HIGHER CORRELATIONS OF DIVISOR SUMS RELATED TO PRIMES III: SMALL GAPS BETWEEN PRIMES

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1 Proc London Math Soc C 2007 London Mathematical Society doi:02/plms/pdm02 HIGHER CORRELATIONS OF DIVISOR SUMS RELATED TO PRIMES III: SMALL GAPS BETWEEN PRIMES D A GOLDSTON and C Y YILDIRIM Abstract We use divisor sums to approximate prime tuples and moments for primes in short intervals By connecting these results to classical moment problems we are able to prove that, for any η>0, a positive proportion of consecutive primes are within + η times the average spacing between primes 4 Authors note This paper was written in 2004, prior to the solution, in [8], of the problem considered here In [8] it is shown that Δ = 0 While the main result in Theorem has now been superseded, we believe the method used here is both of interest and future utility in other applications In particular, the work of Green and Tao [2] on arithmetic progressions of primes makes use of Proposition of this paper Introduction Finding mathematical proofs for easily observed properties of the distribution of prime numbers is a difficult and often humbling task, at least for the authors of this paper The twin prime conjecture is a famous example of this, but we are concerned here with the much more modest problem of proving that there are arbitrarily large primes that are unusually close together Statistically this means that we seek consecutive primes whose distance apart is substantially less than the average distance between consecutive primes Let p n denote the nth prime; then by the prime number theorem the average gap size p n+ p n between consecutive primes is log p n Thus we define pn+ p n Δ = lim inf, n log p n so that Δ is the smallest number for which there will be infinitely many gaps between consecutive primes of size less than Δ + ɛ times the average size It is empirically evident that Δ=0, 2 but at the time of writing this has not been proved Up to now three different unconditional methods have been invented which provide non-trivial estimates for Δ The Hardy Littlewood and Bombieri Davenport method In the mid-920s Hardy and Littlewood used the circle method to obtain a conditional result which in 965 Bombieri and Davenport [] both improved and made unconditional This approach can be interpreted as a second moment method using a truncated divisor sum as an approximation of Λn, the von Mangoldt function see the introduction in [9] The method proves that Δ 2 3 Received 8 November 2005; revised 7 March 2007; published online 6 July Mathematics Subject Classification N05 primary, P32 secondary The research of Goldston was supported by NSF; that of Yıldırım was supported by TÜBİTAK

2 654 D A GOLDSTON AND C Y YILDIRIM In the unpublished paper Partitio Numerorum VII, Hardy and Littlewood proved, assuming the Generalized Riemann Hypothesis, that Δ 2 3 In 940 Rankin [20] refined the method of Hardy and Littlewood to show that Δ + 4Θ/5, where Θ is the supremum of the real parts of all the zeros of all Dirichlet L-functions In particular, if the Generalized Riemann Hypothesis Θ = 2 is assumed, this gives Δ The Erdős method By the prime number theorem we have Δ Erdős [4] in 940 was the first to prove unconditionally that Δ < He used the sieve upper bound for primes differing by an even number k, ΛnΛn + k B + ɛskn, 4 n N where Sk is the singular series and B is a constant By this bound there can not be too many pairs of primes with the same difference, and therefore the distribution function for prime gaps must spread out from the average This method gives the result Δ 2B 5 The value B = 4 of Bombieri and Davenport [] see also [] or[3] or B =35 of Bombieri, Friedlander and Iwaniec [2], or even slightly smaller values may be used here The value B =35 only holds for k not too large as a function of N in 4, but this is acceptable for 5 3 The Maier method In 988 Maier [7] found certain rather sparsely distributed intervals where there are e γ more primes than the expected number, and therefore within these intervals the average spacing is reduced by a factor of e γ Hence Δ e γ = In contrast to the first two methods, this method does not produce a positive proportion of small prime gaps These three methods may be combined to obtain improved results Huxley [5, 6] combined the first two methods making use of a weighted version of the first method to find that Δ using B =4, Δ using B =35, 7 and Maier combined his method with Huxley s method with B = 4 to obtain Δ e γ = This last result is the best result known up to now, and as we have seen uses all three of the previously known methods For several years we have been developing tools for dealing with higher correlations of short divisor sums which approximate primes Our first results appeared in [9], and, with considerable help from other mathematicians, we have greatly simplified and improved on these results in [0] In the former paper we had an application to small gaps between primes based on approximating a third moment In particular, we recovered the result 3 The method is based on the same approximation that underlies the method of Bombieri and Davenport, but it detects small prime gaps in a different way In this paper we extend that argument to all moments and obtain the limit of this method Let πn denote the number of primes less than or equal to N

3 HIGHER CORRELATIONS OF DIVISOR SUMS III 655 Theorem Let r be any positive integer For any fixed λ> r 2 2 and N sufficiently large, we have r πn 9 p n N p n+r p n λ log p n In particular, for any fixed η>0 and all sufficiently large N>N 0 η, a positive proportion of gaps p n+ p n for p n N are less than 4 + η log N, and Δ 4 0 Our results depend on the level of distribution of primes in arithmetic progressions, and Theorem makes use of the Bombieri Vinogradov theorem If for primes up to N the level of distribution in arithmetic progressions is assumed to be N ϑ ɛ for any ɛ>0, then Theorem holds with λ> r ϑ/2 2 Hence, assuming the Elliott Halberstam conjecture that ϑ = holds, we obtain the improved result that lim inf n pn+r p n log p n r 2 2, and, in particular, Δ = < 2 There are several improvements that can be made in our results First, we can incorporate Maier s method into our method This is a straightforward adaptation of the argument Maier used to combine his method with Huxley s result, although the result is complicated by the need to prove our propositions in the next section when they are summed over arithmetic progressions Second, and more significantly, we have found in joint work with J Pintz better approximations for prime tuples than those used in this paper, and these lead to significantly stronger results These results will appear in future papers This paper is organized as follows In Section 2 we present our method and state the two main propositions needed in the proof In Section 3 we prove some lemmas which are used in the later sections In Sections 4 and 5 we prove the propositions In Section 6 we examine an optimization problem related to the Poisson distribution which is used in the proof of Theorem, and finally in Section 7 we prove Theorem Notation We will take ɛ to be any sufficiently small positive number whose value can be changed from equation to equation, and similarly C, c, and c will denote small fixed positive constants whose value may change from equation to equation We will let A denote a large positive constant which may be taken as large as we wish, but is fixed throughout the paper For a finite set A we let A denote the number of elements in A We will sometimes write A = A k if A = k For a vector H we denote the number of components by H A dash in a summation sign indicates that all the summation variables are relatively prime to each other, and further any sum without a lower bound on the summation variables will have the variables start with the value Empty sums will have the value zero, and empty products will have the value We will make use of the Iverson notation that, for a statement P,[P]isif P is true and is 0 if P is false Acknowledgements We are indebted to Andrew Granville and Kannan Soundararajan who have greatly clarified and simplified our method We have made extensive use of their results here We have also benefited from ideas of Enrico Bombieri, Brian Conrey, Percy Deift, David Farmer, John Friedlander, Roger Heath-Brown, Hugh Montgomery, Yoichi Motohashi, Michael Rubinstein, Peter Sarnak, and Jimena Sivak We have used work of Dashiell Fryer,

