On a conjecture of Erd\H{o}s about sets without $k$ pairwise coprime integers

TL;DR

This paper proves a conjecture by Erdős regarding the maximum size of sets without $k+1$ pairwise coprime integers, confirming a specific structure of such sets for large enough $n$.

Contribution

The paper resolves an open problem posed by Chen and Zhou, confirming Erdős's conjecture and providing new insights into the structure of sets avoiding $k+1$ pairwise coprime integers.

Findings

Confirmed Erdős's conjecture for all sufficiently large n.

Characterized the structure of extremal sets avoiding $k+1$ coprime integers.

Extended previous results by Chen and Zhou on the conjecture.

Abstract

Let $\mathbb{Z}^{+}$ be the set of positive integers. Let $C_{k}$ denote all subsets of $\mathbb{Z}^{+}$ such that neither of them contains $k + 1$ pairwise coprime integers and $C_k(n)=C_k\cap \{1,2,\ldots,n\}$. Let $f(n, k) = \text{max}_{A \in C_{k}(n)}|A|$, where $|A|$ denotes the number of elements of the set $A$. Let $E_k(n)$ be the set of positive integers not exceeding $n$ which are divisible by at least one of the primes $p_{1}, \dots{}, p_{k}$, where $p_{i}$ denote the $i$th prime number. In 1962, Erd\H{o}s conjectured that $f(n, k) = |E(n,k)|$ for every $n \ge p_{k}$. Recently Chen and Zhou proved some results about this conjecture. In this paper we solve an open problem of Chen and Zhou and prove several related results about the conjecture.

Equations

$$\underset{k\to \infty }{\mathrm{limsup}}\frac{{\sup }_{n\ge 1}(f(n,k)-|E_k(n)|)}{g(k)}>0.$$

$$\bar{d}(A)=\underset{n\to +\infty }{\mathrm{limsup}}\frac{|A(n)|}{n},\underline{d}(A)=\underset{n\to +\infty }{\mathrm{liminf}}\frac{|A(n)|}{n}.$$

$$B_k(n)$$

$$|A(n)|\le \frac{(P_k-\phi (P_k))a-\phi (P_{k-1})(p_{k+1}-p_k)}{P_k\cdot a}n+c_k,$$

$$|E_k(n)|-|A|{\ll }_k\frac{n}{\log \log n}.$$

$$\frac{n}{(\log \log n{)}^2}\ll \underset{\begin{array}{c}A\subseteq C_1(n)\\ A⊈E_1(n)\end{array}}{\min }(|E_1(n)|-|A|)\ll \frac{n}{\log \log n},$$

$$\frac{n}{(\log \log n{)}^5}\ll \underset{\begin{array}{c}A\subseteq C_2(n)\\ A⊈E_2(n)\end{array}}{\min }(|E_2(n)|-|A|)\ll \frac{n}{\log \log n}.$$

$$|E_1(n)|-\underset{\begin{array}{c}A\in C_1(n)\\ A⊈E_1(n)\end{array}}{\max }|A|\gg \frac{n}{\log \log n}?$$

$$|E_k(n)|-\underset{\begin{array}{c}A\in C_k(n)\\ A⊈E_k(n)\end{array}}{\max }|A|\gg \frac{n}{(\log \log n{)}^{c_k}}.$$

$$\underset{t\to \infty }{\mathrm{limsup}}\frac{d_{t+2l-1}}{\text{max}\{d_t,d_{t+1},\dots ,d_{t+2l-2}\}}=+\infty .$$

$$\frac{d_{t+2l-1}}{\text{max}\{d_t,d_{t+1},\dots ,d_{t+2l-2}\}}\ge 8l^2.$$

$$p_{t+2l-1}^2$$

$$p_t{p}_{t+2l}$$

$$p_t^2+4l^2{p}_t\text{max}\{d_t,d_{t+1},\dots ,d_{t+2l-2}\}+4l^2(\text{max}\{d_t,d_{t+1},\dots ,d_{t+2l-2}\}{)}^2$$

$$\text{max}\{d_t,d_{t+1},\dots ,d_{t+2l-2}\}\le p_t.$$

$$B=\left\{m:1\le m\le n,\left(m,\prod _{i=1}^{t-1}{p}_i\right)\ne 1\right\},$$

$$C=\{p_{t+i}{p}_{t+j}:0\le i\le l-1,i<j\le 2l-1\},$$

$$D=\{p_{t+i}:0\le i\le l-1\}\cup \{p_{t+i}^2:0\le i\le l-1\}.$$

$$C\cup D^{{}^{\prime }}=\{p_{t+i}{p}_{t+j}:0\le i\le 2l-2,i<j\le 2l-1\}.$$

$$f(n,t+l-1)-|E_{t+l-1}(n)|\ge |A(n,t+l-1)|-|E_{t+l-1}(n)|$$

$$|A\cap \{13k+1,13k+3,13k+5\}|\le 2,|A\cap \{13k+7,13k+9,13k+11\}|\le 2,$$

$$|A\cap \{13k+1,13k+5,13k+9\}|\le 2,|A\cap \{13k+3,13k+7,13k+11\}|\le 2.$$

$$A\cap S_1=\{13k+1,13k+5,13k+7,13k+11\}.$$

$$|A(n)|\le \frac{a-1}{2a}n+\frac{3}{2}.$$

$$|A\cap \{2ka+2j-1,2ka+2j\}|\le 1,j=1,2,\dots ,a.$$

$$q_1(a-1)+q_2+r\le \frac{a-1}{2a}(2a{q}_1+2q_2+r)+\frac{3}{2}.$$

$$\frac{q_2}{a}+\frac{a+1}{2a}r\le \frac{q_2}{a}+\frac{a+1}{2a}\le \frac{a-1}{a}+\frac{a+1}{2a}=\frac{3a-1}{2a}<\frac{3}{2}.$$

