INTEGERS WITH A DIVISOR IN (y, 2y]

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1 INTEGERS WITH A DIVISOR IN y, 2y] KEVIN FORD Abstract We determine, up to multiplicatie constants, how many integers n x hae a diisor in y, 2y] Introduction Let Hx, y, z) be the number of integers n x which hae a diisor in the interal y, z] In the author s paper [3], the correct order of growth of Hx, y, z) was determined for all x, y, z In particular, x ) Hx, y, 2y) 3 y x), log y) δ log log y) 3/2 where + log log 2 δ = = log 2 In this note we proe only the important special case ), omitting the parts of the argument required for other cases In addition, we present an alternate proof, dating from 2002, of the lower bound implicit in ) This proof aoids the use of results about uniform order statistics required in [3], and instead utilizes the cycle lemma from combinatorics Although shorter and technically simpler than the argument in [3], this method is not useful for counting integers with a prescribed number of diisors in y, 2y] We mention here one of the applications of ), a 50-year old problem of Erdős [], [2]) known colloquially as the multiplication table problem Let Ax) be the number of positie integers n x which can be written as n = m m 2 with each m i x Then x Ax) log x) δ log log x) 3/2 This follows directly from ) and the inequalities ) x x x H 4, 4, Ax) H 2 k 0 x 2 k, x 2 k+, More on the history of estimations of Hx, y, z), further applications and references may be found in [3] Heuristic argument For breity, let τn, y, z) be the number of diisors of n in y, z] Write n = n n, where n is composed only of primes 2y and n is composed only of primes Date: July 7, Mathematics Subject Classification: Primary N25; Secondary 62G30 Research supported by National Science Foundation grants DMS and DMS x 2 k )

2 2 KEVIN FORD > 2y For simplicity, assume n is squarefree and n y 00 Assume for the moment that the set Dn ) = {log d : d n } is uniformly distributed in [0, log n ] If n has k prime factors, then the expected alue of τn, y, 2y) should be about 2k log 2 2k This is precisely log n log y when k k 0 + O), where k 0 := Using the fact eg Theorem 08 of [5]) that the log log y log 2 number of n x with n haing k prime factors is of order x log y log log y) k, k! we obtain a heuristic estimate for Hx, y, 2y) of order x log y k k 0 +O) log log y) k k! xlog log y)k 0 k 0! log y x log y) δ log log y) /2 This is slightly too big, and the reason stems from the uniformity assumption about Dn ) In fact, for most n with about k 0 prime factors, the set Dn ) is far from uniform, possessing many clusters of diisors and large gaps between clusters This substantially decreases the likelihood that τn, y, 2y) The numbers log log p oer p n are well-known to behae like random numbers in [0, log log 2y] Consequently, if we write n = p p k, j log log y where p < p 2 < < p k, then we expect log log p j k 0 = j log 2 + O) for each j Large deiation results from probability theory see Smirno s theorem in 4; also see Ch of [5]) tell us that with high probability there is a j for which log log p j j log 2 c log log y, where c is a small positie constant Thus, the 2 j diisors of p p j will be clustered in an interal of logarithmic length about log p j 2 j e c log log y On a logarithmic scale, the diisors of n will then lie in 2 k j translates of this cluster A measure of the degree of clustering of the diisors of an integer a is gien by La) = measl a), L a) = d a[ log 2 + log d, log d) The probability that τn, y, 2y) should then be about Ln )/ log y Making this precise leads to the upper and lower bounds for Hx, y, 2y) gien below in Lemmas 2 and 32 The upper bound for La) gien in Lemma 3 iii) below quantifies how small La) must be when there is a j with log log p j considerably smaller than j log 2 What we really need to count is n for which n has about k 0 prime factors and Ln ) log n This roughly corresponds to asking for log log p j j log 2 O) for all j The anologous problem from statistics theory is to ask for the likelihood than gien k 0 random numbers in [0, ], there are k 0 x + O) of them which are x, uniformly in 0 x In section 4, Lemma 4, we will see that this probability is about /k 0 / log log y and this leads to the correct order ) Notation: Let τn) be the number of positie diisors of n, and define ωn) to be the number of distinct prime diisors of n Let P + n) be the largest prime factor of n and let P n) be the smallest prime factor of n Adopt the notational conentions P + ) = 0 and P ) = Constants implied by O, and are absolute The notation f g means f g and g f

