On primitive sets of squarefree integers

Transcription

1 On primitive sets of suarefree integers R. Ahlswede and L. Khachatrian Fakultät für Mathematik Universität Bielefeld Postfach Bielefeld and A. Sárközy * Eötvös Loránd University Department of Algebra and Number Theory H 088 Budapest, Rákóczi út 5 Hungary * Research partially supported by Hungarian National Foundation for Scientific Research, Grant No. T This paper was written while A. Sárközy was visiting the Universität Bielefeld.

2 Introduction The set of positive integers is denoted by N while the set of positive suarefree integers is denoted by N. For A N we write SA) = a A a. A set A N is said to be primitive, if there are no a A, a A with a a, a a. The family of the primitive sets A N is denoted by P, and P denotes the family of the primitive sets A consisting of suarefree integers, i.e., with A P, A N. A subscript N indicates if we restrict ourselves to integers not exceeding N, so that N N = {, 2,...,N}; N N denotes the set of the positive) suarefree integers not exceeding N; P N is the family of the primitive subsets of {, 2,...,N}; P N denotes the family of the primitive sets selected from the suarefree integers not exceeding N. The number of distinct prime factors of n is denoted by wn), while Ωn) denotes the total number counted with multiplicity) of prime factors of n: Ωn) = p nα. α Here p α n denotes that p α n but p α+ n.) It is well known and easy to prove see, e.g., [8, p. 244]) that ) ) max A = N [N/2] = A P N 2 + o) N..) Behrend [2] proved that max SA) < c A P N log ) /2 for some absolute constant c and all N 3 and Erdős, Sárközy and Szemerédi [6] determined the value of the best possible constant in the following sense: max SA) = + o) ) as N..2) A P N 2π log ) /2 Erdős [3] proved that a A a log a < c 2 for all A P with / A.3) where c 2 is an absolute constant, and he conjectured see [3]) that the sum in.3) is maximal if A is the set of the primes: a A a log a p p log p for all A P with / A..4) 2

3 This conjecture has not been proved yet; Erdős and Zhang have several partial results. In this paper our goal is to study the suarefree analogs of the problems in.) and.2), i.e., to estimate max A and max SA)..5) A P N A P N 2 The results Since lim N N N / N N = ζ2) = 6 π 2, i.e., roughly speaking the proportion of integers being suarefree is 6, thus one expects that π 2 the maxima in.5) are less than the ones in.) and.2) by a factor 6 + o). In case π 2 of the second maximum these rough heuristics lead to the following conjecture: Conjecture. We have max SA) = + o) ) 6 as N. 2.) A P N π 2 2π log ) /2 This conjecture was raised by Pomerance and Sárközy in [0]. Indeed, we will prove this conjecture in the following more general form: Theorem. Let Q = {, 2,...} = {p α,p α 2 2,...} with p < p 2 <...) be a set of powers of distinct primes with SQ) <. 2.2) Then we have max SA) = + o) ) A P N a for a A, ) as N 2.3) 2π log ) /2 Note that here Q = is allowed and, indeed, in this special case we obtain theorem.2) of Erdős, Sárközy and Szemerédi. Choosing we obtain Corollary. 2.) holds, i.e., Conjecture is true. Q = {2 2, 3 2, 5 2,...,p 2,...}, 2.4) Another important special case is when Q consists of the primes not exceeding a fixed number K: 3

