Maximal sets of numbers not containing k + 1 pairwise coprime integers

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1 ACTA ARITHMETICA LXXII. 995) Maximal sets of numbers not containing k + airwise corime integers by Rudolf Ahlswede and Levon H. Khachatrian Bielefeld). Introduction. We continue our work of [], in which an old conjecture of Erdős [5] was disroved. There also some cases were settled in the ositive and related questions were investigated. For further related work we refer to [9] [2], and [5]. While restating now the conjecture of Erdős in its original form and its general form of [8], we also introduce our notation and some basic definitions. Here we follow [] as closely as ossible. N denotes the set of ositive integers and P = {, 2,...} = {2, 3, 5,...} denotes the set of all rimes. For two numbers u, v N we write u v iff u divides v, u, v) stands for the largest common divisor of u and v, [u, v] is the smallest common multile of u and v. The numbers u and v are called corimes if u, v) =. We are articularly interested in the sets.) N s = and { u N : u, s.2) N s n) = N s, n, ) } i = where for i j, i, j equals {i, i +,..., j}. Erdős introduced in [5] and also in [6] [8], [0]) fn, k, s) as the largest integer r for which an.3) A N s n), A = r, exists with no k + numbers in A being corimes. Certainly the set.4) En, k, s) = {u N s n) : u = s+i v for some i = 0,,..., k } does not have k + corimes. The case s =, in which we have N n) =, n, is of articular interest. [77]

2 78 R. Ahlswede and L. H. Khachatrian Conjecture. fn, k, ) = En, k, ) for all n, k N. It seems that this conjecture of Erdős aeared for the first time in rint in his aer [5] of 962. General Conjecture. fn, k, s) = En, k, s) for all n, k, s N. Erdős mentions in [8] that he did not succeed in settling the case k =. We focus on this secial case by calling it Conjecture 2. Notice that fn,, s) = En,, s) for all n, s N. En,, s) = {u N n) : s u;,..., s u}. Whereas in [] Conjecture was disroved for k = 22, Conjecture 2 was almost settled with the following result. Theorem 2 []). For every s N and n s i/ s+ s ), fn,, s) = En,, s) and the otimal configuration is unique. After the resentation of these results on his 80th birthday at a conference in his honour Erdős conjectured that with finitely many excetions Erdős sets are otimal or, in other terminology, that for every k N, fn, k, ) En, k, ) occurs only for finitely many n. We call this Conjecture. Analogously we seak of Conjecture 2 which is settled in the affirmative by Theorem 2 of []) and of the General Conjecture, which is established in this aer. Actually the main ste is the roof of Conjecture. It can easily be extended to the general case with a bulk of notation. To simlify notation we write in the case s =, Nn) N n), fn, k) fn, k, ) and En, k) En, k, ). We climbed the mountain to Conjecture in 3 stes by going through a series of weaker conjectures of increasing strength: Conjecture A. The infinite Erdős set E, k) = {m i : i k, m N} has maximal lower) density among subsets of N without k + corimes.

3 Conjecture B. Sets not containing airwise corime integers 79 lim fn, k) En, n k) = for every k N. A few more definitions and known facts are needed. For A N we define An) = A, n and A is the cardinality of A. We call da = lim inf n An) /n the lower and da = lim su n An) /n the uer asymtotic density of A. If da = lim n An) /n exists, then we call da the asymtotic density of A. Erdős sets can be nicely described in terms of sets of multiles. The set of multiles of A is MA) = {m N : a m for some a A} and the set of non-multiles of A is NA) = N\MA). Thus En, k) = M{,..., k }), n and also for any finite A = {a,......, a t } N and a = t a i, NA), a the set of integers in, a not divisible by any member of A. Already Dirichlet knew that t ) NA), a = a ai if the elements of A are airwise relatively rime. For general A, by inclusion-exclusion, t NA), a = a + ) a i [a i<j i, a j ]... and therefore.5) dna) = t a i + i<j [a i, a j ] The main results. It is convenient to introduce the family Sn, k, s) of all subsets of N s n) no k + elements of which are airwise relatively rime. In case s = we also write Sn, k) and S, k) in the unrestricted case n =. Theorem A. su A S,k) da = de, k) = k i ).

