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,…,an are integers, we denote their least common multiple by [a0,…,an]. Consider two coprime positive integers u0 and r, and put uk=u0+kr, Lk=[u0,…,uk]. 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 Ln⩾u0rα+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 Ln⩾u0rr+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⟧, uk=u0+kr and Ln=[u0,…,un]. Then, for any n⩾r(r+1),
(1)Ln⩾u0rr+1(r+1)n.

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
(2)[m2+c,(m+1)2+c,…,n2+c]⩾2n.

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 Ln is a multiple of (u0⋯un/n!). This is can be proved by considering a suitable partial fraction expansion (cf. [11]) or by considering the integral ∫01xu0/r-1(1-x)ndx (cf. [13]).

For ℓ∈⟦0,n⟧, with a slight modification of notation as in [11], we put Ln,ℓ=[un-ℓ,…,un] and
(3)Bn,ℓ:=un-ℓ⋯unℓ!.
Clearly, we have Ln=Ln,n and for any ℓ∈⟦0,n⟧, Ln⩾Ln,ℓ. This result can be restated as

Lemma 3.
For any ℓ∈[0,n], one can find a positive integer An,ℓ such that Ln,ℓ=An,ℓBn,ℓ.

Our modification aims to emphasize the estimate of the terms An,ℓ and Bn,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 Bn,ℓ+1=(un-ℓ-1/(ℓ+1))Bn,ℓ, thus
(4)Bn,ℓ+1⩽Bn,ℓ⟺u0+(n-(ℓ+1))r⩽ℓ+1⟺ℓ+1⩾nr+u0r+1.

We will put ℓn=min{⌊un/(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 Bn,ℓn⩾u0(r+1)n.

Proof.
We can just proceed by mathematical induction.

If n=0, then ℓn=0 and B0,0=u0, and it holds. In fact, if n⩽u0, then we have ℓn=n and
(5)Bn,n=u0(u01+r)⋯(u0n+r)⩾u0(1+r)n.

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

It is obvious that ℓn⩽ℓn+1⩽ℓn+1 and un/(r+1)-1⩽ℓn⩽un/(r+1).

If ℓn=ℓn+1, then (n+1)r+u0⩽(ℓn+1)(r+1) and un-ℓn=u0+(n-ℓn)r⩽ℓn+1.

Thus,
(6)un+1un-ℓn=un-ℓn+(ℓn+1)run-ℓn=1+ℓn+1un-ℓnr⩾1+r.

Hence,
(7)Bn+1,ℓn+1=un+1un-ℓn·Bn,ℓn⩾u0(r+1)n+1.

If ℓn+1=ℓn+1, then
(8)Bn+1,ℓn+1=un+1ℓn+1·Bn,ℓn⩾u0(r+1)n+1.

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

We can complete our proof now. Suppose that n⩾r(r+1). Then,
(9)ℓn=⌊unr+1⌋⩾⌊r2+u0r+1⌋⩾r2.

By considering the first r multiples of r, we have rr+1 | ℓn!. Since (r,u0)=1, we deduce that for all k∈⟦0,n⟧, (r,uk)=1. Writing the result of Lemma 3 for ℓ=ℓn as
(10)ℓn!·Ln,ℓn=An,ℓn·un-ℓn⋯un,
we conclude that rr+1 | An,ℓn.

Using the Lemma 4, we obtain
(11)Ln⩾Ln,ℓn=An,ℓnBn,ℓn⩾u0rr+1(r+1)n,
which is our conclusion.

Consider the case n=r2+r-1. If u0>r, then we still have ℓn⩾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 ℓn0′=r2 and n0-ℓn0′=r-1 so that Bn0,ℓn0′⩾u0(r+1)n0.

On the one hand, when n=r2-r-1, we have
(12)ℓn=⌊u0+r(r2-r-1)r+1⌋=⌊r2(r+1)-2r(r+1)+r+u0r+1⌋=r2-2r+⌊r+u0r+1⌋=r2-2r+1.