3 INTEGERS WITH A DIVISOR IN y, 2y] 3 We shall make frequent use of the following estimate, which is a consequence of the Prime Number Theorem with classical de la Valée Poussin error term For certain constants c 0, c, 2) p = log log x + c 0 + Oe c log x ) x 2) p x We also need the standard siee bound eg [4]; Theorem 06 and Exercise 02 of [5]) 3) {n x : P n) > z} x log z and Stirling s formula k! 2πkk/e) k x 2z 4) 2 Lower bound In this section we proe the lower bound implicit in ) The first step is to bound Hx, y, 2y) in terms of a sum of La)/a Next, sums of La)/a are related ia the Cauchy- Schwarz inequality to sums of a function W a) which counts pairs of diisors of a which are close together With a strategic choice of sets of a to aerage oer, the problem is reduced to the estimation of a certain combinatorial sum This is accomplished with the aid of a tool closely related to the so-called cycle lemma Lemma 2 If 3 y x, then Hx, y, 2y) x log 2 y La) a a y /8 Proof Let y 0 be a sufficiently large constant If 3 y y 0, then Hx, y, 2y) x xl) log 2 y since L) = log 2 If y y 0, consider integers n = apb x with a y /8, all prime factors of b are > 2y or in [y /4, y 3/4 ], and p is a prime with logy/p) L a) The last condition implies that τap, y, 2y) In particular, y 7/8 y/a < p 2y Thus, each n has a unique representation in this form Fix a and p and note that x/ap) x/2y 9/8 ) 2 y7/8 If x/ap) 4y, 3) implies that the number of b x with P b) > 2y is x If ap ap log y x/ap) < 4y, then the number of b x composed of two prime factors in ap y/4, y 3/4 ] is likewise x Hence ap log y Hx, y, 2y) x log y a a y /8 logy/p) L a) Since L a) is the disjoint union of interals of length log 2 and p y 7/8, for each a we hae by repeated application of 2) logy/p) L a) p La) log y p

4 4 KEVIN FORD Lemma 22 For any set A of positie integers, La) a A a 6 a A where a A ) 2 τa) a W a) a W a) = {d, d ) : d a, d a, log d/d log 2} Proof Since τa) log 2 = τa, e u, 2e u ) du, by the Cauchy-Schwarz inequality, ) 2 τa) ) 2 log 2) 2 = τa, e u, 2e u ) du a a a A a A ) La) ) τ 2 a, e u, 2e u ) du a a a A Partition the real numbers into interals I j of length log 2 and let k j be the number of diisors of a in I j Then τ 2 a, e u, 2e u ) du log 2) k j + k j+ ) 2 4log 2) kj 2 4log 2)W a) j j We apply Lemma 22 with sets A of integer haing particular distribution of prime factors Partition the primes into sets D, D 2,, where each D j consists of the primes in an interal λ j, λ j ], with λ j λ 2 j More precisely, let λ 0 = 9 and inductiely define λ j for j as the largest prime so that 2) log 2 p λ j <p λ j For example, λ = 2 and λ 2 = 7 By 2), we hae and thus for some absolute constant K, a A log log λ j log log λ j = log 2 + Oe c log λj ), 22) 2 j K log λ j 2 j+k j 0) For a ector b = b,, b J ) of non-negatie integers, let A b) be the set of square-free integers a composed of exactly b j prime factors from D j for each j Lemma 23 Assume b = b,, b J ) Then a A b) W a) a 2 log 2)b + +b J b! b J!, J 2 j+b + +b j j=