4 656 D A GOLDSTON AND C Y YILDIRIM an undergraduate student at San Jose State University in the MARC program, who worked on properties of Laguerre polynomials needed in our method The first-named author also thanks the American Institute of Mathematics where much of the collaboration mentioned above took place In a recent preprint [2] Ben Green and Terence Tao proved a landmark result on arithmetic progressions of primes One tool they used was the current Proposition from an earlier not widely distributed preprint of this paper They corrected an oversight in our original proof which we have incorporated into our Lemma 2 and the proof of Proposition 2 Approximating prime tuples Our approach for finding small gaps between primes is to compute approximations of the moments for the number of primes in short intervals, and this computation uses short divisor sums to approximate prime tuples Given a positive integer h, let H = {h,h 2,,h k }, with 0 h,h 2,,h k h distinct integers, 2 and let ν p H denote the number of distinct residue classes modulo p that the elements of H occupy We define the singular series SH = k ν ph 22 p p p If SH 0 then H is said to be admissible ThusH is admissible if and only if ν p H <pfor all p Letting Λn denote the von Mangoldt function, define Λn; H =Λn + h Λn + h 2 Λn + h k 23 The Hardy Littlewood prime tuple conjecture [4] states that for H admissible, Λn; H =N SH+o, as N 24 n N This is trivially true if H is not admissible We approximate Λn as in our earlier work [6] by using the truncated divisor sum and then approximate Λn; H by Λ R n = μd log R d, 25 d n d R Λ R n; H =Λ R n + h Λ R n + h 2 Λ R n + h k 26 For convenience we also define Λ R n; H =ifh = Our method is founded on the following two propositions which allow us to obtain information about primes Suppose H and H 2 are both sets of distinct positive integers that are less than or equal to h, with H = k and H 2 = k 2, and let k = k + k 2 We always assume k Proposition Let H = H H 2 and r = H H 2 IfR = on /k log R H /k and h R A for any large constant A>0, then we have for R, N, Λ R n; H Λ R n; H 2 =N SH+o k log R r 27 n N

5 HIGHER CORRELATIONS OF DIVISOR SUMS III 657 Proposition 2 Let H = H H 2, r = H H 2, and h 0 h Let H 0 = H {h 0 }, and r 0 = r if h 0 Hand r 0 = r + if h 0 HIfR k N /2k log N Bk for a sufficiently large positive constant Bk, and h R /2k, then we have for R, N, Λ R n; H Λ R n; H 2 Λn + h 0 =N SH 0 +o k log R r0 28 n N If the Elliott Halberstam conjecture is assumed, then equation 28 holds for R k N /k ɛ with any ɛ>0 The restriction on the size of R in Proposition 2 can be improved in the situation when h 0 H H 2 Ifweletk = k H {h 0 } H 2 {h 0 }, then we see that the reduction of cases at the start of the proof of Proposition 2 implies that Proposition 2 holds in the range R k N /2k log N Bk except in the trivial case when k = 2 and k = 0 where the result holds for R N In the case of the Elliott Halberstam conjecture, k can also be replaced by k in the bound for R We actually prove both propositions with the error term o k replaced by a series of lower order terms, which however are not needed in any of our applications If we take H 2 = in Proposition, we have, for R = on /k and h R A for any given constant A>0, that for R, N, Λ R n; H =N SH+o k, 29 n N in agreement with the Hardy Littlewood prime tuple conjecture 24 In applying these propositions it is critical to have some form of positivity in the argument For example, in the special case when H 2 =, Proposition 2 takes the form, for R N /2k log N Bk, { N SH+o k log R if h 0 H, Λ R n; HΛn + h 0 = N SH 0 +o k 20 if h 0 H, n N which would seem to exhibit that our approximation detects primes However, since Λ R n; H is not non-negative, it is impossible to conclude anything about primes from 20 alone On the other hand, consider instead the special case of Proposition 2 where H = H 2 = H which gives, on taking H = k, for R N /4k log N Bk, { Λ R n; H 2 N SH+o k log R k+ if h 0 H, Λn + h 0 = N SH 0 +o k log R k 2 if h 0 H n N The restriction on the size of R here makes it impossible to conclude from 2 that any given tuple H will contain two or more primes, but Granville and Soundararajan found a simple argument which uses the non-negativity of Λ R n; H 2 to prove from 2 that Δ To prove their result, we need a formula of Gallagher that as h, SH =h k + O k h k /2+ɛ 23 h,h 2,,h k h distinct Granville unpublished and Montgomery and Soundararajan [9] have recently proved more precise results, but these are not needed here We fix k ; the argument works equally well for any k, and we can take k = if we wish Suppose now that R = N /4k log N Bk, h log N

6 658 D A GOLDSTON AND C Y YILDIRIM By differencing, equation 2 continues to hold when the sum on the left-hand side is over N<n 2N, and therefore we have 2N Λn + h 0 Λ R n; H 2 n=n+ Also by Proposition, h 0 h knshlog R k+ + 2N n=n+ h 0 h h 0 h i, i k Λ R n; H 2 NSHlog R k, NSH 0 log R k and therefore we find, on summing over all distinct tuples h,h 2,,h k h and applying 23, that, for ρ a fixed number and h, N, 2N Λn + h 0 ρ log N Λ R n; H 2 Since n=n+ h 0 h h,h 2,,h k h distinct Nh k log R k k log R + h ρ log N Nh k log R k h ρ 4 log N Λ R n dn log R n ɛ, we see that the contribution in the sum above from terms where n + h 0 is a prime power is N /2+ɛ which is negligible, and therefore we may restrict the sum over h 0 to terms where n + h 0 is prime The right-hand side above is positive if h>ρ 4 log N, which implies with this restriction on h that there is a value of n, with N<n 2N, such that logn + h 0 >ρlog N h 0 h n+h 0 prime If ρ>, this implies that for N sufficiently large there are at least two terms in this sum, and thus by taking ρ + we obtain 22 The proof of Theorem is a refinement of the above argument, where we detect primes by the square of the linear combination of tuple approximations k a 0 + a j Λ R n; H j 24 j= h,h 2,,h j h distinct Here the a j are available to optimize the argument While it is possible to use 24 directly, we have chosen in the proof of Theorem to first approximate moments, which highlights the Poisson model which the prime numbers are thought to satisfy This method also has the advantage of simplifying the combinatorics that occur in the problem The moment method leads to an optimization problem which is familiar in the theory of orthogonal polynomials, the solution of which was provided to us by Enrico Bombieri and Percy Deift The final result that we obtain depends on the asymptotics of the smallest zero of a certain sequence of Laguerre polynomials; these results are obtained by Sturm comparison type theorems and have appeared in the literature; Michael Rubinstein first pointed these out to us