4 Corollary 2. If K 2, then we have max! A P N a, Q p = for all a A p K SA) = + o) ) p K ) as N. p 2π log ) /2 Moreover, we can prove that if Q is finite, then Q need not consist of prime powers, it suffices to assume coprimality: Theorem. Let Q = {,..., t } be a finite set of pairwise coprime positive integers: i, j ) = for i < j t. Then 2.3) holds. Since the proof of Theorems and are similar and, indeed, the proof of the latter is slightly easier, we will prove only Theorem here. On the other hand, the proof of a general theorem, which includes both theorems as special cases, could be more troublesome.) It comes perhaps as a surprise that the heuristics at the beginning of Section 2 fails in case of the first maximum in.5). Indeed, an integer m N N resp. N N ) is called maximal in N N resp. N N ) if m m for all m m, m N N resp. N N ). Let MN) resp. M N)) denote the set of all maximal integers in N N resp. N N ). Clearly MN) P N,M N) P N and MN) = {n N N : N2 } < n N and M N) = {n N N : N2 } { < n N n N N : 2 n and N 3 < n N 2 { n NN : 6 n and N 5 < n N } 3 { n NN : p k n and N < n N }... p k+ p k p k is the k th prime). Hence MN) = N N 2 = maxa PN A by.)) and simple calculation shows that ) M N) = 2 3 k 2 p + )p 2 + )...p k + ) + o) } NN 0, 5676+o)) NN max A. A P N We will prove that this maximum is much greater but can be estimated surprisingly well. Theorem 2. For N > N o we have i) max A > 0, N. A P π 2 N 4

5 ii) max A < 0, N. A P π 2 N Actually, we can show by the same method but with more computation that for large N we have 0, N < max A < 0, N.) π 2 A P π 2 N More generally, one might like to count the elements a of A with certain weights fa) where fx) is a nice function we will return to this uestion in a subseuent paper). Theorem corresponds to the weight fa) =, while in case of Theorem 2 the weighting is fa) = a for all a N). If Erdős s conjecture.4) is right, then the heuristics above also fails in case of the weighting fa) =, since then we have a log a max A P / A a A a log a = max A P / A a A a log a = p p log p. One might like to study what happens for other weightings. Another important special case is fa) = with 0 < σ < σ = 0 and σ = correspond to the special cases fa) =, a σ resp. fa) = studied above). a We introduce for 0 σ Fσ) = lim N max A P N a A ) / ) max, a σ A P N a σ a A if the limit exists and conjecture that this is the case for σ = 0. Then by Theorem 2 F0) > 6 π 2. Moreover we have the following: Conjecture 2. i) For all 0 < σ < the limit Fσ) exists. ii) Fσ) > 6 π 2. iii) lim σ Fσ) = 6 π 2. iv) Fσ) is decreasing in 0, ). Finally, we draw attention to our paper [], where we studied normalized weights in the cases σ = 0, for prefix free and also suffix free) sets of numbers rather than primitive sets and obtained several conclusive results. In Theorem 2 there the function EN) = max prefix free A N N a A a ) / is bounded by constants from below and above for all large N. Here lim EN) is also N conjectured to exist. If so, what is its value? 5 a N N a

6 3 Auxiliary results In order to prove Theorem, we need several lemmas. Lemma. For x we have and p x p α x p = log log x + c 3 + O ) log x ) p = log log x + c α 4 + O. log x In the second sum the summation is over all prime powers not exceeding x.) These two euivalent) formulas can be found in any book on prime number theory see, e.g., [9, p. 20]). Lemma 2. Write Px,k) = { n : n x, Ωn) = k }. 3.) Then uniformly for x 3, k N we have n < c log x 5 log log x) /2. n Px,k) Proof: This follows from [4] and [5] see also [] and [2]) and as pointed out by the referee also directly from Lemma and the ineuality ak ea, uniformly for all a > and all k! a /2 k N. Lemma 3. For any f with fx) > 0 for x > x 0 and we have uniformly for k N, satisfying n Px,k) where Px,k) is defined by 3.)). Proof: See [4]. lim fx)log log x x) /2 = 0, log log x fx) < k < log log x + fx), 3.2) n = + o) ) log x as x 2π log log x) /2 Lemma 4. Define Q = {, 2,...} as in Theorem so that 2.2) holds), write Bx,k,Q) = { n : n Px,k), n for Q }, 3.3) 6

