Consider a sequence of positive integers in arithmetic progression u_k = u₀ + kr with (u₀, r) = 1. Denote the least common multiple of u₀, …, u_n by L_n. We show that if n⩾r² + r, then L_n⩾u₀r^r+1(r + 1), and we obtain optimum result on n in some cases for such estimate. Besides, for quadratic sequences m² + c, (m + 1)² + c, …, n² + c, we also show that the least common multiple is at least 2ⁿ when m ⩽ ⌈n/2⌉, 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 n positive integers is asymptotically upper bounded by (e + ε) ⁿ and lower bounded by (e − ε) ⁿ for any prefixed ε. As for effective uniform estimate, Hanson [2] obtained the upper bound 3ⁿ about forty years ago by considering Sylvester series of one. Nair [3] gave 2ⁿ 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 a₀, …, a_n are integers, we denote their least common multiple by [a₀, …, a_n]. Consider two coprime positive integers u₀ and r, and put u_k = u₀ + kr, L_k = [u₀, …, u_k]. A recent result of Hong et al. (cf. [8, 9]) shows that for any positive integers a, α, r, and n such that a⩾2, α, r⩾a, n⩾2αr, we have L_n⩾u₀r^α+a−2(r + 1) ⁿ.

Recently, Wu et al. [10] improved the Hong-Kominers lower bound. A special case of Hong-Kominers result tells us that if n⩾r(r + 3) (or n⩾r(r + 4)) if r is odd (or even), then L_n⩾u₀r^r+1(r + 1) ⁿ. In this note, we find that the Hong-Kominers lower bound is still valid if n ∈ [r(r + 1), r(r + r₀)) with r₀ = 3 or 4 if r is odd or even. That is, we have the following result.

Theorem 1. Let n, u₀, r ∈ ℕ with (u₀, r) = 1. One puts for any k ∈ ⟦0, n⟧, u_k = u₀ + kr and L_n = [u₀, …, u_n]. Then, for any n⩾r(r + 1),

()

Furthermore, if u₀ > r or u₀ < min {3, r}, the same estimate holds when n = r² + r − 1.

In 2007, Farhi [11] showed that [1² + 1, 2² + 1, …, n² + 1]⩾0.32(1.442) ⁿ. 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 c, m, n ∈ ℕ be such that 0 < m < n. Suppose that m ⩽ ⌈n/2⌉. Then, one has

()

This theorem improves the result in [11].

2. Proof of the First Theorem

Let x, y ∈ ℝ with y ≠ 0. We say that x is a multiple of y if there is an integer z such that x = yz. As usual, ⌊x⌋ denotes the largest integer not larger than x, and ⌈x⌉, denotes the smallest integer not smaller than x.

We will introduce the two following results. The first is a known result which tells us that L_n is a multiple of (u₀ ⋯ u_n/n!). This is can be proved by considering a suitable partial fraction expansion (cf. [11]) or by considering the integral (cf. [13]).

For ℓ ∈ ⟦0, n⟧, with a slight modification of notation as in [11], we put L_n,ℓ = [u_n−ℓ, …, u_n] and

()

Clearly, we have L_n = L_n,n and for any ℓ ∈ ⟦0, n⟧, L_n⩾L_n,ℓ. This result can be restated as

Lemma 3. For any ℓ ∈ [0, n], one can find a positive integer A_n,ℓ such that L_n,ℓ = A_n,ℓB_n,ℓ.

Our modification aims to emphasize the estimate of the terms A_n,ℓ and B_n,ℓ that will give us some improvement. If n⩾u₀, then we see that the first term u₀ does not play an important role as it will be a factor of the term u₀ + u₀r = u₀(1 + r). Hence, the behaviour should be different when n is large. It will be more interesting to give a control over the last ℓ terms.