4 80 R. Ahlswede and L. H. Khachatrian Theorem B. lim n fn, k) = for every k N. En, k) Theorem. For every k N there is an nk) such that fn, k) = En, k) for all n > nk) and the otimal set is unique. After the examle of [] this is the strongest statement one can hoe for. A key tool in the roof of Theorem is a combinatorial result of indeendent interest. For a subfamily A ) [m] l, that is, a set of l-element subsets of an m-element set, the lower) shadow A is defined by { ) } [m] A = B : B A for some A A l and the uer) shadow of B [m] l ) is { ) } [m] δb = A : B A for some B B. l With any function g : A R + we associate the function h : A R +, where hb) = max A δ{b} A ga). Theorem 2. Let A ) [m] l have the roerty that no k + elements of A are disjoint. Then for any function g : A R + and its associated function h : A R + defined as above) hb) ga). k In articular, B A A A A k A. Even though Theorem A now follows from Theorem, we give our original roof, because it is much simler than that of Theorem, which is based on Theorem 2. It also shows how the ideas develoed. The original roof of Theorem B is not based on Theorem 2, but since it is rather technical, it is not resented in this aer. It should be mentioned, however, that Theorem B imlies su da = A S,k) su da = A S,k) su da. A S,k) Finally, we remark that insection of our methods and roofs shows that they also aly to the general case of fn, k, s) for s >. Only some extra notation is needed. Therefore we just state the results.

5 Sets not containing airwise corime integers 8 Theorem. For every k, s N there exists an nk, s) such that for all n nk, s), and the otimal set is unique. En, k, s) = fn, k, s) 3. Reduction to left comressed sets. The oeration ushing to the left is frequently used in extremal set theory, but to our surrise it seems not to be as oular in combinatorial number theory, erhas because its usefulness is less obvious. Anyhow, our first but not only) idea is to exloit it. Definition. A N s is said to be left comressed if for any a A of the form and any l of the form a = i ra, a, r ) =, s l < r, l, a ) =, it follows that a = i l a A as well. For any n N { } we denote the family of all left comressed sets from Sn, k, s) by Cn, k, s). Lemma. For n N, max A = max A = fn, k, s). A Sn,k,s) A Cn,k,s) P r o o f. For any A Sn, k, s) and s l < r we consider the artition of A, where A = A A 0, A = {a A : a = i ra i ), a, r l ) = ; i la A}, A 0 = A \ A. Define A = {u N s : u = i l a, a, l r ) =, i ra A } and notice that by our definitions A N s n). Consider now A = A A ) \ A and observe that A = A and also that A Sn, k, s). Finitely many iterations of this rocedure to rimes s l < r give the result. The oeration which led from A to A can be denoted by L s,l,r. This is a left ushing oeration: A = L s,l,r A).

6 82 R. Ahlswede and L. H. Khachatrian Moreover, by countably many left ushing oerations one can transform every A S, s) into a left comressed set A such that 3.) An) A n) and therefore also 3.2) da da, da da. For the left comressed sets C, k) in S, k) we have thus shown the following. and Lemma 2. su db = su db B S,n) B C,n) su B S,n) db = su B C,n) db. Next we mention two useful observations. Any otimal B Sn, k, s), that is B = fn, k, s), is an uset : 3.3) B = MB) N s n) and it is also a downset in the following sense: 3.4) b B, b = q α... qα t t, α i b = q... q t B. Finally, we introduce for any B N the unique rimitive subset P B) which has the roerties 3.5) b, b 2 P B) b b 2 and B MP B)). We know from 3.4) that for an otimal B Sn, k, s), P B) consists only of squarefree integers. R e m a r k. We could use also the following concet of left comressedness: Definition 2. A N s is left comressed if for any a A of the form a = α i i a, α i, a, i ) =, it follows that for any j, s j < i, in case α i 2, and in case α i =, a = j α i i a A, a = j a A if a, j ) =. While the two definitions are different in general, it can be easily seen that if the considered set A N s is also an uset and a downset, then both definitions of left comressedness coincide.