5 INTEGERS WITH A DIVISOR IN y, 2y] 5 Proof Let B = b + + b J and for j 0 let b j = i j b j Let a = p p B, where 23) p b j +,, p b j D j j J) and the primes in each interal D j are unordered Since W p p B ) is the number of pairs Y, Z {,, B} with 24) log p i log p i log 2, i Y i Z we hae 25) a A b) W a) a b! b J! Y,Z {,,B} p,,p B 23),24) p p B When Y = Z, 2) implies that the inner sum on the right side of 25) is log 2) B, and there are 2 B such pairs Y, Z When Y Z, let I = max[y Z) Y Z)] With all the p i fixed except for p I, 24) implies that U p I 4U for some number U Let EI) be defined by b EI) < I b EI), ie p I D EI) By 2), a A b) U p I 4U p I D EI) p I maxlog U, log λ EI) ) 2 EI) Thus, by 2) the inner sum in 25) is 2 EI) log 2) B With I fixed, there correspond 2 B I+ 4 I = 2 B+I pairs Y, Z By 25), [ ] W a) 2 log 2)B B + 2 I EI) 2 log 2)B J 2 j 2 I, a b! b J! b! b J! and the claimed bound follows a A b) I= j= b j <I b j Now suppose that M is a sufficiently large positie integer, b i = 0 for i < M, and b j Mj for each j By 22), τa) J a = ) 2k b j! p p 2 p bj j=m p D j 26) Let 2 k J j=m 2 log 2)k 2b M! b J! log log y k = log 2 log 2 b j! p 2 D j p 2 p b j λ j ) bj p bj D j p bj {p,,p bj } 2M, J = M + k

6 6 KEVIN FORD Let B be the set of ectors b,, b J ) with b i = 0 for i < M and b + + b J = k Let B be the set of b B with b j minmj, MJ j + )) for each j M If b B and a A b), then by 22), J J M log a b j log λ j M l + )2 J+K l < log y 8 j=m if M is large enough, as 2 J+ 2 M log y Put J 27) fb) = 2 M h+b M + +b h By Lemma 23, a A b) W a) a h=m l=0 2 log 2)k + 2 M fb) ) b M! b J! 2 log 2)k b M! b J! fb) since fb) /2 By Lemmas 2 and 22, plus 26), we hae for large y x2 log 2)k 28) Hx, y, 2y) log 2 y b b B M! b J!fb) Obsere that the product of factorials is unchanged under permutation of b M,, b J Roughly speaking, fb) gb) := max 2 b M )+ +b j ) j Note that b M ) + + b J ) = k J M + ) = 0 Gien real numbers z,, z k with zero sum, there is a cyclic permutation z of the ector z = z,, z k ) all of whose partial sums are 0: let i be the index minimizing z + + z i and take z = z i,, z k, z,, z i ) In combinatorics, this fact is know as the cycle lemma Thus, there is a a cyclic permutation b of b with gb ) = Thus, we expect that /fb ) will be /k on aerage oer b and that /fb) /k on aerage oer b B This is essentially what we proe next; see 20) below Lemma 24 For positie real numbers x,, x r with product X, let x r+i = x i for i Then r r ) [ ] x +j x h+j max, X), min, X) j=0 h= Proof Put y 0 = and y j = x x j for j The sum in question is r r ) y h+j r y j = y j y +j + + y r+j Since y r = X, j=0 h= j=0 y +j + + y r+j = Xy y j ) + y +j + + y r [min, X)y y r ), max, X)y y r )]