Note that B_n,ℓ+1 = (u_n−ℓ−1/(ℓ + 1))B_n,ℓ, thus

()

We will put ℓ_n = min {⌊u_n/(r + 1)⌋, n}. The following lemma tells us that keeping the first smaller terms can increase at least the power n 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 n = 0, then ℓ_n = 0 and B_0,0 = u₀, and it holds. In fact, if n ⩽ u₀, then we have ℓ_n = n and

()

When n⩾u₀, we have (nr + u₀)/(r + 1) ⩽ (nr + n)/(r + 1) ⩽ n and ℓ_n = ⌊(nr + u₀)/(r + 1)⌋. It remains to derive the result for the case n + 1 from that of n⩾u₀.

It is obvious that ℓ_n ⩽ ℓ_n+1 ⩽ ℓ_n + 1 and u_n/(r + 1) − 1 ⩽ ℓ_n ⩽ u_n/(r + 1).

If ℓ_n = ℓ_n+1, then (n + 1)r + u₀ ⩽ (ℓ_n + 1)(r + 1) and .

Thus,

()

Hence,

()

If ℓ_n+1 = ℓ_n + 1, then

()

In either case, the principle of mathematical induction assures the result.

We can complete our proof now. Suppose that n⩾r(r + 1). Then,

()

By considering the first r multiples of r, we have r^r+1 | ℓ_n!. Since (r, u₀) = 1, we deduce that for all k ∈ ⟦0, n⟧, (r, u_k) = 1. Writing the result of Lemma 3 for ℓ = ℓ_n as

()

we conclude that

Using the Lemma 4, we obtain

()

which is our conclusion.

Consider the case n = r² + r − 1. If u₀ > r, then we still have ℓ_n⩾r².

Supposing now that r⩾2 and u₀ ⩽ min {2, r − 1}, we shall prove that, for n₀ = r² + r − 1, it is still possible to choose and so that .

On the one hand, when n = r² − r − 1, we have

()

Thus, n − ℓ_n = r − 2 and

()

On the other hand, we write

()

It suffices to show that for any s ∈ ⟦1, r − 1⟧,

()

Equation (15) is equivalent to

()

By expanding, it remains to verify for any integer r⩾s + 1⩾2 that

()

With simple computation, we obtain for any real r, s > 0 with r⩾s + 1,

()

which allow to establish (15).

Equation (16) is equivalent to

()

Such inequality holds for any positive integer r when u₀ = 1 or 2.

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 c, m, n ∈ ℕ be such that 0 < m < n. Put

()

Then,

()

Proof. We shall denote x^ı = cos (log x) + ısin(log x).

Consider the integral of complex-valued function of a real variable

()

Firstly, by integrating by parts (n − m) 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 0 ⩽ k ⩽ n − m and x_k, y_k ∈ ℤ. On the other hand, it is easy to see that the number obtained by integrating by parts (n − m) times is not zero. So, its modulus is not smaller than 1, and we conclude that

()

Now, we write furthermore

()

Suppose now that m ⩽ m^′ : = ⌈n/2⌉, then and n = 2m^′ or 2m^′ − 1.

The following Stirling estimate

()

allows us to obtain

()

If n = 2m^′, then

()

As the function

is increasing and its value at x = 4 is ⩾1.08, we conclude that for any integer m^′⩾4

()

If n = 2m^′ − 1, then we can conclude similarly for m^′⩾5

()

In brief, for any integer m, n with n⩾8 and m ⩽ ⌈n/2⌉, we just establish that

()

The case n = 7 can be checked directly as .

For the case n = 6, we have

()

For the case n = 5, we have

()

It is easy to check the cases n = 2, 3, and 4 as it involves only two terms, and we can make our final conclusion.

Acknowledgment

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

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Note on the Lower Bound of Least Common Multiple

Abstract

1. Introduction

2. Proof of the First Theorem

3. Proof of the Second Theorem

Acknowledgment

References

References

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Note on the Lower Bound of Least Common Multiple

Abstract

1. Introduction

2. Proof of the First Theorem

3. Proof of the Second Theorem

Acknowledgment

References

References

Related

Information