REFINEMENTS OF SELBERG S SIEVE

REFINEMENTS OF SELBERG S SIEVE BY SARA ELIZABETH BLIGHT A issertation submitte to the Grauate School New Brunswick Rutgers, The State University of New Jersey in partial fulfillment of the requirements for the egree of Doctor of Philosophy Grauate Program in Mathematics Written uner the irection of Henryk Iwaniec an approve by New Brunswick, New Jersey May, 2010

ABSTRACT OF THE DISSERTATION Refinements of Selberg s Sieve by Sara Elizabeth Blight Dissertation Director: Henryk Iwaniec This thesis focuses on refinements of Selberg s sieve as well as new applications of the sieve Sieve methos are aresse in four ways First, we look at lower boun sieves We will construct new lower boun sieves that give us non-trivial lower bouns for our sums The lower boun sieves we construct will give better results than those previously known Secon, we create an upper boun sieve an use it to boun the number of primes to improve Selberg s version of the Brun-Titchmarsh Theorem We improve a constant in the boun of the number of primes in an arbitrary interval of fixe length Thir, we construct an upper boun sieve to improve the large sieve inequality in special cases Sieve methos allow us to improve this well-known boun of exponential sums Finally, we inclue some notes on the use of successive approximations to give a choice of an upper boun sieve that minimizes the main term an the remainer term simultaneously ii

Acknowlegements I woul like to thank my avisor, Professor Iwaniec, for his invaluable guiance over the years I have learne so much from his classes an from working with him on research Also, I woul like to thank the rest of my committee, Professors Chamizo, Miller, an Weibel for their help an feeback uring this process All of the Rutgers faculty have been very helpful, especially Professors Cohen, Greenfiel, an Nussbaum I want to thank my friens an family for their support My fellow grauate stuents have mae the grauate experience very enjoyable In particular, I want to thank Beth, Emilie, an Leigh for their wonerful frienship Finally, I woul like to thank my family for always listening, supporting me uring the tough times, an celebrating with me uring the goo times iii

Deication This thesis is eicate to my family an friens, especially to my Mom for her invaluable avice of starting with a fresh piece of paper iv

Table of Contents Abstract ii Acknowlegements iii Deication iv 1 Introuction 1 2 Sifting Limit 5 21 Introuction 5 22 Various Lower Boun Sieves 6 221 Beta Sieve 6 222 Diamon-Halberstam Sieve 8 223 Selberg Sieve 9 23 Selberg s Choice 12 24 Choice for Small Sifting Dimension 14 241 Choice of F t 17 25 Further Generality 19 251 Choice of Λ 21 252 Lower Boun Sieve with Three Primes 23 253 Analysis of T F s 25 26 New Sifting Limit Results 28 27 Analysis of Choice of Sieve 29 28 Appenix 30 3 Brun-Titchmarsh Theorem 35 31 Introuction 35 v

32 Improvement of Brun-Titchmarsh Theorem 36 321 Preliminaries 36 322 Brun-Titchmarsh Theorem 38 4 Large Sieve Inequality 42 41 Introuction 42 411 Improvement of Large Sieve 43 42 Proof of Theorem 45 421 Preliminary Analysis 47 422 Choice of g 49 423 Conclusion 49 5 Notes on Successive Approximations 52 51 Introuction 52 511 Property of S + 53 52 Preliminaries 55 53 Solution to System of Equations 59 6 Conclusion 67 61 Further Research 67 Vita 69 vi

1 Chapter 1 Introuction In sifting theory, a quantity we are intereste in is how many of the numbers n x have no prime factors p less than a parameter z If we take P z to be the prouct of all primes p z, then this means that we want to count how many numbers n x are such that n, P z = 1 In summation form, this is Sx, z = n x 1 11 n,p z=1 Although this is an interesting problem by itself, we woul like to look at an even more general problem Instea of looking at the sum 11, we look at the weighte sum below SA, x, z = n x a n 12 n,p z=1 where A = a n is a sequence of non-negative real numbers Now we woul like to estimate this sum We woul like to remove the conition n, P z = 1 from the summation One way we can o this is by using the Mobiüs function We recall that 1 if m = 1, µ = 13 0 otherwise Therefore, m SA, x, z = n x = n x a n a n n,p z n µ µ Unfortunately, the Mobiüs function oes not have nice asymptotics, so this is not the best approach for estimations Therefore, we look at more general functions If we

2 want an upper boun for SA, x, z, we take a sequence of real numbers Λ + = λ + such that Then m λ + SA, x, z = 1 if m = 1, 0 if m 1 n x a n n λ + We note that the two conitions we impose on λ + are equivalent to the conitions that λ + 1 = 1 an m λ+ 0 for all m If these conitions are satisfie an λ+ = 0 for > D, then Λ + = λ + is calle an upper boun sieve of level D With a clever choice of λ +, we will be able to construct goo upper bouns for SA, x, z Now we woul like to construct a lower boun for SA, x, z Trivially, we know that the sum is non-negative because each a n 0 We woul like to be able to construct non-trivial lower bouns as well If we want a lower boun, we take a sequence of real numbers Λ = λ such that m λ = 1 if m = 1, 0 if m 1 Then SA, x, z n x a n n λ We note that the two conitions we impose on λ are equivalent to the conitions that λ 1 = 1 an m λ 0 for all m 1 If these conitions are satisfie an λ = 0 for > D, then Λ = λ is calle a lower boun sieve of level D For now, we let SΛ = n x a n n λ 14 where Λ = λ is a general sieve of level D, either an upper boun sieve or a lower boun sieve Changing the orer of summation, we have SΛ = λ n x n 0 a n = λ A x

