Let $n$ be a positive integer and $f(x)$ be a polynomial with nonnegative
integer coefficients. We prove that ${\rm lcm}_{\lceil n/2\rceil \le i\le n}
\{f(i)\}\ge 2^n$ except that $f(x)=x$ and $n=1, 2, 3, 4, 6$ and that $f(x)=x^s$
with $s\ge 2$ being an integer and $n=1$, where $\lceil n/2\rceil$ denotes the
smallest integer which is not less than $n/2$. This improves and extends the
lower bounds obtained by Nair in 1982, Farhi in 2007 and Oon in 2013.Comment: 6 pages. To appear in Comptes Rendus Mathematiqu

Hong, Shaofang

Luo, Yuanyuan

Qian, Guoyou

Wang, Chunlin

English

arXiv

Shaofang Hong

Yuanyuan Luo

Guoyou Qian

Chunlin Wang

Comptes Rendus Mathématique

C. R. Acad. Sci. Paris, Ser. I 351 (2013) 781–785Contents lists available at ScienceDirectC. R. Acad. Sci. Paris, Ser. Iwww.sciencedirect.comNumber theoryUniform lower bound for the least common multiple ofa polynomial sequence ✩Une borne inférieure uniforme pour le plus petit commun multiple d’unesuite polynomialeShaofang Hong a,b, Yuanyuan Luo a, Guoyou Qian c, Chunlin Wang aa Mathematical College, Sichuan University, Chengdu 610064, PR Chinab Yangtze Center of Mathematics, Sichuan University, Chengdu 610064, PR Chinac Center for Combinatorics, Nankai University, Tianjin 300071, PR Chinaa r t i c l e i n f o a b s t r a c tArticle history:Received 28 August 2013Accepted after revision 9 October 2013Available online 26 October 2013Presented by Jean-Pierre SerreLet n be a positive integer and f (x) be a polynomial with nonnegative integer coefficients.We prove that lcmn/2in{ f (i)}  2n , except that f (x) = x and n = 1,2,3,4,6 and thatf (x) = xs , with s  2 being an integer and n = 1, where n/2 denotes the smallest integer,which is not less than n/2. This improves and extends the lower bounds obtained byM. Nair in 1982, B. Farhi in 2007 and S.M. Oon in 2013.© 2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.r é s u m éSoit n un entier  1 et f (x) un polynôme à coefficients entiers  0. Nous démontrons que,à l’exception de certains cas explicites, on a ppcmn/2in{ f (i)}  2n , où n/2 dénote leplus petit entier  n/2. Ceci améliore, et étend, les bornes inférieures obtenues par M. Nairen 1982, B. Farhi en 2007 et S.M. Oon en 2013.© 2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.1. IntroductionThe least common multiple of consecutive positive integers was first studied by Chebyshev, who made an importantprogress for the proof of the prime number theorem. Actually, Chebyshev [3] introduced the function ψ(x) := ∑pkx log p =log lcm1ix{i}, where x > 0 is a real number. From Chebyshev’s work, one can derive that the prime number theorem isequivalent to the statement: ψ(n) = log lcm(1, . . . ,n) ∼ n as n tends to infinity. Since then, the least common multiple ofsequences of integers became popular. Bateman, Kalb and Stenger [2] gave an asymptotic formula of log lcm1in{b + ai}as n tends to infinity, where a  1 and b  0 are coprime integers. Hong, Qian and Tan [11] got an asymptotic formula ofthe least common multiple of a sequence of products of linear polynomials. Qian and Hong [14] investigated the asymptoticbehavior of the least common multiple of any consecutive arithmetic progression terms. Further, Farhi and Kane [6] andHong and Qian [10] obtained some results on the least common multiple of consecutive arithmetic progression terms.