SPARSE VARIANCE FOR PRIMES IN ARITHMETIC PROGRESSION J. BRUΜDERN AND T. D. WOOLEY∗ Abstract. An analogue of the Montgomery-Hooley asymptotic formula is established for the variance of the number of primes in arithmetic progressions, in which the moduli are restricted to the values of a polynomial. 1. Introduction The distribution of primes is a fundamental problem in the theory of numbers. Our concern here is with the error terms that arise in the prime number theorem for arithmetic progressions. Let π denote Euler’s totient, and deο¬ne the relevant quantity πΈ(π₯; π, π) for real π₯ ≥ 2, and coprime natural numbers π and π, through the equation ∑ π₯ + πΈ(π₯; π, π). log π = π(π) π≤π₯ π≡π mod π Montgomery [8] obtained an asymptotic formula for the variance π (π₯, π¦) = π ∑ ∑ π≤π¦ β£πΈ(π₯; π, π)β£2 π=1 (π,π)=1 that was promptly reο¬ned by Hooley [5], and now takes the shape π (π₯, π¦) = π₯π¦ log π¦ + ππ₯π¦ + π(π₯1/2 π¦ 3/2 + π₯2 (log π₯)−π΄ ). (1) Here, the coeο¬cient π is a certain real constant, the exponent π΄ is a preassigned positive real number, and the relation (1) holds uniformly for 1 ≤ π¦ ≤ π₯. In this range, the asymptotic formula (1) implies the upper bound π (π₯, π¦) βͺ π₯π¦ log π₯ + π₯2 (log π₯)−π΄ (2) which is known as the Barban-Davenport-Halberstam theorem. If one singles out an individual progression π modulo π from the mean π (π₯, π¦), then one recovers from the bound (2) the Siegel-Walο¬sz theorem, the latter asserting that the estimate πΈ(π₯; π, π) βͺ π₯(log π₯)−π΄/2 1991 Mathematics Subject Classiο¬cation. 11N13, 11P55. Key words and phrases. Primes in progressions, Barban-Davenport-Halberstam theorem. ∗ The authors thank Shandong University for providing a creative atmosphere during the period while this paper was conceived, and the Hausdorο¬ Research Institute for very good working conditions while the paper was completed. The second author is supported in part by a Royal Society Wolfson Research Merit Award. 1 2 J. BRUΜDERN AND T. D. WOOLEY holds uniformly in π and π. Although it seems diο¬cult to improve this last bound, the asymptotic formula for π (π₯, π¦) shows that in mean square, the average size of πΈ(π₯; π, π) is about the square-root of (π₯ log π₯)/π. One might ask whether conclusions similar to those described above are possible for thinner averages, for example with the moduli π restricted to the values of a polynomial. This was studied recently by Mikawa and Peneva [7], who obtained a result analogous to the Barban-Davenport-Halberstam theorem. More precisely, let π ∈ β[π‘] denote an integer-valued polynomial of degree π ≥ 2, with positive leading coeο¬cient and derivative π ′ . Then, there exist real numbers π¦1 such that the inequalities π (π¦) ≥ 2 and π ′ (π¦) ≥ 1 hold for all π¦ ≥ π¦1 . Denote the smallest such π¦1 with π¦1 ≥ 1 by π¦0 (π ). We may now deο¬ne ππ (π₯, π¦) = ∑ π¦0 (π )<π≤π¦ ′ π (π) π (π) ∑ β£πΈ(π₯; π (π), π)β£2 . (3) π=1 (π,π (π))=1 In this notation, the aforementioned estimate of Mikawa and Peneva asserts that whenever π΄ ≥ 1 is a real number, then there is a π΅ ≥ 1 such that, uniformly for π (π¦) ≤ π₯(log π₯)−π΅ , one has ππ (π₯, π¦) βͺ π₯2 (log π₯)−π΄ . (4) ′ We note here that Theorem 2 of [7] relates to a variant of ππ (π₯, π¦), with the weight π (π) in (3) replaced by π −1 π(π (π)). However, as is easily checked, the bound (4) above is equivalent to the conclusion in [7]. Our purpose in this paper is to establish a version of the Montgomery-Hooley asymptotic formula for ππ (π₯, π¦). Theorem. Let π be an integer-valued polynomial of degree π ≥ 2, with positive leading coeο¬cient. Then, whenever π (π¦) ≤ π₯, one has ( ) ππ (π₯, π¦) = π₯π (π¦) log π (π¦) + πΆ(π )π₯π (π¦) + π π₯1/2 π (π¦)3/2 + π₯2 (log π₯)−π΄ , wherein πΆ(π ) denotes a certain real number depending only on π . A new approach1 to the Montgomery-Hooley theorem was developed by Goldston and Vaughan [3] and Vaughan [10]. In contrast with previous work that depended on the additive theory of primes, the evaluation of π (π₯, π¦) is now linked to the problem of ο¬nding an asymptotic formula for the number of solutions of the equation π1 − π2 = βπ in primes π1 , π2 not exceeding π₯, and natural numbers β and π with π ≤ π¦. This may be viewed as a ternary additive problem that is well within the competence of the circle method. Moreover, the primes enter the scene only through an application of the Siegel-Walο¬sz theorem. Vinogradov’s estimate for exponential sums over primes is not required since the minor arc analysis can be performed primarily with the product βπ. Following these patterns, we are led to consider the equation π1 − π2 = βπ (π), and again, a treatment by the circle method is possible. A detailed discussion of this equation is to be found in §§3 and 4, following a preparatory analysis in the next section. Reο¬ecting a similar problem in the work of Vaughan [10], the treatment of the major arcs invokes certain auxiliary averages of multiplicative functions that are supplied in §§5 and 6, thereby completing the proof of the theorem. There is an extensive literature relating to improvements of the error term in the asymptotic formula (1) that are subject to the truth of the Riemann hypothesis for Dirichlet 1For historical comments on the genesis of this method, the curious reader is directed to p. 118 of [3]. SPARSE VARIANCE FOR PRIMES IN ARITHMETIC PROGRESSION 3 πΏ-functions (see, in particular, Hooley [5], [6], Friedlander and Goldston [2], Goldston and Vaughan [3] and Vaughan [11]). Likewise, in the conclusion of our theorem above, the term π₯2 (log π₯)−π΄ may be replaced conditionally by π₯2−πΏ(π) , where πΏ(π) is a positive real number the size of which is determined by the maximal savings available for Weyl sums involving a degree π polynomial. We spare the reader any details. Our methods are rather ο¬exible. One may replace the primes with other sequences of interest, and functions can be substituted for the polynomial π with growth more rapid than any polynomial. For example, by working along the lines of BruΜdern and Perelli [1], it should be possible to deduce an asymptotic formula for ππ (π₯, π¦) when π (π) = [ exp((log π)πΎ )], and πΎ is any positive number smaller than 3/2. 2. Two preparatory steps Our initial advance on the analysis of the variance ππ (π₯, π¦) involves the isolation of a term accessible to the circle method, and this is the objective of this section. Fix the value of π΄ ≥ 1 for the rest of the paper, and choose a permissible value of π΅ = π΅(π΄) so that the asymptotic relation (4) holds uniformly for π (π¦) ≤ π₯(log π₯)−π΅ . An inspection of [7] reveals that one may take π΅ = 24π π΄. When π₯ is suο¬ciently large, there is a unique solution π¦ to the equation π (π¦) = π₯(log π₯)−π΅ that we denote by π¦1 = π¦1 (π₯). As a truncated analogue of the sum ππ (π₯, π¦) deο¬ned in (3), we deο¬ne ππ′ (π₯, π¦) = ∑ ′ π (π) π¦1 <π≤π¦ π (π) ∑ β£πΈ(π₯; π (π), π)β£2 . (5) π=1 (π,π (π))=1 Then in view of the deο¬nition (3) and the estimate (4), it is easily veriο¬ed that ππ (π₯, π¦) = ππ′ (π₯, π¦) + π(π₯2 (log π₯)−π΄ ). (6) We have removed small values of π from ππ (π₯, π¦) mainly for technical convenience. To do so, we had to invoke the work of Mikawa and Peneva [7], but it would be possible, and perhaps more coherent, to cover small π by our method. However, a critical step for small π would depend on the ideas of section 2.2 in [7], and that we would have