3 where A x = n x a n 15 n 0 In orer to treat the summation, we shall assume some asymptotics of the partial sums A x We write A x = gx + ra, where gx is the expecte main term an ra, is an error term which we think of as being small In the main term, X n x a n so g is the ensity of the masses a n attache to n 0mo If we think of ivisibility by istinct primes as inepenent events, we are le to assume that g is a multiplicative function with 0 < gp < 1 if p P z an gp = 0 otherwise Given these asymptotics of A x, we have SΛ where = X gλ + λ ra, = XV D, z + RA, D V D, z = gλ, RA, D = λ ra, In this thesis, we will aress sieve methos in four ways First, we will look at lower boun sieves A funamental problem in sieve theory is the sifting limit problem We wish to construct lower boun sieves that give non-trivial lower bouns for our sums Given a particular asymptotic for our ensity function g, the problem is to fin a lower boun sieve that gives a non-trivial lower boun Details of the problem are escribe in chapter 2 Selberg establishe the asymptotic result for this problem He also preicte a value for the sifting limit in general This is still an open problem In this thesis, we construct a lower boun sieve that gives better results than those previously known in many cases Secon, we will create an upper boun sieve an use it to boun the number of primes to improve Atle Selberg s version of the Brun-Titchmarsh Theorem We will

4 improve a constant in the boun of the number of primes in an arbitrary interval of fixe length Thir, we will construct an upper boun sieve to improve the large sieve inequality in special cases Sieve methos will allow us to improve this well-known boun of exponential sums Finally, we make some notes on the use of successive approximations to give an upper boun sieve that simultaneously minimizes the main term an remainer term We provie useful lemmas to solve systems of equations prevalent in sieve theory

5 Chapter 2 Sifting Limit 21 Introuction From chapter 1, we know that if Λ = λ is a lower boun sieve of level D, then where an SA, x, z = a n XV D, z + RA, D n x n,p z=1 V D, z = RA, D = gλ λ ra, Since a n 0 for all n, we know trivially that SA, x, z 0 We woul like to fin a choice of Λ that gives a nontrivial lower boun To o this, we first make a couple efinitions Definition 211 Let gp be a multiplicative function supporte on square free numbers with 0 < gp < 1 for p P z an gp = 0 for p P z an κ > 0 is a number which satisfies gp log p = κ log x + O1 21 p x Further assume for any w 2 with w < z that w p z Then κ is calle the sifting imension 1 gp 1 log z κ log w

6 To control the size of gp, we assume gp 2 log p < p In this chapter, we will be examining a quantity β κ, known as the sifting limit for a sifting imension κ A precise efinition of sifting limit is very complicate For such a efinition, the reaer is referre to Selberg s Lectures on Sieves [4, Section 14] Loosely speaking, for a sifting imension κ an a lower boun sieve Λ of level D, by the sifting limit β κ,λ we mean the minimum of log D/ log z for which V D, z > 0 for log D/ log z > β κ,λ an V D, z 0 for log D/ log z β κ,λ By the sifting limit β κ for a sifting imension κ, we mean the greatest lower boun of the β κ,λ over all possible lower boun sieves Λ of level D For ease of notation, in this thesis we enote β κ,λ by β κ when the sieve Λ is unerstoo Selberg propose that the sifting limit is 2κ He was able to prove this result asymptotically as κ approache infinity For 1/2 < κ < 1, Iwaniec an Rosser constructe a sieve with β κ < 2κ However, at this time, a lower boun sieve with a sieving limit of 2κ has not been foun for κ > 1 The sifting limit problem is to fin lower boun sieves that give β κ 2κ for each κ > 1 There are many types of lower boun sieves, each with its own avantages an isavantages We will mention a few of these sieves an the corresponing sifting limits Then we will give an improvement on Selberg s lower boun sieve, which will provie significant improvement of the sifting limit for κ 3 22 Various Lower Boun Sieves There are many choices of lower boun sieves that provie goo sifting limits Here we will focus on the beta-sieve, the Diamon-Halberstam sieve, an the Selberg sieve 221 Beta Sieve The beta sieve was create by Iwaniec an Rosser The sieve works very well when the sifting imension is small We let β κ enote the sifting limit for sifting imension κ

7 Using formulas B9, 1142, an 1157 of [2], we establishe the following numerical values of β κ κ β κ 05 10000000000 055 10340771100 06 11042161305 065 11922077070 07 12912892849 075 13981115251 08 15107489225 085 16279798714 09 17489723058 095 18731283112 1 20000000000 105 21292406269 11 22605745188 115 23937776845 12 25286648100 125 26650802364 13 28028915201 135 29419847168 14 30822608556 145 32236332483 15 33660254038 2 48339865967 Asymptotically, the beta sieve gives β κ cκ where c = 3591 is the number which solves the equation c/e c = e We also note that β κ < 2κ if 1 2 < κ < 1

8 222 Diamon-Halberstam Sieve The Diamon-Halberstam sieve works well for slightly larger sifting imension We have the following values of sifting limits [1, p227] κ β κ 10 2000000 15 3115821 20 4266450 25 5444068 30 6640859 35 7851463 40 9072248 45 10300628 50 11534709 55 12773074 60 14014644 65 15258588 70 16504285 75 17751146 80 18998853 85 20247056 90 21495510 95 22744013 100 23992408 The Diamon-Halberstam sieve is an infinite iteration of the Ankeny-Onishi sieve[1] Therefore, it is believe that the sifting limit β κ cκ as κ, where c = 2445

9 223 Selberg Sieve Finally, we turn to the Selberg sieve The Selberg sieve oes not give goo sifting limits when the sifting imension is very small However, asymptotically, β κ 2κ which is better than any other sieve By making ifferent choices an better estimates, we have been able to moify the Selberg sieve to give goo sifting limits for small κ These sifting limits are smaller than the sifting limits of the other sieves for κ 3 In this section, we explain Selberg s approach In the next section, we will explain the moifications With straightforwar calculations, we see that if Λ + is an upper boun sieve of level D 1 an Λ is a lower boun sieve of level D 2, then Λ = Λ + Λ is a lower boun sieve of level D 1 D 2, efine by n λ = n Applying this lower boun sieve, we have SA, x, z = n x n,p z=1 a n n x λ + a n n n λ λ = n x a n n λ n λ + For Λ, Selberg chose λ 1 = 1, λ p = 1 for p z an λ = 0 otherwise, so that Λ is a lower boun sieve of level z Then, n λ = 1 p n p P z This is a crue lower boun sieve However, with a goo choice of Λ +, the overall choice of Λ = Λ Λ + is still goo For Λ +, Selberg chose his Λ 2 sieve, which is the convolution 1 of Λ with itself, Λ + = ΛΛ with Λ = {ρ } That is he chose λ + such that m λ + = m ρ 2

