On cycles in the coprime graph of integers^{1}

Paul Erdős Mathematics Institute Hungarian Academy of Sciences Reáltanoda u. 13-15, H-1053 Budapest, Hungary

Gabor N. Sarkozy Computer Science Department Worcester Polytechnic Institute Worcester, MA 01609 gsarkozy@cs.wpi.edu

Submitted: June 20, 1996; Accepted: December 2, 1996

Dedicated to Herbert S. Wilf on the occasion of his 65^{th} birthday

Abstract

In this paper we study cycles in the coprime graph of integers. We denote by f(n, k) the number of positive integers m ≤ n with a prime factor among the first k primes. We show that there exists a constant c such that if A ⊂ {1, 2, …, n} with |A| > f(n, 2) (if 6|n then f(n, 2) = \frac{2}{3}n), then the coprime graph induced by A not only contains a triangle, but also a cycle of length 2l + 1 for every positive integer l ≤ cn.

1 Introduction

1.1 Notations and definitions

Let (a, b) denote the greatest common divisor and [a, b] the least common multiple of integers a, b. Consider the coprime graph on the integers. This is the graph whose vertex set is the set of integers and two integers a, b are connected by an edge if and only if (a, b) = 1. Let A ⊂ {1, 2, …, n} be a set of positive integers. The coprime graph of A, denoted by G(A), is the induced coprime graph on A. A_{(m,u)} denotes the integers a_{i} ∈ A, a_{i} ≡ u (mod m). ϕ(n) denotes Euler’s function, ω(n) denotes the number of distinct prime factors of n and μ(n) is the Moebius function.

V (G) and E(G) denote the vertex set and edge set of a graph G. K_{n} is the complete graph on n vertices, C_{n} is the cycle on n vertices and K(m, n) denotes the complete bipartite graph between U and V, where |U| = m, |V | = n. H is a subgraph of G, denoted by H ⊂ G, if V (H) ⊂ V (G) and E(H) ⊂ E(G).

1Mathematics Subject Classification 11B75,05C38

1.2 The coprime graph of integers

Recently the investigation of various graphs on the integers has received significant attention (see e.g. “Graphs on the integers” in [10]). The most popular graph seems to be the coprime graph, altough there are many attractive problems and results concerning the divisor graph ([7], [8], [12] and [14]). Several reasearchers studied special subgraphs of the coprime graph. Perhaps the first problem of this type was raised by the first author in 1962 ([6]): What is largest set A ⊂ {1, 2, …, n} such that K_{k} ⊄ G(A)? Of course, the set of m ≤ n which have a prime factor among the first k − 1 primes is such a set (let us denote the cardinality of this set by f(n, k − 1)), and the first author conjectured that this set gives the maximum. For k = 2, 3 this is trivial, and for k = 4 it was recently proved by Szabó and Tóth [15]. However, the conjecture recently was disproved by Ahlswede and Khachatrian [1]. They also gave some positive results in [2] and [3].

Another interesting question is what conditions guarantee a perfect matching in the coprime graph. Newman conjectured more than 25 years ago, that if I_{1} = {1, 2, …, n} and I_{2} is any interval of n consecutive integers, then there is a perfect coprime matching from I_{1} to I_{2}. This conjecture was proved by Pomerance and Selfridge [13] (see also [4]). Note that the statement is not true if I_{1} is also an arbitrary interval of n consecutive integers. Example: I_{1} = {2, 3, 4} and I_{2} = {8, 9, 10}, any one-to-one correspondance between I_{1} and I_{2} must have at least one pair of even numbers in the correspondance.

In this paper we are going to investigate another natural question of this type (also initiated by the first author), namely what can we say about cycles in G(A). The case of even cycles is not hard from earlier results (at least for not too long cycles). In fact, if l ≤ ⌊\frac{1}{10} log log n⌋, for the largest set A ⊂ \{1, 2, …, n\} with C_{2l} ⊄ G(A), we have |A| = f(n, 1) + (l − 1) = ⌊\frac{1}{2}n⌋ + (l − 1). This cardinality can be obtained by taking all the even numbers and the first l − 1 odd primes, then obviously C_{2l} ⊄ G(A). The upper bound follows from the following theorem in [9]: If n ≥ n_{0}, |A_{(2,1)}| = s > 0, |A| > f(n, 1) and r = min\{s, ⌊\frac{1}{10} log log n⌋\}, then K(r, r) ⊂ G(A).