Thus, n-ℓn=r-2 and
(13)ur-2⋯ur2-r-1(r2-2r+1)!⩾u0(r+1)r2-r-1.

On the other hand, we write
(14)Bn0,ℓn0′=ur-1⋯ur2+r-1r2!=ur-2⋯ur2-r-1(r2-2r+1)!·ur2-rur2ur-2(r2-r+1)·∏s=1r-1ur2-r+sur2+r-s(r2-2r+s+1)(r2-(s-1)).

It suffices to show that for any s∈⟦1,r-1⟧,
(15)ur2-r+sur2+r-s(r2-2r+s+1)(r2-(s-1))⩾(r+1)2,(16)ur2-rur2ur-2(r2-r+1)⩾(r+1)2.

Equation (15) is equivalent to
(17)ur2-r+sur2+r-s⩾(r2-2r+s+1)(r2-(s-1))(r+1)2.

By expanding, it remains to verify for any integer r⩾s+1⩾2 that
(18)fs(r):=2u0r3-(4s-1)r2+2s(s-1)r+s2+u02-1⩾0.

With simple computation, we obtain for any real r,s>0 with r⩾s+1,
(19)fs(s+1)=2(u0-1)s3+6(u0-1)s2 +2(3u0-2)s+u02+2u0⩾0, fs′(r)⩾6r2-2(4s-1)r+2s(s-1) ⩾2(r-s)(3r-s+1)⩾0,
which allow to establish (15).

Equation (16) is equivalent to
(20)(u0+(r2-r)r)(u0+r3) ⩾(u0+(r-2)r)(r2-r+1)(r+1)2
or
(21)(2-u0)r4+(u0-1)r3+(1-u0)r2 +(2-u0)r+u02-u0⩾0.

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

Finally, we still have
(22)Bn0,ℓn0′⩾u0(r+1)n0.
and the condition ℓn0′⩾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
(23)Lm,n′:=[m2+c,(m+1)2+c,…,n2+c].

Then,
(24)Lm,n′⩾∏k=mnk2+c(n-m)!.

Proof.
We shall denote xı=cos(logx)+ısin(logx).

Consider the integral of complex-valued function of a real variable
(25)∫01xm-1+cı(1-x)n-mdx.

Firstly, by integrating by parts (n-m) times,
(26)∫01xm-1+cı(1-x)n-mdx =xm+cı(1-x)n-mm+cı|01 +n-mm+cı∫01xm+cı(1-x)n-m-1dx =(n-m)!∏k=mn-1(k+cı)∫01xn-1+cıdx =(n-m)!∏k=mn(k+cı).

Secondly, by expanding
(27)∫01xm-1+cı(1-x)n-mdx =∫01∑k=0n-m(-1)k(n-mk)xm-1+k+cıdx =∑k=0n-m(-1)k(n-mk)m+k+cı.

On one hand, put such complex number in Cartesian form, and after multiplying it by Lm,n′, we get a linear combination of xk+ykcı with integer coefficients, where 0⩽k⩽n-m and xk,yk∈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
(28)Lm,n′⩾∏k=mnk2+c(n-m)!.

Now, we write furthermore
(29)Lm,n′⩾n!(m-1)!(n-m)!=m(nm).

Suppose now that m⩽m′:=⌈n/2⌉, then Lm,n′⩾Lm′,n′⩾m′(nm′) and n=2m′ or 2m′-1.

The following Stirling estimate
(30)(ne)n2πn⩽n!⩽(ne)n2πne1/(12n)
allows us to obtain
(31)(2nn)⩾22nnπe-1/(6n).

If n=2m′, then
(32)m′(2m′m′)⩾22m′m′πe-1/(6m′).

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′⩾4(33)m′(2m′m′)⩾22m′.

If n=2m′-1, then we can conclude similarly for m′⩾5(34)m′(2m′-1m′)⩾(2m′-1)(2(m′-1)m′-1)⩾22m′-1m′-1/2m′πe-1/(6m′)⩾22m′-1.

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

The case n=7 can be checked directly as 4(74)⩾27.

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

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

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.