10 where {ρ } is another sequence of real numbers with ρ 1 = 1 an ρ = 0 for > D/z = Y Then Λ + is an upper boun sieve of level D/z an Λ = Λ Λ + is a lower boun sieve of level D He kept the choice of ρ open Applying these choices, we see SA, x, z = n x a n 1 2 1 ρ p n n p P z = XV D, z + RA, D where V D, z = gλ, RA, D = λ r A, in the notation of chapter 1 We note that with the above efinitions, λ = λ + p λ + + λ+ /p Rewriting λ + in terms of ρ an noting that is squarefree, an manipulating the result, we see that Then V D, z = 1 λ = [ 1, 2 ]= ρ 1 ρ 2 [p, 1, 2 ]= ρ 1 ρ 2 g[ 1, 2 ]ρ 1 ρ 2 g[p, 1, 2 ]ρ 1 ρ 2, p 2 2 where p, 1, 2 run inepenently over ivisors of P z, p prime In orer to look at the first sum, we first efine the multiplicative function h by hp = gp 1 gp We note that since 0 < gp < 1 for p P z, we have hp > 0 for all p P z an h > 0 for all Since gp = 0 for p P z, hp = 0 for p P z Then we have g[ 1, 2 ]ρ 1 ρ 2 = h 1 2 1 1 m 0mo gmρ m 2 In orer to treat the secon sum, we nee to make some more efinitions We efine g p = g[p, ]/gp an h p as h p = if p an h p = h otherwise

11 Finally, we efine G p := h p 1 g p mρ m 2 m 0mo By expaning the square an simplifying, we fin G p = g[p, m 1, m 2 ] ρ m1 ρ m2 gp m 1 m 2 Therefore, p gpg p = gp p = m 1 2 h p 1 g p mρ m m 0mo m 2 ρ m1 ρ m2 g[p, m 1, m 2 ], which is the secon sum in our expression for V D, z Rewriting g p an h p in terms of g an h, we fin that p P z gpg p = p gp h m 0mo z gmρ m + 1 hp m 0mo p z gmρ m 2 Finally, V D, z = 1 h p m 0mo z gp h gmρ m 2 m 0mo z gmρ m + 1 hp m 0mo p z gmρ m 2 Since the sum m 0mo z gmρ m is prevalent, we make a change of variables to simplify the expression Selberg chose Making this substitution, we fin V D, z = y = µ h hy 2 z m 0mo p gmρ m 22 2 gph y y p 23

12 We note that the original variables ρ can be foun in terms of the new variables y by Mobiüs inversion For the normalization, we note that ρ 1 = 1 means that 1 = hy Also, the support of ρ being Y is equivalent to the support of y being Y This is the point where two ifferent paths may be taken The first path is ieal for large sifting imension because it illumines the asymptotics Selberg was able to achieve However, in orer to clearly see the asymptotics, some estimates are mae which worsen the result for small sifting imension The secon path is more computationally heavy, but yet gives better results for small sifting imension The work of this thesis expans upon the secon path to give even more precise results for small sifting imension We will explore these paths in the following sections 23 Selberg s Choice In section 223, we gave the set-up of Selberg s sifting limit argument We now continue with an explanation of his work Since hp gp, equation 23 gives V D, z hy 2 hpy y p 2 24 p Rewriting this, we fin V D, z hy 2 hpy y p 2 where p = h y2 y /p y 2 p = h{y 2 l} l = p y /p y 2 Selberg then chose { } log Y/ min 1, if 1 Y y = J 1 log z 0 otherwise, where J = h Y

13 We recall that Y = D/z We note that with this choice of y, l = 0 except for x/z < < xz In this range, l = p y /p y 2 J 2 p log p 2 J 2 log p 2 log = J log z log z log z 25 p By making these crue estimates, we lose some of our precision The precision will not matter for the main term of the asymptotic result, but it oes effect the result for small sifting imension We have Y/z<<Y z h{y 2 2 l} J Y/z<<Y h log log z 26 Also, Therefore, Y/z h{y 2 2 l} = J Y/z h 27 J 2 V D, z J 2 Y/z Y/z = = h{y 2 l} h Y/z<<Y z h log Y z log z h h Definition 231 We efine V z by We efine Y/z<<Y z log Y z2 log z V z 1 = h log log z Y/z h h log Y z log z Y/z<<Y z h IX, z = h X h Y/z<<Y z h

14 Then J 2 V D, z so J 2 V D, zv z V z 1 1 log Y z2 IY/z, z log z log Y z2 IY/z, zv z log z From Opera e Cribro by Frielaner an Iwaniec [2, p111], we have 2eκ t/2 IX, zv z e κ t if t = 2 log X/ log z > 2κ We note that s = log D/ log z = 2log Y/ log z+1 Therefore, log Y/ log z = s 1/2 Letting X = Y/z, we have t = s 3 Therefore, if s > 2κ + 3, Hence, Thus, V D, z > 0, if 2eκ s 3/2 IY/z, zv z e κ s 3 J 2 V D, zv z 1 s + 3 2e κ s + 3 2e κ 2eκ s 3/2 < 1 s 3 2eκ s 3/2 s 3 assuming s > 2κ + 3 This occurs when s > 2κ + 2 2κ log κ + log κ + 9 This provies an upper boun for the sifting limit in the case of sifting imension κ 24 Choice for Small Sifting Dimension In the previous section, we mae some estimates to give a clear asymptotic We now use better bouns to achieve goo sifting limits for small sifting imension In equation 23 we choose our variables y as follows log J 1 F if 1 Y y = log Y 0 otherwise 28 where F is a general continuous, piecewise smooth function which will be picke to optimize results for each sifting imension We efine α = log z/ log Y = 2/s 1