The case of odd cycles is more interesting. As it was mentioned above to guarantee a triangle we need at least f(n, 2) + 1 = ⌊\frac{n}{2}⌋ + ⌊\frac{n}{3}⌋ − ⌊\frac{n}{6}⌋ + 1 (= \frac{2}{3}n + 1 if 6|n) numbers. Somewhat surprisingly this cardinality already guarantees the existence of odd cycles of “almost” every length. More precisely:

Theorem 1. There exist contants c, n_{0} such that if n ≥ n_{0}, A ⊂ {1, 2, …, n} and |A| > f(n, 2), then C_{2l+1} ⊂ G(A) for every positive integer l ≤ cn.

It would be interesting to determine here the best possible value of the constant c. Perhaps it is c = \frac{1}{6}. One may obtain this trivial upper bound for 6|n by taking all the even numbers and the first \frac{n}{6} + 1 odd numbers for A; clearly C_{2l+1} ⊄ G(A) for l > \frac{n}{6}.

In the proof we will distinguish two cases depending on the size of |A_{(6,1)}| + |A_{(6,5)}|. The theorem will be an immediate consequence of the following two theorems:

Theorem 2. There exist constants c_{1}, c_{2}, n_{1} such that if n ≥ n_{1},

|A_{(6,1)}| = s_{1}, |A_{(6,5)}| = s_{2}, 1 ≤ s_{1} + s_{2} ≤ c_{1}n,

and

|A| > f(n, 2),

then C_{2l+1} ⊂ G(A) for every positive integer l ≤ c_{2}n.

Theorem 3. For every ϵ > 0, there exist constants c_{3} = c_{3}(ϵ) and n_{2} = n_{2}(ϵ) such that if n ≥ n_{2},

|A_{(6,1)}| = s_{1}, |A_{(6,5)}| = s_{2}, s_{1} + s_{2} ≥ ϵn,

and

|A| > f(n, 2),

then C_{2l+1} ⊂ G(A) for every positive integer l ≤ c_{3}n.

2 Proofs

2.1 Proof of Theorem 2

We may assume without loss of generality that s_{1} = max(s_{1}, s_{2}). We take an arbitrary 1 ≤ l ≤ c_{2}n positive integer. The rough outline of the construction of a C_{2l+1} in G(A) is the following: First we will pick a number a ∈ A_{(6,1)} with relatively large ϕ(a) and the remaining 2l numbers will be chosen alternately from A_{(6,2)} and A_{(6,3)}.

We need the following lemma:

Lemma 1. The number of integers 1 ≤ k ≤ n satisfying \frac{ϕ(k)}{k} < \frac{1}{t} is less than

n exp(− exp c_{4}t)

(where exp z = e^{z}), uniformly in t > 2.

This lemma can be found in [5]. We apply Lemma 1 with

t = \frac{1}{c_{4}} log log \frac{2n}{s_{1}}  (1)

(t > 2 holds for small enough c_{1}). Then the number of integers 1 ≤ k ≤ n for which \frac{ϕ(k)}{k} < \frac{1}{t} (where t is defined by (1)) is less than \frac{s_{1}}{2} . Hence there exists an integer a ∈ A_{(6,1)} satisfying \frac{ϕ(a)}{a} ≥ \frac{1}{t}.

The number of those integers u, for which

0 < 6u + 2 ≤ n and (6u + 2, a) = 1

hold, is given by the following sieve formula

∑_{d|a} μ(d)g_{1}(n, d),  (2)

where g_{1}(n, d) denotes the number of those integers v, for which

6v + 2 ≤ n and d|6v + 2.

(i.e. 6v ≡ −2 (mod d).) It is clear from (a, 6) = 1, that

|g_{1}(n, d) − \frac{n − 2}{6d}| ≤ 1.

We also use the following lemma:

Lemma 2 (see [11], page 394). There exists an n_{3} such that if n ≥ n_{3} then

ω(n) < 2\frac{log n}{log log n}.

This lemma implies that in (2), for large enough n, the number of terms is

2^{ω(a)} < 2^{2 \frac{log n}{log log n}} .

Indeed, if a < n_{3} this is trivial, and if a ≥ n_{3}, then

ω(a) < 2\frac{log a}{log log a} ≤ 2\frac{log n}{log log n},

since the function g(u) = 2\frac{log u}{log log u} is increasing if u is large enough (see [11], page 394).