7 Sets not containing airwise corime integers 83 Besicovitch has shown in the thirties see [3]) that MA) need not have a density for general A. Erdős [4] has given a characterisation for sets A for which dma) exists. Here we have the following Conjectures. The set of multiles MA) of any left comressed set A in the sense of Definition 2) ossesses asymtotic density. We conjecture this even for left comressed sets in the sense of Definition. Moreover, we think that even a stronger statement is true: For any left comressed set A in the sense of Definitions or 2, da exists. 4. Proof of Theorem A. We remind the reader of the abbreviations fn, k), En, k), Nn), Sn, k), Cn, k) for fn, k, ), En, k, ), N k), Sn, k, ), and Cn, k, ) res. We also introduce 4.) On, k) = {B Sn, k) : B = fn, k)}. By the remarks at the end of Section 3 we know that for A On, k) we have roerties I). I) a) P A) N, the set of squarefree numbers, b) A = MP A)) Nn). We also know from Lemma that c) On, k) Cn, k). For infinite sets A N we choose the lower asymtotic density da as a measure and define 4.2) O, k) = {A S, k) : da = su db}, B S,k) which is not automatically non-emty. C, k) are the left comressed sets in S, k). Again it suffices to look at A C, k) with the roerties a) P A) N, b) A = MP A)). Sets of multiles have been studied intensively in the thirties cf. Halberstam and Roth [3]). Let P A) = {a, a 2,...}, where the elements are written in the usual lexicograhical or alternatively in natural) order. It is easy to show see [3]) that 4.3) dmp A)) = b i),

8 84 R. Ahlswede and L. H. Khachatrian where 4.4) b i) = a i j<i [a j, a i ] +... is the density of the set B i) of those integers in MP A)) which are divisible by a i and not by a, a 2,..., or a i. We can say more about b i) if we use the rime number factorization of the squarefree numbers a i. Lemma 3. Let a i = q... q r, q <... < q r and q j P for j =,..., r. Then { i) B i) = n N : n = q α... qα r r q with α j, q, ) } =, ii) db i) = b i) = q )... q r ) q r q r ). P r o o f. Since A is left comressed and P A) is written in lexicograhical order, q is of the described form and i) holds. To verify ii) just observe that from.6), db i) = α j = q r = q r q α... qα r r ) ) q r α = q α... ) α r = q )... q r ). We are now ready to rove Theorem A. Suose to the contrary that there exists an A S, k) with k ) da > j. j= We know already that we can assume A C, k), P A) N, MP A)) = A and that P A) = {a, a 2,...} is in lexicograhical order. We have k ) b i) > j j= and hence for a suitable ma) also m k ) b i) > j j= q α r r for m ma).

9 Sets not containing airwise corime integers 85 We can therefore consider A = M{a,..., a m }), because A S, k) and still m k ) 4.5) da = da = b i) > j. Write P A ) = {a,..., a m } in the form j= 4.6) P A ) = R... R t, where R s is the set of all a j s with greatest rime factor + a j ) = s. Notice that in case t > k by left comressedness we necessarily have t A and also t R t, because otherwise A S, k). Hence where 4.7) τr s ) = dmp A )) = a=q...q r s R s q <...<q r < s m b i) = t τr s ), s= q )... q r ) s ) s i ). We now consider R t = {a l, a l+,..., a m } for some l m. We have m 4.8) τr t ) = b i). By the igeon-hole rincile there exists a subset R t = {a i,..., a ir } R t such that r 4.9) b ij) τr t) ai and,..., a ) i r >. t t t j= Now we relace the set A by the set A = MR... R t R t ), where { } R t aij = : a ij R j. t One readily verifies that A C, k). We now estimate da from below. The contribution of every element a ij / t R t to MR... R t R t )\MR... R t ) are the elements in the form u = q β... qβ r r q, where a ij = q... q r t, β j, and q, t i) =. The density of this set of integers equals b ij) t ) = i q )... q r ) i=l