7 HIGHER CORRELATIONS OF DIVISOR SUMS III Lemmas We recall the Riemann zeta-function defined for Res > by ζs = p s 3 p The zeta-function is analytic everywhere except for a simple pole with residue at s =, and therefore ζs 32 s is an entire function We need to use a classical zero-free region result By [23, Theorem 3 and 38] there exists a small positive constant C such that ζσ + it 0 in the region for all t, and further σ C log t +2 ζσ + it σ +it log t +2, log t +2, 34 ζσ + it in this region Let c denote the contour s = c + it, with <t<, and let L denote the contour given by C s = + it 35 log t Lemma We have, for R 2 and c>0, R s 2πi c ζ + s s 2 ds =+O e c log R, 36 and, for any fixed constant B, and L log s +2 B R s ds s 2 / log R log s +2 B R s ds s 2 e c log R, 37 log R 38 Proof We first prove 37 The integral to be bounded is, for any w 2, B log t +2 R C/ log t +2 t + C 2 dt w log w B R C/ logt+2 log t B dt + 0 w t 2 dt wlog w B e C log R/ log w log wb +, w and on choosing log w = 2 C log R we see that this is C log R B/2 e C log R/2 e c log R, which proves 37 To prove 36, we note that by the second bound in 34 the integrand in 36 vanishes as t in the region to the right of L, and therefore we can move the contour c to the left

8 660 D A GOLDSTON AND C Y YILDIRIM to L, pass the simple pole at s = 0 with residue, and obtain R s R s ds =+ 2πi c ζ + s s2 2πi L ζ + s s 2 ds On L we deduce from 34 that /ζ + s log t + 2, and therefore we may use the estimate 37 to obtain 36 Finally, the left-hand side of 38 is B log t +2 log R 2 dt + t / log R t >/ log R t 2 dt log R Lemma 2 Let f R s,s 2 be analytic in the strip B σ,σ 2 b for some positive constants B and b, and suppose also that, for any ɛ>0, f R s,s 2 e ɛ log R in this strip as t, t 2 ForR 2 and 0 <c,c 2 c, let UR = ζ + s + s 2 R s+s2 2πi 2 f R s,s 2 c 2 c ζ + s ζ + s 2 s 2 s 2 ds ds Then UR =f R 0, 0 log R + C R + O e c log R, 30 where C R = f R 0, 0 + ds f R s, s s 2 2πi L 2 ζ + sζ ss 4, 3 where L 2 is defined in 32 and 33 below Proof One would expect to proceed by moving both contours to the left to L There is, however, a complication because the integrand now contains the function ζ + s + s 2 which necessitates that also s + s 2 be restricted to the region to the right of L if we wish to use the bounds in 34 This was pointed out to us by J Sivak and also Y Motohashi This problem was handled in similar ways in [2] and in [7] We follow here our method in [8] Let V = e log R 32 and define the contours, for j = or 2, 4 L j { c 4 j } c j = = + it: <t<, log V log V { 4 j } c L j = log V + it: t 4 j V, { } L j = 4 j c log V + it: t 4 j V, H j = { σ j ± i4 j V : σ j 4 j c log V Now, by the bound for f R and 34, the integrand of UR is e ɛ log R log t + 2 log t max, s + s 2 R σ+σ2 s 2 s 2 2 } 33 provided s, s 2, and s + s 2 are on or to the right of L Thusifs and s 2 are to the right of L 2, this integrand vanishes as t or t 2, and we may shift the contours c and c 2 to L and L 2, respectively, without changing the value of UR Next, we truncate these contours so that they may be replaced with L and L 2 ; the error introduced by this is, since

9 s + s 2 5c/6 log V on these contours, HIGHER CORRELATIONS OF DIVISOR SUMS III 66 log V e ɛ log R R 5c/6 log V log t +2 2 c/6 log V +it 2 dt log t 2 V/6 t 2 dt log V 4 log e c R V 5c/6 ɛ Hence UR = ζ + s + s 2 R s+s2 2πi 2 f R s,s 2 L 2 L ζ + s ζ + s 2 s 2 s 2 ds ds 2 + O e c log R 34 2 To replace the s -contour along L with the contour along L, we consider the rectangle formed by L, H and L which contains poles of the integrand as a function of s at s =0 and s = s 2 Hence we see that UR = ζ + s + s 2 R s+s2 Res f R s,s 2 ds 2πi L s 2 =0 ζ + s ζ + s 2 s 2 s ζ + s + s 2 R s+s2 Res f R s,s 2 ds 2πi L s 2 = s 2 ζ + s ζ + s 2 s 2 s ζ + s + s 2 R s+s2 2πi 2 f R s,s 2 L 2 L H ζ + s ζ + s 2 s 2 s 2 ds ds 2 + O e c log R 2 = f R 0,s 2 Rs2 2πi L 2 s 2 ds 2 + ds 2 f R s 2,s 2 2 2πi L 2 ζ s 2 ζ + s 2 s ζ + s + s 2 R s+s2 2πi 2 f R s,s 2 L H ζ + s ζ + s 2 s 2 s 2 ds ds 2 + O e c log R 2 L 2 = I + I 2 + I 3 + O e c log R 35 Here the contours along L and H are oriented clockwise To evaluate I, we consider the rectangle formed by L 2, H 2 and L 2 which contains a double pole at s 2 = 0, and obtain, by 37 of Lemma and the bound for f R and 34, I = f R 0, 0 log R + f R s 2 0, 0 + 2πi L 2 H 2 f R 0,s 2 Rs2 s 2 2 ds 2 = f R 0, 0 log R + f 0, 0 + O e c log R 36 s 2 Since I 2 is included in C R, to complete the proof of Lemma 2 we only need to show that I 3 e c log R This is done in the same way as we handled I ; we consider the rectangle formed by L 2, H 2 and L 2, which however in this case contains a simple pole at s 2 = 0, and obtain I 3 = f R s, 0 Rs 2πi L 2 H 2 s 2 ds ζ + s + s 2 R s+s2 + f R s,s 2 L H L 2 H 2 ζ + s ζ + s 2 s 2 s 2 ds ds By 37 of Lemma and the bound for f R and 34 both of these integrals are e c log R which completes the proof Let 4 Proof of Proposition κ = H H 2, r = H H 2, k = H + H 2 = k + k 2, 4

10 662 D A GOLDSTON AND C Y YILDIRIM and therefore 0 r, k,k 2 κ and κ = k r 42 Next, without loss of generality, we take H =h,h 2,,h k, H = {h,h 2,,h k }, H 2 = {h k+,h k+2,,h k }, H H 2 = {h,h 2,,h r }, h = h k, h 2 = h k,, h r = h k r+ = h κ+, H := H H 2 = {h,h 2,,h κ } Here r = 0 when H H 2 = and the fourth and fifth lines in 43 may be removed With this notation we have S k H, H 2 = n N Λ R n; H Λ R n; H 2 43 k Let = n N = d,d 2,,d k R d i n+h i, i k d,d 2,,d k R i= μd i log R d i k μd i log R d i i= n N d i n+h i, i k 44 D k =[d,d 2,,d k ], 45 the least common multiple of d,d 2,,d k The sum over n above is zero unless d i,d j h j h i, for i<j k, in which case the sum runs through a unique residue class modulo D k, and we have = N + O 46 D k Hence S k H, H 2 =N n N d j n+h j, j k d,d 2,,d k R d i,d j h j h i, i<j k D k j= k μd j log R + OR k d j = NT k H, H 2 +OR k 47 We next decompose the d i into relatively prime factors Let Pk be the set of all non-empty subsets of the set of k elements {, 2,,k} This is just the power set with the empty set removed For B Pk, we let P B k denote the set of all members of Pk for which B is a subset Thus for example if k = 4 then P {,2} 4 = {{, 2}, {, 2, 3}, {, 2, 4}, {, 2, 3, 4}} Since the d i are squarefree we can decompose them into the relatively prime factors d i = a ν, for i k, 48 ν P {i} k