need to duplicate in any case. In the interest of brevity, we therefore prefer to work temporarily with ππ′ (π₯, π¦). In the next step, we open the square in (5). Let ∑ π(π₯; π, π) = log π π≤π₯ π≡π mod π and Φπ (π§, π¦) = π ′ (π) . π(π (π)) π§<π≤π¦ ∑ (7) Then it follows from (5) that ππ′ (π₯, π¦) = π1 − 2π₯π2 + π₯2 Φπ (π¦1 , π¦), (8) 4 J. BRUΜDERN AND T. D. WOOLEY where π1 = ∑ π (π) π¦1 <π≤π¦ π2 = π (π) ∑ ′ π=1 (π,π (π))=1 π ′ (π) π(π (π)) <π≤π¦ π (π) ∑ ∑ π¦1 π(π₯; π (π), π)2 , π(π₯; π (π), π). π=1 (π,π (π))=1 By the Prime Number Theorem, one ο¬nds that π (π) ∑ ∑ π(π₯; π (π), π) = log π = π₯ + π(π₯(log π₯)−3π΄ ). π≤π₯ πβ€π (π) π=1 (π,π (π))=1 Consequently, using only straightforward estimates, one discovers that whenever π (π¦) ≤ π₯, one has π2 = π₯Φπ (π¦1 , π¦) + π(π₯(log π₯)2−3π΄ ). (9) Similarly, one may derive the relation π (π) ∑ π(π₯; π (π), π)2 = ∑ ∑ (log π1 )(log π2 ). π1 ≤π₯ π2 ≤π₯ π1 β€π (π) π2 β€π (π) π1 ≡π2 mod π (π) π=1 (π,π (π))=1 Here we separate the diagonal terms with π1 = π2 from the oο¬-diagonal ones. Note that the conditions π1 β£π (π) and π1 ≡ π2 mod π (π) imply that π1 = π2 . We may therefore ignore the constraint ππ β€ π (π) (π = 1, 2) when considering the oο¬-diagonal terms. In this way, by symmetry, we derive the formula π (π) ∑ π(π₯; π (π), π)2 = ∑ (log π1 )(log π2 ). π1 <π2 ≤π₯ π1 ≡π2 mod π (π) π≤π₯ πβ€π (π) π=1 (π,π (π))=1 ∑∑ (log π)2 + 2 We next reintroduce the terms with πβ£π (π), and sum over π. Then, much as in the analysis leading to the relation (9), we ο¬nd that ∑ ∑ π1 = 2π0 + π ′ (π) (log π)2 + π(π₯(log π₯)2 ), (10) π≤π₯ π¦1 <π≤π¦ where π0 = ∑ π¦1 <π≤π¦ π ′ (π) ∑∑ (log π1 )(log π2 ). (11) π1 <π2 ≤π₯ π1 ≡π2 mod π (π) On applying partial summation within (10), and substituting the outcome together with (8) and (9) into (6), we obtain our ο¬nal formula ∑ ππ (π₯, π¦) = 2π0 − π₯2 Φπ (π¦1 , π¦) + π (π¦) (log π)2 + π(π₯2 (log π₯)−π΄ ). (12) π≤π₯ SPARSE VARIANCE FOR PRIMES IN ARITHMETIC PROGRESSION 5 3. The circle method In this section we apply the circle method to evaluate the sum π0 deο¬ned in equation (11). With this objective in view, we rewrite the congruence condition π1 ≡ π2 mod π (π) in (11) in the form π2 − π1 = βπ (π), with β ∈ β€. In addition, we observe that the supplementary condition β ≥ 1 is equivalent to the summation condition π1 < π2 imposed in the second summation of (11). Consequently, on writing ∑ ∑ ππ (πΌ) = π ′ (π) π(πΌβπ (π)), (13) π¦1 <π≤π¦ π (πΌ) = ∑ β≤π₯/π (π) (log π)π(πΌπ), (14) π≤π₯ we may use orthogonality to infer from (11) that ∫ 1 ππ (πΌ)β£π (πΌ)β£2 ππΌ. π0 = 0 We analyse this integral by splitting the range of integration into major and minor arcs. The dissection depends on the absolute constant π· ≥ 1 that arises in Lemma 1 below. Let πΆ = 4π3 π·π΅, and write π = (log π₯)πΆ . Let π denote the union of the pairwise disjoint intervals β£πΌ−π/πβ£ ≤ π/π₯ with 1 ≤ π ≤ π ≤ π and (π, π) = 1. Let πͺ = [π/π₯, 1+π/π₯]βπ. The contribution from the minor arcs is controlled with the aid of the bound sup β£ππ (πΌ)β£ βͺ π₯(log π₯)−2π΄ (15) πΌ∈πͺ that we establish in a moment. Equipped with (15) and the elementary estimate ∫ 1 ∑ β£π (πΌ)β£2 ππΌ = (log π)2 βͺ π₯ log π₯, 0 π≤π₯ we infer that π0 = ππ + π(π₯2 (log π₯)1−2π΄ ), (16) where ∫ ππ = ππ (πΌ)β£π (πΌ)β£2 ππΌ. (17) π The major arc contribution ππ will be discussed in §4. In the remainder of this section, we brieο¬y sketch a proof of (15). This bound of Weyl’s type is certainly familiar territory for workers in the ο¬eld, but a suitable reference does not seem to be available. The argument below is streamlined so as to conο¬rm (15) with little eο¬ort, but when one seeks for the strongest possible bounds, one can do rather better, in particular when π