Nontrivial lower bounds for the least common multiple of some finite sequences of integers


 Sheila Lloyd
 4 years ago
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1 J. Numbe Theoy, 15 (007), p Nontivial lowe bounds fo the least common multiple of some finite sequences of integes Bai FARHI Abstact We pesent hee a method which allows to deive a nontivial lowe bounds fo the least common multiple of some finite sequences of integes. We obtain efficient lowe bounds (which in a way ae optimal) fo the aithmetic pogessions and lowe bounds less efficient (but nontivial) fo quadatic sequences whose geneal tem has the fom u n = an(n + t) + b with (a, t, b) Z 3, a 5, t 0, gcd(a, b) = 1. Fom this, we deduce fo instance the lowe bound: lcm{1 +1, +1,..., n +1} 0.3(1.44) n (fo all n 1). In the last pat of this aticle, we study the intege lcm(n, n+1,..., n+ ) ( N, n N \ {0}). We show that it has a diviso d n, simple in its dependence on n and, and a multiple m n, also simple in its dependence on n. In addition, we pove that both equalities: lcm(n, n+1,..., n+) = d n, and lcm(n, n + 1,..., n+) = m n, hold fo an infinitely many pais (n, ). MSC: 11A05. Keywods: Least common multiple. 1 Intoduction and notations In this aticle, [x] denotes the intege pat of a given eal numbe x. Futhe, we say that a eal x is a multiple of a nonzeo eal y if the atio x/y is an intege. log lcm{1,...,n} The pime numbes theoem (see e.g. []) shows that lim n + = n 1. This is equivalent to the following statement: ε > 0, N = N(ε) / n N : (e ε) n lcm{1,..., n} (e + ε) n. 1
2 Concening the effective estimates of the numbes lcm{1,..., n} (n 1), one has among othes, two main esults. The fist one is by Hanson [1] which shows (by using the development of the numbe 1 in Sylveste seies) that lcm{1,..., n} 3 n fo all n 1. The second one is by Nai [3] which poves (simply by exploiting the integal 1 0 xn (1 x) n dx) that one has lcm{1,..., n} n fo all n 7. In this, we pesent a method which allows to find a nontivial lowe bounds fo the least common multiple of n consecutive tems (n N ) of some sequences of integes. We obtain efficient lowe bounds (which in a way ae optimal) fo the aithmetical pogessions (see Theoem 5). Besides, we also obtain less efficient lowe bounds (but nontivial) fo the quadatic sequences whose geneal tem has the fom: u n = an(n + t) + b with (a, t, b) Z 3, a 5, t 0, gcd(a, b) = 1 (see Coollay 10). Ou method is based on the use of some identities elated to the sequences which we study. Moe pecisely, let (α i ) i I be a given finite sequence of nonzeo integes. We see an identity of type 1 i I α i β i = 1 whee β γ i (i I) and γ ae nonzeo integes. If lcm{β i, i I} is bounded (say by a eal constant R > 0), one concludes that lcm{α i, i I} γ (see Lemma 1). It emains to chec R whethe this late estimate is nontivial o not. Howeve, the point is that looing fo identities of the above types is not easy. Theoem stems fom concete and inteesting example of such identities. Though, it is not liewise that we can find othe nontivial applications, than the ones pesented hee, fo that specific example. In ode to have nontivial lowe bounds of least common multiple fo othe families of finite sequences, it could be necessay to see fo new identities elated to those sequences. In the last pat of this aticle, we study the least common multiple of some numbe of consecutive integes, lage than a given positive intege. In Theoem 11, we show that the intege lcm{n, n + 1,..., n + } (n N, N) has a diviso d n, simple in its dependence on n and and a multiple m n, simple in its dependence on n. In addition, we pove that d n, and m n, ae optimal in that sense that the equalities lcm{n,..., n + } = d n, and lcm{n,..., n + } = m n, hold fo infinitely many pais (n, ). Moe pecisely, we show that both equalities ae satisfied at least when (n, ) satisfies some conguence modulo! (see Theoem 1).