11 HIGHER CORRELATIONS OF DIVISOR SUMS III 663 where a ν is the product of all the primes that precisely divide all the d i for which i ν, and none of the other d i This decomposition is unique and the 2 k factors a ν are pairwise relatively prime to each other We next denote by DH the divisibility conditions d i,d j = ν P {i,j} k a ν hj h i, for i<j k, 49 and have T k H, H 2 = d,d 2,,d k R DH ν Pk μa ν ν k log R 40 a ν d j j= We now apply the formula, for c>0, { c+i x s 2πi c i s 2 ds = 0 if 0 <x, log x if x, 4 and have where T k H, H 2 = 2πi k c k c 2 c F s,s 2,,s k = a ν,ν Pk DH F s,s 2,,s k ν Pk k j= R sj s j 2 ds j, 42 μa ν ν a ν +τ ν, 43 and τ ν = j ν s j 44 Y Motohashi has pointed out to us that we could also define F s,s 2,,s k = p ɛ μp ɛ μp ɛ k p ɛ p ɛ k p ɛ s ++ɛ k s k where the subscript ɛ indicates that we sum over the values 0 ɛ j, for j k, with the conditions p ɛi,p ɛj h i h j, for i<j k Using this definition would simplify some of the calculations in this section We next consider the divisibility conditions DH The variables a ν indexed by the singleton sets ν = {j}, with j k, are not constrained by these divisibility conditions, and therefore can contain any prime as a factor Further, if r, then h j h i = 0 for j = k i + and i r Thus these constraints drop out of DH and the unconstrained variables are both the singleton sets ν = {j}, with j k, and also the doubleton sets ν = {i, k i +}, with i r If r = 0 there are none of these doubleton sets The remaining a ν are constrained by at least one of the divisibility relations, and therefore must divide some h j h i so that they can only contain prime factors less than or equal to h We therefore see that we can write

12 664 D A GOLDSTON AND C Y YILDIRIM F s,,s k as the Euler product, for σ j > 0, j k, F s,,s k = k p + f Hp; s,s 2,,s k +sj p h j= k r p + +sj p +sj+s, 45 k j+ p>h j= j= where ν f H p; s,s 2,,s k = 46 +τν p ν Pk, ν 2 p h j h i for all i,j ν Factoring out the dominant zeta-factors we write r ζ + s j + s k j+ F s,s 2,,s k =G H s,s 2,,s k ζ + s j ζ + s k j+ and proceed to analyze G H Let Δ:= j= i<j κ κ j=r+ ζ + s j, 47 h j h i h k2, 48 so that this product is over all the non-zero differences of h i and h j for i<j k Here of course Δ is not the same function as in the first section From the discussion above 45, r f H = unless p Δ, and therefore where Let G H s,s 2,,s k = k p Δ j= k j= p +sj+s k j+ j= p +sj + f Hp; s,s 2,,s k p +sj r hs,s 2,,s k = k p Δ j= j= s = k p +sj j= p +sj+s k j+ p +sj r j= + r hs,s 2,,s k, 49 p +sj+s k j+ j= p +sj+s k j+ 420 k minσ j, 0 42 j= Taking σ j 5, for j k, wehave hs,s 2,s k k +O k p 6/5 k, 422 p

13 and thus in this region we have HIGHER CORRELATIONS OF DIVISOR SUMS III 665 G H s,s 2,,s k k p Δ k exp ak +O k p s p s p k 2 log2h k exp akk 2 log2h s p k 2 log2h p k exp bklog2h s log log log 2h, 423 where the sum which was originally over p Δ has been majorized by replacing these primes by the primes 2, 3,,p m with m = νδ and using the fact that 2 3 p m Δ and 48 to see by the prime number theorem that p m k 2 log2h By this bound and 47 wesee that if r then F has simple poles at s i + s k i+ = 0, for i r By34, for s i with i k, and s i + s k i+ with i r, to the right of L, F s,s 2,,s k k exp bklog2h s log log log 2h k r log 2 t j +2 j= i= max, 424 s i + s k i+ We are now ready to begin the evaluation of T k H, H 2 By 424 we see that the integrand in 42 goes to zero as any one of the variables t j We will first move successively the contours c j, for r + j κ, tol; by47 these correspond to the cases where the integrand has only a simple pole at s j =0Ifr = κ there are none of these terms and we skip ahead to 428 Thus, moving c r+ to L and passing a simple pole at s r+ = 0 we obtain T k H, H 2 = 2πi k + 2πi k k j= c j j r+ G Hs,s 2,,s k sr+=0 κ j=r+2 R sj ζ + s j s 2 ds j j r ζ + s j + s k j+ R sj+s k j+ ζ + s j ζ + s k j+ s 2 j s k j+ 2 ds j ds k j+ k j= j= j r+ c j L F s,s 2,,s k k j= R sj s j 2 ds j 425 We bound the second multiple integral on the right by moving all the contours c j, with j r +,to/ log R which leaves the value of the integral unchanged If s j and s k j+ are on / log R, then ζ + s j + s k j+ ζ + s j ζ + s k j+ log R log2 + s j + s k j+ log2 + s j log2 + s k j+ 426

14 666 D A GOLDSTON AND C Y YILDIRIM In the multiple integral s = σ r+ C/log 2 for σ r+ on L, and we take a fixed C< 2 log 2 Then by 47, 424, 426 and Lemma, the second term in 425 is k exp bklog2h /2 κ r log 2 s +2 R s ds / log R s 2 log R r 2r log 2 s +2 R s ds s 2 log 2 s +2 R s ds s 2 / log R k exp bklog2h /2 log R κ+2r e c log R k e k c log R, 427 where we used log 2h log R for the last line We continue this process, next moving c r+2 tolin the first term, and estimating the secondary term as above, and so on successively through the contours up to c κ Hence we conclude that T k H, H 2 =U r + O k e c k log R, 428 where r U r = 2πi 2r G s,s 2,,s r,s κ+,s κ+2,s k and j= c k j+ c j r j= L ζ + s j + s k j+ R sj+s k j+ ζ + s j ζ + s k j+ s j 2 s k j+ 2 ds j ds k j+, 429 G s,s 2,,s r,s κ+,s κ+2,,s k = G H s,s 2,,s r, 0, 0,,0,s κ+,s κ+2,,s k 430 We will now prove that r U r = G H 0, 0,,0log R r + A j Hlog R r j + O k e c k log R, 43 j= where the A j H are explicitly computable arithmetic functions which for h R A with any A>0 satisfy the bound A j H k log log 2h bk 432 We will also prove at the end of this section that G H 0, 0,,0 = SH 433 From these results Proposition follows The multiple integral in U r would decouple into a product of double integrals evaluated in Lemma 2 if G were a constant, but since this is not the case, we need to apply Lemma 2 inductively To do this we need estimates for the partial derivatives of G H Let a =a,a 2,,a k, and define D a G H = a a2 a k s a s a 2 2 G s a H s,s 2,,s k 434 k k We have, for σ j > c k, with j k, D a G H k log log 2h b k exp bklog2h s log log log 2h 435