15 We note that Selberg s choice of y from the previous section correspons to: 1 if 0 t 1 α, F t = 29 1 α 1 t if 1 α < t 1 Now, we have J 2 V D, z = Y log hf 2 log Y h p P z [ log gp F F log Y ] log p 2 log Y We now treat the sum over primes in the secon line We assume that F is a continuous piecewise smooth function with F 0 F u 2 u for 0 u 1 We now apply Lemma 283 with [ log p log log Φ = F F log Y log Y log Y + log p ] 2 log Y Therefore, we have p P z α = κ [ log gp F F log Y 0 ] log p 2 log Y F v F v + u 2 u u + O log log z log z where α = log z/ log Y an v = log / log Y We apply Lemma 284 with Φv = F 2 v κ α 0 F v F u + v 2 u u For the contribution from the error term of Olog log z/ log z, we note that h = V z 1 Then we have c 1 V zj 2 V D, z = κ +O 1 F v 2 fv/α 0 1 α 0 c F v F u + v 2 u 0 u fv/α 1 log log z log z where c 1 = e γκ Γκ + 1, γ is Euler s constant, f is given by 216 an V z = p P z 1 + hp 1

16 Then we have Then c 1 V zj 2 V D, z = α c V zj 2 V D, z = κ +O κ +O 1 1 F v 2 fv/α 0 1 α 0 c 0 1 α 0 0 F v F v + u 2 u u fv/α 1 log log z log z F 2 vf v/αv 0 V z 1 α F v F u + v 2 1 u f v/αuv log log z log z Applying the conition that F v = 0 for v > 1, we have α c V zj 2 V D, z = κ κ κ +O 1 0 1 F 2 vf v/αv 1 v 1 α 0 1 α α 0 1 1 α 0 α 1 v V z 1 α Definition 241 For α = 2/s 1, we efine T F s = κ κ κ 1 0 1 F 2 vf v/αv 1 v 1 α 0 1 α α 0 1 1 α With this efinition, we have 0 α 1 v F v F u + v 2 1 u f v/αuv F v F u + v 2 1 u f v/αuv F v 2 1 u f v/αuv log log z log z F v F u + v 2 1 u f v/αuv F v F u + v 2 1 u f v/αuv F v 2 1 u f v/αuv Proposition 242 Let F be a continuous piecewise smooth function with F 0 F u 2 u for all 0 u 1 an F v = 0 for v > 1 Assume T F s as efine above is positive Then there is some z 0 such that if z > z 0, then V D, z is also positive

17 Proof As z, the error term above approaches zero an the main term is positive as state 241 Choice of F t We have V D, z positive if T F s is positive Now we wish to examine ifferent choices for the function F We woul like to fin the minimum of s such that T F s is positive for some function F To o so, we will look at various families of functions We recall that the requirement on F is that it is continuous, piecewise smooth an satisfies F 0 F u 2 u for all 0 u 1 For a particular function F, we let β κ F be the minimum s such that T F s is positive in the case of sifting imension κ For ease of notation we will enote β κ F by β κ where F is unerstoo We note that F t = 1 t m + c for any m > 1/2 an constant c satisfies the conitions on F With the choice of F t = 1 t, we have β 5 < 1076 β 4 < 87499 β 35 < 781 With the choice of F t = 1 t 07 we have β 3 < 66125 All of these sifting limits are smaller than the sifting limits given by the Diamon- Halberstam sieve We note that all computations were one using Maple math software We woul also like to compare this result to the sieve given in section 23 To fin asymptotics of the sifting limit, Selberg chose y corresponing to the choice of F t given in 29 The sifting limit using F t = 1 t is much better than this choice of Selberg For example for κ = 3, the function F t = 1 t gives us β 3 < 675 while Selberg s choice gives us β 3 < 724

18 Although this choice of Selberg is not the best choice for small sifting imension, his choice oes give us some insight into the problem He chose a piecewise efine function for F with a break at t = 1 α We note that this is a natural choice ue to the limits of integration in the efinition of T F s We now follow this example, but keep the efinition of F very general to allow us more freeom We efine F 1 t if 0 t 1 α F t = F 1 1 α F 2 1 α F 2t if 1 α < t 1 210 where F 1 t an F 2 t are continuous, piecewise monotonic functions such that F 1 0 F 1 w 2 w an F 2 0 F 2 w 2 w for 0 w 1 Then F t is a continuous piecewise monotonic function that satisfies F 0 F w 2 w for 0 w 1 We also note that the simple case of a single function F follows when F 1 = F 2 Using this efinition of F t, we fin that T F s is: T F s = 1 α 0 + F 11 α 2 F 2 1 α 2 κ F 11 α 2 F 2 1 α 2 κ F 11 α 2 F 2 1 α 2 κ κ κ 1 α α F 1 v 2 f v/αv 1 1 α 1 F 2 v 2 f v/αv α 1 α 1 v 1 1 v 1 α 1 2α 1 α v 1 α 1 α v 1 2α 0 1 2α α 0 0 0 F 2 v 2 1 u f v/αuv F 2 v F 2 u + v 2 1 u f v/αu F 1 v F 11 α 2 F 2 1 α F 1 2u + v F 1 v F 1 u + v 2 1 u f v/αuv F 1 v F 1 u + v 2 1 u f v/αuv u f v/αuv This oes give us more improvements For κ = 3, we chose F 1 t = 1 t 083 an F 2 t = 1 t 057 With this choice, we have: β 3 < 6576 With the choice of F t = 1 t 07, we ha β 3 < 66125