3 Results.1 Basic Results Lemma 1 Let (α i ) i I and (β i ) i I be two finite sequences of nonzeo integes such that: 1 = 1 α i β i γ i I fo some nonzeo intege γ. Then, the intege lcm{α i, i I}.lcm{β i, i I} is a multiple of γ. Theoem Let (u ) N be a stictly inceasing sequence of nonzeo integes. Then, fo any positive intege n, the intege: { } lcm {u 0,..., u n }.lcm (u i u j ) ; j = 0,..., n is a multiple of the intege (u 0 u 1... u n ).. Results about the aithmetic pogessions Theoem 3 Let (u ) N be a stictly inceasing aithmetic pogession of nonzeo integes. Then, fo any nonnegative intege n, the intege lcm{u 0,..., u n } is a multiple of the ational numbe: u 0... u n n! (gcd{u 0, u 1 }) n. Theoem 4 (Optimality of Theoem 3) Let (u ) N be a stictly inceasing aithmetic pogession of nonzeo integes such that u 0 and u 1 ae copime. Then, fo any positive intege n which satisfies: u 0 u n 0 mod(n!), we have: lcm{u 0,..., u n } = u 0... u n. n! Theoem 5 Let (u ) N be an aithmetic pogession of integes whose diffeence and fist tem u 0 ae positive and copime. Then: 3
4 1) Fo any n N, we have: lcm{u 0,..., u n } u 0 ( + 1) n 1. Besides, if n is a multiple of ( + 1), we have: lcm{u 0,..., u n } u 0 ( + 1) n. ) Fo any n N, we have: lcm{u 0,..., u n } ( + 1) n 1. 3) Fo any n N, we have: lcm{u 0,..., u n } n n + 1 { ( + 1) n 1 + ( 1) n 1}. 4) Fo any n N satisfying n u 0 3+1, we have: lcm{u 0,..., u n } 1 ( + 1) n 1+ u 0. π The following Conjectue impoves the pats 1) and ) of Theoem 5. Besides, the fist pat of this Theoem ensues its validity in the paticula case whee the intege n is a multiple of ( + 1). Conjectue 6 In the situation of Theoem 5, we have fo any n N: lcm{u 0,..., u n } u 0 ( + 1) n. The two following Theoems study the optimality of the pat 4) of Theoem 5. Theoem 7 The coefficient 3 affected to which appeas in the condition n u of the pat 4) of Theoem 5 is optimal. Theoem 8 1) The optimal absolute constant C fo which the assetion: Fo any aithmetic sequence (u ) as in Theoem 5 and fo any nonnegative intege n satisfying n u 0 3+1, we have: lcm{u 0,..., u n } C ( + 1) n 1+ u 0 is tue, satisfies: 1 π C 3. 4
5 ) Moe geneally, given n 0 N, the optimal constant C(n 0 ) (depending uniquely on n 0 ) fo which the assetion: Fo any aithmetic sequence (u ) as in Theoem 5 and fo any intege n satisfying n max{n 0, u }, we have: lcm{u 0,..., u n } C(n 0 ) ( + 1) n 1+ u 0 is tue, satisfies: 1 π C(n 0) < 4(n 0 + 4) n Comments: i) The lowe bound poposed by Conjectue 6 is optimal on the exponent n of (+1). Indeed, fo any positive intege n and fo any aithmetic pogession (u ) as in Theoem 5, we obviously have: lcm{u 0,..., u n } u 0 u 1... u n u 0 (max{u 0, n}) n ( + 1) n. Fo any given positive eal ε, we can choose two abitay positive integes u 0 and n and a positive intege, which is copime with u 0 and sufficiently lage as to have ( + 1) ε > (max{u 0, n}) n. The aithmetic pogession (u ), with fist tem u 0 and diffeence, will then satisfy: lcm{u 0,..., u n } < u 0 ( + 1) n+ε. ii) A simila agument to that of the above pat i) shows that the exponent (n 1) of ( + 1) which appeas in the lowe bound of the pat ) of Theoem 5 is optimal. iii) Fo small values of n accoding to, the lowe bound of the pat 3) of Theoem 5 implies the one of the pat ) of the same Theoem. Moe pecisely, it can be checed that the necessay and sufficiently condition fo the holding of this impovement is n n , that is n f(), whee n 1 n 1 1 f is a eal function which is equivalent to 1 log as tends to infinite. iv) Unde the additional assumptions 7 u 0 and n u (esp. u 0 and n u ), the lowe bound of the pat 4) of Theoem 5 implies the one of the pat 1) (esp. )) of the same Theoem up to the multiplicative constant (esp. 1 ). π π (Notice that the function x x(x + 1) u 0 x is deceasing on the inteval [7, u 0 ], then if 7 u 0, we have ( + 1) u 0 > u 0 ). 5