15 HIGHER CORRELATIONS OF DIVISOR SUMS III 667 To obtain these estimates, we logarithmically differentiate G H to see that G H k G H log p s p s p k 2 log2h The sum above is bounded by k 2 log p log2h s k 2 log2h s logk 2 log2h, p p k 2 log2h and 435 follows in this case by 423 By the product rule, further partial derivatives will satisfy the above bound with the sum having log p replaced by log p ck, which only changes the value of b k in435 We first consider the case r = in 429 By Lemma 2 applied with f R = G we see by 423 that the conditions of the lemma are satisfied and therefore U = G H 0, 0,,0 log R + A + O k e c k log R, where A = G 0, 0,,0 + ds G s,s s k 2πi L 2 ζ + s ζ s s 4 It remains to prove that A satisfies 432 By 435 the first term in A satisfies this bound In the integral term we move the contour L 2 to the imaginary axis with a semicircle of radius δ =/ logk 2 log2h to the right of the double pole at s = 0 Using 34 and 423 wesee that the part of the integral over the contour on the imaginary axis is bounded by k log log 2h bk logt log log 2h bk δ t 4 dt k δ 3 k log log 2h b k, and the integral over the contour on the semicircle is bounded by log log 2h3bk k πδ δ 2 k log log 2h b k, which completes the proof for r = For the general case of 429, we move all the contours to / log R and apply Lemma 2 for the double integral over s and s k to obtain U r = 2πi 2r 2 / log R / log R r j=2 G log R + G 2 + O r e c s=0 r s k =0 log R ζ + s j + s k j+ R sj+s k j+ ζ + s j ζ + s k j+ s k j+ 2 s j 2 ds j ds k j+ = U r log R + U r + O r e c r log R, 436 where the error term was estimated using 426 asin427 Here G 2 s 2,s 3,,s r,s κ+,s κ+2,,s k = G 0,s 2,s 3,,s r,s κ+,s κ+2,,s k, 0 s k + ds G s,s 2,,s r,s κ+,,s k,s 2πi L 2 ζ + s ζ s s 4, and therefore U r is of the same form as U r with G replaced by a partial derivative of G or an absolutely convergent integral of G with respect to the variable s when s k = s As we saw in the case r =, both of these terms satisfy 432 and 435 We now apply Lemma 2

16 668 D A GOLDSTON AND C Y YILDIRIM for the double integral over s 2 and s k, and continue this process until all the variables are exhausted We thus arrive at 43 and the bound 432 follows by 435 and the argument used in the case r = It remains to prove 433 By 49 and 420 wehave G H 0, 0,,0 = k p + f Hp;0, 0,,0 κ κ κ, p p p p Δ where, by 46, f H p;0, 0,,0 = p Therefore by 22 we need to prove that ν Pk, ν 2 p h j h i for all i,j ν ν Pk, ν 2 p h j h i for all i,j ν p Δ ν ν = k ν p H 437 If ν p H =q, then h,h 2,,h k must fall into q distinct residue classes, say r,r 2,,r q mod p Let M p l ={m: h m r l mod p}, for l q Thus given p, the sets M p l, with l q, give a disjoint partition of the set {, 2,,k}, and therefore q M p l = k 438 l= The conditions p h j h i hold if and only if h i and h j are in the same residue class modulo p and thus if and only if i and j are in M p l for some l Hence the ν Pk which will satisfy p h j h i for all i, j ν are precisely the subsets of M p l with at least two elements We conclude, using 438, that M p = ν Pk, ν 2 p h j h i for all i,j ν q {ν : ν M p l, ν 2} l= ν = = = = = ν M p ν q l= ν M pl, ν 2 q l= M pl j=2 ν j Mp l j q + Mp l l= q M p l q l= = k ν p H

17 HIGHER CORRELATIONS OF DIVISOR SUMS III Proof of Proposition 2 We first reduce the proof to the special case when h 0 H= H H 2 Let SH, H 2,h 0 = Λ R n; H Λ R n; H 2 Λn + h 0 5 n N Since trivially Λ R n dn log R, we see that, for i =, 2, Λ R n, H i dn log R ki, 52 and since dn n ɛ and, in Proposition 2, R N /2,wehave SH, H 2,h 0 = Λ R n; H Λ R n; H 2 Λn + h 0 +OR +ɛ = R<n N R<n N n+h 0 prime Λ R n; H Λ R n; H 2 Λn + h 0 +ON /2+ɛ, 53 where we have removed the prime powers in the last line If n + h 0 is a prime greater than R then its only divisor less than or equal to R is d =, and therefore Thus, if h 0 H H 2, SH, H 2,h 0 = log R 2 if h 0 H H 2, SH, H 2,h 0 = log R and if h 0 H H 2, SH, H 2,h 0 = Λ R n + h 0 Λn + h 0 =Λn + h 0 log R R<n N n+h 0 prime R<n N n+h 0 prime R<n N n+h 0 prime Λ R n; H {h 0 }Λ R n; H 2 {h 0 } logn + h 0 +ON /2+ɛ ; Λ R n; H {h 0 }Λ R n; H 2 logn + h 0 +ON /2+ɛ ; Λ R n; H Λ R n; H 2 logn + h 0 +ON /2+ɛ In these sums we may once again include the terms that are less than or equal to R and the prime powers if we wish with the same error term, and therefore in each situation we have reduced the proof to the case when h 0 is distinct from the other h i Henceforth we can therefore take h 0 H 54 Proceeding to the proof, we have SH, H 2,h 0 = d,d 2,,d k R k μd i log R d i i= n N d i n+h i, i k Λn + h 0 55 By the Chinese Remainder theorem, the sum will run through an arithmetic progression modulo D k provided d i,d j h j h i, for i<j k, and will be empty otherwise As in 49 we denote these conditions by DH Using Iverson notation, we let x ψx; q, a := Λn =[a, q =] + Ex; q, a, 56 n x n amod q φq