Thus

∑ ; \begin{matrix}u : 6u + 2 ≤ n, \\ (6u + 2, a) = 1 \end{matrix} 1 ≥ \frac{n − 2}{6} ∑_{d|a} \frac{μ(d)}{d} − 2^{2 \frac{log n}{log log n}} =

= \frac{n − 2}{6} ∏_{p|a} (1 − \frac{1}{p}) − 2^{2 \frac{log n}{log log n}} =

= \frac{n − 2}{6} \frac{ϕ(a)}{a} − 2^{2 \frac{log n}{log log n}} ≥ \frac{n − 2}{6t} − 2^{2 \frac{log n}{log log n}} ≥ \frac{n}{7t}

for sufficiently large n.

Therefore

∑ ; \begin{matrix}u : 6u + 2 ≤ n, \\ (6u + 2, a) = 1, \\ 6u + 2 ∈ A_{(6,2)} \end{matrix} 1 ≥ ∑ ; \begin{matrix}u : 6u + 2 ≤ n, \\ (6u + 2, a) = 1 \end{matrix} 1 − ∑ ; \begin{matrix}u : 6u + 2 ≤ n, \\ 6u + 2 ∉ A_{(6,2)} \end{matrix} 1 ≥

≥ \frac{n}{7t} − (s_{1} + s_{2}) ≥ \frac{n}{8t}.

Once again applying Lemma 1 with

t′ = \frac{1}{c_{4}} log log \frac{2n}{\frac{n}{8t}},

(so t′ < t) there are at least \frac{n}{16t} integers in the form 6u + 2 satisfying

6u + 2 ≤ n, (6u + 2, a) = 1, 6u + 2 ∈ A_{(6,2)}

and

\frac{ϕ(6u + 2)}{6u + 2} ≥ \frac{1}{t′} > \frac{1}{t}.

We choose b_{1} arbitrarily from these integers.

Applying Lemma 1 with

t′′ = \frac{1}{c_{4}} log log \frac{1}{c_{1}} ,

the number of integers 1 ≤ k ≤ n for which \frac{ϕ(k)}{k} < \frac{1}{t′′} is at most c_{1}n. Therefore the number of integers 1 ≤ k ≤ n for which

k ∈ A_{(6,2)} and \frac{ϕ(k)}{k} ≥ \frac{1}{t′′}  (3)

is at least

\{1, 2, …, n\}_{(6,2)} − (s_{1} + s_{2}) − c_{1}n ≥ ⌊\frac{n − 2}{6}⌋ + 1 − 2c_{1}n ≥ \frac{n}{7},

if c_{1} is small enough. We choose b_{2}, b_{3}, …, b_{l} as arbitrary numbers from the numbers satisfying (3). Let b_{l+1} = a.

Put e_{i} = [b_{i}, b_{i+1}] for all 1 ≤ i ≤ l. The number of those integers u, for which

6u + 3 ≤ n and (6u + 3, e_{i}) = 1

is again clearly the following:

∑_{d|e_{i}} μ(d)g_{2}(n, d),  (4)

where g_{2}(n, d) denotes the number of those integers v, for which

6v + 3 ≤ n and d|6v + 3

(i.e. 6v ≡ −3 (mod d)). Since (6, a) = 1, and 2|b_{i}, but 3 |̸b_{i} for all 1 ≤ i ≤ l, it is easy to see that

g_{2}(n, d) = \{ \begin{matrix}0 & if 2|d \\ \frac{n−3}{6d} + ϵ & otherwise, \end{matrix}

where |ϵ| ≤ 1.

Furthermore, in (4) for large enough n the number of terms is

2^{ω(e_{i})} < 2^{2 \frac{log(n^{2})}{log log(n^{2})}} < 2^{4 \frac{log n}{log log n}} ,

where we again used Lemma 2, e_{i} ≤ n^{2} and the fact that g(u) = 2\frac{log u}{log log u} is increasing if u is large enough.

Therefore

∑ ; \begin{matrix}u : 6u + 3 ≤ n \\ (6u + 3, e_{i}) = 1 \end{matrix} 1 = ∑ ; \begin{matrix}d|e_{i}, \\ 2 |̸d \end{matrix} μ(d)g_{2}(n, d) ≥

≥ \frac{n − 3}{6} ∏ ; \begin{matrix}p|e_{i} \\ 2 |̸p \end{matrix} (1 − \frac{1}{p}) − 2^{4 \frac{log n}{log log n}} ≥

≥ \frac{n − 3}{6} ∏_{p|e_{i}} (1 − \frac{1}{p}) − 2^{4 \frac{log n}{log log n}} ≥

≥ \frac{n − 3}{6} ∏_{p|b_{i}} (1 − \frac{1}{p}) ∏_{p|b_{i+1}} (1 − \frac{1}{p}) − 2^{4 \frac{log n}{log log n}} .