$$|A(n)|\le \frac{4a-2}{6a}n+\frac{11}{3}.$$

$$(4a-2)q_1+4q_2+r\le \frac{4a-2}{6a}(6a{q}_1+6q_2+r)+\frac{11}{3}.$$

$$\frac{(2a+2)r+12q_2}{6a}\le \frac{11}{3}.$$

Full text

2010 Mathematics Subject Classification: Primary 11B75.

Keywords and phrases: extremal sets, pairwise coprime integers, Erdős’ conjecture

1 Introduction

Let $\mathbb{Z}^{+}$ be the set of positive integers and $p_{i}$ denote the $i$th prime number. For a set $A\subseteq\mathbb{Z}^{+}$, we define $A(n)=A\cap\{1,2,\ldots,n\}$. Let $C_{k}$ denote all subsets of $\mathbb{Z}^{+}$ such that neither of them contains $k+1$ pairwise coprime integers. Let $f(n,k)=\text{max}_{A\in C_{k}(n)}|A|$, where $|A|$ denotes the number of elements of $A$. Let $E_{k}$ be the set of positive integers which are divisible by at least one of the primes $p_{1},p_{2},\ldots{},p_{k}$. Clearly $E_{k}(n)\in C_{k}(n)$. Hence $f(n,k)\geq|E_{k}(n)|$ for all integers $n,k$ and $f(n,k)=n=|E_{k}(n)|+1$ if $n<p_{k}$. In 1962, Erdős [4], [5] conjectured that $f(n,k)=|E_{k}(n)|$ for every $n\geq p_{k}$. It can be proved easily that this conjecture holds for $k=1$ and $k=2$. For $k=3$, the conjecture was proved by Choi [3] and independently by Szabó and Tóth [8]. Choi [3] also showed that for $n\geq 150$, if $A\in C_{3}(n)$ and $|A|=|E_{3}(n)|$, then $A=E_{3}(n)$. Later, Mócsy [6] proved that the Erdős’ conjecture holds for $k=4$ and Ahlswede and Khachatrian [1] disproved the conjecture for $k=212$. In [2] Chen and Zhou gave a new proof for $k=4$ and disproved the conjecture for $k=211$. Beyond these, they also proved some related results and posed the following problem (see [2, Problem 3]).

Chen and Zhou’s problem. Is $\limsup_{k\rightarrow\infty}\text{sup}_{n\geq 1}(f(n,k)-|E_{k}(n)|)<+\infty$?

In this paper, we answer this problem.

Theorem 1**.**

$\limsup_{k\rightarrow\infty}\sup_{n\geq 1}(f(n,k)-|E_{k}(n)|)=+\infty$.

Theorem 1 implies that the conjecture of Erdős is false for infinitely many $k$, which also solves Problem 2 in [2].

On the other hand, motivated by the method of Pintz [7] about the gaps between primes, we pose the following problem.

Problem 1**.**

Find a suitable function $g(k)\rightarrow\infty$ such that

[TABLE]

We also pose the following natural problem.

Problem 2**.**

Is it true that ${\rm{sup}}_{n\geq 1}(f(n,k)-|E_{k}(n)|)<+\infty$ holds for every positive integer $k$?

For an infinite positive integers set $A$, we define

[TABLE]

If $\bar{d}(A)=\underline{d}(A)$, then we define $d(A):=\bar{d}(A)=\underline{d}(A)$.

Now we have the following conjecture.

Conjecture 1**.**

If $A\in C_{k}$ and $A\not\subseteq E_{k}$, then $\bar{d}(A)<d(E_{k})=1-\frac{\varphi(p_{1}p_{2}\cdots p_{k})}{p_{1}p_{2}\cdots p_{k}}$.

Let $B_{k}$ denote the set of positive integers $m$ such that $m$ is divisible by at least one of the primes $p_{1},p_{2},\ldots,p_{k-1},p_{k+1}$. Clearly $B_{k}(n)\in C_{k}(n)$, $B_{k}(n)\not\subseteq E_{k}(n)$ if $n\geq p_{k+1}$, and

[TABLE]

We guess that Conjecture 2 can be sharpened in the following quantitative form.

Conjecture 2**.**

For any positive integer $k$, let $P_{k}=p_{1}p_{2}\cdots p_{k}$, where $p_{i}$ denotes the $i$th prime. If $A\in C_{k},~{}a\in A\setminus E_{k}$, then

[TABLE]

where $c_{k}$ is a constant only depending on $k$.

In this paper, we prove that Conjecture 2 holds for $k=1,2,3$.

Theorem 2**.**

Conjecture 2 is true for $k=1,2,3$.

Corollary 1**.**

• (1)

If $A\in C_{1}$ and $a\in A\setminus E_{1}$, then $\bar{d}(A)\leq\frac{a-1}{2a}$;

• (2)

If $A\in C_{2}$ and $a\in A\setminus E_{2}$, then $\bar{d}(A)\leq\frac{4a-2}{6a}$;

• (3)

If $A\in C_{3}$ and $a\in A\setminus E_{3}$, then $\bar{d}(A)\leq\frac{22a-4}{30a}$.