7 We hae 29) where Let x i = 2 +b M +i INTEGERS WITH A DIVISOR IN y, 2y] 7 b b B M! b J!fb) S 0 S 0 = b B S j) = b B b j >Mj S 2 j) = M j<k/m S j) b M! b J!fb), b B b J+ j >Mj b M! b J!fb), b M! b J!fb) for i k Then x x k = and fb) = x + x x x x 2 x k j<k/m By Lemma 24 and the multinomial theorem, 20) S 0 = k k ) x +j x h+j b M! b J! k b B j=0 h= S 2 j), = kk k! To bound S j), apply Lemma 24 with x i = 2 b j+i for i J j and note that X = x x J j = 2 j+ M b M b j < Write b = b M,, b j, b j+,, b J ), whose sum of components is k b j Ignoring the terms with h j in 27), using Lemma 24 and the multinomial theorem, we find S j) b j >Mj J j 2kk k! kk k! Hence, if M 2 then 2) b j! b j >Mj b j >Mj 2 Mj)! b i j b i! 2 M j+b M + +b j J j k ) k b j b j!k b j )! b j! M j<k/m S j) kk 0k! J j i=0 J j ) x +i x h+i h=

8 8 KEVIN FORD The estimation of S 2 j) is similar Let x i = 2 +b M+i X = x x J M+ j = 2 j b J j+ b J for i J M + j, so that Put b = b J j+ and let b = b M,, b J j, b J j+2,, b J ), whose sum of components is k b Then, ignoring the terms with h > J j in 27), we hae If M is large enough, then 22) S 2 j) = b>mj By 29), 20), 2) and 22), b! b 2 j J M + j i J j+ b i! b>mj 2 j 2 b k + j)k k k! b! b>mj kk 2 j e j 2 Mj k! Mj)! j 2 b b! S 2 j) kk 0k! 2 b j+b J j+2+ +b J J M + j k + j 2) k b k b)! b b B M! b J!fb) kk 2k! The lower bound in ) for large y now follows from 28) and Stirling s formula If y y 0 for some fixed constant y 0, the lower bound in ) follows from Hx, y, 2y) x 3 Upper bound, part I In this section, we proe the upper bound implicit in ), except for the estimation of some integrals which will be dealt with in section 4 As with the lower bound argument, we begin by bounding Hx, y, 2y) in terms of a sum inoling La) Using a relatiely simple upper bound for La) proed in Lemma 3 below, the sums inoling La) are bounded in terms of particular multiariate integrals The estimates for these integrals in section 4 allow us then to complete the proof Lemma 3 We hae i) La) minτa) log 2, log 2 + log a); ii) If a, b) =, then Lab) τb)la); iii) If p < < p k, then Lp p k ) min 0 j k 2k j logp p j ) + log 2)

9 INTEGERS WITH A DIVISOR IN y, 2y] 9 Proof Part i) is immediate, since L a) is the union of τa) interals of length log 2, all contained in [ log 2, log a) Part ii) follows from L ab) = d b{u + log d : u L a)} Combining parts i) and ii) with a = p p j and b = p j+ p k yields iii) Remarks More generally, we may define for non-negatie real numbers x,, x k the quantity ) k k L x; η) = η + ε i x i, ε i x i ε,,ε k {0,} Then Lx; η) = measl x; η) measures the distribution of the subset sums of x,, x k The proof of Lemma 3 iii) gies Lemma 32 If 3 y x, then where 3) St) = i= i= L x; η) min 0 j k 2k j x + + x j + η) Hx, y, 2y) x max y t x St), P + a) t µ 2 a)= La) a log 2 t/a + P + a)) Proof First, we reduce the problem to estimating H x, y, z), the number of squarefree integers n x with τn, y, z) Write n = n n, where n is squarefree, n is squarefull and n, n ) = The number of n x with n > log y) 4 is x n n >log y) 4 x log y) 2 Assume now that n log y) 4 For some f n, n has a diisor in y/f, 2y/f], hence 32) Hx, y, 2y) ) ) x x, y, 2y + O n f f log y) 2 H n log y) 4 f n Next, we show that for 3 y x 3/5, 33) H x, y, 2y ) H 2 x, y, 2y ) x S2y ) + Sx /y ) ) Let A be the set of squarefree integers n x 2, x ] with a diisor in y, 2y ] Put z = 2y, y 2 = x 4y, z 2 = x y If n A, then n = m m 2 with y i < m i z i i =, 2) For some j {, 2} we hae p = P + m j ) < P + m 3 j ) Write n = abp, where P + a) < p < P b) and b > p Since τap, y j, z j ), we hae p y j /a By 3), gien a and p, the number of possible b is x ap log p x ap log Q,