is small. This is of no relevance for us, but an analysis of ππ (π₯, π¦) under GRH is sensitive to this comment. Let π¦0 ≤ π§0 ≤ π§, and consider the weighted Weyl sum ∑ π(π½; π§0 , π§) = π ′ (π)π(π½π (π)). π§0 <π≤π§ 6 J. BRUΜDERN AND T. D. WOOLEY Lemma 1. There is an absolute constant π· ≥ 1 with the property that whenever π½ ∈ β, and π and π are coprime natural numbers with β£ππ½ − πβ£ ≤ 1/π, one has )1/(π·π3 ) ( . π(π½; π§0 , π§) βͺ π§ π (log π§) (π + π§ π β£ππ½ − πβ£)−1 + π§ −1 + (π + π§ π β£ππ½ − πβ£)π§ −π Proof. When π is a natural number, let π½π ,π (π ) denote the number of integral solutions of the simultaneous equations π ∑ (π₯ππ − π¦ππ ) = 0 (1 ≤ π ≤ π), π=1 with 1 ≤ π₯π , π¦π ≤ π (1 ≤ π ≤ π ). Then, by Theorem 7.4 of Vaughan [9], for example, there exists an absolute constant π· ≥ 1 such that whenever π ≥ 21 π·π 3 , one has π½π ,π (π ) βͺ π 2π −π(π+1)/2 . Applying Theorem 5.2 of Vaughan [9] in conjunction with partial summation, we deduce that 3 π(π½; π§0 , π§) βͺ π§ π (log π§)(π−1 + π§ −1 + ππ§ −π )1/(π·π ) . The conclusion of the lemma now follows from a familiar transference principle (see, for example, Exercise 2 of Chapter 2 of Vaughan [9]). β‘ We are now ready to deduce (15). Let π¦(β) be the upper bound for π determined by the simultaneous conditions π ≤ π¦ and π (π) ≤ π₯/β. Then π (π¦(β)) = min{π (π¦), π₯/β}. Reversing the order of summation in (13) yields ∑ ππ (πΌ) = π(πΌβ; π¦1 , π¦(β)). (18) β≤(log π₯)π΅ Suppose that πΌ ∈ πͺ. For any integer β with 1 ≤ β ≤ (log π₯)π΅ , we apply Dirichlet’s theorem to ο¬nd coprime integers π = π(β) and π = π(β) with 1 ≤ π ≤ π¦(β)π−2/3 and β£πΌβπ − πβ£ ≤ π¦(β)−π+2/3 . The validity of the simultaneous conditions βπ ≤ π and β£πΌ − π/(βπ)β£ ≤ π/π₯ would imply that πΌ ∈ π, whence for each β, at least one of these inequalities fails. If βπ > π, then π > (log π₯)πΆ−π΅ , and an application of Lemma 1 with π½ = πΌβ yields 3 3 π(πΌβ; π¦1 , π¦(β)) βͺ π₯(log π₯)(π−1 + π¦(β)−2/3 )1/(π·π ) βͺ π₯(log π₯)1+(π΅−πΆ)/(π·π ) . If βπ ≤ π but β£πΌ − π/(βπ)β£ > π/π₯, meanwhile, then in like manner Lemma 1 gives 3 3 π(πΌβ; π¦1 , π¦(β)) βͺ π₯(log π₯)π−1/(π·π ) βͺ π₯(log π₯)1−πΆ/(π·π ) . Summing over natural numbers β with β ≤ (log π₯)π΅ , we deduce (15) from (18). 4. The major arcs We move on to analyse the major arc contribution ππ to π0 , and subsequently incorporate the conclusions of the previous section in order to obtain an asymptotic formula for π0 . For πΌ ∈ π, one may write πΌ = π½ + π/π for some π½ ∈ β with β£π½β£ ≤ π/π₯, and coprime integers π and π with 1 ≤ π ≤ π ≤ π. Then, by (14) and Lemma 3.1 of Vaughan [9], one has π(π) π½(π½) + π(π₯(log π₯)−5πΆ ), π (πΌ) = π(π) SPARSE VARIANCE FOR PRIMES IN ARITHMETIC PROGRESSION 7 where π½(π½) = ∑ π(π½π). π≤π₯ The trivial upper bound ππ (πΌ) βͺ π₯ log π₯ (which is available from (13)) and (17) now suο¬ce to conο¬rm that ∫ π/π₯ π ∑ π(π)2 ∑ ππ = ππ (π½ + π/π)β£π½(π½)β£2 ππ½ + π(π₯2 (log π₯)−πΆ ). 2 π(π) −π/π₯ π=1 π≤π (π,π)=1 We would like to extend the sum over all natural numbers π here. It is convenient to do this in two steps. In the ο¬rst of these steps, we extend the range for π above to π ≤ π₯1/(2π) . For this, we note that the intervals β£πΌ − π/πβ£ ≤ π/π₯ with 1 ≤ π ≤ π ≤ π₯1/(2π) and (π, π) = 1 are pairwise disjoint. When π and π are coprime natural numbers with 1 ≤ π ≤ π and π < π ≤ π₯1/(2π) , and β£π½β£ ≤ π/π₯, one has π½ + π/π ∈ πͺ. Therefore, on recalling (15), for any such π½, π and π, one has β£ππ (π½ + π/π)β£ ≤ sup β£ππ (πΌ)β£ βͺ π₯(log π₯)−2π΄ . πΌ∈πͺ Combining this estimate with the elementary inequality ∫ 1/2 ∫ π/π₯ 2 β£π½(π½)β£2 ππ½ = [π₯], β£π½(π½)β£ ππ½ ≤ −1/2 −π/π₯ we deduce that ∑ ππ = π≤π₯1/(2π) ∫ π/π₯ π π(π)2 ∑ ππ (π½ + π/π)β£π½(π½)β£2 ππ½ + π(π₯2 (log π₯)1−2π΄ ). 