✩ The work was supported partially by National Science Foundation of China Grant #11371260, by the Ph.D. Programs Foundation of Ministry of Educationof China Grant #20100181110073 and by Postdoctoral Science Foundation of China Grant #2013M530109.E-mail addresses: sfhong@scu.edu.cn, s-f.hong@tom.com, hongsf02@yahoo.com (S. Hong), yuanyuanluoluo@163.com (Y. Luo), qiangy1230@163.com,qiangy1230@gmail.com (G. Qian), wdychl@126.com (C. Wang).1631-073X/$ – see front matter © 2013 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.http://dx.doi.org/10.1016/j.crma.2013.10.005782 S. Hong et al. / C. R. Acad. Sci. Paris, Ser. I 351 (2013) 781–785Effective bounds for the least common multiple of integer sequences are given by several authors. Hanson [7] provedthat lcm1in{i} < 3n for any integer n  1. Nair [12] showed that lcm1in{i}  2n for any integer n  7. Lower bounds ofthe least common multiple of finite arithmetic progression have been investigated by Farhi [4,5], Hong and Feng [8], Hongand Kominers [9] and Wu et al. [15]. For the quadratic case, some results are also achieved. Farhi [5] provided a nontriviallower bound for lcm1in{i2 + 1}. Oon [13] improved Farhi’s lower bound by proving that lcm1in{i2 + c}  2n with cbeing a positive integer.In this paper, we find surprisingly that 2n is the uniform lower bound for the least common multiple of polynomialsequences of nonnegative integer coefficients. That is, we have the following result.Theorem 1.1. Let n  1 be an integer and f (x) be a polynomial of nonnegative integer coefficients. Then lcmn/2in{ f (i)}  2nexcept that f (x) = x and n = 1,2,3,4,6 and that f (x) = xs with s  2 being an integer and n = 1, where n/2 stands for thesmallest integer, which is not less than n/2.In particular, we have the following interesting result.Corollary 1.2. Let n  1 be an integer and f (x) be a polynomial of nonnegative integer coefficients. Then lcm1in{ f (i)}  2n exceptthat f (x) = x and n = 1,2,3,4,6 and that f (x) = xs with s  2 being an integer and n = 1.Evidently, if we take f (x) = x, then Corollary 1.2 becomes Nair’s lower bound [12]. If one picks f (x) = x2 + c, thenTheorem 1.1 reduces to Oon’s result [13].The paper is organized as follows. In Section 2, we present some basic facts that are needed in the proof of our mainresult. Consequently, in Section 3, we prove three results about the least common multiple, and then show Theorem 1.1 asthe conclusion of this paper.2. PreliminariesIn this section, we show three lemmas that can be proved with a little effort and are needed in the proof of Theorem 1.1.Recall that a complex number is called an algebraic integer if it is a root of a monic polynomial of integer coefficients (see,for example, [1]).Lemma 2.1. Let s  1 be an integer and f (x) = ∑si=0 ai xi ∈ Z[x] be a polynomial of degree s. If α1, . . . ,αs are s roots of f (x), thenas(∏j∈{1,...,s}\{i} α j) is an algebraic integer for each integer i with 1  i  s.Proof. Clearly, Lemma 2.1 is true if s = 1. We let s  2 in what follows. Write βi := as(∏ j∈{1,...,s}\{i} α j) for each integer iwith 1  i  s. If at least two of α1, . . . ,αs are zero, then βi = 0 for each integer i with 1  i  s. So Lemma 2.1 holds in thiscase. If exactly one of α1, . . . ,αs is zero, saying αt = 0 for some integer