6 v) Now, we chec that if u 3 0 and n u 0 3+1, the lowe bound of the pat 4) of Theoem 5 implies (up to a multiplicative constant) the one of Conjectue 6. Indeed, if u 3 0 and n u 0 3+1, the decease of the function x x(x + 1) u 0 1 x on the inteval [1, + [ implies: ( + 1) u 0 1 u ( 3 0 u ) 1 > u 3 0 which gives (by using the lowe bound of the pat 4) of Theoem 5): lcm{u 0,..., u n } 3π u 0( + 1) n. Moe geneally, fo any given eal ξ 3, if we suppose 1u ξ 0 and n u then the decease of the function x x(x + 1) u 0 x ξ+ 1 on u 0 ( ) 1 the inteval [1, + [ implies: ( + 1) ξ+ 1 u0 u 0 ξ ξ + 1 > u 0 ξ which gives (by using the lowe bound of the pat 4) of Theoem 5): lcm{u 0,..., u n } 1 πξ u 0( + 1) n+ξ 3 Rema that if ξ > 3, this lowe bound is stonge than the one of Conjectue 6..3 Results about the quadatic sequences Theoem 9 Let u = (u ) N be a sequence of integes whose geneal tem has the fom: u = a( + t) + b ( N), with (a, t, b) Z 3, a 1, t 0 and gcd{a, b} = 1. Also let m and n (with m < n) be two nonnegative integes fo which none of the tems u (m n) of u is zeo. Then the intege lcm{u m,..., u n } is a multiple of the ational numbe:. A u (t, m, n) := { u 0...u n (n)! if (t, m) = (0, 0) (m + t 1)! um...un (n+t)! othewise. Coollay 10 Let u = (u ) N be a sequence of integes as in the above Theoem and n be a positive intege. Then, if the (n + 1) fist tems u 0,..., u n of the sequence u ae all nonzeo, then we have: lcm{u 0,..., u n } { ( b a ) n if t = 0 4 ( b a ) n if t 1. t t 4 6
7 Rema. It is clea that the lowe bound of Coollay 10 is nontivial only if a 5. Such as it is, this coollay cannot thus give a nontivial lowe bound fo the numbes lcm{1 + 1, + 1,..., n + 1} (n 1). But we ema that if 3 is an intege, it gives a nontivial lowe bound fo the last common multiple of consecutive tems of the sequence ( n + 1) n 1 which is a subsequence of (n + 1) n. So we can fist obviously bound fom below lcm{1 + 1, + 1,..., n + 1} by lcm{ + 1, + 1,..., + 1} (with := [ n ]), then use Coollay 10 to bound fom below this new quantity. We obtain in this way: ( ) lcm{1 + 1, + 1,..., n ( ) n 1 + 1} > = 8 { ( } n. 4 4 ) This gives (fo any choice of 3) a nontivial lowe bound fo the numbes lcm{1 + 1, + 1,..., n + 1} (n 1). We easily veify that the optimal lowe bound coesponds to = 5, that is: lcm{1 + 1, + 1,..., n + 1} 0, 3(1, 44) n ( n 1)..4 Results about the least common multiple of a finite numbe of consecutive integes The following Theoem is an immediate consequence of Theoems 3 and 4. Theoem 11 Fo any nonnegative intege and any positive intege n, the intege lcm{n, n + 1,..., n + } is a multiple of the intege n ( ) n+. Futhe, if the conguence n(n + ) 0 mod(!) is satisfied, then we have pecisely: ( ) n + lcm{n, n + 1,..., n + } = n. The following esult is independent of all the esults peviously quoted. It gives a multiple m n, of the intege lcm{n, n + 1,..., n + } ( N, n N ) which is optimal and simple in its dependance on n. Theoem 1 Fo any nonnegative intege and any positive intege n, the intege lcm{n, n+1,..., n+} divides the intege n ( ) {( n+ lcm ) ( 0, ) ( 1,..., )}. Futhe, if the conguence n mod(!) is satisfied, then we have pecisely: ( ) {( ) ( ) ( )} n + lcm{n, n + 1,..., n + } = n lcm,,...,
8 3 Poofs Poof of Lemma 1. In the situation of Lemma 1, we have: lcm{α i, i I}.lcm{β i, i I} γ = lcm{α i, i I}.lcm{β i, i I} α j β j j I = lcm{α i, i I}. lcm{β i, i I}. α j β j j I 1 This last sum is clealy an intege because fo any j I, the two numbes lcm{α i,i I} α j and lcm{β i,i I} β j ae integes. Lemma 1 follows. Poof of Theoem. Theoem follows by applying Lemma 1 to the identity: n j=0 1 u j. 1 (u i u j ) = 1, u 0 u 1... u n which we obtain by taing x = 0 in the decomposition to simple elements of the 1 ational faction x (x+u 0 )(x+u 1 )...(x+u n ). Poof of Theoem 3. By eplacing if necessay the sequence (u n ) n by the u sequence with geneal tem v n := n ( n N), we may assume that u gcd{u 0,u 1 } 0 and u 1 ae copime. Unde this hypothesis, we have to show that the intege lcm{u 0,..., u n } is a multiple of the ational numbe u 0...u n (fo any n N). n! Let n be a fixed nonnegative intege. Fom Theoem, the intege lcm{u 0,..., u n } is a multiple of the ational numbe lcm { u 0... u n (u i u j ) ; 0 j n Let denotes the diffeence of the aithmetic sequence (u ). We have fo any (i, j) N : u i u j = (i j), then fo any j {0,..., n}: (u i u j ) = (i j) }. = n {( j)(1 j)( j)... ( 1)}. {1.... (n j)} = n ( 1) j j!(n j)!. 8