2 π(π) π=1 −π/π₯ (19) (π,π)=1 Before extending the range for π even further, we complete the integral in (19) to β£π½β£ ≤ 21 . In this range for π½, one has π½(π½) βͺ β£π½β£−1 , so that another application of the trivial bound ππ (πΌ) βͺ π₯ log π₯ in combination with straightforward estimates yields the upper bound ∫ π ∑ π(π)2 ∑ β£ππ (π½ + π/π)π½(π½)2 β£ ππ½ 2 π(π) π=1 1/(2π) π≤π₯ 1 (π,π)=1 π/π₯≤β£π½β£≤ 2 ∑ 1 ∫ 1/2 π½ −2 ππ½ βͺ π₯2 (log π₯)2−πΆ . βͺ π₯(log π₯) π(π) π/π₯ π≤π₯ Hence, we may indeed extend the integration in (19) to [− 21 , 21 ] with the introduction of acceptable errors. However, by orthogonality and (13), one has ∫ 1/2 ∑ ∑ ∑ ∑ ( πβπ (π) ) 2 ′ ππ (π½ + π/π)β£π½(π½)β£ ππ½ = π (π) π π −1/2 π ≤π₯ π ≤π₯ π¦ <π≤π¦ β≤π₯/π (π) 1 1 2 π2 −π1 =βπ (π) = ∑ π¦1 <π≤π¦ π ′ (π) ∑ β≤π₯/π (π) ( πβπ (π) ) π ([π₯] − βπ (π)). π 8 J. BRUΜDERN AND T. D. WOOLEY In the last sum, we may replace [π₯] by π₯, introducing an error bounded by π(π₯ log π₯) into this identity. We summarize and inject these results into (19) to infer that ππ = π0 + π(π₯2 (log π₯)−π΄ ), where π0 = ∑ π≤π₯1/(2π) (20) ∑ π(π)2 ∑ ′ π (π) ππ (βπ (π))(π₯ − βπ (π)), π(π)2 π¦ <π≤π¦ 1 in which we have written ππ (π) = β≤π₯/π (π) π ∑ π=1 (π,π)=1 ( ππ ) π π for Ramanujan’s sum. Recall that π¦(β) was deο¬ned through the equation π (π¦(β)) = min{π (π¦), π₯/β}. In the deο¬nition of π0 , we exchange the order of summation of β and π and then sort π according to residue classes modulo π. Then π ∑ π(π)2 ∑ ∑ ∑ π0 = π (βπ (π)) π ′ (π)(π₯ − βπ (π)). (21) π 2 π(π) 1/(2π) π=1 β≤π₯/π (π¦1 ) π≤π₯ π¦1 <π≤π¦(β) π≡π mod π For small π, Euler’s summation formula readily yields an asymptotic formula for the innermost sum, with a main term independent of π. It is then possible to sum ππ (βπ (π)) over π. This process will largely disentangle the sum in (21). Two lemmata make this analysis precise. Lemma 2. With π₯, π¦ and β in the ranges as above, and uniformly in π, one has ∫ ∑ 1 π (π¦(β)) ′ (π₯ − βπ‘) ππ‘ + π(π₯π¦ π−1 + π¦ 2π−1 β). π (π)(π₯ − βπ (π)) = π π (π¦1 ) π¦1 <π≤π¦(β) π≡π mod π Proof. Observe ο¬rst that by Euler’s summation formula, if π : [π, π ] → β is continuously diο¬erentiable, then ∫ ∑ 1 π π(π) = π(π‘) ππ‘ + πΈ, (22) π π π<π≤π π≡π mod π where the implicitly deο¬ned real number πΈ satisο¬es the bound ∫ π β£πΈβ£ ≤ β£π ′ (π‘)β£ ππ‘ + β£π(π)β£ + β£π(π )β£. π For the application at hand, we take π(π‘) = π ′ (π‘)(π₯ − βπ (π‘)), and put π = π¦1 and π = π¦(β). Note, in particular, that π₯ − βπ (π‘) = π(π₯) for all relevant choices of π‘, and hence π(π‘) βͺ π¦ π−1 π₯ for π ≤ π‘ ≤ π . Also, one has π ′ (π‘) = π ′′ (π‘)(π₯ − βπ (π‘)) − βπ ′ (π‘)2 . Then it follows that πΈ βͺ π₯π¦ π−1 + π¦ 2π−1 β. (23) Finally, by substitution one obtains ∫ π¦(β) ∫ π (π¦(β)) ′ π (π‘)(π₯ − βπ (π‘)) ππ‘ = (π₯ − βπ ) ππ. (24) π¦1 π (π¦1 ) SPARSE VARIANCE FOR PRIMES IN ARITHMETIC PROGRESSION The conclusion of the lemma follows by collecting together (22), (23) and (24). 9 β‘ In advance of the statement of the next lemma, for each natural number π, we deο¬ne π(π) to be the number of solutions of the congruence π (π) ≡ 0 mod π. Lemma 3. For any β ∈ β, the sum π 1∑ ππ (βπ (π)) π€β (π) = π π=1 is a multiplicative function of π. When π is a prime, one has { π − 1, when πβ£β, π€β (π) = π(π) − 1, when π β€ β. Proof. Since ππ (π) is multiplicative as a function of π (see, for example, Theorem 272 of Hardy and Wright [4]), it suο¬ces to apply the Chinese Remainder Theorem to the sum over π in order to conο¬rm the multiplicativity of π€β (π). Next, when π is a prime, we apply the deο¬nition of ππ (π) and orthogonality to obtain the relation π π−1 1 ∑ ∑ ( πβπ (π) ) π π€β (π) = = card{1 ≤ π ≤ π : βπ (π) ≡ 0 mod π} − 1, π π=1 π=1 