10 0 KEVIN FORD where Q = maxp + a), y j /a) Since a has a diisor in y j /p, z j /p], we hae logy j /p) L a) or log2y j /p) L a) Since L a) is the disjoint union of interals of length log 2 with total measure La), by repeated use of 2) we obtain logcy j /p) L a) p Q p La) log Q c =, 2), and 33) follows Write x 2 = x/n, y = y/f Each n x 2 / log 2 y, x 2 ] lies in an interal 2 r+ x 2, 2 r x 2 ] for some integer 0 r 5 log log y Applying 33) with x = 2 r x 2 for each r gies H x 2, y, 2y ) x 2 log 2 y + r 2 r x 2 S2y ) + S2 r x 2 /y ) ) x 2 max y t x2 St) Here we used the fact that St) L) = log 2 Finally, log 2 t log 2 t n τn )/n = O) and the lemma follows from 32) Define 34) T P, Q) = P + a) P a Q,µ 2 a)= La) a If a t /2 or P + a) > t /3000, then log 2 t/a + P + a)) log 2 t Otherwise, e eg < P + a) e eg for some integer g satisfying 0 g log log t 5 Thus we hae T t, ) 35) St) log 2 t + e 2g T exp{e g }, t /2 ) for a fixed large integer g 0 Further define T k P, Q) = g 0 g log log t 5 P + a) P, a Q ωa)=k, µ 2 a)= La) a We next bound T k P, Q) in terms of a mutiariate integral Heuristically, p z /p log log z, hence by partial summation we expect for nice functions f that ) log log p f,, log log p k log log P log log P log log P ) k fξ) dξ p p k p < <p k P Lemma 33 Suppose P is large and Q Let = Then where U k ) = log log P log 2 T k P, Q) e log Q log P 2 log log P ) k U k ), 0 ξ ξ k 0 ξ ξ k min 0 j k 2 j 2 ξ ξ j + ) dξ and suppose k 0

11 Proof Let α = We hae log P Also 36) p P p α = p P INTEGERS WITH A DIVISOR IN y, 2y] T k P, Q) Q α p + O α p P P + a) P ωa)=k La) a α ) log p = log log P + O) p In a similar manner to how we constructed the sets D j in section 2, we find that there is an absolute constants K so that the following holds for all P : the interal [2, P ] may be partitioned into subinterals E 0,, E +K with = and for each j, and log 2 p / log P p E j log log P log 2 37) p E j = log log p j + K) log 2 Consider a = p p k, p < < p k P and define j i by p i E ji i k) Put l i = log log p i By Lemma 3 iii) and 37), log 2 La) 2 k min 0 g k 2 g 2 l lg + ) 2 k+k F j), where F j) = min 0 g k 2 g 2 j jg + ) Let J denote the set of ectors j satisfying 0 j j k + K Then T k P, Q) Q α 2 k+k j J F j) p < <p k p i E ji i k) p p k ) α Let b j be the number of primes p i in E j for 0 j + K Using the hypothesis that k 0, the sum oer p,, p k aboe is at most +K ) bj log 2) k b j=0 j! p α b 0! b +K! p E j = + K) log 2) k dξ where e 0K log 2) k Rj) Rj) dξ, Rj) = {0 ξ ξ k : j i + K)ξ i j i + i}