9 Hence: { } lcm (u i u j ) ; 0 j n = lcm { n ( 1) j j!(n j)! ; 0 j n } = n lcm {j!(n j)! ; 0 j n} = n n! (because each intege j!(n j)! divides n! and fo j = 0 o n, we have j!(n j)! = n!). Thus the intege lcm{u 0,..., u n } is a multiple of the ational numbe u 0...u n. But n n! ou hypothesis u 0 copime with u 1 implies that is copime with all tems of the sequence (u ), which implies that n is copime with the poduct u 0... u n. By the Gauss lemma, we finally conclude that the intege lcm{u 0,..., u n } is a multiple of the ational numbe u 0...u n as equied. n! Poof of Theoem 4. We need the following peliminay Lemma: Lemma. Let n be a positive intege and x and y be two integes satisfying: x y 0 mod(n) and xy 0 mod(n!). Then x and y ae multiples of n. Poof. We distinguish the following fou cases: If n = 1: In this case, the esult of Lemma is tivial. If n is pime: In this case, since x = x(x y) + xy, we have x 0 mod(n), but since n is supposed pime, we conclude that x 0 mod(n) and then that y = x (x y) 0 mod(n). If n = 4: In this case, we have x y 0 mod(4) and xy 0 mod(4) and we have to show that x and y ae multiples of 4. Let us ague by contadiction. Then, since x y mod(4), we have: Eithe x y 1, 3 mod(4) which implies xy 1 mod(4) and contadicts the conguence xy 0 mod(4). O x y mod(4) which implies xy 4 mod(8) and contadicts the conguence xy 0 mod(4) again. Thus the Lemma holds fo n = 4. If n 5 and n is not pime: In this case, it is easy to see that the intege (n 1)! is a multiple of n, so that the intege n! is a multiple of n. We thus have x y 0 mod(n) and xy 0 mod(n ). Let us ague by contadiction. Suppose that one at least of the two integes x and y is not a multiple of n. To fix the ideas, suppose fo instance that x 0 mod(n). Then, thee exists a pime numbe p dividing n such that v p (x) < v p (n). But since xy 0 mod(n ), we have v p (xy) v p (n ), that is v p (x) + v p (y) v p (n). This implies that v p (y) v p (n) v p (x) > v p (x) (because v p (x) < v p (n)). Thus, the padic valuations of the integes x and y ae distinct. Then we have: v p (x y) = min(v p (x), v p (y)) = v p (x) < v p (n), which 9
10 contadicts the fact that (x y) is a multiple of n. The Lemma is poved. Retun to the poof of Theoem 4: The case n = 1 is tivial. Next, we assume that n. Fom Theoem 3, the intege lcm{u 0,..., u n } is a multiple of the ational numbe u 0...u n. To pove n! Theoem 4, it emains to pove that u 0...u n is also a multiple of lcm{u n! 0,..., u n }, which means that u 0...u n is a multiple of each of integes u n! 0,..., u n. Since u 0 u n is assumed a multiple of n!, the numbe u 0...u n is obviously a multiple of each n! of integes u 1,..., u n 1. To conclude, it only emains to pove that this same numbe u 0...u n is a multiple of u n! 0 and u n, which is equivalent to pove that the two integes u 1... u n and u 0... u n 1 ae multiples of n!. We fist pove that u 0 and u n ae multiples of n. Denoting the diffeence of the aithmetic sequence (u ), we have u n u 0 = n 0 mod(n) and u 0 u n 0 mod(n!) (by hypothesis). This implies (fom the above Lemma) that u 0 and u n effectively ae multiples of n. We now pove that the two integes u 1... u n and u 0... u n 1 ae multiples of n!. Fo any 1 n 1, we have: u = u 0 + mod(u 0 ), then: u 1... u n 1 (1.)(.)... ((n 1).) mod(u 0 ) (n 1)! n 1 mod(u 0 ). It follows that: u 1... u n 1 u n (n 1)!u n n 1 mod(u 0 u n ). Since u n is a multiple of n and (by hypothesis) u 0 u n is a multiple of n!, the last conguence implies that u 1... u n 1 u n is a multiple of n!. Similaly, fo any 1 n 1, we have: u n = u n mod(u n ), then: u n 1... u 1 ( (n 1).)... ( 1.) mod(u n ) ( 1) n 1 (n 1)! n 1 mod(u n ). It follows that: u 0 u 1... u n 1 ( 1) n 1 (n 1)!u 0 n 1 mod(u 0 u n ). Since u 0 is a multiple of n and (by hypothesis) u 0 u n is a multiple of n!, the last conguence implies that u 0... u n 1 is also a multiple of n!. This completes the poof of Theoem 4. Poof of Theoem 5. Fo any intege {0,..., n}, the intege lcm{u 0,..., u n } is obviously a multiple of the intege lcm{u,..., u n } and fom Theoem 3, this last intege is a multiple of the ational numbe u...u n. It follows that fo any (n )! {0,..., n}, we have: lcm{u 0,..., u n } u... u n (n )!. (1) 10