10 86 R. Ahlswede and L. H. Khachatrian and hence b ij) = t )b ij). Therefore, using 4.9) we have t da τr s ) + t ) τr t) t t > τr s ) = da, s= s= because t > t. We notice that P A ) R... R t R t and hence max a P A ) + a) t. Continuing this rocedure we arrive after finitely many stes at the set M{,..., k }) and by 4.5) at the statement that its density k / i ) must be bigger than itself. This roves that max B S,k) db = de, k). 5. A finite version of Lemma 3. We now work in Nn) and need sharer estimates on cardinalities than just bounds on densities. It suffices to consider A Cn, k) On, k). We know that P A) = {a <... < a m } N and that A = MP A)) Nn). Define B i) n) = {u Nn) : a i u and a j u for j =,..., i } and write m 5.) A = B i) n). Lemma 4. Let a i = q... q r and q <... < q r with q j P. Then { i) B i) n) = u Nn) : u = q α... qα r r T, α i, T, ) } =. B i) n) ii) lim = n n q )... q r ) q r ). q r q r iii) For every ε > 0, every h N and every a i = q... q r, q < < q r h, there exists an nh, ε) such that for n > nh, ε) we have ε)n ) < B i) n) q )... q r ) q r < + ε)n ). q )... q r ) P r o o f. i) immediately follows from the facts that A is comressed, an uset and a downset. ii) We know that dn m = ) m

11 for m N and hence Sets not containing airwise corime integers 87 B i) n) lim = n n α i = q α... qα r r q r q )... q r ) ) q r ). iii) follows from ii), because the constant number of sequences converges uniformly. 6. Combinatorial result for shadows and a roof of Theorem 2. For A ) [m] l and B [m] l ) the lower shadow A and the uer shadow δb were defined in Section 2. We begin with a secial case of Theorem 2. Lemma 5. Let A ) [m] l have the roerty that no k + of its members are airwise disjoint. Then A k A. P r o o f. The standard left ushing oeration reserves the no k + disjoint -roerty and only can decrease the shadow. We can assume therefore that A is left-comressed. We distinguish two cases. C a s e : m k + )l. Counting airs A; B) with B A in two ways we get A l m l + A l k + )l l + A = k A. C a s e 2: m k + )l. We consider the following artition of, m : I =, k, I 2 = k +, 2k +,..., I j = j )k + ), jk + ),..., I l = l )k + ), lk + ), I l+ = lk + ), m. First we show that for every A A there exists an index j, j l, for which 6.) A I... I j ) = j. To see this, assume that this does not hold for some A A. Then necessarily A I l+, because otherwise A I... I l ) = l since A = l. Therefore we must have A I... I l ) l and a fortiori A I... I l ) l 2, A I... I l 2 ) l 3,..., A I I 2 ), A I = 0. However, since A is also left comressed, we can then choose k + elements from A including A) which are airwise disjoint. This contradicts our assumtion on A.

12 88 R. Ahlswede and L. H. Khachatrian Now, for every A A define j A, j A l, as the largest index j for which 6.) holds. This can be used to artition A into disjoint subsets: ḷ 6.2) A = A i, where A i = {A A : j A = i}. Some of the subsets may be emty. Consider now the shadows A i i l) and their subshadows A i = {B A i : B I... I i ) = i }. It follows immediately from the definition of the A i that 6.3) A i A i2 = for all i i 2. Moreover, using left comressedness of A it can be shown easily that 6.4) A = l A i. In the light of 6.2) 6.4) it suffices to show that 6.5) A i k A i for i =,..., l. We look therefore for fixed i at the intersections and artition A i as follows: 6.6) A i =. U i = {A I... I i ) : A A i } U U i A U i, A U i = {A A i : A I... I i ) = U}. Also, we introduce the intersections V i = {B I... I i ) : B A i } and artition A i as follows: A i =. A i ) V, 6.7) V V i A i ) V = {B A i : B I... I i ) = V }. Now counting for the -oeration airs again in two ways we get the inequality i A U i ik + ) i )) A i ) V ik A i ) V. U U i V V i V V i Together with 6.6) and 6.7) it imlies 6.5). The next result is of a more general structure. It enables us to get immediately Theorem 2 from Lemma 5. Let G = V, W, E) be a biartite grah.