12 2 KEVIN FORD Obserer that, in Rj), there are b s numbers ξ j satisfying s + K)ξ i s + for each s, and Vol{0 x x b } = /b! Since 2 j i 2 +K)ξ i 2 K 2 ξ i for ξ Rj), we hae Hence and the lemma follows F j) 2 K j J min 0 g k 2 g 2 ξ ξg + ) F j) dξ 2 K U k ) Rj) Estimating U k ) is the most complex part of the argument The next lemma will be proed in section 4 Lemma 34 Suppose k, are integers with 0 k 0 Then U k ) + k 2 k + )!2 k + ) Notice that the bound in Lemma 34 undergoes a change of behaior at k = Lemma 35 Suppose P is large and Q Then Proof Let = 38) and 39) k 0 k By Lemma 3 i), T P, Q) e log Q log P log P ) 2 δ log log P ) 3/2 log log P and γ = e log Q log P log 2 By Lemmas 33 and 34, T k P, Q) γ T k P, Q) γ k 0 k k 0 T k P, Q) k ) log log P ) k 2 k k + )! k) 2 + )2 log log P ) k k + )! k 0 P + a) P µ 2 a)=,ωa)=k a Q k 0 γ k 0 2 k log 2 a 2 k Q / log P 2 k k! p P P + a) P µ 2 a)=,ωa)=k p / log P ) k γ γ2 log log P ) + )! 2 log log P ) + )! a / log P

13 k 0 INTEGERS WITH A DIVISOR IN y, 2y] 3 By 36), the sum on p is log log P + O) Thus, 2 log log P + O))0 2 log log P ) 30) T k P, Q) γ γ 0)! + )! Finally, T 0 P, Q) = 0 if Q > and T 0 P, ) = L) = log 2 Recalling the definition of and combining 38), 39), 30) with Stirling s formula completes the proof By Lemma 35 and 35), St) log t) 2 δ log t) 2 log log t) 3/2 + log t) δ log log t) 3/2 g 0 g log log t 5 e 2g log t)/2eg )+g2 δ) g 3/2 The desired upper bound for Hx, y, 2y) now follows from Lemma 32 4 Upper bound, part II The goal of this section is to proe Lemma 34, and thus complete the proof of the upper bound in ) Let Y,, Y n be independent, uniformly distributed random ariables in [0, ] Let ξ be the smallest of the numbers Y i, let ξ 2 be the next smallest, etc, so that 0 ξ ξ n The numbers ξ i are the order statistics for Y,, Y n Then k!u k ) is the expectation of the random ariable X = min 0 j k 2 j 2 ξ ξ j + ) Heuristically, we expect that 4) EX E min j k 2 j+ξ j, so we need to understand the distribution of min j k ξ j j Let Q k u, ) be the probability that ξ i i u for eery i In the special case = k, Smirno in 939 showed that Q k x k, k) e 2x2 for each fixed x The corresponding probability estimate for two-sided bounds on the ξ i was established by Kolmogoro in 933 and together these limit theorems are the basis of the Kolmogoro-Smirno goodness-of-fit statistical tests In the next lemma, we proe new, uniform estimates for Q k u, ) The remainder of the section is essentially deoted to proing 4) The details are complicated, but the basic idea is that if 2 j 2 ξ ξ j ) is much large than 2 ξ j j, then for some large l, the numbers ξ j l,, ξ j are all ery close to one another As shown below in Lemmas 43 and 44, this is quite rare Lemma 4 Let w = u + k Uniformly in u 0 and w 0, we hae Q k u, ) u + )w + )2 k