11 The idea consists in choosing as a function of n, and u 0 in ode to optimize the lowe bound (1), that is to mae the quantity u...u n maximal. (n )! Let (v ) 0 n denotes the finite sequence of geneal tem: v := u...u n. We (n )! have the following intemediate Lemma: Lemma. The sequence (v ) 0 n eaches its maximum value at { [ ] } n u0 0 := max 0, Poof. Fo any {0,..., n 1}, we have: n, hence: u 0 + v +1 v v +1 v n u n u = u +1...u n (n 1)! / u...u n (n )! = n u = [ ] n u This pemits us to detemine the vaiations of the finite sequence (v ) 0 n accoding to the position of n compaed to u 0. If n < u 0, the sequence (v ) 0 n is deceasing and it thus eaches its maximum value at = 0. In the othe case i.e n u 0, the sequence (v ) 0 n is inceasing until the intege [ ] n u then it deceases, so it eaches its maximum value at = [ n u 0 ] The Lemma follows. The following intemediay lemma gives an identity which pemits to bound fom bellow v by simple expessions (as function as u 0, and n) fo the paticula values of which ae athe close to the intege 0 of the above Lemma. Lemma. Fo any {0,..., n}, we have: v = n +1 1 u 0 x+ 0 1 (1 x) n dx. () Poof. Fo any 0 n, we have: v := u... u n (n )! = u (u + )... (u + (n )) (n )! u = n +1 ( u + 1 )... ( u + n ) (n )! = n +1 Γ ( u + n + 1 ) Γ ( ) u.γ (n + 1) = n +1 β ( u, n + 1 ), 11
12 whee Γ and β denote the Eule s functions. The identity () of Lemma follows fom the well nown integal fomula of the βfunction. The Lemma is poved. Because of some technical difficulties concening the lowe bound of the ighthand side of () fo = 0, we ae led to bound fom below this side fo othe values of which ae close to 0. So, we obtain the lowe bounds of the pats 1) and 4) of Theoem 5 by bounding fom below v fo = [ n 1 + 1] and +1 fo the neaest intege to the eal n+ u 0 espectively. Futhe, we obtain the +1 emaining pats ) and 3) of Theoem 5 by anothe method which doesn t use the identity (). We fist pove the pats 1) and 4) of Theoem 5. Poof of the pat 1) of Theoem 5: Let 1 := [ n 1 + 1]. Using the identity (), we ae going to get a lowe +1 bound fo v 1 which depends on u 0, and n. The intege 1 satisfies n 1 < +1 1 n = n+. We thus have: n 1+1 (n 1) (3) and fo any eal x [0, 1]: x 1+ u 0 1 (1 x) n 1 x n u 0 1 (1 x) (n 1) +1, which gives: 1 0 x 1+ u 0 1 (1 x) n 1 dx 1 0 {x(1 x) } n 1 +1 x u 0 1 dx. (4) By studying the function x x(1 x), we may show that fo any eal x [0, 1], we have: x(1 x). Substituting this into the ighthand side of (4), (+1) +1 we deduce that: 1 0 x 1+ u 0 1 (1 x) n 1 dx (n 1) +1 ( + 1). (5) n 1 u 0 By combining the two elations (3) and (5), we finally obtain: n x 1+ u 0 1 (1 x) n 1 dx u 0 ( + 1) n 1. Then the fist lowe bound of the pat 1) of Theoem 5 follows fom the elations () and (1). If n is a multiple of (+1), the second lowe bound of the pat 1) of Theoem 5 follows by taing in the above poof instead of 1 the intege = n +1. 1
13 Poof of the pat 4) of Theoem 5: The paticula case n = 0 of the pat 4) of Theoem 5 follows fom the fact that the function x x(x + 1) u 0 1 x is deceasing on the inteval [1, + [. Next, we suppose that n 1. The hypothesis n u means that the eal n+ u 0 is geate than o equal to 1. Since this same eal n+ u 0 is less than o equal to n + 1 (because n 1), then thee exists an intege {0,..., n} satisfying: It follows that: and that fo any eal x ]0, 1[: 1 n + u n +1 (n 1)+u (6) x + u 0 1 (1 x) n x (n 1)+u 0 (+1) 1 (1 x) (n 1)+u = {x(1 x) } (n 1)+u 0 (+1) 1 x(1 x) ( ) (n 1)+u 0 (+1) 1 ( + 1) (+1) x(1 x) (because x(1 x) Consequently: 1 0 x + u 0 1 (1 x) n dx fo any x [0, 1]). (+1) +1 ( ( + 1) (+1) ) (n 1)+u 0 (+1) 1 0 dx x(1 x). Since 1 0 dx = π, we deduce that: x(1 x) 1 x + u (1 x) n dx π (n 1)+u0 +1 By combining the two elations (6) and (7), we finally obtain: ( + 1) n 1+ u 0. (7) n x + u 0 1 (1 x) n 1 ( + 1) n 1+ u 0 π and we conclude the lowe bound of the pat 4) of Theoem 5 by using the identity () and the lowe bound (1). 13
14 We obtain the two emaining pats ) and 3) of Theoem 5 by using the same idea which consists to bound fom below v = u...u n fo some paticula values (n )! of {0,..., n}. The only diffeence with the last pats 1) and 4) poved above is that hee such paticula values ae not explicit, we just show thei existence by using the following Lemma: Lemma. Let x be a eal and n be a positive intege. Then: 1) thee exists an intege (1 n) such that: ( ) n x n +1 x(x + 1) n 1. ) Thee exists an odd intege l (1 l n) such that: ( ) n l x n l+1 n l n + 1 x { (x + 1) n 1 + (x 1) n 1}. Poof. The fist pat of Lemma follows fom the identity: n ( ) n x n +1 = nx(x + 1) n 1 (8) =1 which can ) be poved by deiving with espect to u the binomial fomula u x n = (u + x) n and then by taing u = 1 in the obtained fomula. n ( n =0 The second pat of Lemma follows fom the identity: ( n 1 n odd ) x n +1 = 1 nx { (x + 1) n 1 + (x 1) n 1} (9) which follows fom (8) by emaing that: ( ) { n n x n +1 = 1 ( ) n x n +1 + ( 1) n 1 n odd The Lemma is poved. =1 n ( } n )( x) n +1. =1 14