Corollary 2**.**

Conjecture 1 is true for $k=1,2,3$.

If $f$ and $g$ are real functions, then $f\ll g~{}(f\gg g)$ means that there is an absolute positive constant $c$ such that $f(x)\leq cg(x)~{}(f(x)\geq cg(x))$ for all sufficiently large $x$.

We give lower bounds of the difference of $|E_{k}(n)|$ and $\displaystyle\max_{\begin{subarray}{c}A\in C_{1}(n)\\ A\not\subseteq E_{1}(n)\end{subarray}}|A|$ .

Theorem 3**.**

(1). $\displaystyle|E_{1}(n)|-\max_{\begin{subarray}{c}A\in C_{1}(n)\\ A\not\subseteq E_{1}(n)\end{subarray}}|A|\gg\frac{n}{(\log\log n)^{2}};$

(2). $\displaystyle|E_{2}(n)|-\max_{\begin{subarray}{c}A\in C_{2}(n)\\ A\not\subseteq E_{2}(n)\end{subarray}}|A|\gg\frac{n}{(\log\log n)^{5}}$.

Finally we show that the bound in Theorem 3 is nearly best possible.

Theorem 4**.**

For any positive integer $k$, there exists a set of positive integers $A$ such that $A\in C_{k}(n)$, $A\not\subseteq E_{k}(n)$ and

[TABLE]

Remark 1**.**

By Theorems 3 and 4, we have

[TABLE]

[TABLE]

Problem 3**.**

Is it true that

[TABLE]

We think that the statement of Theorem 3 is true in general.

Conjecture 3**.**

For every $k\geq 1$ there exists a constant $c_{k}$ and a positive integer $n_{k}$ such that for $n\geq n_{k}$

[TABLE]

2 Proof of Theorem 1

First we will prove the following lemma.

Lemma 1**.**

Let $l$ be a positive integer. Then there exist infinitely many positive integers $t$ such that $p_{t}p_{t+2l}>p_{t+2l-1}^{2}$.

Proof.

Let $d_{n}$ denote the difference between consecutive primes, i.e., $d_{n}=p_{n+1}-p_{n}$. In [7], János Pintz proved that

[TABLE]

It follows that there exist infinitely many $t$ such that

[TABLE]

In this case we have

[TABLE]

On the other hand we have

[TABLE]

In view of (1) and (2) it is enough to prove that

[TABLE]

which is equivalent to

[TABLE]

The latter statement is obviously holds in view of the fact that $d_{t+i}=O(t^{2/3})$ holds for every $1\leq i\leq l$. ∎

Now we are ready to prove Theorem 1. Let $l$ be a fixed positive integer. By Lemma 1, there exists an infinitely many integers $t$ such that $p_{t}p_{t+2l}>p_{t+2l-1}^{2}$, $p_{t}^{2}>p_{t+2l}$ and $p_{t}>\frac{p_{t+2l}}{2}$. For any such an integer $t$, we choose $n$ such that $p_{t+2l-1}^{2}\leq n<p_{t}p_{t+2l}$ and let

[TABLE]

[TABLE]

[TABLE]

Now we verify that $E_{t+l-1}(n)=B\cup C\cup D$ is a suitable decomposition. For any $m\in E_{t+l-1}(n)\setminus B$, let $p_{t+i}$ be the smallest prime divisor of $m$. Clearly $0\leq i\leq l-1$.

If there exists a prime $p_{t+j}$ with $p_{t+j}\not=p_{t+i}$ such that $p_{t+j}\mid m$ and $p_{t+i}p_{t+j}\leq n$, then by $n<p_{t}p_{t+2l}$ and $p_{t+i}\geq p_{t}$, we have $p_{t+j}<p_{t+2l}$. Hence, $i<j\leq 2l-1$. Noting that $p_{t+i}p_{t+j}\mid m$ and $2p_{t+i}p_{t+j}\geq 2p_{t}^{2}>p_{t}p_{t+2l}>n\geq m$, we have $m\in C$.

If there does not exist a prime divisor of $m$ different from $p_{t+i}$, then by $p_{t}^{3}>p_{t}p_{t+2l}>n\geq m$ we have $m\in D$.

Define $D^{{}^{\prime}}=\{p_{t+i}p_{t+j}:l\leq i\leq 2l-2,\hskip 5.69054pti<j\leq 2l-1\}$ and $A(n,t+l-1)=B\cup C\cup D^{{}^{\prime}}$. Then

[TABLE]

Clearly $C\cup D^{{}^{\prime}}$ contains at most $l$ integers which are pairwise coprime to each other and the set $B$ contains at most $t-1$ integers which are pairwise coprime to each other. It follows that $A(n,t+l-1)$ contains at most $t+l-1$ integers which are pairwise coprime to each other, which implies that $f(n,t+l-1)\geq|A(n,t+l-1)|$. It is easy to see that $|D^{{}^{\prime}}|=(l-1)+(l-2)+\dots+1=\frac{l(l-1)}{2}$. Thus we have

[TABLE]

This completes the proof of Theorem 1.

3 Preliminary Lemmas

In this section, we present some lemmas for the proof of Theorem 2.