Hence for i = 1 we have

∑ ; \begin{matrix}u : 6u + 3 ≤ n, \\ (6u + 3, e_{1}) = 1, \\ 6u + 3 ∈ A_{(6,3)} \end{matrix} 1 ≥ \frac{n − 3}{6t′t′′} − 2^{4 \frac{log n}{log log n}} − (s_{1} + s_{2}) ≥ \frac{n}{7tt′′} − (s_{1} + s_{2}) ≥ \frac{n}{8tt′′} .

For i = l we get

∑ ; \begin{matrix}u : 6u + 3 ≤ n, \\ (6u + 3, e_{l}) = 1, \\ 6u + 3 ∈ A_{(6,3)} \end{matrix} 1 ≥ \frac{n − 3}{6t′′t} − 2^{4 \frac{log n}{log log n}} − (s_{1} + s_{2}) ≥ \frac{n}{7t′′t} − (s_{1} + s_{2}) ≥ \frac{n}{8t′′t}.

Finally for 1 < i < l

∑ ; \begin{matrix}u : 6u + 3 ≤ n, \\ (6u + 3, e_{i}) = 1, \\ 6u + 3 ∈ A_{(6,3)} \end{matrix} 1 ≥ \frac{n − 3}{6(t′′)^{2}} − 2^{4 \frac{log n}{log log n}} − (s_{1} + s_{2}) ≥ \frac{n}{7(t′′)^{2}} − (s_{1} + s_{2}) ≥ \frac{n}{8(t′′)^{2}} .

Thus it is not hard to see that if c is small enough then we may choose a different f_{i} for each 1 ≤ i ≤ l such that

(b_{i}, f_{i}) = (b_{i+1}, f_{i}) = 1 and f_{i} ∈ A_{(6,3)}.

Then

a, b_{1}, f_{1}, b_{2}, f_{2}, …, b_{l}, f_{l}, a

is a C_{2l+1} in G(A) completing the proof of Theorem 2.

2.2 Proof of Theorem 3

Here we assume that s_{2} ≥ \frac{ϵ}{2}n, the case s_{1} ≥ \frac{ϵ}{2}n is similar. Denote by P_{r} the product of the primes not exceeding r. The rough outline of the proof will be the following: First we find 3 positive integers j_{1}, j_{2} and j_{3} such that (j_{1}, j_{2}) = (j_{1}, j_{3}) = (j_{2}, j_{3}) = 1 and |A_{(P_{r},j_{i})}| is relatively large for each i = 1, 2, 3. Then if 1 ≤ l ≤ c_{3}n, to construct a C_{2l+1} in G(A) first we pick a number a ∈ A_{(P_{r},j_{1})} and then the remaining 2l numbers will be chosen alternately from A_{(P_{r},j_{2})} and A_{(P_{r},j_{3})}.

We will need the following lemma Lemma 3 . For every σ > 0 and δ > 0, there exists an r_{0} = r_{0}(σ, δ) such that if r ≥ r_{0}, n ≥ n_{4}(σ, δ, r) and u = 1, 2, …, P_{r}, then for all but σ \frac{n}{P_{r}} integers k satisfying

1 ≤ k ≤ n, k ≡ u (mod P_{r}),

we have

α(k) = ∏ ; \begin{matrix}p|k \\ p > r \end{matrix} (1 − \frac{1}{p}) > 1 − δ.

This lemma can be found in [9].

Now we prove Theorem 3. Let r denote a positive integer for which r ≥ r_{0}(\frac{ϵ}{8}, \frac{ϵ}{8}). We evidently have

∑_{i=1}^{\frac{1}{6} P_{r}} (|A_{(P_{r},6i−1)}| + |A_{(P_{r},6i)}| + … + |A_{(P_{r},6i+5)}|) =

= |A_{(6,0)}| + |A_{(6,1)}| + … + |A_{(6,4)}| + 2|A_{(6,5)}| = |A| + |A_{(6,5)}| > f(n, 2) + s_{2} > \frac{2}{3}n − 2 + \frac{ϵ}{2}n.