18 670 D A GOLDSTON AND C Y YILDIRIM and have n N d i n+h i, i k Λn + h 0 =[DH] ψn + h 0 ; D k,a ψh 0,D k,a =[DH]ψN; D k,a+oh log N, 57 where a is an integer satisfying the congruence relations a h 0 h j mod d j, for j k The term ψn; D k,a has a non-zero main term if a, D k =, which is equivalent to d j,h j h 0 =, for j k, 58 and, if a, D k > then ψn; D k,a log N 2 ;thus k SH, H 2,h 0 =N μd i log R φd k d i d,d 2,,d k R DH d j,h j h 0=, j k + O d,d 2,,d k R i= k μ 2 d i log R d i i= max amod D k a,d k = EN; D k,a + O R k hlog N 2 = N T k H, H 2,h 0 +OE k +O R k hlog N 2 59 We handle the error term E k with the Bombieri Vinogradov theorem First, we have E k log R k μ 2 D k max EN; D k,a log R k d,d 2,,d k R q R k μ 2 q max amod q a,q= amod D k a,d k = EN; q, a q=d k d,d 2,,d k R Given q, the number of ways to write q = D k that is, write q as the least common multiple of k squarefree numbers is bounded by dq k, since each of the k numbers in the least common multiple must be a divisor of q Applying Cauchy s inequality, we have E k log R k μ 2 qdq k max EN; q, a amod q q R k a,q= 2k log R k dq q q max EN; q, a 2 q R k q R k We now use the estimate amod q a,q= dn k k xlog x 2k n x and the trivial estimate EN; q, a N log N/q for q N to conclude that E k k log R 4k +k N log N max EN; q, a q R k amod q a,q= 50 By the Bombieri Vinogradov theorem the last sum over q is N/log N A for any A>0 provided that R k N /2 log N B, 5

19 HIGHER CORRELATIONS OF DIVISOR SUMS III 67 where B = BA We conclude under this condition that E k k Nlog N 4k +k+/2 A/2 = o k N 52 if A>24 k + k + 2 To complete the proof of the proposition we will now prove that, for R k N and h R /2k, T k H, H 2,h 0 =T k+ H {h 0 }, H 2 +O k e c k log R, 53 which by 428, 432, and 433 completes the proof To prove 53, we have T k+ H {h 0 }, H 2 = = = d 0,d,,d k R d i,d j h j h i, 0 i<j k d,d 2,,d k R d i,d j h j h i, i<j k d,d 2,,d k R DH [d 0,D k ] k μd j log R d j k μd j log R d j= j d 0 R d 0,d j h j h 0, j k μd 0 [d 0,D k ] log R d 0 k μd j log R T H {h 0 }, H 2 54 d j= j On letting g =d 0,D k where d 0 = gd, we see that [d 0,D k ]=D k d and d,d k =Thus T H {h 0 }, H 2 = d R d,m= gd R g D k g,d j h j h 0, j k d,d k = = D k μgd d D k g R g D k g,d j h j h 0, j k μg log R gd d R/g d,d k = μd d log R/g d For log m log R we have by the prime number theorem or see [9, Lemma 2] μd d log R d = m φm + O e c log R 55 Applying this and dropping the redundant condition g R since g k j= g, d j h k R when h R /k, we see that T H {h 0 }, H 2 = φd k g D k g,d j h j h 0, j k ddk μg+o e c logr/h k D k We now claim that, using Iverson notation, μg =[d j,h j h 0 =, j k] g D k g,d j h j h 0, j k One way to see this is through the decomposition of the d i into relative factors 48 from which we see that we can write g = ν Pk b ν, where b ν a ν with the b ν pairwise relatively

20 672 D A GOLDSTON AND C Y YILDIRIM prime to each other Then the sum becomes We conclude that k j= ν P {j} k b ν d j,h j h 0 μb ν =[d j,h j h 0 =, j k] T H {h 0 }, H 2 = φd k [d j,h j h 0 =, j k]+o k and on substituting this result into 54 wehave k T k+ H {h 0 }, H 2 = φd k d,d 2,,d k R Dk d j,h j h 0=, j k + O k log R k e c log R/h k The first term is T k H, H 2,h 0 and, by 50, dd k dq D k q d,d 2,,d k R q R k dqk+ q q R k Thus the error term is which proves 53 ifh R /2k k log R 2k+ ddk e c logr/h k, D K μd j log R d j= j d,d 2,,d k R q=d k k log R 2k+ +k e c logr/h k k e c k log R, dd k D k 6 Optimization of a quadratic form related to the Poisson distribution The content of this section and the proof given here was provided to us by E Bombieri and P Deift The final tool we need for our proof of Theorem is an optimization procedure related to the Poisson distribution Let X be a Poisson random variable with expected value λ, defined by the discrete probability density function λ λj pj = ProbX = j =e, for j =0,, 2, 6 j! We define an inner product with respect to this density function by fx,gx = fjgjpj = e λ fjgj λj j! 62

21 HIGHER CORRELATIONS OF DIVISOR SUMS III 673 The kth moment of the Poisson distribution is defined by μ k λ =Ex k = x k, = e λ j k λ j j! = e λ λ d k λ j dλ j! = e λ λ d k e λ 63 dλ More explicitly, we have k { } k μ k λ = λ ν, 64 ν ν= where { ν k } denotes the Stirling numbers of the second type, defined to be the number of ways to partition a k-set that is, a set with k elements into ν non-empty subsets not counting the order of the subsets It is easy to see that { k ν } = ν ν=0 { k ν } + { } k ν since the last element in our k-set either is put into its own singleton set or is put into one of the ν subsets which contain some of the earlier elements To prove 64 we use the identity j { } k j ν! = j k 66 ν ν This identity arises from counting the number of partitions of a k-set into at most j sets, where the order of these sets is counted On one hand there are j choices for where to place each of the k elements, so this number is j k, while on the other hand, if ν of these j sets are non-empty, then there are ν! { ν k } such partitions and ν j ways to choose the ν non-empty sets Rewriting 66 in the form j k j j! = ν=0 { } k ν j ν!, multiplying by λ j e λ and summing over j, we obtain, by 63, μ k λ =e λ = e λ = k ν=0 j k λ j λ j j! j { k ν} j ν! ν=0 { k ν} λ ν, by interchanging the j and ν summations, which proves 64 Our method for finding small gaps between primes leads us to define a second bilinear form given by fx,gx ρ = x ρ, fxgx = j ρfjgjpj 67 65

22 674 D A GOLDSTON AND C Y YILDIRIM where ρ is a real number This is not an inner product because it is not necessarily nonnegative Letting a =a 0,a,a 2,,a k, consider and the associated quadratic form P a x = k a i x i, 68 i=0 where we define Q = Q a λ, ρ = P a x,p a x ρ = a i a j x ρ, x i+j 0 i,j k = a i a j μ i+j+ λ ρμ i+j λ 0 i,j k = a i a j c i+j, 69 0 i,j k c m = c m λ, ρ =μ m+ λ ρμ m λ 60 The optimization problem we need to solve is to maximize Q over all vectors normalized by a k = when ρ>0 is fixed The solution involves the generalized Laguerre polynomials defined for α> by n n + α x L α n x = ν ν n ν ν! 6 ν=0 The zeros of the Laguerre polynomials are real, positive, and simple see [22, Chapter 6] We denote the smallest zero of L n α x byx n, α The solution of our problem is obtained in the following proposition Proposition 3 For each k and ρ>kfixed, we have, for 0 <λ<x k +,ρ k, ρ k λ max Q aλ, ρ = k +!λ k L k+ a k = L ρ k k λ Thus, for each k and ρ>k, 62 inf {λ >0: Q a λ, ρ > 0,a k =} = x k +,ρ k 63 The proof of this proposition will ultimately reduce to evaluating the determinant c 0 c c 2 c k c c 2 c 3 c k+ D k = det c 2 c 3 c 4 c k+2 = det [c i+j ] i=0,,2,,k 64,,2,,k c k c k+ c k+2 c 2k The solution of the optimization problem can be obtained by choosing a so that P a x is orthogonal to all lower degree polynomials with respect to, ρ Thus we consider the k equations P a x,x i ρ =0, for i =0,, 2,,k, 65 and prove the following lemma