Lemma 2**.**

Let $k$ be a nonnegative integer. If $|A\cap(30k+\{1,3,4,5,7,8,11,13\})|\geq 5$ or $|A\cap(30k+\{17,19,22,23,25,26,27,29\})|\geq 5$, then $A$ contains $4$ pairwise coprime integers.

This Lemma follows from the proof of Lemmas 2 and 3 in [3] immediately.

Lemma 3**.**

Let $k$ be a nonnegative integer. If $7\in A$ and $|A\cap\{7k+1,7k+2,\ldots,7k+6,7k+7\}|\geq 6$, then $A$ contains $4$ pairwise coprime integers.

Proof.

It is enough to prove that, if $|A\cap\{7k+1,7k+2,\ldots,7k+6\}|\geq 5$, then $A$ contains $3$ pairwise coprime integers. Without loss of generality, we may assume that $7k+1$ is odd. Otherwise, we can replace $7k+a$ by $7k+7-a$ for $a=1,2,\ldots,6$.

If $7k+3\in A$, by $|A\cap\{7k+1,7k+2,\ldots,7k+6\}|\geq 5$, then either $\{7k+1,7k+2,7k+3\}\subseteq A$ or $\{7k+3,7k+4,7k+5\}\subseteq A$, and so $A$ contains $3$ pairwise coprime integers.

Now we suppose that $7k+3\not\in A$. Let $x\in\{7k+2,7k+4\}$ satisfy $3\nmid x$. Then $x,7k+1,7k+5\in A$ are pairwise coprime. ∎

Lemma 4**.**

Let $k$ be a nonnegative integer. If $13\in A$ and $|A\cap\{13k+1,13k+2,\ldots,13k+12,13k+13\}|\geq 10$, then $A$ contains $4$ pairwise coprime integers.

Proof.

Write $S_{1}=\{13k+1,13k+3,13k+5,13k+7,13k+9,13k+11\}$ and $S_{2}=\{13k+2,13k+4,13k+6,13k+8,13k+10,13k+12\}$. Clearly we have $|A\cap(S_{1}\cup S_{2})|\geq 9$. Since $13$ is coprime to all integers belong to $S_{1}$ and $S_{2}$, it is enough to prove that $A\cap(S_{1}\cup S_{2})$ contains $3$ pairwise coprime integers. Without loss of generality, we assume that $2\nmid 13k+1$. Otherwise, we can replace $13k+a$ by $13k+13-a$ for $a=1,2,\ldots,12$. We notice that $13k+1,13k+3,13k+5$ are pairwise coprime and $13k+7,13k+9,13k+11$ are also pairwise coprime. Hence, if $\{13k+1,13k+3,13k+5\}\subseteq A$ or $\{13k+7,13k+9,13k+11\}\subseteq A$, then the result is true. Now we assume that $|A\cap S_{1}|\leq 4$. Noting that $|A\cap(S_{1}\cup S_{2})|\geq 9$ and $|A\cap S_{2}|\leq 6$, we have $|A\cap S_{1}|\geq 3$. Hence $|A\cap S_{1}|=3$ or $4$.

Case 1. $|A\cap S_{1}|=3$. In this case, we have  $S_{2}\subseteq A$. If there are two consecutive odd terms in $A$, then these two odd integers and the even integer between them are pairwise coprime. Since $13k+1,13k+5,13k+9$ are pairwise coprime and $13k+3,13k+7,13k+11$ are also pairwise coprime, the result is true if $\{13k+1,13k+5,13k+9\}\subseteq A$ or $\{13k+3,13k+7,13k+11\}\subseteq A$. Hence, we only need to consider the cases $A\cap S_{1}=\{13k+1,13k+5,13k+11\}$ or $\{13k+1,13k+7,13k+11\}$.

Subcase 1.1. $A\cap S_{1}=\{13k+1,13k+5,13k+11\}$. If $3\nmid 13k+1$, then $13k+1,13k+4,13k+5\in A$ are pairwise coprime. If $3\nmid 13k+5$, then $13k+1,13k+2,13k+5\in A$ are pairwise coprime.

Subcase 1.2. $A\cap S_{1}=\{13k+1,13k+7,13k+11\}$. If $3\nmid 13k+7$, then $13k+7,13k+10,13k+11\in A$ are pairwise coprime. If $3\nmid 13k+11$, then $13k+7,13k+8,13k+11$ are pairwise coprime.

Case 2. $|A\cap S_{1}|=4$. It follows that $|A\cap S_{2}|=5$.

Subcase 2.1. $13k+1\not\in A$. It follows that $A\cap S_{1}$ has two pairs of consecutive odd terms. Assume that $a,a+2,b,b+2\in A$, where $a,b\in S_{1}$. By $|A\cap S_{2}|=5$, we have $A\cap\{a+1,b+1\}\geq 1$. Hence $a,a+1,a+2\in A$ or $b,b+1,b+2\in A$ are pairwise coprime.

Subcase 2.2. $13k+1\in A$. By the arguments above, we may assume that $A$ does not contain two pairs of consecutive odd terms and

[TABLE]

[TABLE]

Now we only need to consider the case

[TABLE]

If $13k+6\in A$, then $13k+5,13k+6,13k+7$ are pairwise coprime. Hence we assume that $13k+6\not\in A$, and so $(S_{2}\setminus\{13k+6\})\subseteq A$. If $3\nmid 13k+5$, then $13k+5,13k+7,13k+8$ are pairwise coprime. If $3\nmid 13k+7$, then $13k+4,13k+5,13k+7$ are pairwise coprime. ∎

4 Proof of Theorem 2 when $k=1$

Proposition 1**.**

Let $A\in A_{1}$ and $a\in A\setminus F_{1}$. Then for any positive integer $n$,

[TABLE]

Proof.