15 Poof of the pats ) and 3) of Theoem 5: We have fo any {1,..., n}: v := u... u n (n )! ()(( + 1))... (n) (n )! ( ) n = n +1. These lowe bounds of v (1 n) implie (by using the above Lemma) that thee exist an intege {1,..., n} and an odd intege l {1,..., n} fo which we have: v ( + 1) n 1 and v l n n + 1 { ( + 1) n 1 + ( 1) n 1}. We conclude by using Relation (1). This completes the poof of Theoem 5. Poof of Theoem 7. Let us ague by contadiction. Then, we can find a ational numbe a > 3 (with a, b ae positive integes) fo which we have fo any b aithmetic pogession (u ) with positive diffeence, satisfying the hypothesis of Theoem 5 and fo any nonnegative intege n u 0 a 1: b lcm{u 0,..., u n } 1 ( + 1) n 1+ u 0. π We intoduce a nonnegative paamete δ and the aithmetic pogession (u ) (depending on δ) with fist tem u 0 := abδ + 1 and diffeence := b δ. The integes u 0 and ae copime because they veify the Bézout identity (1 abδ)u 0 + a δ = 1. The sequence (u ) thus satisfies all the hypotheses of Theoem 5. Since the intege n = 1 satisfies n u 0 a 1 = 1, we must have: b lcm{u 0, u 1 } 1 u 0 ( + 1). (10) π Futhe, we have and lcm{u 0, u 1 } = u 0 u 1 = (abδ + 1) ( (ab + b )δ + 1 ) = O(δ ) 1 u 0 ( + 1) = 1 π π b δ(b δ + 1) a b + 1 b δ ( ) = O δ a b + 1. But since a + 1 >, The elation (10) cannot holds fo δ sufficiently lage. b Contadiction. Theoem 7 follows. Poof of Theoem 8. Let us pove the assetion 1) of Theoem 8. The fact that the constant C of this assetion is geate than o equal to 1 is an π immediate consequence of the pat 4) of Theoem 5. In ode to pove the uppe bound C 3, we intoduce a paamete δ N and the aithmetic 15
16 sequence (u ) (depending on δ), with fist tem u 0 := 3δ + and diffeence := δ + 1. The integes u 0 and ae copime because they veify the Bézout identity u 0 3 = 1. So, this sequence (u ) satisfies all the hypotheses of Theoem 5. Since u = 0, we must have fo any nonnegative intege n: lcm{u 0,..., u n } C ( + 1) n 1+ u 0, in paticula (fo n = 0): u 0 C ( + 1) u 0 1, hence: u 0 C. u 0 ( + 1) 1 Since this last uppe bound holds fo any δ N, we finally deduce that: C lim δ + u 0 ( + 1) u 0 1 3δ + = lim δ+1 δ + δ + 1(δ + ) δ+1 = 3 as equied. Now, let us pove the assetion ) of Theoem 8. Let n 0 be a fixed nonnegative intege. As above, the lowe bound C(n 0 ) 1 is an immediate consequence of the pat 4) of Theoem 5. In ode to pove the uppe bound π of Theoem 8 fo the constant C(n 0 ), we choose an intege n 1 such that n n 1 n and that (n 1 + 1) is pime (this is possible fom the Betand postulate). Then, we intoduce a paamete δ N which is not a multiple of (n 1 + 1) and the aithmetic pogession (u ) (depending on δ), with fist tem u 0 := 3δn 1! and diffeence := δn 1! + n These integes u 0 and ae copime. Indeed, a common diviso d 1 between u 0 and divides 3 u 0 = 3(n 1 + 1), thus it divides gcd{u 0, 3(n 1 + 1)} = 3gcd{δn 1!, n 1 + 1}. Futhe, the fact that (n 1 + 1) is pime implies that (n 1 + 1) is copime with n 1!, moeove since δ is not a multiple of (n 1 + 1), the intege (n 1 + 1) also is copime with δ. It follows that (n 1 + 1) is copime with the poduct δn 1!. Hence d divides 3. But since 3 divides δn 1! (because n 1 3) and 3 doesn t divide n (because n is a pime numbe 5) then 3 cannot divide the sum δn 1! + (n 1 + 1) =, which poves that d 3. Consequently d = 1, that is u 0 and ae copime effectively. The sequence (u ) which we have intoduced thus satisfies all the hypotheses of Theoem 5. Since n 1 max{n 0, u } (because n 1 n and u = 3n 1 < 0), then we must have lcm{u 0,..., u n1 } C(n 0 ) ( + 1) n 1 1+ u 0. This gives: C(n 0 ) lcm{u 0,..., u n1 } ( + 1) n 1 1+ u 0. Now, since u 0 is a multiple of n 1!, we have fom Theoem 4: lcm{u 0,..., u n1 } = u 0...u n1 n 1. Hence:! u 0... u n1 C(n 0 ) n 1!. ( + 1) n 1 1+ u 0 16
