Note on the Lower Bound of Least Common Multiple

Abstract

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 2^𝑛 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 3^𝑛 about forty years ago by considering Sylvester series of one. Nair [3] gave 2^𝑛 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), 𝐿_𝑛⩾𝑢₀𝑟^𝑟+1(𝑟+1)^𝑛. (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 𝑚²+𝑐,(𝑚+1)²+𝑐,…,𝑛²+𝑐⩾2^𝑛. (2)

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 ∫¹₀𝑥^{𝑢0/𝑟−1}(1−𝑥)^𝑛𝑑𝑥 (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 𝐵_𝑛,ℓ𝑢∶=^𝑛−ℓ⋯𝑢_𝑛.ℓ! (3) 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 𝐵_𝑛,ℓ+1⩽𝐵_𝑛,ℓ⟺𝑢₀+((⩾𝑛−ℓ+1))𝑟⩽ℓ+1⟺ℓ+1𝑛𝑟+𝑢₀.𝑟+1 (4)

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 𝐵_𝑛,ℓ𝑛⩾𝑢₀(𝑟+1)^𝑛.

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 𝐵_𝑛,𝑛=𝑢₀𝑢₀1⋯𝑢+𝑟⁰𝑛+𝑟⩾𝑢₀(1+𝑟)^𝑛. (5)
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 𝑢_{𝑛−ℓ𝑛}=𝑢₀+(𝑛−ℓ_𝑛)𝑟⩽ℓ_𝑛+1.
Thus, 𝑢_𝑛+1𝑢_{𝑛−ℓ𝑛}=𝑢_{𝑛−ℓ𝑛}+ℓ_𝑛+1𝑟𝑢_{𝑛−ℓ𝑛}ℓ=1+^𝑛+1𝑢_{𝑛−ℓ𝑛}𝑟⩾1+𝑟. (6)
Hence, 𝐵_{𝑛+1,ℓ𝑛+1}=𝑢_𝑛+1𝑢_{𝑛−ℓ𝑛}⋅𝐵_𝑛,ℓ𝑛⩾𝑢₀(𝑟+1)^𝑛+1. (7)
If ℓ_𝑛+1=ℓ_𝑛+1, then 𝐵_{𝑛+1,ℓ𝑛+1}=𝑢_𝑛+1ℓ_𝑛+1⋅𝐵_𝑛,ℓ𝑛⩾𝑢₀(𝑟+1)^𝑛+1. (8)
In either case, the principle of mathematical induction assures the result.

We can complete our proof now. Suppose that 𝑛⩾𝑟(𝑟+1). Then, ℓ_𝑛=𝑢_𝑛⩾𝑟𝑟+1²+𝑢₀𝑟+1⩾𝑟². (9)

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 ℓ_𝑛!⋅𝐿_𝑛,ℓ𝑛=𝐴_𝑛,ℓ𝑛⋅𝑢_{𝑛−ℓ𝑛}⋯𝑢_𝑛, (10) we conclude that 𝑟^𝑟+1|𝐴_𝑛,ℓ𝑛.

Using the Lemma 4, we obtain 𝐿_𝑛⩾𝐿_𝑛,ℓ𝑛=𝐴_𝑛,ℓ𝑛𝐵_𝑛,ℓ𝑛⩾𝑢₀𝑟^𝑟+1(𝑟+1)^𝑛, (11) 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 ℓ^′_𝑛0=𝑟² and 𝑛₀−ℓ^′_𝑛0=𝑟−1 so that 𝐵_{𝑛0,ℓ′𝑛0}⩾𝑢₀(𝑟+1)^𝑛0.