7 and define fx) as in Lemma 3. Then uniformly for k satisfying 3.2) we have n = + o) ) ) log x as x. 3.4) 2π log log x) /2 n Bx,k,Q) Proof: Fix an ε > 0, and let H be a positive integer with < ε; 3.5) by 2.2), such a number H exists. Write >H Q = { : Q, H} and Q 2 = { : Q,H < x}. Clearly we have Bx,k,Q) Bx,k,Q ), and if n Bx,k,Q ) but n / Bx,k,Q), then we have n Px,k) and n for some Q 2. It follows that n n n Bx,k,Q) n Bx,k,Q ) n. 3.6) 2 n Px,k) n By the exclusion inclusion principle we have n = n + ) j j n Bx,k,Q ) n Px,k) = n Px,k) n + j ) j i,..., ij Q i < < ij i < < ij H n Px,k) i... ij n i... ij n t Px/ i... ij,k Ω i... ij ) t. By Lemma 3, it follows that for x we have n n Bx,k,Q ) = + o) ) log x 2π log log x) + /2 j = + o) ) H ) ) j i < < ij H logx/ i... ij ) i... ij 2π log logx/ i... ij )) /2 log x 2π log log x) /2. 3.7) 7

8 By 3.5) here we have ) ) = ) ) < H >H < ) exp 2 < >H ) >H + 2 ) ) exp2ε). 3.8) Moreover, by 3.5) and Lemma 2 we have 2 n Px,k) n n = 2 t Px,k) t = 2 t Px/,k Ω)) t < 2 c 5 log x log log x) /2 < c 5 ε log x log log x) /2. 3.9) Since 3.6), 3.7), 3.8) and 3.9) hold for ε > 0 and x > x 0 ε), thus 3.4) follows. Lemma 5. Write z = [log ]. 3.0) Then we have S {n : n N, Ωn) ωn) > 00 log z} ) = o log ) /2 ). Proof: This is the Lemma in [6]. Lemma 6. For n 30 we have p n p < c 6 log log log n. 3.) Proof: Clearly we have p n whence, by the well known ineuality + ) < p p n ϕn) > c 7 p n log log n = n ϕn) 8

9 see, e.g. [9, p. 24]), it follows that On the other hand, we have p n + ) p p n = exp p n = exp p n 3.) follows from 3.2) and 3.3). + ) < c 8 log log n. 3.2) p p) log + p + O p ) > exp p 2 p n p + c ) 4 Completion of the proof of Theorem The lower bound for the maximum in 2.3) is a straightforward conseuence of Lemma 4. Indeed, define again z by 3.0) and let A = BN,z,Q) where BN,z,Q) is defined by 3.3)). Then clearly we have and, by Lemma 4, SA ) = n BN,z,Q) It follows that the maximum in 2.3) is A P N, a for a A, Q n = + o) ) max SA) SA ) = + o) ) A P N a for a A, ) ) 2π log ) /2. 2π log ) /2. 4.) In order to prove that this bound is also an upper bound for maxsa), we will adopt the method of the proof in [6] which is a method of combinatorial flavour and it reminds one of the proof of Sperner s theorem). Consider an A with A P N and such that a for a A, Q. Define A by A = { a : a A, Ωa) ωa) 00 log z } where again, z is defined by 3.0). By 2.2) and Lemma 5, it suffices to prove that SA ) + o) ) ) 9 as N. 4.2) 2π log ) /2

10 Following combinatorial terminology, we will refer to the integers n with Ωn) = k as integers on level k. First we separate the a s of different level, i.e., we write A k = { a : a A, Ωa) = k }. Let r and r 2 denote the occuring smallest, resp. greatest level, so that A = r 2 k=r A k and SA ) = r 2 k=r SA k ). Now we separate the contribution of the A k s with k > z high level a s ; here again z is defined by 3.0)), k = z, resp. k < z low level a s ): SA ) = SA k ) + SA z ) + SA k ) = Σ + Σ 2 + Σ ) z<k r 2 r k<z We will compare Σ and Σ 3 with the sum of the reciprocals of the numbers having z prime factors which can be constructed from the high level a s by dropping prime factors, resp. from the low level a s by adding prime factors without producing a multiple of a ). Indeed, let D denote the set of the integers d such that Ωd) = z and there is an a A k with d a, k > z, and let E denote the set of the integers e such that Ωe) = z, there is no Q with e, and there is an a A k with a e, k < z. It follows from the primitivity of A that D E = D A z = E A z =, 4.4) and clearly we have D BN,z,Q), E BN,z,Q) and A z BN,z,Q). 4.5) By Lemma 4, it follows from 4.4) and 4.5) that SD) + SE) + SA z ) S BN,z,Q) ) = + o) ) We will show that ) 2π log ) /2. 4.6) Σ + o) ) SD) 4.7) and Σ 3 + o) ) SE) + o log ) /2 ). 4.8) 0