19 Another choice of F suggeste by Selberg [5, p482] is F t = 1 t + c where c is a constant epenent on κ With c = 01, we have β 3 < 65206 With c = 007, we have β 4 < 853 By using the piecewise efine F with F 1 t = 1 t 086 an F 2 t = 1 t 098 +01 we achieve β 3 < 651998 which is an improvement over Selberg s metho presente in section 23 25 Further Generality In the previous section, we consiere the sieve Λ Λ + where Λ + was Selberg s Λ 2 sieve an Λ = λ q was given by λ 1 = 1, λ p = 1 for p z, an λ q = 0 otherwise Now, we woul like to consier a more general lower boun sieve Λ We let Λ = λ q supporte on q z an we let Λ + = Λ 2 = ρ 2 be Selberg s Λ 2 sieve in terms of ρ with support D/z = Y Proposition 251 With Λ = Λ Λ 2 = λ, we have V D, z = D gλ = q z q P z λ q gq D/z 2 h µcy c 211 c q where y = µ h m 0mo z gmρ m 212

20 Proof We first write λ in terms of λ q an ρ Therefore, n x a n n λ = a n n x q n = n x = a n 1 Y q P z q n q P z ρ 1 λ q λ q 2 Y 2 P z 1 n 1 P z ρ 2 n ρ 1 q z q P z ρ 2 2 n 2 P z λ q ρ 2 n 0mo [q, 1, 2 ] a n V D, z = D = 1 Y 1 P z gλ ρ 1 2 Y 2 P z ρ 2 q z q P z = λ q gq ρ 1 q z 1 Y q P z 1 P z λ q g[ 1, 2, q] 2 Y 2 P z [1, 2, q] ρ 2 g q Now we turn to the right-han sie of equation 211 We only nee to show that 1 Y 1 P z ρ 1 2 Y 2 P z [1, 2, q] ρ 2 g q Applying the efinition of y c 212, we fin Y h c q µcy c 2 = Y = Y = Y = Y 1 h c q 1 h 1 h 2 h µcy c 1 hc m 0mo z m 0mo z c q m 0mo c z gmρ m gmρ m 2 c q,m 1 2 hc 1 2 gmρ m gq, m

21 Now we expan the square Therefore, Y = Y = m Y z = m Y z = m Y z 2 h µcy c 1 h ρ m ρ m ρ m c q V D, z = m 0mo z n Y n P z n Y n P z n Y n P z D ρ n ρ n gmρ m 1 gq, m gmgn gq, mgq, n m,n n 0mo n P z 1 h gmgn gq, mgq, ngm, n [m, n, q] ρ n g q gλ = q z q P z λ q gq Y gnρ n 1 gq, n 2 h µcy c c q 251 Choice of Λ We now consier specific choices for Λ Previously, we chose λ 1 = 1, λ p = 1 if p P z an λ = 0 otherwise We can instea make the following choice: Lemma 252 Let λ 1 = 1, λ p = 1 for p P z, λ p1 p 2 = 1 for p 2 < p 1 z 1/3, λ p1 p 2 p 3 = 1 for p 3 < p 2 < p 1 z 1/3, an λ = 0 otherwise, where p 1 p 2 p 3 P z Then Λ = {λ } is a lower boun sieve of level z Proof We first note that λ 1 = 1 Also, the sieve is of level z because λ = 0 for z We have n λ n λ

22 where n is the squarefree part of n with all prime ivisors z 1/3 since the rest contributes a non-positive amount Let n = p 1 p m Then m m m m 1 λ = 1 + = 0 1 2 3 3 n The last equality follows from the ientities m m 1 m 1 m = +, an = 3 3 2 2 m 1 2 + m 1 1 have We let Y = D/z Then accoring to Proposition 251 an Lemma 252, we V D, z = hy 2 Y gp 1 p 1 z Y p 1 P z + p 2 <p 1 z 1/3 p 1 p 2 P z p 3 <p 2 <p 1 z 1/3 p 1 p 2 p 3 P z hy y p1 2 gp 1 gp 2 Y gp 1 gp 2 gp 3 Y 2 h y y p1 y p2 + y p1 p 2 y y p1 p 2 p 3 h y p1 y p2 y p3 +y p1 p 2 + y p1 p 3 + y p2 p 3 2 The above is a logical choice of lower boun sieve However, Selberg presente another choice of lower boun sieve that will give more flexibility an in fact better results Lemma 253 Let T be a positive integer Let λ 1 = 1, λ p = 1 for p P z, λ p1 p 2 = 4T 2/T T + 1 for p 2 < p 1 z 1/3, λ p1 p 2 p 3 = 6/T T + 1 for p 3 < p 2 < p 1 z 1/3 an λ = 0 otherwise, where p 1 p 2 p 3 P z Then Λ = {λ } is a lower boun sieve of level z Proof We first note that λ 1 = 1 Also, the sieve is of level z because λ = 0 for z We have n λ n λ

23 where n is the squarefree part of n with all prime ivisors z 1/3 since the rest contributes a non-positive amount Let n = p 1 p m n λ = 1 m + 4T 2 1 T T + 1 m 2 6 T T + 1 m 1T mt m 1 = 0 T T + 1 m 3 since T is an integer an m 1 We o note that if T is not an integer, this conition oes not hol With this choice of Λ for some integer T, we have V D, z = hy 2 Y gp 1 p 1 z Y p 1 P z + 4T 2 T T + 1 6 T T + 1 p 2 <p 1 z 1/3 p 1 p 2 P z p 3 <p 2 <p 1 z 1/3 p 1 p 2 p 3 P z hy y p1 2 gp 1 gp 2 Y gp 1 gp 2 gp 3 Y hy y p1 y p2 + y p1 p 2 2 y y p1p 2p 3 + y p1p 2 h y p1 y p2 y p3 +y p1 p 3 + y p2 p 3 2 252 Lower Boun Sieve with Three Primes In this section, we present the sifting limit arguments for the lower boun sieve Λ = Λ Λ 2 where Λ is given by λ 1 = 1, λ p = 1 for p P z, λ p1 p 2 = 4T 2/T T + 1 for p 2 < p 1 z 1/3 an λ p1 p 2 p 3 = 6/T T + 1 for p 3 < p 2 < p 1 z 1/3, an λ = 0 otherwise, where p 1 p 2 p 3 P z We again make the following choice of y log J 1 F if 1 Y y = log Y 0 otherwise 213 where Y = D/z