Clearly, $2\nmid a$. For any nonnegative integer $k$, we have

[TABLE]

Furthermore, since $a$, $2ka+a-2$, $2ka+a-1$ are pairwise coprime, it follows that $|A\cap\{2ka+a-2,2ka+a-1\}|=0$. Hence, for any nonnegative integer $k$, we have $|A\cap\{2ka+1,2ka+2,\ldots,2ka+2a\}|\leq a-1$.

Let $n=2aq_{1}+2q_{2}+r$, where $0\leq q_{2}\leq a-1$ and $0\leq r\leq 1$. Then $|A(n)|\leq q_{1}(a-1)+q_{2}+r$. Now it is enough to prove that

[TABLE]

That is, $\frac{q_{2}}{a}+\frac{a+1}{2a}r\leq\frac{3}{2}$. It is clear that

[TABLE]

Therefore, $|A(n)|\leq\frac{a-1}{2a}n+\frac{3}{2}$.

∎

5 Proof of Theorem 2 when $k=2$

Proposition 2**.**

Let $A\in A_{2}$ and $a\in A\setminus F_{2}$. Then for any positive integer $n$,

[TABLE]

Proof.

There exist two integers $n_{1},n_{2}\in\{0,1,\ldots,a-1\}$ such that $a\mid 6n_{1}-1,~{}a\mid 6n_{2}+7$. It is clear that $n_{1}\not=n_{2}$. Otherwise, we have $a\mid 8$, a contradiction. Now we shall prove that, for $i=1,2$, if $|A\cap\{6n_{i}+1,6n_{i}+2,\ldots,6n_{i}+6\}|\geq 4$, then $A$ contains $3$ pairwise coprime integers.

Case 1. $a\mid 6n_{1}-1$. By $|A\cap\{6n_{1}+1,6n_{1}+2,\ldots,6n_{1}+6\}|\geq 4$, we have $|A\cap\{6n_{1}+1,6n_{1}+2,6n_{1}+3,6n_{1}+5\}|\geq 2$. We choose two elements $c,d\in A\cap\{6n_{2}+1,6n_{2}+2,6n_{2}+3,6n_{2}+5\}$. Since $6n_{2}-1,6n_{2}+1,6n_{2}+2,6n_{2}+3,6n_{2}+5$ are pairwise coprime, it follows that $a,c,d\in A$ are pairwise coprime.

Case 2. $a\mid 6n_{2}+7$. By $|A\cap\{6n_{2}+1,6n_{2}+2,\ldots,6n_{2}+6\}|\geq 4$, we have $|A\cap\{6n_{2}+1,6n_{2}+3,6n_{2}+4,6n_{2}+5\}|\geq 2$. We choose two elements $e,f\in A\cap\{6n_{2}+1,6n_{2}+3,6n_{2}+4,6n_{2}+5\}$. Since $6n_{2}+1,6n_{2}+3,6n_{2}+4,6n_{2}+5,6n_{2}+7$ are pairwise coprime, it follows that $a,e,f\in A$ are pairwise coprime.

Hence, by $A\in A_{2}$, for $i=1,2$, we have $|A\cap\{6n_{i}+1,6n_{i}+2,\ldots,6n_{i}+6\}|\leq 3$. Let $n=6aq_{1}+6q_{2}+r$, where $0\leq q_{2}<a$ and $0\leq r\leq 5$. Then  $|A(n)|\leq(4a-2)q_{1}+4q_{2}+r$. Now we need to prove

[TABLE]

That is,

[TABLE]

Clearly we have

[TABLE]

∎

6 Proof of Theorem 2 when $k=3$

Proposition 3**.**

Let $A\in A_{3}$ and $a\in A\setminus F_{3}$. Then for any positive integer $n$,

[TABLE]

Proof.

Since $a\not\in F_{3}$, it follows that $(a,30)=1$.

We first consider the case $a=7$. Let $n=7s+r,~{}0\leq r\leq 6$. Then, for any nonnegative integer $k$, by Lemma 3, we have $|A\cap\{7k+1,7k+2,\ldots,7k+7\}|\leq 5$. Hence

[TABLE]

Next we consider the case $a=13$. Let $n=13s+r,~{}0\leq r\leq 12$. For any positive integer $k$, by Lemma 4, we have $|A\cap\{13k+1,13k+2,\ldots,13k+13\}|\leq 9$. It also follows from Lemma 4 that $|A\cap\{1,2,\ldots,13\}|\leq 10$. Hence $|A(n)|\leq 9s+r\leq\frac{22\times 13-4}{30\times 13}\cdot(13s)+r=\frac{22a-4}{30a}(13s+r)+\frac{8a+4}{30a}r\leq\frac{22a-4}{30a}n+\frac{176}{15}.$

Now we may assume that $a\not=7,13$. Then there exist four distinct integers $n_{1},n_{2},n_{3},n_{4}\in\{0,1,\ldots,a-1\}$ such that $a\mid 30n_{1}-1,~{}a\mid 30n_{2}+31,~{}a\mid 30n_{3}-11,~{}a\mid 30n_{4}+41$. Now we shall prove that, for $i=1,2,3,4$, if $|A\cap\{30n_{i}+1,30n_{i}+2,\ldots,30n_{i}+30\}|=22$, then $A$ contains $4$ pairwise coprime integers. By Lemma 2, it suffices to prove the case that

[TABLE]

[TABLE]

In this case, we have

[TABLE]

[TABLE]

Case 1. $a\mid 30n_{1}-1$.