14 4 KEVIN FORD Proof Without loss of generality, suppose k 00, u k/0 and w k If min i k ξ i i u ) < 0, let l be the smallest index with ξ l < l u and write ξ l = l u λ, so that 0 λ Let { R l λ) = Vol 0 ξ ξ l l u λ : ξ i i u } i l ) Then we hae Q k u, ) = k! = k! 0 u+λ l k 0 u+λ l k { l u λ R l λ) Vol R l λ) k l)! ) k l k + w + λ l dλ } ξ l+ ξ k dλ Now suppose that ξ k 2w+2 λ as before, we hae = k u w 2 Then min i k ξ i i u < 0 Defining l and 2w + 2 ) k = k! Vol{ 0 ξ ξ k 2w + 2 } k l w 2 + λ = k! Thus, for any A > 0, we hae Q k u, ) = A 2w + 2 ) k k! + k! 0 u+λ l k w 2+λ Noting that 2 λ λ, we hae ) k l k l w 2 + λ = k l + w + λ 0 u+λ l k w 2+λ 0 R l λ) k l)! k w 2+λ<l k ) k l dλ ) k l R l λ) k + w + λ l dλ k l)! R l λ) k l)! k l [ Ak l w 2 + λ) k l k l + w + λ) k l] dλ = exp w + 2 λ k l { e 2w+2) 2w + 2) + ) k l + w + λ k l ) k l) } w + 2 λ) j + ) j w + λ) j j=2 jk l) j

15 INTEGERS WITH A DIVISOR IN y, 2y] 5 Thus, taking A = e 2w+2, we conclude that Q k u, ) e 2w+2 2w + 2 ) k { } 2w + 2 = exp k + Ow)) { } 2uw + Ou + w 2 + ) = exp j t j+a>0 2uw + Ou + w2 + ) Lemma 42 If t 2, b 0 and t + a + b > 0, then t j u + )w + )2 k ) a + j) j b + t j) t j e 4 t + a + b) t Proof Let C t a, b) denote the sum in the lemma We may assume that a > t, otherwise C t a, b) = 0 The associated complete sum is ealuated exactly using one of Abel s identities [6], p20, equation 20)) 42) t j=0 t j ) a + j) j b + t j) t j = a + ) t + a + b) t ab 0) b If a, put A = max, a) and B = max, b) By 42), C t a, b) C t A, B) A + ) t + A + B) t B 43) 2t + a + b + 3) t 2e 3t ) t+a+b t + a + b) t < e 4 t + a + b) t Next assume a < Since + c/x) x is an increasing function for x >, we hae a + j) j = j ) j + a + ) j j ) j + a + ) t j t Thus, by 43), C t a, b) ) t t + a C t, b) t e 4 t + a)t + b ) t e 4 t + a + b) t ) t = e 4 t + a + b + ) t a + )b t

16 6 KEVIN FORD For breity, write S k u, ) = {ξ : 0 ξ ξ k : ξ i i u i k)}, so that Q k u, ) = k! Vol S k u, ) Lemma 43 Suppose g, k, s, u, Z satisfy 2 g k/2, s 0, k/0, u 0, u + k + Let R be the subset of ξ S k u, ) where, for some l g +, we hae 44) Then l u VolR) ξ l l u + 0s + ))g g 2)!, ξ l g l u s u + )u + k) 2 k + )! Proof Fix l satisfying maxu, g + ) l k Let R l be the subset of ξ S k u, ) satisfying 44) for this particular l We hae VolR l ) V V 2 V 3 V 4, where, by Lemma 4, Thus V = Vol{0 ξ ξ l g l u+ : ξ i i u i} ) l g l u + = Vol{0 θ θ l g : θ i i u ) l g l u + Q l g u, l u + ) = l g )! ) l g l u + u + )g 2 l g)!, V 2 = Vol{ l u s ξ l g ξ l l u+ } = ) g s +, g! V 3 = Vol{ l u ξ l l u+ } =, V 4 = Vol{ξ l+ ξ k : ξ i i u i} ) k l u + l = Q k l0, u + l) k l)! ) k l u + l u + k) 2 k l + )! VolR) s + )g u + )g 2 u + k) 2 g! k k + g)! k + g l g l l u+ i} ) l u + ) l g u + l) k l By Lemma 42 with t = k + g, a = g + u, b = u + k ), the sum on l is ) g e 4 + ) k g k g = k k k g k g + )! 0 k g k k!