24 To make notation easier, we introuce some new variables We let v = log / log Y, u = log p 1 / log Y, w = log p 2 / log Y an x = log p 3 / log Y With this notation, we see that J 2 V D, z = hf v 2 Y gp 1 p 1 z Y h F v F u + v 2 + 4T 2 T T + 1 gp 1 gp 2 p 2 <p 1 z 1/3 Y h 6 gp 1 gp 2 gp 3 T T + 1 p 3 <p 2 <p 1 z 1/3 Y F v + F u + w + v F u + v F w + v F v F u + v 2 F w + v F x + v +F u + w + v +F u + x + v +F w + x + v F u + w + x + v 2 We now apply the same arguments as in section 24 to this expression in orer to analyze the sums We employ the notation that α = log z/ log Y We also let β = α/3 = log z 1/3 / log Y, so that x < w < u β in the secon an thir sums Finally, we note that Y means v 1 Therefore, α c V zj 2 V D, z = κ 1 F v 2 f v/αv 0 1 α 0 0 +κ 2 4T 2 T T + 1 κ 3 6 T T + 1 +O α c F v F u + v 2 f v/αuv 1 β β 0 0 w 1 β β β 0 log log z log z 0 x w F v + F u + w + v F u + v F w + v F v F u + w + x + v F u + v + F u + w + v F w + v + F u + x + v F x + v + F w + x + v 2 v u f α u 2 f v α w w v u w x u w x v We again efine T F s to be the right-han sie of the above expression without the error term Then we have V D, z > 0 for large enough z if s is chosen such that T F s > 0 We now turn to the analysis of T F s

25 253 Analysis of T F s This section escribes the computation of T F s The computations are teious an long, so we only inclue highlights of the computations The results from these computations can be foun in the next section The first step in computing T F s is to apply the conition that F v = 0 for v > 1 We make the following efinitions: B = {0 v 1, 0 u α} B 1 = {u, v B : u + v 1} B 2 = {u, v B : u + v > 1} C = {0 v 1, 0 w β, w u β} C 1 = {u, w, v C : u + v + w 1} C 2 = {u, w, v C : u + v + w 1, u + v 1} C 3 = {u, w, v C : u + v 1, w + v 1} C 4 = {u, w, v C : w + v 1} D = {0 v 1, 0 x β, x w β, w u β} D 1 = {u, w, x, v D : x + v 1} D 2 = {u, w, x, v D : x + v 1, w + v 1} D 3 = {u, w, x, v D : w + v 1, u + v 1, x + w + v 1} D 4 = {u, w, x, v D : x + w + v 1, u + v 1, x + w u} D 5 = {u, w, x, v D : u + v 1, x + w + v 1, x + w u} D 6 = {u, w, x, v D : u + v 1, x + w + v 1, x + u + v 1} D 7 = {u, w, x, v D : x + u + v 1, w + u + v 1} D 8 = {u, w, x, v D : w + u + v 1, x + w + u + v 1} D 9 = {u, w, x, v D : x + w + u + v 1} Then we have T F s +T A κt B + κ 2 4T 2 T T + 1 T C κ 3 6 T T + 1 T D

26 where T C = D 9 T A = 1 0 F v 2 f v α v v u T B = F v F u + v 2 f B 1 α u v v u + F v 2 f B 2 α u v + C 1 F v F u+v F w+v+f u+w+vf C 2 F v F u+v F w+v 2 f + v + F v 2 f C 4 α C 3 F v F w+v 2 f v α u w u w v v α u w u w v v α u w u w v u w u w v v u T D = F v 2 f w x D 1 α u w x v v u + F v F x+v 2 f w x D 2 α u w x v v u + F v F x+v F w+v 2 f w x D 3 α u w x v v u + F v F x+v F w+v+f x+w+v 2 f w x D 4 α u w x v v u + F v F x+v F w+v F u+v 2 f w x D 5 α u w x v v u + F v F x+v F w+v F u+v+f x+w+v 2 f w x D 6 α u w x v 2 F v F x + v F w + v F u + v v u + f w x +F x + w + v + F x + u + v D 7 α u w x v 2 F v F x + v F w + v F u + v v u + f w x +F x + w + v + F x + u + v + F w + u + v D 8 α u w x v 2 F v F x + v F w + v F u + v + F x + w + v v u + f w x +F x + u + v + F w + u + v F x + w + u + v α u w x v Now we can simply use the efinition of F t on the interval 0 t 1 instea of a piecewise efinition The next step is to write the sets B i, C i an D i as unions of sets such that the variables are in intervals For example, B 1 = {1 α v 1, 0 w 1 v} {0 v 1 α, 0 w α}

27 We o not inclue the ecomposition of each set here, since the ecompositions are quite complicate For example, D 8 ecomposes into 22 such sets Ieally, once we have the ecompositions we woul like to be able to input T F s into a math software program to compute its value an leave the choice of F t open Unfortunately, the integrals over the C i an D i are too ifficult for typical math software to hanle Hence, we now make a general choice of F t = 1 t+c where c is a constant to be chosen later With this choice of F, we can now simplify some of the integrals by han using changes of variables an by changing the orer of integration The math software oes not use these techniques We simplify the integrals as much as necessary in orer for the software to be able to process the rest In some cases, we estimate some of the integrals in orer to make computation time feasible In two cases, we use the estimate that ln 1 u 4 ln 2 2u 2 + u for u [0, 1/2] We note that our choice of F t = 1 t + c simplifies some of our computations quite a bit For example, over C 1, F v F u + v F w + v + F u + w + v = 0 Over D 4, F v F x + v F w + v + F x + w + v = 0 Finally, over D 9, F v F x + v F w +v F u+v+f x+w +v+f x+u+v+f w +u+v F x+w +u+v = 0 Applying our efinition of F t, we have T A = 1 0 1 + c v 2 f v α v v T B = uf uv B 1 α v u + 1 + c v 2 f B 2 α u v T C = 1 + c u v w 2 f C 2 v u + wf C 3 α u wv v + 1 + c v 2 f C 4 α v α u w u w v u w u w v

