Consider a sequence of positive integers in arithmetic progression with . Denote the least common multiple of by . We show that if , then , and we obtain optimum result on in some cases for such estimate. Besides, for quadratic sequences , we also show that the least common multiple is at least when , which sharpens a recent result of Farhi.

1. Introduction

Integer sequences in arithmetic progressions constitute a recurrent theme in number theory. The most notable result in this new century is perhaps the existence of arbitrary long sequences of primes in arithmetic progressions due to Green and Tao [1].

The bounds of the least common multiple for the finite sequences in arithmetic progressions also attract some attention. The prime number theorem assures that the least common multiple of the first positive integers is asymptotically upper bounded by and lower bounded by for any prefixed . As for effective uniform estimate, Hanson [2] obtained the upper bound about forty years ago by considering Sylvester series of one. Nair [3] gave as alower bound in a simple proof ten years later in view of obtaining a Chebyshev-type estimate on the number of prime numbers as in a tauberian theorem due to Shapiro [4].

Recently, some results concerning the lower bound of the least common multiple of positive integers in finite arithmetic progressions were obtained by Farhi [5]. Some other results about the least common multiple of consecutive integers and consecutive arithmetic progression terms are given by Farhi and Kane [6] and by Hong and Qian [7], respectively. If are integers, we denote their least common multiple by . Consider two coprime positive integers and , and put , . A recent result of Hong et al. (cf. [8, 9]) shows that for any positive integers ,  and such that , we have .

Recently, Wu et al. [10] improved the Hong-Kominers lower bound. A special case of Hong-Kominers result tells us that if (or ) if is odd (or even), then . In this note, we find that the Hong-Kominers lower bound is still valid if with or 4 if is odd or even. That is, we have the following result.

Theorem 1. Let with . One puts for any and . Then, for any ,
Furthermore, if or , the same estimate holds when .

In 2007, Farhi [11] showed that . Note that Qian et al. [12] obtained some results on the least common multiple of consecutive terms in a quadratic progression. We can now state the second result of this paper.

Theorem 2. Let be such that . Suppose that . Then, one has

This theorem improves the result in [11].

2. Proof of the First Theorem

Let with . We say that is a multiple of if there is an integer such that . As usual, denotes the largest integer not larger than , and , denotes the smallest integer not smaller than .

We will introduce the two following results. The first is a known result which tells us that is a multiple of (). This is can be proved by considering a suitable partial fraction expansion (cf. [11]) or by considering the integral (cf. [13]).

For , with a slight modification of notation as in [11], we put and Clearly, we have and for any , . This result can be restated as

Lemma 3. For any , one can find a positive integer such that .

Our modification aims to emphasize the estimate of the terms and that will give us some improvement. If , then we see that the first term does not play an important role as it will be a factor of the term . Hence, the behaviour should be different when is large. It will be more interesting to give a control over the last terms.

Note that , thus

We will put . The following lemma tells us that keeping the first smaller terms can increase at least the power in the estimate of lower bound in such a way (cf. also [13]).

Lemma 4. One has .

Proof. We can just proceed by mathematical induction.
If , then and , and it holds. In fact, if , then we have and
When , we have and . It remains to derive the result for the case from that of .
It is obvious that and .
If , then and .
If , then
In either case, the principle of mathematical induction assures the result.

We can complete our proof now. Suppose that . Then,

By considering the first multiples of , we have . Since , we deduce that for all , . Writing the result of Lemma 3 for as we conclude that .

Using the Lemma 4, we obtain which is our conclusion.

Consider the case . If , then we still have .

Supposing now that and , we shall prove that, for , it is still possible to choose and so that .

On the one hand, when , we have

Thus, and

On the other hand, we write

It suffices to show that for any ,

Equation (15) is equivalent to

By expanding, it remains to verify for any integer that

With simple computation, we obtain for any real with , which allow to establish (15).

Equation (16) is equivalent to or

Such inequality holds for any positive integer when or .

Finally, we still have and the condition allows us to conclude.

3. Proof of the Second Theorem

We shall start by proving the following lemma.

Lemma 5. Let be such that . Put

Proof. We shall denote .
Consider the integral of complex-valued function of a real variable
Firstly, by integrating by parts times,
Secondly, by expanding
On one hand, put such complex number in Cartesian form, and after multiplying it by , we get a linear combination of with integer coefficients, where and . On the other hand, it is easy to see that the number obtained by integrating by parts times is not zero. So, its modulus is not smaller than 1, and we conclude that

Now, we write furthermore

Suppose now that , then and or .

The following Stirling estimate allows us to obtain

If , then

As the function is increasing and its value at is , we conclude that for any integer

If , then we can conclude similarly for

In brief, for any integer with and , we just establish that

The case can be checked directly as .

For the case , we have

For the case , we have

It is easy to check the cases ,  and as it involves only two terms, and we can make our final conclusion.


The author would like to thank the referees for their helpful suggestions that improved the presentation of this paper.