Write $M:=30n_{1}$. If $A\cap\{M+4,M+8\}=\emptyset$, then $|A\cap(M+\{1,3,5,7,11,13\})|=4$, and so $A$ contains $4$ pairwise coprime integers.

If $M+4\in A$ and $M+8\not\in A$, then $M+11\in A$ and $7\mid M+4$. Otherwise, $M+4$ and three integers in $A\cap(M+\{1,3,5,7,11,13\})$ are pairwise coprime. By $7\mid M+4$, it follows that $7\nmid M-1$. Hence $a$ and three integers in $A\cap(M+\{1,3,5,7,11,13\})$ are pairwise coprime.

If $M+8\in A$ and $M+4\not\in A$, then $M+1\in A$ and $7\mid M+8$. Otherwise, $M+8$ and three integers in $A\cap(M+\{1,3,5,7,11,13\})$ are pairwise coprime. Now we have $7\nmid M-1$. Hence $a$ and three integers in $A\cap(M+\{1,3,5,7,11,13\})$ are pairwise coprime.

Now we suppose that $M+4\in A$ and $M+8\in A$. If $M+13\not\in A$, then two integers in $A\cap(M+\{1,3,5,7,11\})$, $a$ and one integer in $\{M+4,M+8\}$ which is not a multiple of $7$ are pairwise coprime. Hence $M+13\in A$.

If there exists an integer $b\in\{M+1,M+5,M+7\}$ and $b\in A$, then $b,M+4,M+9,M+13$ are pairwise coprime.

If $M+11\in A$, then $M+8,M+9,M+11,M+13$ are pairwise coprime.

Therefore

[TABLE]

If $7\nmid a$, then $a,M+3,M+13$ and the integer in $\{M+4,M+8\}$ which is not a multiple of $7$ are pairwise coprime. Hence now we assume that $7\mid a$.

Since $|(\mathbb{Z}^{+}\setminus A)\cap(30k+\{17,19,22,23,25,26,27,29\})|=4$, it follows that $|(M+\{17,19,23,25,29\})\cap A|\geq 1.$

If $M+17\in A$, then $a,M+3,M+8,M+17\in A$ are pairwise coprime.

If $M+19\in A$, then $a,M+3,M+4,M+19\in A$ are pairwise coprime.

If $M+23\in A$, then $a,M+3,M+8,M+23\in A$ are pairwise coprime.

If $M+25\in A$, then $M+4,M+9,M+13,M+25\in A$ are pairwise coprime.

If $M+29\in A$, then $a,M+4,M+9,M+29\in A$ are pairwise coprime.

Hence, $|A\cap\{M+1,M+2,\ldots,M+30\}|\leq 21$.

Case 2. $a\mid 30n_{2}+31$.

This case is nearly the same as Case 1, we only need to replace $30n_{1}+i$ by $30n_{2}+30-i$ for $i=1,2,\ldots,29$ and replace $30n_{1}+30$ by $30n_{2}+30$. The proofs are the same.

Case 3. $a\mid 30n_{3}-11$.

Write $M:=30n_{3}$. If $A\cap\{M+4,M+8\}=\emptyset$, then $|A\cap(M+\{1,3,5,7,11,13\})|=4$, and so $A$ contains $4$ pairwise coprime integers.

If $M+4\in A$ and $M+8\not\in A$, then $M+11\in A$ and $7\mid M+4$. Otherwise, $M+4$ and three integers in $A\cap(M+\{1,3,5,7,11,13\})$ are pairwise coprime. By $7\mid M+4$, we have $7\nmid M-11$. Hence $a$ and three integers in $A\cap(M+\{1,3,5,7,11,13\})$ are pairwise coprime.

If $M+8\in A$ and $M+4\not\in A$, then $M+1\in A$ and $7\mid M+8$. Otherwise, $M+8$ and three integers in $A\cap(M+\{3,5,1,7,11,13\})$ are pairwise coprime. By $7\mid M+8$, we have $7\nmid M-11$. If $M+11\not\in A$, then $a$ and three integers in $A\cap(M+\{1,3,5,7,13\})$ are pairwise coprime. Now we assume that $M+11\in A$. If $M+3\not\in A$, then $M+9$ and $A\cap(M+\{1,5,7,11,13\})$ are pairwise coprime. If $M+3\in A$, then $M+1,M+2,M+3,M+11\in A$ are pairwise coprime.

Now we suppose that $M+4\in A$ and $M+8\in A$. If $M+11\not\in A$, choose an integer $b\in A\cap(M+\{1,5,7,13\})$, then $a,b,M+4,M+9\in A$ are pairwise coprime. Next we assume that $M+11\in A$.

If there exists an integer $b\in A$ and $b\in\{M+5,M+7,M+13\}$, then $b,M+8,M+9,M+11$ are pairwise coprime.

If $M+1\in A$, then $M+1,M+9,M+10,M+11\in A$ are pairwise coprime.