π and the lemma follows at once. β‘ We inject the asymptotic formula supplied by Lemma 2 into (21). This introduces an error term ∑ π(π)2 ∑ ππ(π)(π₯π¦ π−1 + π¦ 2π−1 β) βͺ π₯2−1/(2π)+π , 2 π(π) 1/(2π) β≤π₯/π (π¦1 ) π≤π₯ so that we now have ∑ π0 = π≤π₯1/(2π) π(π)2 π(π)2 ∑ ∫ π (π¦(β)) π€β (π) (π₯ − βπ‘) ππ‘ + π(π₯2−1/(2π)+π ). (25) π (π¦1 ) β≤π₯/π (π¦1 ) We next exchange the order of summation and complete the sum over π. By Lemma 3, the multiplicative function π€β (π)π(π)2 is non-negative and bounded by π((π, β)π π ). This implies that ∑ π(π)2 π€β (π) = π (β) + π(βπ π₯−1/(2π) ), 2 π(π) 1/(2π) π≤π₯ with π (β) = ∞ ∑ π(π)2 π=1 ( ) β ∏ π(π) − 1 1+ π€β (π) = . π(π)2 π(β) (π − 1)2 (26) πβ€β A trivial estimate for the integral in (25) shows it to be π(π (π¦)π₯), and this suο¬ces here to conclude that ∫ π (π¦(β)) ∑ π0 = π (β) (π₯ − βπ‘) ππ‘ + π(π₯2−1/(2π)+π ). β≤π₯/π (π¦1 ) π (π¦1 ) 10 J. BRUΜDERN AND T. D. WOOLEY The derivative of π‘(π₯ − 12 βπ‘) with respect to π‘ is π₯ − βπ‘, and we have π (π¦(β)) = π (π¦) when β ≤ π₯/π (π¦), and π (π¦(β)) = π₯/β when β > π₯/π (π¦). Hence, if we compute the integral and split the sum over β at π₯/π (π¦) when appropriate, we readily ο¬nd that ( ) (27) π0 = 12 π (π¦1 )2 Θπ (π₯/π (π¦1 )) − π (π¦)2 Θπ (π₯/π (π¦)) + π(π₯2−1/(2π)+π ) where Θπ (π») = ∑ π (β) (π» − β)2 . β β≤π» (28) We summarize the analysis based on the application of the circle method by combining (16), (20) and (27), thereby obtaining the formula 2π0 = π (π¦1 )2 Θπ (π₯/π (π¦1 )) − π (π¦)2 Θπ (π₯/π (π¦)) + π(π₯2 (log π₯)−π΄ ), (29) and thus we enter the opening phase of the endgame. 5. An exercise in multiplicative number theory In view of the relations (29) and (12), the evaluation of ππ (π₯, π¦) is reduced to the deduction of appropriate asymptotic formulae for the functions Φπ and Θπ , deο¬ned in (7) and (28), respectively. This is accomplished largely by a series of exercises in multiplicative number theory. We begin with an analysis of the Dirichlet series associated with the function π (β)/β, restricting attention for the time being to the situation wherein π(2) ≥ 1. An inspection of (26) shows that the series π·(π ) = ∞ ∑ π (π)π−1−π π=1 converges absolutely in the half-plane Re (π ) > 0, and that π (β)/π (1) is multiplicative. Hence, for Re (π ) > 0, one has ∏ π·(π ) = π (1) π·π (π ) (30) π with ( π(π) − 1 π·π (π ) = 1 + 1 + (π − 1)2 )−1 ∑ ∞ π(ππ )−1 π−ππ . (31) π=1 Notice here that since π(π) ≥ 0, the Euler factor π·π (π ) is deο¬ned for all values of π, with the exception of π = 2 in the special case π(2) = 0. It is for this reason that we isolate the latter situation below.2 The Euler product (30) yields an analytic continuation of π·(π ) to a meromorphic function on the half-plane Re (π ) > −2. In order to see this, it will be convenient to write π = π−1−π , and deο¬ne the real number π = π(π) by means of the identity )−1 ( π(π) π(π) − 1 =1− 2 . (32) 1+ 2 (π − 1) π 2We are grateful to the referee for prompting our separate treatment of this case. SPARSE VARIANCE FOR PRIMES IN ARITHMETIC PROGRESSION 11 It is apparent from the relation deο¬ning π(π) that 0 ≤ π βͺ 1, wherein the implicit constant depends only on π . We may rearrange the Euler factors into the shape ( ) π π π π·π (π ) = 1 + 1− 2 , π−1 π 1−π and consequently, ( ) π π 1− , (1 − π)π·π (π ) = 1 + π−1 π ( ) ( ) ( )2 ( ) π π 1−π π π π 1− (1 − π)π·π (π ) = 1 + − 1− , π π π−1 π π−1 π )( ) ( (1 − π/π)(1 − π) 1−π π/π . π·π (π ) = 1 + 1 − π 2 /π2 1 + π/π π−1 On recalling (30), and rewriting π as π−1−π again, the previous formulae transform into the relations π(π + 1)π(π + 2) π·(π ) = π(π + 1)πΈ1 (π ) = π(π + 1)π(π + 2)πΈ2 (π ) = πΈ3 (π ), (33) π (1) π(2π + 4) where we have written ∏( πΈ1 (π ) = 1+ π