13 Sets not containing airwise corime integers 89 Write σs) for the set of vertices adjacent to a vertex s and σs) for the set of vertices adjacent to vertices in S. We assume that σv ) = W. Lemma 6. Suose that for some α R + we have, for every S V, 6.8) S α σs). Then for every function g : V R + and associated function h : W R +, where hb) = max a σb) ga) for all b W, 6.9) ga) α hb). b W a V P r o o f. Let {γ <... < γ r } be the range of g. Then we have the artition V = V... V r, where Clearly, 6.0) V i = {v V : gv) = γ i }, i r. ga) = a V By the definition of h obviously r γ i V i. 6.) hb) = γ r for all b σv r ). We now roceed by induction on r. r = : Here hb) = γ for all b W and hence by 6.8), ga) = γ V γ α W = α hb). b W a V r r: We assume that 6.9) holds for every function g : V R + with r different values. With our g under consideration we associate the function g : V R + defined by { g γi for a V a) = i, i r, γ r for a V r. Denote by h : W R + the usual function corresonding to g. We verify that 6.2) ga) = g a) + γ r γ r ) V r, a V a V 6.3) h b) + γ r γ r ) σv r ). b W hb) = b W

14 90 R. Ahlswede and L. H. Khachatrian From the condition 6.8) and the induction hyothesis alied to g we know that V r α σv r ) and g a) α h b). b W a V These inequalities and 6.2), 6.3) give 6.9). P r o o f o f T h e o r e m 2. Consider G = V, W, E) = A, A, E), where A; B) E iff A B, and A satisfies the hyothesis of Theorem 2 and hence also of Lemma 5. Since every subfamily A A also satisfies this hyothesis, we know that 6.4) A k A. Since A = σa ), 6.4) guarantees 6.8) for α = k. The conclusion 6.9) says now ga) k ha) A A and Theorem 2 is established. A A R e m a r k 2. One might consider instead of the maximal) associated function h an average) associated function h : A R +, where hb) = δb) A ga). Obviously hb) hb) for all B A. A δb) A While for the case m k +)l one can relace h by h in Theorem 2, this is not ossible in general. Examle h cannot be relaced by h in Theorem 2). Choose m = 6, l = 3, and k = and define A = {{, 2, 3}, {, 2, 4}, {, 2, 5}, {, 2, 6}} {{, 3, 4}, {, 3, 5}, {, 3, 6}} {{2, 3, 4}, {2, 3, 5}, {2, 3, 6}}. No two sets in A are disjoint and A is left comressed. Choose { ga) = for A = {, 2, 3}, 0 otherwise and use the notation fc) = C C fc). Then k ga) = ga) > h A), because δ{, 2}) A = δ{, 3}) A = δ{2, 3}) A = 4, and thus h A) = 3 4 < ga) =. 7. A number-theoretical consequence of Theorem 2. We now resent a basic new auxiliary result for every S Cn, k) with Proerties I)

15 Sets not containing airwise corime integers 9 in Section 4. S need not be otimal, that is, it can be in Cn, k)\on, k). Define 7.) S i = {d S : i d, but... i, d) = }. Clearly, 7.2) S i S j = i j) and S =. i S i. Lemma 7. For every k, n N and every S Cn, k) with Proerties I) we have i) S r /k) i r+ S i for every r N, ii) for every α R + and for kα) kα indeendent of n!) kα) S k+i α j k+kα)+ S j. P r o o f. ii) follows from i), so we have to rove i). We consider the set i r+ S i and let, for every l N, 7.3) T l = {d Obviously, 7.4) and for d T l, i r+ } S i : d has exactly l different rimes in its factorization.. i r+ S i =. l 7.5) d = q β... qβ l l, r < q <... < q l, β i. Since S Cn, k), we have 7.6) d i = β i r q β... qβ i i qβ i+ i+... qβ l l S r for i =,..., l. Define 7.7) σd) = {d,..., d l } and σt l ) = T l d T l σd). Since σt l ) S r and σt l ) σt l ) = l l ), to rove i) it is sufficient to show that 7.8) σt l ) k T l for all l N. Let Tl = T l N be the squarefree integers in T l. Then σtl ) = d T l is the set of all squarefree integers of σt l ). σd)