Hence there exists an i_{0} for which

|A_{(P_{r},6i_{0}−1)}| + |A_{(P_{r},6i_{0})}| + … + |A_{(P_{r},6i_{0}+5)}| > \frac{\frac{2}{3}n + \frac{ϵ}{2}n − 2}{\frac{P_{r}}{6}} = \frac{4n}{P_{r}} + 3ϵ\frac{n}{P_{r}} − \frac{12}{P_{r}}.  (5)

Clearly for every u

|A_{(P_{r},u)}| < \frac{n}{P_{r}} + 1.  (6)

We claim that (5) and (6) imply that there exist three integers j_{1}, j_{2} and j_{3} such that

6i_{0} − 1 ≤ j_{1} < j_{2} < j_{3} ≤ 6i_{0} + 5, (j_{1}, j_{2}) = (j_{1}, j_{3}) = (j_{2}, j_{3}) = 1,  (7)

and

|A_{(P_{r},j_{i})}| > \frac{ϵ}{2} \frac{n}{P_{r}} for all i = 1, 2, 3.  (8)

Indeed, if |A_{(P_{r},6i_{0}−1)}| ≤ \frac{ϵ}{2} \frac{n}{P_{r}}, then

|A_{(P_{r},6i_{0})}| + … + |A_{(P_{r},6i_{0}+5)}| > \frac{4n}{P_{r}} + \frac{5ϵ}{2} \frac{n}{P_{r}} − \frac{12}{P_{r}}.  (9)

But then (6) and (9) imply that there exist integers u_{1}, …, u_{5} such that

0 ≤ u_{1} < … < u_{5} ≤ 5,  (10)

and

|A_{(P_{r},6i_{0}+u_{i})}| > \frac{ϵ}{2} \frac{n}{P_{r}} for all i = 1, …, 5,

since otherwise

|A_{(P_{r},6i_{0})}| + … + |A_{(P_{r},6i_{0}+5)}| ≤ 4 (\frac{n}{P_{r}} + 1) + 2\frac{ϵ}{2} \frac{n}{P_{r}} = 4\frac{n}{P_{r}} + ϵ\frac{n}{P_{r}} + 4

would hold in contradiction with (9).

From (10) it follows that the sequence {6i_{0} + u_{1}, …, 6i_{0} + u_{5}} contains a subsequence {j_{1}, j_{2}, j_{3}} of 3 terms which are pairwise relatively prime, proving our claim in this case.

The case when |A_{(P_{r},6i_{0}+5)}| ≤ \frac{ϵ}{2} \frac{n}{P_{r}} is similar. Thus now we may assume that

|A_{(P_{r},6i_{0}−1)}| > \frac{ϵ}{2} \frac{n}{P_{r}} and |A_{(P_{r},6i_{0}+5)}| > \frac{ϵ}{2} \frac{n}{P_{r}} .

In this case we choose j_{1} = 6i_{0} − 1 and j_{3} = 6i_{0} + 5. For j_{2} we choose one from the integers 6i_{0} + 1, 6i_{0} + 2 and 6i_{0} + 3 for which

|A_{(P_{r},j_{2})}| > \frac{ϵ}{2} \frac{n}{P_{r}} .

(there must be one such a j_{2}) and then (7) and (8) clearly hold.

Thus the claim is proved, we have j_{1}, j_{2} and j_{3} satisfying (7) and (8). Let a denote a positive integer for which

a ∈ A_{(P_{r},j_{1})} and ∏ ; \begin{matrix}p|a \\ p > r \end{matrix} (1 − \frac{1}{p}) > 1 − \frac{ϵ}{8}.  (11)

Lemma 2 and the choice of r guarantee that such an a exists.

We are going to estimate from below the number of solutions of

(a, b_{x}) = 1, b_{x} ∈ A_{(P_{r},j_{2})}.  (12)

Assume that p|(a, d) for some d ≡ j_{2}(mod P_{r}). Since (j_{1}, j_{2}) = 1 we clearly have p > r. Denote by h(P_{r}, j, z) the number of those integers d for which d ≤ n, d ≡ j(mod P_{r}) and (z, d) = 1. It is not hard to see that

|h(P_{r}, j_{2}, a) − \frac{n}{P_{r}} ∏ ; \begin{matrix}p|a \\ p > r \end{matrix} (1 − \frac{1}{p})| ≤ 2^{ω(a)} < 2^{2 \frac{log n}{log log n}} .