π−π π(π − 1) ( π(π) 1− π )) , π−π π−2π (1 − π(π)) − π2 (π − 1) π3 (π − 1) π ) ∏( 1 − π(π) . πΈ3 (π ) = 1+ (π − 1)(1 + ππ +2 ) π ∏( πΈ2 (π ) = 1+ (34) ( π(π) 1− π )) , (35) (36) Here π denotes Riemann’s zeta function, and the product πΈπ converges absolutely and locally uniformly in the half-plane Re (π ) > −(π + 1)/2 for π = 1, 2, 3. In particular, the relations recorded in (33) provide the analytic continuation of π·(π ) to the half-plane Re (π ) > −2. In the half-plane Re (π ) ≥ −3/2, meanwhile, the only singularities of π·(π ) are simple poles at π = 0 and π = −1. Returning now to (28), we deduce from one of Perron’s formulae that ∫ 1 2+π∞ π·(π )π» π +2 Θπ (π») = ππ . (37) ππ 2−π∞ π (π + 1)(π + 2) Observe that from (36), one ο¬nds that the function πΈ3 (π ) is holomorphic and uniformly bounded in Re (π ) ≥ −3/2. Invoking only standard estimates for the zeta factors in (33), one can move the line of integration on the right hand side of (37) to Re (π ) = −3/2. The reader may care to inspect the analysis on p. 141 of Goldston and Vaughan [3] for the necessary ideas, though in the present circumstances one should not use the Riemann hypothesis. One may also compare the discussion on p. 791 of Vaughan [10]. If π (π§) is the residue of the integrand in (37) at π = π§, then it follows that 1 Θ (π») 2 π = π (0) + π (−1) + π(π» 1/2 ). (38) 12 J. BRUΜDERN AND T. D. WOOLEY We compute π (0) explicitly by using (33) and (34) along with the familiar expansion 1 π(1 + π ) πΎ = 2 + + π(1). π π π A short calculation then reveals that where π (0) = 21 π(π )π» 2 log π» + Γ0 π» 2 , (39) ) ∏( 1 − π(π)π−1 π(π ) = π (1) 1+ π(π − 1) π (40) and Γ0 = Γ0 (π ) is a certain real number, the precise value of which is not relevant to our subsequent discussion. Another short calculation leads from (26) and (32) to the alternative formulae )( ) ∏( π(π) − 1 1 − π(π)π−1 π(π ) = 1+ 1+ 2 (π − 1) π(π − 1) π ) ∑ ∞ ∏( π(π)2 π(π) π(π) = . (41) = 1+ π(π − 1) ππ(π) π=1 π In the latter form, this constant will play an important role later. A similar technique may be applied to evaluate π (−1). In this instance we use (33) and (35) and then reach the transitional formula π (−1) = −π (1)πΈ2 (−1)π(0)π» log π» + Γ−1 π», (42) where Γ−1 = Γ−1 (π ) is another real number, the precise value of which is again of little importance in what follows. We remark, however, that Γ−1 (π ) could be made just as explicit as π(π ). Observe next that π(0) = −1/2, and so on applying (26), (32) and (35), we ο¬nd that the Euler factors of π (1) and πΈ2 (−1) combine to give the formula )( ) ∏( π(π) − 1 1 − π(π) 1 − π(π)π−1 π (1)πΈ2 (−1) = 1+ − 1+ (π − 1)2 π(π − 1) π(π − 1) π ( ) ( ) ∏ π(π) − 1 π(π) = 1+ 1− 2 = 1. 2 (π − 1) π π We may therefore conclude that π (−1) = 21 π» log π» + Γ−1 π». (43) On substituting (39) and (43) into (38), we obtain the expansion Θπ (π») = π(π )π» 2 log π» + 2Γ0 π» 2 + π» log π» + 2Γ−1 π» + π(π» 1/2 ), which may be incorporated into (29), leading to the asymptotic formula ( ) ( ) π (π¦) π₯ 2 −π₯π (π¦) log − 2Γ−1 π₯π (π¦) 2π0 = π(π )π₯ log π (π¦1 ) π (π¦) ( ) + π π₯1/2 π (π¦)3/2 + π₯2 (log π₯)−π΄ . This completes the evaluation of the expression π0 deο¬ned in equation (11). (44) SPARSE VARIANCE FOR PRIMES IN ARITHMETIC PROGRESSION 13 We turn our attention now to the situation with π(2) = 0. Here, as a consequence of (26), one sees that π (β) = 0 whenever β is odd, and moreover that for all natural numbers π one has )−1 ( π ∏ π (2π) π(π) − 1 = 1+ . π (2) π(2π) (π − 1)2 πβ£π π>2 Deο¬ne { 2, when π = 2, π0 (π) = π(π), when π > 2. In the remainder of this section, we adopt the convention that the decoration of a symbol with a subscript 0 is to be interpreted as indicating that the deο¬nition