16 92 R. Ahlswede and L. H. Khachatrian For an a T l, a = x... x l, x <... < x l, x i P, we consider 7.9) T a) = {d S : d = x β... xβ l l, β i } and for a b σt l ), b = ry... y l, r < y <... < y l, y i P, we consider 7.0) It is clear that 7.) T l =. Ub) = {d S r : d = γ l r y γ... yγ l l, γ i, y γ... yγ l l xγ l T l for some x P}. a T l T a) and σt l ) = Next we observe that for any b σt l ), 7.2) Ub) = max bx/ r T l. b x) T r b σt l ) Ub). and this has brought us into the osition to aly Theorem 2 to the sets A Tl and A σtl ), where is the canonical corresondence between squarefree numbers and subsets. We indicate the corresondence by using small and caital letters such as a A. We define g : A R + by 7.3) ga) = T a). The associated function h : A R + is defined by hb) = Ub). We see from 7.2) that this definition is correct. Theorem 2 therefore yields 7.8) and thus i). 8. Further auxiliary results. We state first the only auxiliary result which is not derived in this aer and is not trivial. It is the weaker version of de Bruijn s strengthening [2] of Buchstab s result [3] that can be found in [3]. Theorem. For the function { 8.) φx, y) = a x : a, ) = } <y there exist ositive absolute constants c, c 2 such that 8.2) c x ) φx, y) c 2 x ) <y <y for all x, y satisfying x 2y 4. Furthermore, the right side inequality in 8.2) remains valid also for x < 2y.

17 We also need Sets not containing airwise corime integers 93 Lemma 8. For ositive constants c, c 2, κ there exists a tc, c 2, κ) such that for t > tc, c 2, κ), c t ) > κ. c 2 P r o o f. Trivial. t Finally, we need a result on bookkeeing. We have two accounts at time 0: x 0 = x and y 0 = y where x, y R +. In any ste i, i, we arbitrarily remove a i, b i with 0 a i x i, 0 b i y i, and add a i 0, b i 0, where a i + b i > βa i + b i ), β >. The new accounts are x i = x i a i + a i, y i = y i b i + b i. Lemma 9. If for some l N the account y l = 0 res. x l = 0) occurs, then we have x l > x + βy res. y l > y + βx). P r o o f. Beginning with accounts x and y at the end the amount y has been removed and transferred to the first account with an increasing factor β. 9. Proof of Theorem. We can assume that as in Section 7 S Cn, k) satisfies Proerties I) and additionally is also otimal, that is, S On, k). Define S i as in 7.) and recall 7.2). Notice also that P S) = P S N ). Equivalent to Theorem is the statement that for large n always 9.) S i =. i k+ Henceforth we assume to the contrary that II) S i for infinitely many n. i k+ Let k 0 N, k 0 > k, be an integer to be secified later. By the disjointness roerty 7.) we can write 9.2) S 0 = S\ i k 0 + From i) in Lemma 7 we know that k 0 i=k+ ) k S i = S i k 0 k k ) k 0 S i i k 0 + i=k+ S i S i ).

18 94 R. Ahlswede and L. H. Khachatrian and hence also that 9.3) S k k 0 S i + γ i=k+ where γ = + k/k 0 k). Let P S 0 ) be the rimitive subset of S 0, which generates S 0. We notice that by the roerties of S, 9.4) P S 0 ) P S), because d P S 0 ) and d d for some d S would by comressedness imly the existence of an e P S 0 ) with e d. Let t be the largest rime occurring in any element of P S 0 ). In other words, t, d) = t for some d P S 0 ) and S i, 9.5) t, d) = for all t > t and all d P S 0 ). By assumtion II) we have t > k. We now consider 9.6) P t S 0 ) = {a P S 0 ) : a, t ) = t }. From Lemma 3i) we know that the contribution of every element a P t S 0 ), a = q... q r t and q <... < q r < t, to MP S 0 )) is the set of integers { 9.7) Ba) = u = q α... qα r r β t Q : α i, β, Q, ) } =. We use the abbreviation 9.8) L t = We also consider the artition 9.9) P t S 0 ) =. a P t S 0 ) Ba). t i k 0 P t i S 0 ), P t i S 0 ) = P t S 0 ) S i. By the igeon-hole rincile for some l, l k 0, 9.0) Ba) L t /k 0 if t > k 0 a P t l S0 ) and for some l, l t, 9.) Ba) L t /t ) if k < t k 0. a P t l S0 )