17 Since this last uppe bound of C(n 0 ) holds fo any δ N which is not a multiple of (n 1 + 1), then we deduce that: C(n 0 ) lim δ + δ 0 mod(n 1 +1) u 0... u n1 n 1!. (11) ( + 1) n 1 1+ u 0 Let us calculate the limit fom the ighthand side of (11). We have: n 1 n 1 u 0... u n1 = (u 0 + ) = {( + 3)n 1!δ + (n 1 + 1)} =0 =0 ( ) n 1 + n 1! n 1+1 ( + 3) =0 δ n 1+1 and: ( + 1) n 1 1+ u 0 = (δn 1! + n 1 + 1) 1/ (δn 1! + n 1 + ) n δn 1! δn 1!+n (δn 1!) n 1+1. Then: u 0... u n1 n 1! ( + 1) n 1 1+ u 0 + n 1 =0 ( + 3) n 1+1 n 1! = (n 1 + 1)(n ) 4 n 1 ( n1 + 1 n 1 ). In the othe wods: u 0... u n1 lim δ + n 1! = (n 1 + 1)(n 1 + 3) ( ) n ( + 1) n 1 1+ u 0 4 n 1 n 1 It is easy to show (by induction on ) that fo any nonnegative intege, we have ( ) +1 < 4. Using this estimate fo = n 1, we finally deduce that: lim δ u 0... u n1 n 1! < (n ( + 1) n 1 1+ u ) n < 4(n 0 + 4) n (because n 1 n 0 + 6). The uppe bound C(n 0 ) < 4(n 0 + 4) n follows by substituting this last estimate into (11). This completes the poof of Theoem 8. Poof of Theoem 9. We fist pove Theoem 9 in the paticula case m = 0. We deduce the geneal case of the same Theoem by shifting the tems of the 17
18 sequence u = (u ). Let u be a sequence as in Theoem 9. The case m = 0: Fom Theoem, the intege lcm{u 0,..., u n } is a multiple of the ational numbe R := u 0... u { n }. (1) lcm (u i u j ) ; j = 0,..., n Now, since we have fo any i, j N: u i u j = {ai(i + t) + b} {aj(j + t) + b} = a(i j)(i + j + t), then: (u i u j ) = = a n = {a(i j)(i + j + t)} (i j). (i + j + t) { a n ( 1) j (n j)!(n+j)! if t = 0 a n ( 1) j (n j)!(n+j+t)! 1 if t 1, ϕ(j,t) j+t whee ϕ(j, t) := 1 if t = 1 and ϕ(j, t) := (j + 1)... (j + t 1) if t. Since (n+t)! (n j)!(n+j +t)! divides (n+t)! (because = ( ) n+t (n j)!(n+j+t)! n j N) and (if t 1) the intege ϕ(j, t) is a multiple of (t 1)! (because ϕ(j,t) = ( ) j+t 1 (t 1)! t 1 N), then the poduct (u i u j ) divides the intege (which does not depend on j): f(t, n) := { a n (n)! if t = 0 a n (n+t)! if t 1. (t 1)! Since j is abitay in {0,..., n}, then the intege lcm{ (u i u j ); j = 0,..., n} divides the intege f(t, n). It follows that the ational numbe R (of (1)) is a multiple of the ational numbe u 0...u n = A u(t,0,n). Consequently, the f(t,n) a n intege lcm{u 0,..., u n } is a multiple of the ational numbe A u(t,0,n). Finally, a n since each tem of the sequence u is copime with a (because gcd{a, b} = 1), we conclude fom the Gauss Lemma that the intege lcm{u 0,..., u n } is a multiple of the ational numbe A u (t, 0, n) as equied. The geneal case (m N): Let us conside the new sequence v = (v ) N with geneal tem: v := u +m = a ( + t ) + b, 18
19 whee a := a, t := m + t and b := am(m + t) + b. Since these integes a, t and b veify a 1, t 0 and gcd{a, b } = gcd{a, b} = 1 obviously, then the sequence v satisfies all the hypotheses of Theoem 9. Thus, fom the paticula case (poved above) of this Theoem, the intege lcm{v 0,..., v n m } = lcm{u m,..., u n } is a multiple of the ational numbe A v (t, 0, n m) = A u (t, m, n) which povides the desied conclusion. Poof of Coollay 10. Fom Theoem 9, the intege lcm{u 0,..., u n } is a multiple of the ational numbe: A u (t, 0, n) := { u 0...u n if t = 0 (n)! (t 1)! u 0...u n if t 1. (n+t)! Let us get a lowe bound fo this last numbe which doesn t depend on the tems of the sequence u. Using the obvious lowe bounds u a( + t) (1 n), we have: u 0... u n then: n n!(n + t)! b{a.1.(1 + t)}{a..( + t)}... {a.n.(n + t)} = ba, t! A u (t, 0, n) b an ( n n ) b a n t ( n+t n ) if t = 0 if t 1 { ( b a ) n if t = 0 4 ( b a n t 4) if t 1 t (because ( ) n n n = 4 n and ( ) n+t n n+t = t 4 n ). The lowe bound of Coollay 10 follows. Poof of Theoem 11. Theoem 11 is only a combination of the esults of Theoems 3 and 4 which we apply fo the aithmetic pogession (u l ) l N with geneal tem u l = l + n (whee n N is fixed). Poof of Theoem 1. Let us pove the fist assetion of Theoem 1. Giving a nonnegative intege and n a positive intege, we easily show that fo any nonnegative intege j, we have: ( ) ( n + n + j 1 )( j )( n + j n = (n + j) j { It follows that the intege lcm n ( )( } n+ j) ; j = 0,..., = n ( ) {( n+ lcm ) 0,..., )} is a multiple of each intege n + j (0 j ). Then it is a multiple of ( lcm{n, n + 1,..., n + }, as equied. 19 ).