17 and the lemma follows INTEGERS WITH A DIVISOR IN y, 2y] 7 To bound U k ), we will bound the olume of the set T k,, γ) = {ξ R k : 0 ξ ξ k, 2 ξ ξ j 2 j γ j k)} Lemma 44 Suppose k,, γ are integers with k 0 and γ 0 Set b = k Then { Y VolT k,, γ)) 2 2b γ k + )!, Y = b if b γ + 5 γ + 5 b) 2 γ + ) if b γ + 4 Proof Let r = max5, b γ) and ξ T k,, γ) Then either 45) ξ j > j γ r j k) or 46) min ξ j j γ ) = ξ j k l l γ [ h, h ] for some integers h r +, l k Let V be the olume of ξ T k,, γ) satisfying 45) If b γ + 5, 45) is not possible, so b γ + 4 and r = 5 By Theorem 4, V Q kγ + 5, ) k! γ + 6)γ + 6 b)2 k + )! If 46) holds, then there is an integer m satisfying 47) m h 3, 2 m < l, ξ 2 l 2 m l γ 2m To see 47), suppose such an m does not exist Then 2 ξ ξ l 2 2 ξ j l/2<j l < 2 2 h 3 2 l γ h+ + 2 l γ, m h 3 Y 2 2b γ k + )! 2 m 2 l γ 2m ) a contradiction Let V 2 be the olume of ξ T k,, γ) satisfying 46) Fix h and m satisfying 47) and use Lemma 43 with u = γ + h, g = 2 m, s = 2m The olume of such ξ is The sum of 2 2m+3 γ + h + )γ + h b)2 k + )! γ + h + )γ + h b)2 2 2m+3 k + )! 20m + 0) 2m 2 m 2)! oer m h 3 is 2 2h Summing oer h r + gies V 2 γ + r + 2)γ b + r + 2)2 2 2r+ k + )! Y 2 2b γ k + )!

18 8 KEVIN FORD Proof of Lemma 34 Assume k, since the lemma is triial when k = 0 Put b = k and define F ξ) = min 0 j k 2 j 2 ξ ξ j + ) For integers m 0, consider ξ R k satisfying 2 m F ξ) < 2 m For j k we hae 2 j 2 ξ ξ j ) max2 j, 2 m 2 j ) 2 m, so ξ T k,, m + ) Hence, by Lemma 44, Next, U k, g) 2 m VolT k,, m + )) k + )! m 0 { b if m b 6 Y m = m + 6 b) 2 m + 2) if m b 5 m 0 2 m Y m = 2 2b m 0 m b 6 b 2 m 2 2b m + m max0,b 5) m 0 2 m Y m 2 2b+t m, m + 6 b) 2 m + 2) 2 m The proof is completed by noting that if b 6, each sum on the right side is b2 b and if b 5, the first sum is empty and the second is 6 b) 2 + b 2 References P Erdős, Some remarks on number theory, Rieon Lematematika 9 955), 45 48, Hebrew English summary) MR 7,460d 2, An asymptotic inequality in the theory of numbers, Vestnik Leningrad Uni 5 960), no 3, 4 49, Russian) MR 23 #A K Ford, The distribution of integers with a diisor in a gien interal, Ann Math 2008), to appear Also aailable on the ArXi, mathnt/ H Halberstam and H-E Richert, Siee methods, Academic Press [A subsidiary of Harcourt Brace Joanoich, Publishers], London-New York, 974, London Mathematical Society Monographs, No 4 MR 54 # R R Hall and G Tenenbaum, Diisors, Cambridge Tracts in Mathematics, ol 90, Cambridge Uniersity Press, Cambridge, 988 MR 90a:07 6 J Riordan, Combinatorial identities, John Wiley & Sons Inc, New York, 968 MR 38 #53 Department of Mathematics, 409 West Green Street, Uniersity of Illinois at Urbana- Champaign, Urbana, IL 680, USA address: ford@mathuiucedu