t ∈ {1, . . . , s}, then βt = (−1)s−1a1 ∈ Z and βi = 0for each integer i with i = t and 1  i  s. Hence Lemma 2.1 is true in this case.Assume now that none of α1, . . . ,αs is zero. Fix an integer i with 1  i  s. Since as = 0, one has βi = as (−1)sa0/asαi =(−1)s a0αi . Therefore, to show that βi is an algebraic integer, it suffices to prove thata0αiis an algebraic integer. From f (αi) = 0,one derives that:as−10αsif (αi) =(a0αi)s+ a1(a0αi)s−1+ · · · + as−1as−20(a0αi)+ asas−10 = 0.This means that a0αi is a root of the integer polynomial g(x) = xs + a1xs−1 + · · · + as−1as−20 x + asas−10 , from which it followsthat a0αi is an algebraic integer. Lemma 2.1 is proved in this case. The proof of Lemma 2.1 is complete. Lemma 2.2. For any positive integer n  7, we have n/2( nn/2) > 2n.Proof. We prove Lemma 2.2 by induction on n. Evidently, n/2( nn/2) > 2n holds for n = 7 and 8. Now let n  7 and weassume that n/2( nn/2) > 2n is true for the n case. Now we consider the n + 1 case. One can easily check that:⌈(n + 1)/2⌉( n + 1(n + 1)/2)={2n/2( nn/2), if n is odd,(2n/2 + 1)( nn/2), if n is even.It then follows that (n + 1)/2( n+1(n+1)/2) > 2n+1. Hence Lemma 2.2 holds for the n + 1 case. Lemma 2.2 is proved. S. Hong et al. / C. R. Acad. Sci. Paris, Ser. I 351 (2013) 781–785 783Lemma 2.3. Let x be an indeterminate and let m and n be positive integers such that m  n. Then we have:n∑k=m(−1)n−k(n − mk − m) n∏j=mj =k(x − j) = (n − m)!. (2.1)Proof. We show (2.1) by induction on n. Obviously, (2.1) is true if n = m. Suppose that (2.1) holds for the n−1 case. Now welet n > m. We prove that (2.1) also holds for the n case. Since (1 − 1)n−m = 0, we have ∑n−1k=m(−1)k−m(n−mk−m) = (−1)n−1−m .So by induction hypothesis, we get that:(n − m)! = (n − m)n−1∑k=m(−1)n−1−k(n − 1 − mk − m) n−1∏j=mj =k(x − j)= (n − m)n−1∑k=m(−1)n−kn − k(n − 1 − mk − m)((x − n) − (x − k)) n−1∏j=mj =k(x − j)=n−1∑k=m(−1)n−k(n − mk − m) n∏j=mj =k(x − j) −(n−1∏j=m(x − j))n−1∑k=m(−1)n−k(n − mk − m)=n−1∑k=m(−1)n−k(n − mk − m) n∏j=mj =k(x − j) +(n − mn − m) n−1∏j=m(x − j)=n∑k=m(−1)n−k(n − mk − m) n∏j=mj =k(x − j).Therefore (2.1) is true for the n case. This finishes the proof of Lemma 2.3. Remark. Lemma 2.3 can be proved using other methods. For example, one can prove it by using the fundamental theoremof algebra.3. Proof of Theorem 1.1In this section, we show Theorem 1.1. We begin with the following lemma.Lemma 3.1. Let s  1 be an integer and f (x) ∈ Z[x] be a polynomial of degree s and with as as its leading coefficient. Then for anytwo positive integers m and n with 1  m  n, we have:lcm(f (m), f (m + 1), . . . , f (n))  1(n − m)!n∏k=m∣∣∣∣ f (k)as∣∣∣∣1s.Proof. If f (k) = 0 for some integer k with m  k  n, then Lemma 3.1 is clearly true. In what follows, we assume thatf (k) = 0 for all integers k with m  k  n.Write f (x) = ∑si=0 ai xi . Suppose that α1, . . . ,αs are s roots of f (x). Then f (x) = as(x − α1) · · · (x − αs). It infers thathk(x) := (−1)s f (k−x) = as ∏si=1(x−(k−αi)) ∈ Z[x] is also a polynomial with the leading coefficient as and k−α1, . . . ,k−αsare s roots of hk(x) for each integer k with m  k  n. So by Lemma 2.1, we know that f (k)k−αi = as∏j∈{1,...,s}\{i}(k − α j) is analgebraic integer for each pair (k, i) with m  k  n and 1  i  s. It follows that lcm( f (m), f (m + 1), . . . , f (n))/(k − αi) isan algebraic integer for each integer i with 1  i  s. Since f (k) = 0 for all m  k  n, we have k − αi = 0 for all pairs (k, i)with 1  i  s and m  k  n. Then letting x = αi(1  i  s) in (2.1), one deduces that:(n − m)!