of that symbol is to be modiο¬ed by replacing π(π) by π0 (π) throughout. We now ο¬nd that π (2) = π0 (1) and ( )−1 ( ) π (2π) ∏ π0 (π) − 1 π = 1+ . π0 (1) (π − 1)2 π−1 πβ£π In particular, it follows that π (2π)/π0 (1) is a multiplicative function of π. With the Dirichlet series π·(π ) deο¬ned as before, we again ο¬nd that π·(π ) converges absolutely in the half-plane Re (π ) > 0. Hence, for Re (π ) > 0, one has π·(π ) = 2 −1−π ∞ ∑ π=1 π (2π)π−1−π = π0 (1)2−1−π ∏ π·π,0 (π ), (45) π with π·π,0 (π ) deο¬ned via (31). We may now execute the argument following (31) to establish that the analogue of (33) holds, save that in present circumstances π·(π )/π (1) is replaced by 21+π π·(π )/π0 (1). It follows, in particular, that the analogue of the asymptotic relation (38) holds. A modest calculation reveals that π 0 (0) = 41 π0 (π )π» 2 log π» + Γ0 π» 2 , (46) where π0 (π ) is deο¬ned via (40), and Γ0 = Γ0 (π ) is a certain (but diο¬erent) real number, the precise value of which is again of no importance in what follows. In addition, as a consequence of the argument underlying (41), we have ) ) ∏( ∏( π0 (π) π(π) π0 (π ) = 1+ =2 1+ = 2π(π ). (47) π(π − 1) π(π − 1) π π>2 So far as π (−1) is concerned, the modiο¬cation to the formula (33) leaves the relation (42) essentially unchanged, save that Γ−1 will need adjustment invisible in the ensuing discussion. We therefore deduce as before that the analogue of (43) holds, whence on substituting (46) and (43) into (38) we conclude that Θπ (π») = 21 π0 (π )π» 2 log π» + 2Γ0 π» 2 + π» log π» + 2Γ−1 π» + π(π» 1/2 ). Incorporating this, as before, into (29), and recalling (47), we arrive at the asymptotic formula (44). In this way, we have established the asymptotic relation (44) in all circumstances. 14 J. BRUΜDERN AND T. D. WOOLEY 6. Another exercise in multiplicative number theory Our treatment of the sum Φπ deο¬ned in equation (7) takes the convolution ∑ π(π)2 π = π(π) π(π) πβ£π as the starting point. By reversing the order of summation, we obtain ∑ π¦0 (π )<π≤π§ ∑ π(π)2 π (π) = π(π (π)) π(π) π≤π (π§) ∑ π¦0 (π )<π≤π§ π (π)≡0 mod π ) ∑ π(π)2 ( π(π) π§ + π(π(π)) . 1= π(π) π π≤π (π§) Since π(π) ≤ π for primes π large enough in terms of π , we now deduce from (41) that ∑ π¦0 (π )<π≤π§ π (π) = π(π )π§ + π((log π§)π ). π(π (π)) This relation may be summed by parts against the diο¬erentiable factor π ′ (π‘)/π (π‘). For large π‘, one has π ′ (π (π‘)/π (π‘)) βͺ π‘−2 , ππ‘ and consequently we are led to the asymptotic formula ( ) ∑ π ′ (π) (log π§)π = π(π ) log π (π§) + π˜(π ) + π , π(π (π)) π§ π¦0 (π )<π≤π§ where π˜(π ) is a real number, the value of which, yet again, plays no role in our conclusions. Substituting this formula into the deο¬nition (7), we infer that whenever π₯1/π (log π₯)−π΅ ≤ π¦ ≤ π₯1/π , one has ( ) π (π¦) Φπ (π¦1 , π¦) = π(π ) log + π(π₯−1/π (log π₯)π΅+π ). (48) π (π¦1 ) We now substitute (44) and (48) into the asymptotic relation (12), thereby deducing that ( ) ∑ ππ (π₯, π¦) = π₯π (π¦) log π (π¦)+π (π¦) (log π)2 − π₯ log π₯ π≤π₯ − 2Γ−1 π₯π (π¦) + π(π₯1/2 π (π¦)3/2 + π₯2 (log π₯)−π΄ ). By the Prime Number Theorem and partial summation, we therefore arrive at the asymptotic formula ππ (π₯, π¦) = π₯π (π¦) log π (π¦) − (2Γ−1 + 1)π₯π (π¦) + π(π₯1/2 π (π¦)3/2 + π₯2 (log π₯)−π΄ ), and this conο¬rms the conclusion of our theorem with πΆ(π ) = −(2Γ−1 + 1). SPARSE VARIANCE FOR PRIMES IN ARITHMETIC PROGRESSION 15 References [1] J. BruΜdern and A. Perelli, Goldbach numbers in sparse sequences. Ann. Inst. Fourier (Grenoble) 48 (1998), 353–378. [2] J. B. Friedlander and D. A. Goldston, Variance of distribution of primes in residue classes. Quart. J. Math. Oxford Ser. 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