On the one hand, when 𝑛=𝑟²−𝑟−1, we have ℓ_𝑛=𝑢₀+𝑟𝑟²−𝑟−1=𝑟𝑟+1₂(𝑟+1)−2𝑟(𝑟+1)+𝑟+𝑢₀𝑟+1=𝑟²−2𝑟+𝑟+𝑢₀𝑟+1=𝑟²−2𝑟+1. (12)

Thus, 𝑛−ℓ_𝑛=𝑟−2 and 𝑢_𝑟−2⋯𝑢_{𝑟2−𝑟−1}𝑟²−2𝑟+1!⩾𝑢₀(𝑟+1)^{𝑟2−𝑟−1}. (13)

On the other hand, we write 𝐵_{𝑛0,ℓ′𝑛0}=𝑢_𝑟−1⋯𝑢_{𝑟2+𝑟−1}𝑟²!=𝑢_𝑟−2⋯𝑢_{𝑟2−𝑟−1}𝑟²!⋅𝑢−2𝑟+1^𝑟2−𝑟𝑢_𝑟2𝑢_𝑟−2𝑟²⋅−𝑟+1_{𝑟−1𝑠=1}𝑢_{𝑟2−𝑟+𝑠}𝑢_{𝑟2+𝑟−𝑠}𝑟²−2𝑟+𝑠+1𝑟²−().𝑠−1 (14)

It suffices to show that for any 𝑠∈⟦1,𝑟−1⟧, 𝑢_{𝑟2−𝑟+𝑠}𝑢_{𝑟2+𝑟−𝑠}𝑟²−2𝑟+𝑠+1𝑟²−()⩾()𝑠−1𝑟+1²,𝑢_𝑟2−𝑟𝑢_𝑟2𝑢_𝑟−2𝑟²⩾()−𝑟+1𝑟+1². (15)(16)

Equation (15) is equivalent to 𝑢_{𝑟2−𝑟+𝑠}𝑢_{𝑟2+𝑟−𝑠}⩾𝑟²−2𝑟+𝑠+1𝑟²−(𝑠−1)(𝑟+1)². (17)

By expanding, it remains to verify for any integer 𝑟⩾𝑠+1⩾2 that 𝑓_𝑠(𝑟)∶=2𝑢₀𝑟³−(4𝑠−1)𝑟²+2𝑠(𝑠−1)𝑟+𝑠²+𝑢²₀−1⩾0. (18)

With simple computation, we obtain for any real 𝑟,𝑠>0 with 𝑟⩾𝑠+1, 𝑓_𝑠(𝑠+1)=2𝑢₀−1𝑠³+6𝑢₀−1𝑠²+23𝑢₀−2𝑠+𝑢²₀+2𝑢₀𝑓⩾0,_′𝑠(𝑟)⩾6𝑟²()()()−24𝑠−1𝑟+2𝑠𝑠−1⩾2𝑟−𝑠)(3𝑟−𝑠+1⩾0, (19) which allow to establish (15).

Equation (16) is equivalent to 𝑢₀+𝑟²−𝑟𝑟𝑢₀+𝑟³⩾𝑢₀+(𝑟−2)𝑟𝑟²−𝑟+1(𝑟+1)² (20) or 2−𝑢₀𝑟⁴+𝑢₀−1𝑟³+1−𝑢₀𝑟²+2−𝑢₀𝑟+𝑢²₀−𝑢₀⩾0. (21)

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

Finally, we still have 𝐵_{𝑛0,ℓ′𝑛0}⩾𝑢₀(𝑟+1)^𝑛0. (22) and the condition ℓ^′_𝑛0⩾𝑟² 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 𝐿^′_𝑚,𝑛∶=𝑚²+𝑐,(𝑚+1)²+𝑐,…,𝑛²+𝑐. (23)
Then, 𝐿^′_𝑚,𝑛⩾∏^𝑛_𝑘=𝑚𝑘²+𝑐()!.𝑛−𝑚 (24)