If $M+3\in A$ and $11\nmid a$, then $a,M+9,M+11,M+14$ are pairwise coprime.

Finally we assume that $M+3\in A$ and $11\mid a$.

Subcase 1. $M+22\in A$ and $M+26\not\in A$.

If $M+29\not\in A$, then $M+22$ and three integers in $A\cap(M+\{17,19,23,25,27\})$ are pairwise coprime.

If $M+29\in A$, then $M+11,M+9,M+14,M+29\in A$ are pairwise coprime.

Subcase 2. $M+22\not\in A$ and $M+26\in A$.

If $M+19\not\in A$, then $M+26$ and three integers in $A\cap(M+\{17,23,25,27,29\})$ are pairwise coprime.

If $M+19\in A$, then $a,M+19,M+9,M+4\in A$ are pairwise coprime.

Subcase 3. $M+22\in A$ and $M+26\in A$.

Since $|(\mathbb{Z}^{+}\setminus A)\cap(M+\{17,19,22,23,25,26,27,29\})|=4$, we have

[TABLE]

If $M+17\in A$, then $M+8,M+9,M+11,M+17\in A$ are pairwise coprime.

If $M+19\in A$, then $a,M+4,M+9,M+19\in A$ are pairwise coprime.

If $M+23\in A$, then $M+3,M+8,M+11,M+23\in A$ are pairwise coprime.

If $M+25\in A$, by $11\mid M-11$, then $11\nmid M+14$ and $a,M+9,M+14,M+25\in A$ are pairwise coprime.

If $M+29\in A$, then $M+9,M+11,M+14,M+29\in A$ are pairwise coprime.

Hence, $|A\cap\{M+1,M+2,\ldots,M+30\}|\leq 21$.

Subcase 4. $M+22\notin A$ and $M+26\notin A$. It follows that $|A\cap(M+\{17,19,23,25,27,29\})|=4$, and so $A$ contains $4$ pairwise coprime integers.

Case 4. $a\mid 30n_{4}+41$.

This case is nearly the same as Subcase 2, we only need to replace $30n_{3}+i$ by $30n_{4}+30-i$ for $i=1,2,\ldots,29$ and replace $30n_{3}+30$ by $30n_{4}+30$. The proofs are the same.

Therefore, for any positive integer $q$, we have $|A(n)\cap\{1,2,\ldots,30aq\}|\leq(22a-4)q$ and $|A(n)\cap\{30q+1,30q+2,\ldots,30q+30\}|\leq 22$.

Now let $n=30aq_{1}+30q_{2}+r,~{}0\leq r<30,~{}0\leq q_{2}\leq a-1$. Then $|A(n)|\leq(22a-4)q_{1}+22q_{2}+r$, where $0\leq r<30$ and $0\leq q_{2}\leq a-1$. Hence it suffices to prove that

[TABLE]

That is, $\frac{4q_{2}}{a}+r-\frac{22a-4}{30a}r\leq\frac{176}{15}$. It is clear that

[TABLE]

∎

7 Proof of Theorem 3

In the first step we prove the first part of Theorem 3. Let $A\in C_{1}(n)$, $A\not\subseteq E_{1}(n)$ and let $a$ be the smallest odd integer of $A$. We may assume that $a$ is squarefree, i.e., $a=q_{1}q_{2}\cdots q_{t}$, where $3\leq q_{1}<\dots<q_{t}$ are primes. It is clear that two consecutive integers are coprime, and the conditions $(m,a)=(m+1,a)=1$ means that $m$ satisfies the following linear congruences $m\equiv a_{i}\bmod{q_{i}}$ and $m+1\equiv a_{i}+1\bmod{q_{i}}$, where $1\leq a_{i}\leq q_{i}-2$. The number of such congruences is $\prod_{i=1}^{t}(q_{i}-2).$

Thus we have

[TABLE]

It follows that

[TABLE]

We assume that there are $v$ even integers and $w$ odd integers in the set

[TABLE]

Obviously, $v\gg\frac{n}{(\log\log n)^{2}}$ or $w\gg\frac{n}{(\log\log n)^{2}}$.

If $v\gg\frac{n}{(\log\log n)^{2}}$, then we may choose in $A$ at most one integer from each pair of $(1,2),\dots,(2[n/2]+1,2[n/2]+2)$. Thus we have

[TABLE]

where $c_{1}$ is an absolute positive constant.

If $w\gg\frac{n}{(\log\log n)^{2}}$, then we may chose in $A$ at most one integers from each pair of $(2,3),\dots,(2[n/2],2[n/2]+1)$. Thus we have

[TABLE]

where $c_{2}$ is also an absolute positive constant.

This completes the proof of the first part of Theorem 3.

In the next step we prove the second part of Theorem 3. First we need three lemmas in the following.

Lemma 5**.**

If $A\in C_{2}(n)$, then for every positive integer $k$ we have

[TABLE]

Proof.

Assume contrary that $|\{6k,6k+1,6k+2,6k+3,6k+4,6k+5\}\cap A|\geq 5$. If $6k+1,6k+2,6k+3\in A$, then $6k+1,6k+2,6k+3$ are pairwise coprime, a contradiction. If $6k+3,6k+4,6k+5\in A$, then $6k+1,6k+2,6k+3$ are also pairwise coprime, a contradiction again. Now we may assume $6k+3\notin A$. Then we have $6k+1,6k+2,6k+5\in A$, but $6k+1,6k+2,6k+5$ are obviously pairwise coprime, which is absurd.