19 Sets not containing airwise corime integers 95 Basic transformation. We consider for this l corresonding to t the set of squarefree numbers) 9.2) P S 0 ) = P S 0 )\P t S 0 )) R t ls 0 ), where 9.3) R t ls 0 ) = {u N : u t P t l S 0 )}. It can haen that P S 0 ) is not rimitive, however, always P S 0 ) Sn, k)! We state the main result for P S 0 ) as Proosition. For suitable n > nk), 9.4) M P S 0 )) Nn) > S 0 + γ L t. P r o o f. For an a Rl ts0 ), a = q... q r, q <... < q r < t, we consider the set { Da) = v Nn) : v = q α... ) } qα r r T, T, =. t Since t was the biggest rime which occurred in P S 0 ), we observe that 9.5) MP S 0 )\P t S 0 )) Da) = for a R t ls 0 ). Moreover, Da) Da ) = for a, a R t ls 0 ), a a. Hence, in the light of 9.0) and 9.), to show 9.4) it is sufficient to rove that for n > nk) and Ba t ) { = u Nn) : u = q α... q α r r β t T, α i, β and T, we have 9.6) Da) > { γk0 Ba t ) if t > k 0, γt ) Ba t ) if t k 0. t ) } =, Three cases in roving 9.6). We always have a = q... q r, q <... < q r < t. C a s e : n/a t ) 2 and t > tc, c 2, k 0 ). Using the right side of the Theorem in Section 8, which is valid without restrictions, we get n Ba t ) c 2 α i, β q α... qα r r β ) 9.7) t t < c 2 n ) q )... q r ) t. t

20 96 R. Ahlswede and L. H. Khachatrian For Da) we have Da) D a) = { u Nn) : u = q... q r T, T, ) } =, t t and since n/q... q r ) 2 t, we can aly the left side of the Theorem to get Da) > D a) c n ) 9.8) q... q r t t = c n ). q... q r t Comaring 9.7) and 9.8) we get Da) Ba t ) > c q )... q r ) t c 2 q... q r ) c c 2 t t > κ = γk 0, where in the last ste we used Lemma 8. Thus we roved 9.6) in this case. C a s e 2: n/a t ) 2 and t tc, c 2, k 0 ). First let us secify k 0 and hence γ. We choose k 0 so large that 9.9) k+i > γk + i ) = + k ) k + i ) for all i N. k 0 k This is of course ossible. Next we choose ε > 0 such that ε 9.20) k+i > γk + i ). + ε Let nε) be a ositive integer so that for n > nε) we can aly Lemma 4iii). So we have Ba t ) < + ε)n ), q )... q r ) t ) t Da) > ε)n ) q )... q r ) t = ε)n q )... q r ) t ), t and hence by 9.20), Da) Ba t ) This establishes 9.6) in this case. > ε + ε t > γt ). t

21 Sets not containing airwise corime integers 97 C a s e 3: n/a t ) < 2. In this case Ba t ) consists of only one element, namely q... q r t. Let now t N satisfy 9.2) t > k0 ) γk 0 and let 9.22) n > t. Notice that in our case necessarily t t, because a t < t and t > t would imly 2a t < 2 < < n by 9.22)) t t and this contradicts our case 2a t > n. Now by 9.2), t t > k0 ) γk 0 and since q k0 we get finally q γk 0 < t. Therefore Da) {q... q r, q 2 q 2... q r,..., q γk 0 q 2... q r, q q 2... q r t }, Da) > γk 0, and again 9.6) holds. k 0, γ, and ε are already fixed and deend only on k. Then for { } 9.23) nk) = max, nε) k0 ) γk 0 and n > nk), 9.6) holds in all three cases and the roof of the Proosition is comlete. Final iterative rocedure and its accounting. We have already noticed that P S 0 ) may not be rimitive. Moreover, M P S 0 )) may even not be left comressed. Let now S Nn) be any set which is obtained from M P S 0 )) by left ushing and is left comressed. We know that 9.24) S Cn, k), S M P S 0 )) Nn) and therefore we know from the Proosition that 9.25) S > S 0 + γ L t. We notice that a, k0 ) > for every a S and the last rime t which occurs as a factor of any rimitive element of P S ) is less than t. If S En, k), then we reeat the whole rocedure and get an S 2 for which S 2 > S + γ L t, where L t is defined analogously to L t with resect to the largest rime t occurring in a member of P S ).