20 Now, in ode to pove the second assetion of Theoem 1, we intoduce the sequence of maps (g ) N of N into N which is defined by: g (n) := n(n + 1)... (n + ) lcm{n, n + 1,..., n + } ( N, n N ). Let us show that (g ) satisfies the induction elation: g (n) = gcd{!, (n + )g 1 (n)} ( (, n) N ). (13) Fo any pai of positive integes (, n), we have: g (n) := n(n + 1)... (n + ) lcm{n, n + 1,..., n + } = n(n + 1)... (n + ) lcm {lcm{n, n + 1,..., n + 1}, n + } = n(n + 1)... (n + ) lcm{n,n+1,...,n+ 1}.(n+) gcd{lcm{n,n+1,...,n+ 1},n+} n(n + 1)... (n + 1) = gcd {lcm{n, n + 1,..., n + 1}, n + } lcm{n, n + 1,..., n + 1} = gcd {n(n + 1)... (n + 1), (n + )g 1 (n)}. Then, the elation (13) follows by emaing that the poduct n(n + 1)... (n + 1) is a multiple of! (because n(n+1)...(n+ 1) = ( ) n+ 1! N) and that g (n) divides! (accoding to Theoem 11). Now, giving a nonnegative intege, by eiteating the elation (13) seveal times, we obtain: g (n) = gcd{!, (n + )g 1 (n)} = gcd{!, (n + )( 1)!, (n + )(n + 1)g (n)}. = gcd{!, (n + ) ( 1)!, (n + )(n + 1) ( )!,..., (n + )(n + 1) (n + l)g l 1 (n)} fo any positive intege n and any nonnegative intege l 1. In paticula, fo l = 1, since g 0 1, we have fo any positive intege n: g (n) = gcd{!, (n + ) ( 1)!, (n + )(n + 1) ( )!,..., (n + )(n + 1) (n + 1).0!} (14) 0
21 Now, if n is a given positive intege satisfying the conguence n mod(!), we have: n + 1 mod(!), (n + )(n + 1) ( 1)! mod(!),..., (n + )(n + 1) (n + 1) ( 1)! mod(!); consequently, the elation (14) gives: Hence: g (n) = gcd {!, 1!( 1)!,!( )!,...,!0!}.! g (n)! = gcd {0!!, 1!( 1)!,...,!0!} { }! = lcm 0!!,! 1!( 1)!,...,!!0! {( ) ( ) ( )} = lcm,,...,. 0 1 But on the othe hand, accoding to the definition of g (n), we have:! lcm{n, n + 1,..., n + } = g (n) n ( ). n+ We thus conclude that: ( ) {( ) n + lcm{n, n + 1,..., n + } = n lcm, 0 ( ),..., 1 ( )} which gives the second assetion of Theoem 1 and completes this poof. Open Poblem. By using the elation (13), we can easily show (by induction on ) that fo any nonnegative intege, the map g which we have intoduced above is peiodic of peiod!. In othe wods, the map g ( N) is defined modulo!. Then, fo fixed in N, it is sufficient to calculate g (n) fo the! fist values of n (n = 1,...,!) to have all the values of g. Consequently, the elation (13) is a pactical mean which pemits to deteminate step by step all the values of the maps g. By poceeding in this way, we obtain: g 0 (n) g 1 (n) 1 (obviously), { { 1 if n is odd g (n) = if n is even, g 6 if n 0 mod(3) 3(n) =,... etc. othewise This calculation point out that the smallest peiod of the map g 3 is equal to 3( 3!). This lead us to as the following inteesting open question: Giving a nonnegative intege, what is the smallest peiod fo the map g? 1
22 Refeences [1] D. Hanson. On the poduct of the pimes. Canad. Math. Bull, 15 (197), p [] G. H. Hady & E. M. Wight. The theoy of numbes. Oxfod Univ. Pess, London, 5th ed, (1979). [3] M. Nai. On Chebyshevtype inequalities fo pimes. Ame. Math. Monthly, 89 (198), n, p
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