23 HIGHER CORRELATIONS OF DIVISOR SUMS III 675 Lemma 3 If D k 0for a given λ, then there is an explicitly obtained vector a with a k =which satisfies 65 and for which Proof Q a λ, ρ = D k D k 66 We take a k = Equation 65 is equivalent to the equations k a j c i+j =0, for i =0,, 2,,k 67 If a satisfies these equations, then with δ ij denoting the Kronecker delta, we have k k Q = a i a j c i+j = = i=0 k a i δ ik i=0 k a j c i+j k a j c j+k 68 On rewriting 67 in the form c 0 a 0 + c a + c 2 a c k a k = c k, c a 0 + c 2 a + c 3 a c k a k = c k+, c k a 0 + c k a + c k+ a c 2k 2 a k = c 2k, we have by Cramer s rule see [24] that these equations have the solution 69 a j = Dj+ k, for j =0,,,k, 620 D k provided that D k 0, where D i k is the determinant with the ith column of D k replaced by the column c k,c k+,,c 2k Thus 68 gives with this choice Q = D k k D j+ k c k+j + D k c 2k 62 On the other hand, if we expand D k into its cofactor expansion along the bottom row, we see that k D k = k+j D k+,j+ c k+j, where the minor D i,j is the determinant of the matrix where the ith row and jth column of D k is removed From 64 we see that D k+,j+ = k j D j+ k, for j k, D k+,k+ = D k, where the factor k j results from shifting the last column of D k by k j places to the left Hence we conclude that which is the required result Q = D k D k, 622

24 676 D A GOLDSTON AND C Y YILDIRIM Our next lemma evaluates D k Lemma 4 We have D k = k! 2! 3! k! λ kk /2 L k ρ k λ 623 Proof where We first claim that D k = k E k 624 μ 0 μ μ 2 μ k μ μ 2 μ 3 μ k+ μ 2 μ 3 μ 4 μ k+2 E k = det = det μ i+j ρ j μ k μ k μ k+ μ 2k ρ ρ 2 ρ k i=0,,2,,k,,2,,k, 625 for if in E k we multiply the lth column by ρ and subtract this from the l + th column for l =, 2,,k we obtain μ 0 c 0 c c k μ c c 2 c k μ 2 c 2 c 3 c k+ E k = det, μ k c k c k c 2k and using the cofactor expansion along the bottom row gives E k = k D k We now introduce the differential operators D = d dλ, Clearly we have the relations δ = λd = λ d dλ, Δ=δ + λ = λ d + λ 626 dλ k k δ k = λ k D k + a j λd j, Δ k = δ k + b j λδ j 627 j= where a j λ and b j λ are polynomials of degree j in λ Now by 64 and 65 wehave μ k =Δμ k, and in general, μ k =Δ i μ k i, for 0 i k, μ k =Δ k 628 From this we see that μ 0 μ μ 2 μ k Δμ 0 Δμ Δμ 2 Δμ k Δ 2 μ 0 Δ 2 μ Δ 2 μ 3 Δ 2 μ k E k = det = det Δi μ j ρ j Δ k μ 0 Δ k μ Δ k μ 2 Δ k μ k ρ ρ 2 ρ k i=0,,2,,k,,2,,k 629 By the second relation in 627 we can replace Δ i by δ i and a linear combination of lower powers of δ, which can be eliminated by row operations Thus we can replace Δ by δ in the above determinant without affecting its value, and then by the first relation in 627 and row

25 HIGHER CORRELATIONS OF DIVISOR SUMS III 677 operations we can replace δ i by λ i D i which on removing the factors of λ in each row gives μ 0 μ μ 2 μ k Dμ 0 Dμ Dμ 2 Dμ k E k = λ kk /2 D 2 μ 0 D 2 μ D 2 μ 3 D 2 μ k det D k μ 0 D k μ D k μ 2 D k μ k ρ ρ 2 ρ k = λ kk /2 det Di μ j 630 We next need the relation ρ j i=0,,2,,k,,2,,k qq q h +=h! q = h q=0 h [ ] h j h q j, 63 j where [ ] h j are the Stirling numbers of the first type, although we do not need to use any properties of these numbers Then we have, by 63, λ h = e λ λ h e λ λ q+h = e λ = qq q h + λq q! q! e λ q=0 q=0 h [ ] = h j h q j λq j q! e λ = h [ ] h j h μ j j λ 632 Thus, using column operations we have λ λ 2 λ k D Dλ Dλ 2 Dλ k D 2 D 2 λ D 2 λ 2 D 2 λ k E k = λ kk /2 det D k D k λ D k λ 2 D k λ k ρ ρ ρ! 2! k! 2 k D i λ j = λ kk /2 det ρ j! 633 j i=0,,2,,k,,2,,k Expanding along the bottom row we see that D i λ j k det ρ j! = j k h h! i=0,,2,,k,,2,,k h=0 ρ det[d h i λ j ] i=0,,,k 634,,,k;j h We will show below that F h = det[d i λ j k ] i=0,,,k =!2! k! λ,,,k;j h h k h 635

26 678 D A GOLDSTON AND C Y YILDIRIM which then gives on retracing our steps k D k = k λ kk /2! 2! k! k h ρ k h! h h h=0 λ k h = k λ kk /2! 2! k!l ρ k k λ, which proves Lemma 4 We prove 635 by the following argument shown to us by Wasin So We consider the complete upper triangular matrix λ λ 2 λ k D Dλ Dλ 2 Dλ k M =[D i λ j ] i=0,,,k = D 2 D 2 λ D 2 λ 2 D 2 λ k,,,k D k D k λ D k λ 2 D k λ k [ ] j = i! λ j i i i=0,,,k,,,k Observe that det M =!2!k!, and further that M = TP, where T =[δ ij i!] i=0,,,k and P =,,,k Now by the matrix inverse formula using minors we have M = det M [ i+j D j,i ] i=0,,,k,,,,k where F h occurs in this matrix as the minor D k,h Further, M = P T, [ ] j λ j i i i=0,,,k,,,k where [ ] [ ] T j = δ ij, P = λ j i j i, i! i=0,,,k i i=0,,,k,,,k,,,k where we used the identity k j s s = i δ ij s i s=0 From this last relation we have, letting M =[ m ij ], F h = k+h k+h k h k det M m hk =! 2! k! λ k h, k! h as desired Proof of Proposition 3 Let a be the solution for 65 found in Lemma 3, which exists for any λ where D k 0, and let b be any other k-vector with b k = Then P b a x isa polynomial of degree k or less, and by the orthogonality property 65, Q b λ, ρ = P a x+p b a x,p a x+p b a x ρ = P a x,p a x ρ + P b a x,p b a x ρ = Q a λ, ρ+q b a λ, ρ