11 4.2) would follow from 4.3), 4.6), 4.7) and 4.8) and this, together with 4.), would complete the proof of the theorem. Thus it remains to prove 4.7) and 4.8). First we will prove 4.7). For z k r 2 define the sets U k, F k by the following recursion: Let U r2 = and F r2 = A r2 = A r2 U r2 ). If z < k r 2 and U k, F k have been defined, then let U k = { u : u N, Ωu) = k, there is f F k with u f } and F k = A k U k. 4.9) Then clearly we have D = U z. 4.0) Moreover, by the primitivity of A, for all z k r 2 we have A k U k =. 4.) If z k < r 2 and u U k, then there is a number f with u f, f F k+. Then we have Ωu) = k and Ωf) = k +, so that f is of the form f = up. By f F k+ we have f a for some a A and thus Ωf) ωf) Ωa) ωa) 00 log z. Moreover, again by f F k+ we have Ωf) = k +. It follows that f has at least representations in form up so that SU k ) p N p = u U k ωf) Ωf) 00 log z = k + 00 log z p N up By Lemma, 4.9) and 4.), it follows that k + 00 log z) f F k+ f = k + 00 log z)sf k+). SU k ) > k + 00 log z z + c 0 SAk+ ) + SU k+ ) ) 4.2) for z k < r 2. For N large enough clearly we have { k + 00 log z if k > z log z > 200 log z z + c 0 for all k z. z 4.3) It follows from 4.2) and 4.3) by a simple induction argument that SU z ) > which, by 4.0), proves 4.7). ) 200 log z 200 log z SA k ) = + o) ) Σ z z<k r 2

12 Now we will prove 4.8). The proof is similar to the proof of 4.7), but while there in each step we moved one level lower by dropping a prime factor, here we move one level higher by adding a prime. Moreover, here a little difficulty arises from the facts that, first, we have to restrict ourselves to integers v with v for Q, and, secondly, also integers v slightly greater than N appear. For r k z define the sets V k, G k by the following recursion: Let V r = and G r = A r = A r V r ). If r k < z and V k, G k have been defined, then let V k+ denote the set of the positive integers v that can be represented in the form v = pg with g G k and a prime p such that p g, p / Q and p N /z2. Moreover, let By the primitivity of A, for all r k z we have G k+ = A k+ V k+. 4.4) A k V k =. 4.5) Clearly, if r k z and g G k, then we have Ωg) = k, and if Q, then g. It follows from these facts that if v V z G z and v N, then we have v E. However, V z also contains numbers v greater than N. Such a v is the product of an a A and at most z r z primes, each of them N /z2. Thus we have v N N /z2 ) z = N +/z. 4.6) It follows that and whence Here we have V z E {n : N < n N +/z } SE) SV z ) N<n N +/z n = O ) z N<n N +/z n. 4.7) = o log ) /2 ). 4.8) It remains to give a lower bound for SV z ). If r < k z and v V k, then v is of the form v = pg with g G k and a prime p such that p g,p / Q,p N /z2, 4.9) and for all g G k and p satisfying 4.9), we have v = pg V k. Since here Ωv) = k, thus v has at most k representations in this form. It follows that SV k ) = v V k v k g G k p g,p/ Q p N /z2 pg = k g g G k p g,p/ Q p N /z2 p. 4.20) 2