22 98 R. Ahlswede and L. H. Khachatrian By iteration we get an S i Cn, k) with 9.26) S i > S i + γ L t i and again in analogy to the first ste we define S i j k S i = j=i S i j ) k 0 j=k+ S i j and the artition and also the sets R ti l Si ). It is clear that the rocedure is finite, i.e. there exists an m N for which 9.27) k 0 j=k+ ) S m j =, S m En, k). Now we do the accounting via Lemma 9. The integers x, y are here k, k 0 x = x 0 = S j y = y0 = S j j= and β = γ >. Furthermore, k x i = and so a i = k L t i j= j= S i j, y i = ) S i, j bi = k 0 j=k+ a i + b i = L t i and a + b = j=k+ S i j, k 0 L t i a R ti l j=k+ ) S i, j Da) count the new elements in the ith ste. We know from the Proosition that a + b > γa i + b i ) and from 9.27) that y m = 0. Hence, by Lemma 9, 9.28) because En, k) x m = S m > x + γy k k 0 k 0 = S j + S j + γ ) γ = + j=0 k k 0 k, S = k j=k+ j= k 0 S j + j=k+ j=k+ S j + S j S, j k 0 + S j,

23 and Sets not containing airwise corime integers 99 k 0 j=k+ S j k 0 k k j k 0 + However, 9.28) says that En, k) > S, which contradicts the otimality of S. Therefore II) must be false and Theorem is roved. R e m a r k 3. For fixed k, s and every n let Hn, k, s) Sn, k, s) be a set with S j. Hn, k, s) = max{ B : B Sn, k, s), B En, k, s)}. We know from the counterexamle in [] that En, k, s) Hn, k, s) < 0 is ossible and that En, k, s) Hn, k, s) > 0 for all n > nk, s) uniqueness). However, by the method of roof of Theorem one can derive for all k, s N. lim En, k, s) Hn, k, s) ) = n References [] R. Ahlswede and L. H. Khachatrian, On extremal sets without corimes, Acta Arith ), [2] N. G. de Bruijn, On the number of uncancelled elements in the sieve of Eratosthenes, Indag. Math ), [3] A. A. B u c h s t a b, Asymtotische Abschätzungen einer allgemeinen zahlentheoretischen Funktion, Mat. Sb. N.S.) ), [4] P. Erdős, On the density of some sequences of integers, Bull. Amer. Math. Soc ), [5], Remarks in number theory IV, Mat. Laok 3 962), [6], Extremal roblems in number theory, in: Theory of Numbers, Proc. Symos. Pure Math. 8, Amer. Math. Soc., Providence, R.I., 965, [7], Problems and results on combinatorial number theory, Chat. 2 in: A Survey of Combinatorial Theory, J. N. Srivastava et al. eds.), North-Holland, 973. [8], A survey of roblems in combinatorial number theory, Ann. Discrete Math ), [9] P. E r d ő s and A. S á r k ö z y, On sets of corime integers in intervals, Hardy Ramanujan J ), 20. [0] P. Erdős, A. Sárközy and E. Szemerédi, On some extremal roerties of sequences of integers, Ann. Univ. Sci. Budaest. Eötvös 2 969), [],,, On some extremal roerties of sequences of integers, II, Publ. Math. Debrecen ), [2] R. Freud, Paul Erdős, 80 A Personal Account, Period. Math. Hungar. 26 2) 993), [3] H. Halberstam and K. F. Roth, Sequences, Oxford University Press, 966, Sringer, 983. [4] R. R. Hall and G. Tenenbaum, Divisors, Cambridge Tracts in Math. 90, 988.

24 00 R. Ahlswede and L. H. Khachatrian [5] C. S z a b ó and G. T ó t h, Maximal sequences not containing 4 airwise corime integers, Mat. Laok ), in Hungarian). FAKULTÄT FÜR MATHEMATIK UNIVERSITÄT BIELEFELD POSTFACH BIELEFELD, GERMANY Received on and in revised form on )