From this and (11) we get for large n

h(P_{r}, j_{2}, a) > \frac{n}{P_{r}} ∏ ; \begin{matrix}p|a \\ p > r \end{matrix} (1 − \frac{1}{p}) − 2^{2 \frac{log n}{log log n}} >

> (1 − \frac{ϵ}{8}) \frac{n}{P_{r}} − 2^{2 \frac{log n}{log log n}} > (1 − \frac{ϵ}{4}) \frac{n}{P_{r}} .  (13)

Denoting the number of solutions of (12) by X, we have by (6), (8) and (13)

X ≥ |A_{(P_{r},j_{2})}| − ∑ ; \begin{matrix}d ≤ n \\ d ≡ j_{2}(mod P_{r}) \\ (a, d) > 1 \end{matrix} 1 =

= |A_{(P_{r},j_{2})}| − ( ∑ ; \begin{matrix}d ≤ n \\ d ≡ j_{2}(mod P_{r}) \end{matrix} 1 − h(P_{r}, j_{2}, a)) >

> \frac{ϵ}{2} \frac{n}{P_{r}} − (\frac{n}{P_{r}} + 1) + (1 − \frac{ϵ}{4}) \frac{n}{P_{r}} = \frac{ϵ}{4} \frac{n}{P_{r}} − 1 > \frac{ϵ}{5} \frac{n}{P_{r}} .  (14)

Therefore using Lemma 2 and (14), if c_{3} is small enough, then we can choose integers b_{1}, b_{2}, … b_{l} satisfying

(a, b_{i}) = 1, b_{i} ∈ A_{(P_{r},j_{2})},

and

∏ ; \begin{matrix}p|b_{i} \\ p > r \end{matrix} (1 − \frac{1}{p}) > (1 − \frac{ϵ}{8}) for all 1 ≤ i ≤ l.  (15)

Put e_{i} = [b_{i}, b_{i+1}] for all 1 ≤ i ≤ l where b_{l+1} is defined to be a.

Let us denote the number of solutions of

(e_{i}, f_{y}) = 1, f_{y} ∈ A_{(P_{r},j_{3})}

by Y_{i}. From (7), if g ≡ j_{3}(mod P_{r}) and p|(e_{i}, g), then p > r. Again we have

|h(P_{r}, j_{3}, e_{i}) − \frac{n}{P_{r}} ∏ ; \begin{matrix}p|e_{i} \\ p > r \end{matrix} (1 − \frac{1}{p})| ≤ 2^{ω(e_{i})} < 2^{2 \frac{log(n^{2})}{log log(n^{2})}} <

< 2^{4 \frac{log n}{log log n}} .

We obtain from this and (15) for large enough n that

h(P_{r}, j_{3}, e_{i}) > \frac{n}{P_{r}} ∏ ; \begin{matrix}p|e_{i} \\ p > r \end{matrix} (1 − \frac{1}{p}) − 2^{4 \frac{log n}{log log n}} ≥

≥ \frac{n}{P_{r}} ∏ ; \begin{matrix}p|b_{i} \\ p > r \end{matrix} (1 − \frac{1}{p}) ∏ ; \begin{matrix}p|b_{i+1} \\ p > r \end{matrix} (1 − \frac{1}{p}) − 2^{4 \frac{log n}{log log n}} >

> (1 − \frac{ϵ}{8})^{2} \frac{n}{P_{r}} − 2^{4 \frac{log n}{log log n}} > (1 − \frac{ϵ}{4}) \frac{n}{P_{r}} .  (16)

(6), (8) and (16) yield that

Y_{i} ≥ |A_{(P_{r},j_{3})}| − ( ∑ ; \begin{matrix}g ≤ n \\ g ≡ j_{3}(mod P_{r}) \end{matrix} 1 − h(P_{r}, j_{3}, e_{i})) >

> \frac{ϵ}{2} \frac{n}{P_{r}} − (\frac{n}{P_{r}} + 1) + (1 − \frac{ϵ}{4}) \frac{n}{P_{r}} = \frac{ϵ}{4} \frac{n}{P_{r}} − 1 > \frac{ϵ}{5} \frac{n}{P_{r}} .

Hence we can choose an integer h_{i} satisfying

(b_{i}, h_{i}) = (b_{i+1}, h_{i}) = 1 and h_{i} ∈ A_{(P_{r},j_{3})}

for all 1 ≤ i ≤ l. Then

a, b_{1}, h_{1}, b_{2}, h_{2}, …, b_{l}, h_{l}, a

form a C_{2l+1} in G(A) and this completes the proof of Theorem 3.

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[14] E. Saias, Etude du graphe divisoriel, I, Periodica Math. Hung., to appear.

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