∏nk=m(k − αi)=n∑(−1)k−m(n − mk − m)1k − αi . (3.1)k=m784 S. Hong et al. / C. R. Acad. Sci. Paris, Ser. I 351 (2013) 781–785Multiplying both sides of (3.1) by lcm( f (m), . . . , f (n)), we obtain that:Ai := (n − m)! lcm(f (m), . . . , f (n)) n∏k=m1k − αiis a nonzero algebraic integer, and so is the product A := ∏si=1 Ai . But one can easily derive that:A = ((n − m)!)s(lcm( f (m), . . . , f (n)))s n∏k=masf (k), (3.2)which implies that A is a nonzero rational number. Thus A is a nonzero rational integer and so |A|  1. This, togetherwith (3.2), concludes the desired result. The proof of Lemma 3.1 is complete. Lemma 3.2. Let f (x) be a polynomial of degree 2 and of nonnegative integer coefficients. Then for any integer m  2, we havelcm( f (m − 1), f (m))  (m(m−1))22m−1 .Proof. Since for any integer m  2, we have gcd( f (m − 1), f (m))|( f (m) − f (m − 1)). This infers that gcd( f (m − 1), f (m)) f (m) − f (m − 1). Write f (x) = a2x2 + a1x + a0 ∈ Z[x], where a0,a1  0 and a2  1. Then we have:lcm(f (m − 1), f (m)) = f (m) f (m − 1)gcd( f (m − 1), f (m)) f (m) f (m − 1)f (m) − f (m − 1)= (a2m2 + a1m + a0)(a2(m − 1)2 + a1(m − 1) + a0)(2m − 1)a2 + a1 m2(m − 1)(a2(m − 1) + a1)(2m − 1)a2 + a1 m2(m − 1)(a2(m − 1) + a1)(2m − 1)a2 + 2m−1m−1 a1= (m(m − 1))22m − 1as desired. This concludes the proof of Lemma 3.2. Lemma 3.3. Let a and b be coprime positive integers. Then lcm(a,a + b,a + 2b) = a(a + b)(a + 2b) or 12 a(a + b)(a + 2b).Proof. Since a and b are coprime, we have gcd(a,a +b)| gcd(a, (a +b)−a) = gcd(a,b) and hence gcd(a,a +b) = 1. Similarly,one has gcd(a + b,a + 2b) = 1 and gcd(a,a + 2b)| gcd(a,2b). So gcd(a,a + 2b) = 1 or 2. Then the desired result followsimmediately from the following well-known identity:lcm(a,a + b,a + 2b) = a(a + b)(a + 2b)gcd(a,a + b,a + 2b)gcd(a,a + b)gcd(a + b,a + 2b)gcd(a,a + 2b) .So Lemma 3.3 is proved. We are now in a position to show Theorem 1.1.Proof of Theorem 1.1. Since f (x) is a polynomial with nonnegative integer coefficients, we may let f (x) = asxs +as−1xs−1 +· · ·+a1x+a0 ∈ Z[x], where ai  0 and as  1. Then for any integer n  7, by Lemmas 3.1 and 2.2 and noting that f (k)  asks ,we have:lcmn/2in{f (i)}∏nk=n/2 | f (k)as |1s(n − n/2)! ∏nk=n/2 k(n − n/2)! = n/2(nn/2)> 2n.So it remains to check that Theorem 1.1 is true for all positive integers n  6 in the following. First we consider the casen = 1. If f (x) has at least two terms or as  2, then lcm( f (n/2), . . . , f (n)) = f (1) = ∑si=0 ai  2. Now let 2  n  6. Wedivide the proof into the following three cases.Case 1. s  3. Then lcm( f ( n2 ), . . . , f (n))  asns  ns  n3 > 2n for each integer n with 2  n  6. So Theorem 1.1 is truein this case.Case 2. s = 2. Then lcm( f ( n2 ), . . . , f (n))  f (n)  a2n2  n2  2n for each integer n with 2  n  4. On the other hand,by Lemma 3.2, we have lcm( f ( n2 ), . . . , f (n))  lcm( f (n − 1), f (n))  (n(n−1))22n−1  2n for n = 5,6. Theorem 1.1 is proved inthis case.S. Hong et al. / C. R. Acad. Sci. Paris, Ser. I 351 (2013) 781–785 785Case 3. s = 1. First let a0 = 0, a1 = 1 and n = 5. Then f (x) = x. Hence lcm n2 in{ f (i)} = lcm 52 i5{i} = lcm(3,4,5) =60 > 25 as required. Now let a0 = 0 and a1  2. Then f (x) = a1x. Denote Ln := lcm n2 in{i}. It is well known thatLn  2n−1. Since a1  2 and a0 = 0, one has lcm n2 in{ f (i)} = a1Ln  2Ln  2n , as claimed.Finally, let a0  1, a1  1 and gcd(a0,a1) = d. One may write a1 = da and a0 = db for some coprime positive integers aand b. If n = 2 and 3, then we have:lcm n2 in{f (i)} = lcm( f (n − 1), f (n)) = f (n − 1) f (n)gcd( f (n − 1), f (n))= f (n − 1) f (n)gcd( f (n) − f (n − 1), f (n)) =(a1(n − 1) + a0)(a1n + a0)gcd(a1,a1n + a0)= d(a(n − 1) + b)(an + b)  n(n + 1) > 2nas required. If 4  n  6, then by Lemma 3.3, one has:lcm n2 in{f (i)} lcm(f (n − 2), f (n − 1), f (n)) = d · lcm(a(n − 2) + b,a(n − 1) + b,an + b) 12d(a(n − 2) + b)(a(n − 1) + b)(an + b)  12(n − 1)n(n + 1) > 2n,as desired.This completes the proof of Theorem 1.1. AcknowledgementsThe authors would like to thank the anonymous referee for very helpful comments and suggestions that improved thispresentation.References[1] S. Alaca, K.S. Williams, Introductory Algebraic Number Theory, Cambridge University Press, Cambridge, 2004.[2] P. Bateman, J. Kalb, A. Stenger, A limit involving least common multiples, Amer. Math. Monthly 109 (2002) 393–394.[3] P.L. Chebyshev, Memoire sur les nombres premiers, J. Math. Pures Appl. 17 (1852) 366–390.[4] B. Farhi, Minoration non triviales du plus petit commun multiple de certaines suites finies d’entiers, C. R. Acad. Sci. Paris, Ser. I 341 (2005) 469–474.[5] B. Farhi, Nontrivial lower bounds for the least common multiple of some finite sequences of integers, J. Number Theory 125 (2007) 393–411.[6] B. Farhi, D. Kane, New results on the least common multiple of consecutive integers, Proc. Amer. Math. Soc. 137 (2009) 1933–1939.[7] D. Hanson, On the product of the primes, Canad. Math. Bull. 15 (1972) 33–37.[8] S. Hong, W. Feng, Lower bounds for the least common multiple of finite arithmetic progressions, C. R. Acad. Sci. Paris, Ser. I 343 (2006) 695–698.[9] S. Hong, S.D. Kominers, Further improvements of lower bounds for the least common multiple of arithmetic progressions, Proc. Amer. Math. Soc. 138(2010) 809–813.[10] S. Hong, G. Qian, The least common multiple of consecutive arithmetic progression terms, Proc. Edinb. Math. Soc. 54 (2011) 431–441.[11] S. Hong, G. Qian, Q. Tan, The least common multiple of a sequence of products of linear polynomials, Acta Math. Hung. 135 (2012) 160–167.[12] M. Nair, On Chebyshev-type inequalities for primes, Amer. Math. Monthly 89 (1982) 126–129.[13] S.M. Oon, Note on the lower bound of least common multiple, Abstr. Appl. Anal. (2013), Art. ID 218125, 4 p.[14] G. Qian, S. Hong, Asymptotic behavior of the least common multiple of consecutive arithmetic progression terms, Arch. Math. 100 (2013) 337–345.[15] R. Wu, Q. Tan, S. Hong, New lower bounds for the least common multiples of arithmetic progressions, Chin. Ann. Math., Ser. B (2013),http://dx.doi.org/10.1007/s11401-013-0001-y.

Uniform lower bound for the least common multiple of a polynomial sequence

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Uniform lower bound for the least common multiple of a polynomial
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Uniform lower bound for the least common multiple of a polynomial sequence

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