27 HIGHER CORRELATIONS OF DIVISOR SUMS III 679 In general by 67 and 69 for any c 0, assuming ρ>0 is fixed, we have Q c λ, ρ = j ρp c j 2 pj = ρ P c x 2, + O c λ < 0, for 0 <λ λ 0 c,ρ, where λ 0 c,ρ is a small positive constant depending on c Thus Q b λ, ρ Q a λ, ρ for 0 <λ<λ 0 c, proving that Q a is maximal at least for small enough λ This will continue to be true for larger λ as long as Q c < 0 for any k -vector c, and therefore as long as the maximal Q for k -vectors is negative By 54 of Szegö [22] we have d dx L n α x = L n α+ x, and therefore we see that the sequence {L k ρ k } of Laguerre polynomials has the property that the negative of the derivative of a term is the previous term Thus the negative derivative of the Laguerre polynomial in the numerator in 62 is the Laguerre polynomial in the denominator Further, in this sequence of Laguerre polynomials the polynomials are all decreasing functions up to their first positive zero, and hence the sequence of smallest positive zeros x k, ρ k is a decreasing sequence Starting with the trivial case when k = we see successively that the Q k with a satisfying 65 will be maximal for 0 <λ<x k +,ρ k This completes the proof of Proposition 3 Our next result evaluates the smallest positive zero x n, α asymptotically as n Lemma 5 Let L α n x, with α>, denote the Laguerre polynomials The zeros of L α n x are real, positive, and simple Let x n, α denote the smallest zero of L α n x If α = βn n and lim n βn/n =A>0, then x n, α lim = A n n Proof The properties of L α n x may be found in [22] by Szegö Equation 636 isa special case of [3, Theorem 44] A simple proof may be obtained by using the same argument as that found in [8] where a result corresponding to 636 for Jacobi polynomials is proved using Sturm comparison theory By [22, 52], the differential equation n +α +/2 u + + α2 x 4x 2 u = has u = e x/2 x α+/2 L n α x as a solution Let H n x := n +α +/2 x + α2 4x 2 4 = x2 4n +2α + x +α 2 4x 2,

28 680 D A GOLDSTON AND C Y YILDIRIM and denote the smaller root of the quadratic in the numerator by x n Then by the Sturm comparison argument in [8], and noting that lim n α/n =A, we have x n, α x n lim = lim n n n n = lim n = 2 + lim n = A 2 2n + α + 4n 2 +2α +2+4nα +4n n α n α 4+4 lim n n 7 Gaps between primes In this section we prove Theorem We want to examine statistically the number of primes in the interval n, n + h] for N<n 2N with N In this range the average distance between consecutive primes is log N, and thus we will take h to be a multiple of this length We therefore let ψx = n x Λn, 7 ψn, h =ψn + h ψn, 72 h = λ log N, 73 and in this paper we assume that λ 74 The model for our method is due to Gallagher [5], who proved that if the Hardy Littlewood conjecture 24 holds uniformly for h log N then one can asymptotically evaluate all the moments for the number of primes in intervals of length h Thus assuming 24, Gallagher proved that M k N, h, ψ := Nlog N k 2N n=n+ ψn, h k μ k λ, 75 as N, where μ k λ is the Poisson moment from 63 and 64 In order to obtain unconditional results we make use of our approximation Λ R n; H, where H is the set formed by the distinct numbers among h,h 2,,h k Taking N<n 2N, we first need to approximate ψn, h k = Λn + h Λn + h 2 Λn + h k h,h 2,,h k h =+o h,h 2,,h k h log N k H Λn; H 76 To define our approximation, we extend the definition of Λ R n; H in26 to vectors or lists H =h,h 2,,h k by letting Thus our approximation of ψn, h k is Λ R n; H := log R k H Λ R n; H 77 ψ R k n, h := h,h 2,,h k h Λ R n; H 78

29 HIGHER CORRELATIONS OF DIVISOR SUMS III 68 For convenience we also define ψ 0 R n, h = We next define the approximate moments, letting k = i + j, M ij R = Nlog R k 2N n=n+ and note that M 00 R = We also need the mixed moments M ij R = Nlog R k+ 2N n=n+ ψ R i n, hψ R j n, h, 79 ψ R i n, hψ R j n, hψn, h, 70 for which we note by the prime number theorem that M 00 R λ/θ = μ λ/θ, in accord with 7 and 73 below Using Propositions and 2 we will prove asymptotically that these approximate moments are also Poisson moments with an increased expected value involving the truncation level R Define θ by R = N θ 7 Proposition 4 As N we have, for k = i + j and for any fixed 0 <θ</k, λ M ij R =+o k μ k, 72 θ and for any fixed 0 <θ</2k, M ij R =+o k μ k+ λ θ 73 Proof By differencing, Propositions and 2 continue to hold unchanged when we sum for N<n 2N We first extend Proposition for vectors H and H 2 Recalling the notation H which denotes the number of components of the vector H, let k = H + H 2 and H = H H 2, where H i is the set of distinct components of H i Then by 77 and Proposition note that the k in Propositions and 2 is equal to H + H 2 here, we have for R = on /k, 2N n=n+ Λ R n; H Λ R n; H 2 = log R k H H2 2N n=n+ Λ R n; H Λ R n; H 2 = N SH+o log R k H 74 Thus we see that this result depends on k and not the individual values of H and H 2 Hence, letting h,h 2,,h k list the components of H and H 2 in any order, we have M ij R = Nlog R k = h,h 2,,h k h 2N h,h 2,,h k h n=n+ SH+ok log R H, Λ R n; H Λ R n; H 2 provided R = on /k We group terms in this sum according to the number of distinct values ν of h,h 2,,h k, and denote these distinct values by h,h 2,,h ν There are { k ν} ways to partition the k values h i into these ν disjoint sets, and all of these will occur in the sum above

30 682 D A GOLDSTON AND C Y YILDIRIM Hence by 23 we have M ij R = k ν= { } k h ν + o ν k log R ν =+o k μ k λ θ which proves the first part of Proposition 4 The second part is proved identically using Proposition 2 Now consider where S k = S k N, R, λ, ρ = Nlog R 2k+ 2N n=n+ P k ψ R n, h = ψn, h ρ log N P k ψ R n, h 2, 75 k a l ψ l R n, hlog R k l, 76 l=0 and the a l are arbitrary functions of N, R, k, λ, and ρ which are to be chosen to optimize the argument On multiplying out we have S k = Nlog R 2k+ a i a j log R 2k i j Letting = 2N n=n+ 0 i,j k 0 i,j k ψn, h ρ θ log R ψ R i n, hψ R j n, h a i a j M ij 77 λ = λ θ, ρ = ρ θ, 78 we have by Proposition 4, on taking i + j = κ and assuming that 0 <θ</2κ, M ij = M ij R ρ θ M ijr = μ κ+ λ ρμ κ λ+o κ θ κ+ = c κ λ, ρ+o κ θ κ+, 79 using the notation of 60 in the last line To evaluate S k we need to apply these results for 0 κ 2k, all of which will hold if we impose the condition θ< 4k + 4k 720 Thus S k = 0 i,j k a i a j c i+j λ, ρ+o k max a l 2 l k = Q a λ, ρ+o k, 72