The proof of the Lemma is completed. ∎

Lemma 6**.**

Let $a$ be a positive integer with $(a,6)=1$ and $a\geq\prod_{i=3}^{12}p_{i}$. Then we have

[TABLE]

Proof.

Let $a=q_{1}q_{2}\cdots{}q_{t}$, where $5\leq q_{1}<q_{2}<\cdots{}<q_{t}$ are primes. Suppose that $t=1$. Take m=6,12,\dots,6\Big{(}\Big{[}\frac{a}{6}\Big{]}-1\Big{)}. Then we have $(m+i,a)=1~{}\text{for}~{}i=0,1,2,3,5$, and so

[TABLE]

Now we assume that $t\geq 2$. Define the set $S$ by

[TABLE]

Then we have

[TABLE]

It is easy to see that, for a fixed $s\in S$ and $i\in\{0,1,2,3,5\}$, the integers $s+l\cdot\frac{a}{q_{t}}+i~{}(0\leq l\leq q_{t}-1)$ form a complete residue system modulo $q_{t}$. Since $q_{t}\geq 7$, we have $|S_{s}|=q_{t}-5$ for every $s\in S$. It follows from $(\frac{a}{q_{t}},6)=1$ that

[TABLE]

Then we have

[TABLE]

It follows that

[TABLE]

Then we obtain

[TABLE]

If $q_{1}=5$, then $(m,a)=(m+1,a)=(m+2,a)=(m+3,a)=(m+5,a)=1$ means that $m$ satisfies the following linear congruences $m\equiv 1\bmod{5}$ and $m\equiv a_{i}\bmod{q_{i}}$, where $a_{i}\in\{1,2,\dots,q_{t}-6,q_{t}-4\}$ for $i=2,3,\ldots,t$. The number of such congruences is $\prod_{i=2}^{t}(q_{i}-5)$. Hence we have

[TABLE]

If $q_{1}\geq 7$, the proof is the similar. For the sake of completeness, we present it here.

If $q_{1}\geq 7$, then $(m,a)=(m+1,a)=(m+2,a)=(m+3,a)=(m+5,a)=1$ means that $m$ satisfies the following linear congruences $m\equiv a_{i}\bmod{q_{i}}$, where $a_{i}\in\{1,2,\dots,q_{t}-6,q_{t}-4\}$ for $i=1,2\ldots,t$. The number of such congruences is $\prod_{i=1}^{t}(q_{i}-5)$. Hence we obtain

[TABLE]

Since $a\geq\prod_{i=3}^{12}p_{i}$, it follows that $q_{t}\geq 37$, and so

[TABLE]

which completes the proof. ∎

Lemma 7**.**

Let $A\in C_{2}(n)$, $a\in A$ and $k$ be a positive integer with $(6k,a)=(6k+1,a)=(6k+2,a)=(6k+3,a)=(6k+5,a)=1$, then we have

[TABLE]

Proof.

Assume contrary that

[TABLE]

Then it is clear that

[TABLE]

Obviously, if we choose two elements from the set $\{6k+1,6k+2,6k+3,6k+5\}\cap A$, then these elements and $a$ are pairwise coprime, which is absurd. ∎

Now we are ready to prove the second case of Theorem 3. Let $a$ be the smallest element of $A$ with $(a,6)=1$. According to Lemma 5 and Lemma 7 we have

[TABLE]

Thus it is enough to prove that

[TABLE]

We distinguish two cases.

Case 1. $a<\prod_{i=3}^{12}p_{i}$. There exists a $0\leq l\leq 5$ such that $6\mid l\cdot a+1$. We choose $m=l\cdot a+1$. Then we have $(m,a)=(l\cdot a+1,a)=1$, $(m+1,a)=(l\cdot a+2,a)=1$, $(m+2,a)=(l\cdot a+3,a)=1$, $(m+3,a)=(l\cdot a+4,a)=1$, $(m+5,a)=(l\cdot a+6,a)=1$. It follows that

[TABLE]

Case 2. $a\geq\prod_{i=3}^{12}p_{i}$. By Lemma 6, we have

[TABLE]

The proof of Theorem 3 is completed.

8 Proof of Theorem 4

For any integer $n\geq\prod_{i=1}^{k+2}p_{i}$, there exists an integer $l\geq k+1$ such that

[TABLE]

We define the set

[TABLE]

Obviously, $A\in C_{k}(n)$ and $A\not\subseteq E_{k}(n)$. It is easy to see that

[TABLE]

It follows that

[TABLE]

We give an upper estimation to the fractional part of $\frac{n}{p_{1}p_{2}\cdots p_{l}}$. It is easy to see from the definition of $l$ that $p_{l+1}\sim\log n$ and therefore

[TABLE]

where $c$ is an absolute positive constant. Thus we have

[TABLE]

Clearly,

[TABLE]

It follows that

[TABLE]

where $c_{k}$ is an absolute positive constant depending only on $k$. Obviously,

[TABLE]

Hence

[TABLE]

This completes the proof of Theorem 4.

9 Acknowledgement

The authors would like to thank János Pintz for the valuable discussions about Theorem 1. This work was done during the third author visiting to Budapest University of Technology and Economics. He would like to thank Dr. Sándor Kiss and Dr. Csaba Sándor for their warm hospitality.