Proof. We shall denote 𝑥^𝚤=cos(log𝑥)+𝚤sin(log𝑥).
Consider the integral of complex-valued function of a real variable ¹₀𝑥^{√𝑚−1+𝑐𝚤}(1−𝑥)^𝑛−𝑚𝑑𝑥. (25)
Firstly, by integrating by parts (𝑛−𝑚) times, ¹₀𝑥^{√𝑚−1+𝑐𝚤}(1−𝑥)^𝑛−𝑚=𝑥𝑑𝑥_{√𝑚+𝑐𝚤}(1−𝑥)^𝑛−𝑚𝑚+√𝑐𝚤¹₀+𝑛−𝑚𝑚+√𝑐𝚤¹₀𝑥^{√𝑚+𝑐𝚤}(1−𝑥)^{𝑛−𝑚−1}=()!𝑑𝑥𝑛−𝑚∏^𝑛−1_𝑘=𝑚𝑘+√𝑐𝚤¹⁰𝑥^{√𝑛−1+𝑐𝚤}=()!𝑑𝑥𝑛−𝑚∏^𝑛_𝑘=𝑚𝑘+√.𝑐𝚤 (26)
Secondly, by expanding ¹₀𝑥^{√𝑚−1+𝑐𝚤}(1−𝑥)^𝑛−𝑚=𝑑𝑥_{10𝑛−𝑚𝑘=0}(−1)^𝑘𝑘𝑥𝑛−𝑚_{√𝑚−1+𝑘+𝑐𝚤}=𝑑𝑥_{𝑛−𝑚𝑘=0}(−1)^𝑘(_𝑘^𝑛−𝑚)𝑚+𝑘+√.𝑐𝚤 (27)
On one hand, put such complex number in Cartesian form, and after multiplying it by 𝐿^′_𝑚,𝑛, we get a linear combination of 𝑥_𝑘+𝑦_𝑘√𝑐𝚤 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 𝐿^′_𝑚,𝑛⩾∏^𝑛_𝑘=𝑚𝑘²+𝑐()!.𝑛−𝑚 (28)

Now, we write furthermore 𝐿^′_𝑚,𝑛⩾𝑛!()!()!𝑛𝑚𝑚−1𝑛−𝑚=𝑚. (29)

Suppose now that 𝑚⩽𝑚^′∶=⌈𝑛/2⌉, then 𝐿^′_𝑚,𝑛⩾𝐿^′_𝑚′,𝑛⩾𝑚^′𝑛_𝑚^′ and 𝑛=2𝑚^′ or 2𝑚^′−1.

The following Stirling estimate 𝑛𝑒^𝑛√2𝜋𝑛⩽𝑛!⩽𝑛𝑒^𝑛√2𝜋𝑛𝑒^1/(12𝑛) (30) allows us to obtain 𝑛⩾2𝑛2^2𝑛√𝑒𝑛𝜋^−1/(6𝑛). (31)

If 𝑛=2𝑚^′, then 𝑚^′2𝑚^′𝑚^′⩾2^2𝑚′𝑚^′𝜋𝑒^{−1/(6𝑚′)}. (32)

As the function (𝑥/𝜋)𝑒^−1/(6𝑥) is increasing and its value at 𝑥=4 is ⩾1.08, we conclude that for any integer 𝑚^′⩾4𝑚^′2𝑚^′𝑚^′⩾2^2𝑚′. (33)

If 𝑛=2𝑚^′−1, then we can conclude similarly for 𝑚^′⩾5𝑚^′2𝑚^′𝑚−1_′⩾2𝑚^′2𝑚−1^′𝑚−1_′−1⩾2^2𝑚′−1𝑚^′−1/2𝑚^′𝜋𝑒^{−1/(6𝑚′)}⩾2^2𝑚′−1. (34)

In brief, for any integer 𝑚,𝑛 with 𝑛⩾8 and 𝑚⩽⌈𝑛/2⌉, we just establish that 𝐿^′_𝑚,𝑛⩾2^𝑛. (35)

The case 𝑛=7 can be checked directly as 4⁷₄⩾2⁷.

For the case 𝑛=6, we have []⩾[]=()16+𝑐,25+𝑐,36+𝑐25+𝑐,36+𝑐25+𝑐)(36+𝑐()⩾gcd25+𝑐,36+𝑐25×3611⩾64=2⁶. (36)

For the case 𝑛=5, we have []⩾[]⩾9+𝑐,16+𝑐,25+𝑐16+𝑐,25+𝑐16×2525−16⩾2⁵. (37)

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.