28 v u T D = 1 + c v 2 f w x D 1 α u w x v v u + xf w D 2 α u w xv v u + 1 + c x w v 2 f w x D 3 α u w x v v + 2 + 2c 2v x w u 2 f D 5 α v u + 1 + c v u 2 f w x D 6 α u w x v v u + xf w D 7 α u w xv v + 1 + c v u w x 2 f D 8 α u w x u w x v u w x u w x v Once we have the integrals in a simple enough form, we use Maple software to numerically integrate the functions with error less than 10 10 The results of these computations are in the next section 26 New Sifting Limit Results Let Λ 1 = Λ 2 Λ where Λ = {ρ } an Λ = {λ q } We efine λ q by λ 1 = 1, λ p = 1 for p P z, λ p1 p 2 = 4T 2/T T + 1 for p 2 < p 1 z 1/3, λ p1 p 2 p 3 = 6/T T + 1 for p 3 < p 2 < p 1 z 1/3 an λ = 0 otherwise, where T is an integer to be etermine later an p 1 p 2 p 3 P z In the notation of the previous sections, with the choice of F t = 1 t + c, we have the following results: For κ = 2, with c = 02214971799 an T = 16, we have β 2 < 445 For κ = 25 with c = 017 an T = 19, we have β 25 < 5455 For κ = 3 with c = 013 an T = 24 we have β 3 < 6458

29 For κ = 4 with c = 011 an T = 31 we have β 4 < 847 The sieve of Diamon an Halberstam gives a smaller sifting limit for κ = 2 an κ = 25 However, our sifting limit is very close to that of Diamon an Halberstam for κ = 25 In aition, the sifting limits for κ = 3, 4 are much smaller than those given by the Diamon-Halberstam sieve 27 Analysis of Choice of Sieve There are many choices for a lower boun sieve In the previous sections, we have mae the choice of Λ 1 = Λ Λ 2, a convolution of a lower boun sieve an Selberg s Λ 2 upper boun sieve Our original choice of Λ was λ 1 = 1, λ p = 1 for p P z an λ = 0 otherwise This is a pretty weak lower boun sieve an yet our results have been very goo This is ue to the power of Selberg s Λ 2 sieve For example, with this choice of Λ an Λ = {ρ } we have SA, x, z = a n 1 1 n x p n p P z 2 ρ 214 n We woul like SA, x, z to be a goo estimate for the number of n x such that n, P z = 1 If n oes not have any prime factors less than z, then a n is counte with weight 1 If n has one prime factor less than z then it is counte with weight zero The sieve comes into play when n has more than one prime factor less than z In such a case, the lower boun sieve Λ gives a negative weight to a n However, when n is highly composite, the upper boun sieve Λ 2 gives a weight close to zero Therefore, the overall weight for a n is negative but small Therefore, the crueness of the lower boun sieve Λ is counterbalance by the power of the upper boun sieve Λ 2 When we change our Λ to also aress n with up to three prime factors, we saw an improvement in the sieve However, the improvement was relatively small, so it is likely that further improvements to Λ will not have a noticeable effect on the sifting limit results The traeoff between better results an computational ifficulty seems to make a Λ aressing five prime factors inavisable

30 It is possible that a better choice of Λ that still aress three prime factors coul be use So far, we have use the choice of Λ with λ 1 = 1, λ p = 1 for p P z, λ p1 p 2 = 4T 2/T T + 1 for p 2 < p 1 z 1/3, λ p1 p 2 p 3 = 6/T T + 1 for p 3 < p 2 < p 1 z 1/3 an λ = 0, where T is an integer an p 1 p 2 p 3 P z This choice of Λ makes computations feasible However, it oes not aress the case when n has two or three prime factors with one of the factors p > z 1/3 This accounts for a small portion of n, so we o not lose much We can take this case into account though by choosing Λ with λ 1 = 1, λ p = 1 for p P z, λ p1 p 2 = f 1 T for p 2 < p 1 z such that p 2 2 p 1 z an λ p1 p 2 p 3 = f 2 T for p 3 < p 2 < p 1 z with p 2 2 p 1 z, where p 1 p 2 p 3 P z an f 1 T, f 2 T are some functions that make Λ a lower boun sieve of level z 28 Appenix In this appenix, we give some useful general lemmas for sifting limit calculations These lemmas are reference in the previous chapter Throughout this appenix, we let κ > 0 an let gp be a multiplicative function with the three following properties 1 p x gp log p = κ log x + δx where δx is boune for x 2 2 w p z 1 + gp log z log w κ if z > w 2 3 p gp2 log p < We note that these three conitions are satisfie by the multiplicative function g that gives the ensity of the masses a n attache to n 0 mo in sifting theory The three conitions are also satisfie by the multiplicative function h efine by hp = gp/1 gp Lemma 281 We have p x where b is a constant given in the proof 1 gp = κ log log x + b + O log x