13 By 2.2), 4.6) and Lemmas and 6, here the innermost sum is p g,p/ Q p N /z2 p p N /z2 p p g p = log 2 log z + O) c 6 log log +/z + O) > z c log z. 4.2) It follows from 4.4), 4.5), 4.20) and 4.2) that SV k ) > z c log z k for r < k z. For N large enough clearly we have z c log z k SG k ) = z c log z SAk ) + SV k ) ) 4.22) k { if k < z c log z c log z z for all k z. It follows from 4.22) and 4.23) by a simple induction argument that SV z ) > 4.23) ) c log z log z c SA k ) = + o) ) Σ ) z r k<z 4.8) follows from 4.7), 4.8) and 4.24) and this completes the proof of Theorem. 5 Proof of Theorem 2 P = {p,p 2,p 3,...} = {2, 3, 5,...} denotes the set of primes and p k is the k th prime. For given N and k consider the following sets: It is well known and easy to check that T N p,...,p k ) = {m N N : m p... p k } and R N p,...,p k ) = {m N N : m,p... p k ) = }. R N p,...,p k ) lim N N N = k i= ). 5.) p i + More generally, for 0 α < β we have αn m βn : m N, m,p... p k ) = lim N αn n βn : n N = k i= ). 5.2) p i + 3

14 Clearly, every a N N can be uniuely represented as a = a a 2, where a T N p,...,p k ), a 2 R N p,...,p k ). For any primitive set A P N and a 2 R N p,...,p k ) we set Fa 2 ) = { a A : a = a a 2, a T N p,...,p k ) }. Of course some of the Fa 2 ) can be empty. Then A =. a 2 R N p,...,p k ) Fa 2 ), A = a 2 R N p,...,p k ) Fa 2 ). 5.3) For n N, let L k n) be a primitive set of T n p,...,p k ) with maximal cardinality. We observe that in the case n < p k+ the set L k n) is an optimal primitive suarefree set in N n, i.e. max B = L k n). B P n Since A is a primitive set then necessarily we have Fa 2 ) N L k a 2 ) for every a 2 R N p,...,p k ). Therefore by 5.3), for every primitive set A P N ) A N L k holds. 5.4) a 2 R N p,...,p k ) Finding in general the value L k n) seems difficult. However, for small k and arbitrary n this can easily be done. Take k = 3. We have L 3 ) = L 3 2) = L 3 3) = L 3 4) = 2, L 3 3) = L 3 4) = {2, 3} ) L 3 n) = 3 for n 5, L 3 n) = {2, 3, 5} or {6.0, 5} ). Conseuently, for a 2 R N 2, 3, 5) we obtain a 2 L 3 N a 2 ) =, if N 3 < a 2 N, L 3 N a 2 ) = 2, if N 5 < a 2 N 3, and L 3 N a 2 ) = 3, if a 2 N 5. 4

15 Hence by 5.4) we have A N 3 < a 2 N : a 2, 30) =,a 2 N + N 5 < a 2 N 3 : a 2, 30) =,a 2 N 2+ a 2 N 5 : a 2, 30) =,a 2 N 3. Using 5.2) we get 2 A ) 5 3 N N = N N + on) 0, 6389 N N + on). ) ) ) + on) Taking k = 5, by similar calculations details are omitted) we obtain a slightly better bound: This proves the upper bound ii). Proof of i): Consider the following set A 0, 6366 N N + on). A =. a 2 R N 2,3,5) Ga 2 ), where for every a 2 R N 2, 3, 5) Ga 2 ) is defined by {a 2 }, if N < a 3 2 N It can easily be verified that A P N. Simple calculations, using 5.2), yield {2a 2, 3a 2 }, if N < a 5 2 N 3 {2a 2, 3a 2, 5a 2 }, if N 2 Ga 2 ) = < a 2 N 5 N {6a 2, 0a 2, 5a 2 }, if < a N 2 N {30a 2 }, if < a 2 N, if a 2 N A 0, 6328 N N + on). By considering the set R N 2, 3, 5, 7, ), similarly one can construct a primitive set A with a slightly better bound: A 0, 6362 N N + on)

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