper, we are focusing on the the minimum number of colour in
to avoid a monochromatic
k-term a.p. From the known results of
,
and
, see [
20,
21,
22], we have
. Additionally, if
, then
.
In this paper, we give necessary and sufficient conditions for (Theorem 3). We also show that for all (Theorem 4). Finally, we give an upper bound for for any prime and integer (Theorem 5).
2. Main Results
Note that . Therefore, we shall assume that .
Proof. Colour with colours that avoids any monochromatic k-term a.p. and colour with different colours that avoids any monochromatic k-term a.p.. Note that this colouring avoids any monochromatic k-term a.p. in . Hence, the lemma follows. □
The following corollary is a consequence of Lemma 1 by noting that for .
Since , we have the following corollary.
Corollary 2. or .
Corollary 3. and for .
Proof. There is a l-colouring of that avoids any k-term a.p.. Therefore, . If , then by Corollary 2, . This is not possible as any l-colouring of has a monochromatic k-term a.p. Hence, .
Since any l-colouring of has a monochromatic k-term a.p., by Corollary 2, . By first part of the corollary, . Again, by Corollary 2, for . □
For , a family of sets is called a partition of if
- (a)
for ;
- (b)
.
Here, we allow for some j. If each avoids any k-term a.p., the partition is called a free k-term a.p. partition.
Let
. Then, there exists a
l-colouring of
that avoids any monochromatic
k-term a.p. For
, let
Note that is a free k-term a.p. partition of . So, can be considered as the minimum size of a free k-term a.p. partition of .
A subset
is said to have a pure arithmetic progression of length
k (or pure
k-term a.p.), if there exists a
such that
Given a subset
and
, we write
Lemma 2. Let be a partition of such that each does not contain any k-term a.p. Suppose there is a that does not contain any pure -term a.p. Let and for all . Then, is a free k-term a.p. partition of .
Proof. Note that . So, it is not hard to see that is a partition of . Now, if contains a k-term a.p. for some , say , then contains the k-term a.p. , a contradiction. So, we may assume that does not contain any k-term a.p. for all .
Suppose contains a k-term a.p. for some , say . If , then contains the k-term a.p. , a contradiction. So, we may assume that . This implies that contains the pure -term a.p. , a contradiction. □
Theorem 3. Suppose . Then, if and only if there exists a free k-term a.p. partition of such that does not contain any pure -term a.p. for some .
Proof. If there exists a free k-term a.p. partition of and does not contain any pure -term a.p., then by Lemma 2, is a free k-term a.p. partition of where and for all . It follows from Corollary 2 that .
Suppose . Then, there exists a free k-term a.p. partition of . By relabelling if necessary, we may assume that . Let and for . Note that is a free k-term a.p. partition of . If contains a pure -term a.p., say , then is a k-term a.p. in , a contradiction. Hence, does not contain any pure -term a.p. □
Corollary 4. Suppose and . Then, for all .
Proof. Let
be a free
k-term a.p. partition of
. Since
, by Theorem 3, each
contains a pure
-term a.p. Let
Note that is a free k-term a.p. partition of . Therefore, for all . The equality follows from Corollary 2. □
Proof. By Corollary 3,
and
for
. It follows from Corollary 4 that
which is equivalent to
. Since
, we have
□
Let with . We shall write and .
Theorem 4. for all .
Proof. For
and
, let
In particular, when
Note that
is a partition of
. Now,
and
So, .
Clearly, is a free k-term a.p. partition for . Suppose is a free k-term a.p. partition for some and . We claim that is also a free k-term a.p. partition.
Since
,
does not contain
. Suppose
contains a pure
-term a.p.
where
. Note that
for some
. If
, there exists a
such that
a contradiction. So, we may assume that
. If both
and
are in
for some
, then
, a contradiction. So, we may assume that for each
s,
contains at most one element of the form
where
. This implies that
and
, a contradiction. Hence,
does not contain any pure
-term a.p. By Lemma 2,
is a free
k-term a.p. partition. Note that
and
By Lemma 2, is a free k-term a.p. partition.
Suppose
is a free
k-term a.p. partition for some
. We claim that
is also a free
k-term a.p. partition. Note that
Clearly,
does not contain
. Suppose
contains a pure
-term a.p., say
where
. If
, then
, a contradiction. So, we may assume that
. If both
and
are in
for some
, then
, a contradiction. So, we may assume that for each
s,
contains at most one element of the form
where
. Since
, we must have both
and
in
. This implies that
, a contradiction. Hence,
does not contain any pure
-term a.p. Note that
and
By Lemma 2,
is a free
k-term a.p. partition. Note that
and
By Lemma 2, is a free k-term a.p. partition. By double induction, is a free k-term a.p. partition of . In particular, is a free k-term a.p. partition of . Hence, for all . □
Note that Theorem 4 is best possible because when , for all and . Note also that for , , therefore .
The following corollary follows from Corollary 3.
Corollary 6. .
Lemma 3. Let be a prime and . For each , let Suppose for some . If A contains a p-term a.p., , then and the p-term a.p is contained entirely in for some with .
Proof. Suppose
. Then,
This implies that for all i, a contradiction. Hence, and the p-term a.p is contained entirely in . □
For each real number x, let be the smallest integer such that .
Lemma 4. Let be a prime. If , then Proof. For each
, let
and
be defined by
for all
. Note that
is a bijection.
Let
and
be a free
k-term a.p. partition of
. Then,
is a free
k-term a.p. partition of
. Let
for all
and
. Then,
.
Let
where
. Let
Then, is a partition of if and is a partition of if . Let . Then, if and if .
For each
, let
be such that
. Now, for each
, let
Since , there is a such that . If contains a k-term a.p., say , then it contains the p-term a.p. . By Lemma 3, . This implies that contains the k-term a.p. for some . Since , contains the k-term a.p. where and . This is not possible as is a free k-term a.p. partition of . Thus, does not contain any k-term a.p.
For each
, let
Clearly, . If contains a k-term a.p., say , then it contains the p-term a.p. . By Lemma 3, . This implies that contains the k-term a.p. for some . Since , contains the k-term a.p. where and . This is not possible as is a free k-term a.p. partition of . Thus, does not contain any k-term a.p.
Now,
is a free
k-term a.p. partition of
. Hence,
and the lemma follows. □
Theorem 5. Let be a prime. Then, ,and for,
Proof. Clearly,
. By Lemma 4,
Suppose
. Then,
. By Theorem 4,
. Thus,
Suppose
. Then,
. So,
□
Corollary 7. For a fixed , there exists an integer such that for all prime , Proof. Since for , we may assume that . Now, the corollary follows Theorem 5 and by noting that . □
The following corollary follows from Corollaries 3 and 7.
Corollary 8. For a fixed , there exists an integer such that for all prime ,