31 Proof We first write p x gp log p = Lx = κ log x + Rx with Rx 1 We then have p x gp = x 2 log u 1 Lu x x 1 = κ 2 log u log u + Ru 2 log u log x u = κ log 2 u + Ru log u x 2 + 2 x = κ log log x κ log log 2 + Rx log x + 1 Ru + ulog u 2 u Ru x ulog u 2 u Ru 2 ulog u 2 u where the 1 term appears because R2 = log 2/2 We note that the thir term Rx/ log x 1/ log x Aitionally, the final term is O1/ log x Therefore, gp = κ log log x + b + O1/ log x p x where Ru b = 1 κ log log 2 + 2 ulog u 2 u We now give a general theorem from Opera e Cribro [2, Theorem A7] Theorem 282 Y g = W z { cflog Y/ log z + Olog Y 1 } 215 where f is the solution to the ifferential ifference equation: ft = t κ if 0 < t 1, tf t = κft ft 1 if t > 1, 216 W z = 1 + gp, p<z an c = e γκ Γκ + 1 1 With the above theorem, we can prove the following general lemma:

32 Lemma 283 Let Φw be a continuous piecewise smooth function with Φw = Ow Let α = log z/ log Y Then p z log p gpφ = κ log Y α 0 Φw w log log z w + O log z Proof S = log p gpφ log Y p z z log w = Φ gp p w = 1 z 1 Φ log Y log w log Y z log log ww + Φ 1 log w Rw log Y where Rw 1/ log 2w By partial summation, z log w S = Φ log log ww 1 log Y log w z +O Φ Rw z 1 + Rw log w log Y Φ 1 log Y We note that since Φu u, we have Φ log w/ log Y 1/w log Y S = = α 0 α 0 Φw 1 z log w w + O Y 1 + Φw 1 w w + O log log z log z 1 w w log 2w We nee one more final lemma: Lemma 284 Let Φv be a continuous, piecewise smooth function for 0 v 1 with Φv = Φ0 + Ov Then c 1 W z 1 gφ log log Y 1 Y = +O Y 1 Φ log w f log Y 1 log log Y c log Y log w log z where c, W z, an f are as in Theorem 282

33 Proof Define Φ 1 v = Φv Φ0 Then Φ 1 v = Ov Suppose that the result is true for Φ 1 v Then S = 1 Y = Φ0 log gφ log Y 1 Y g + 1 Y gφ 1 v From Theorem 282, we have Y log Y g = cw zf + O log z 1 W z log 2Y Then S log Y = cw zf log z +O log log Y W z log Y Y = cw z +O 1 W z Y log w = cw z Φ 1 1 log Y Y Φ0 + cw z 1 log w Φ0f log z log log Y log Y f log w Φ 1 f log Y log w log z Y log w + cw z Φ 1 f 1 log Y log w log log Y + O W z log z log Y log w log z We have now reuce the question to Φv with Φv = Ov S = = 1 Y Y 1 Φ Y = cw z log gφ log Y log w log Y 1 Φ w log w log Y g f +O W z Φ1 log 2Y + W z log w log z Y 1 1 log 2w log w Φ log Y

34 We note that Φv = Ov an Φv piecewise smooth gives Φ v 1, so S Y log w = cw z Φ 1 log Y W z +O log Y + W z log Y = cw z 1 0 Φvf Y 1 f v α + O log w log z w w log 2w log log Y W z log Y

35 Chapter 3 Brun-Titchmarsh Theorem 31 Introuction A historic problem in analytic number theory is to boun the number of primes in an interval of a fixe length We efine πx to be the number of primes p x If we wish to boun the number of primes in an interval of length y, we want to boun πx+y πx We want to fin some function F y that is inepenent of x such that πx + y πx < F y for all x, when y is sufficiently large A theorem with a boun of this type is calle a Brun-Titchmarsh Theorem name for Viggo Brun an Ewar Charles Titchmarsh Brun first establishe a boun of this type in 1915 with with a constant c 1 > 2 F y = c 1y log y Then Titschmarsh establishe a boun for primes in arithmetic progressions We efine πx; a, k to be the number of primes p x such that p is congruent to a moulo k Then we want a function Gy, k inepenent of x an a such that πx + y; a, k πx; a, k < Gy, k for y sufficiently large Titchmarsh prove this for Gy, k = c 2 y φk log y/k with a constant c 2 > 2, where φ is Euler s totient function

36 Various improvements of these bouns have been mae over the years Our focus is on Selberg s improvement using sieve methos In section 22 of his Lectures on Sieves [4], Selberg establishe the following two bouns: 2y πx + y πx < log y + 28 πx + y; a, k πx; a, k < 2y φk log y/k + 28 Following Selberg s notes, we are able to moify his constructions an create a computer program to establish the improve bouns: 2y πx + y πx < log y + 28168 2y πx + y; a, k πx; a, k < φk log y/k + 28168 These bouns will be proven using sieve methos in the next section 32 Improvement of Brun-Titchmarsh Theorem 321 Preliminaries In orer to establish the bouns we esire, we first introuce a function θ q, following Selberg s work Definition 321 For q prime, we efine Q to be the prouct of all primes p q Then we efine θ q = max a,x x p q 1 1 p a n a+x n,q=1 1 We note that the maximum exists by perioicity We also note that p q 1 p 1 = φq/q In his Lectures on Sieves [4, Section 22], Selberg prove the following Lemma regaring θ q : Lemma 322 Selberg With θ q as efine in 321, we have that θ q grows faster than any power of q Also θ 1 = 1, θ 2 = 1, θ 3 = 4/3, θ 5 = 28/15 an θ 7 = 106/35

37 We can exten this result further by computing values of θ q for larger q Selberg compute the above values, an with a simple computer program we compute the following values Lemma 323 B With θ q as efine in 321, we have that θ 1 = 1, θ 2 = 1, θ 3 = 4/3, θ 5 = 28/15, θ 7 = 106/35, θ 11 = 388/77, θ 13 = 7102/1001, θ 17 1087759 an θ 19 1696824 Proof To etermine θ q we may assume 0 a Q an 0 x Q ue to the perioicity mo Q Therefore, θ q can always be foun by inspecting a finite number of cases In this way, we are able to compute the values in the lemma For the proof of the growth of θ q, we refer the reaer to section 22 of Selberg s Lectures on Sieves [4, Lemma] Selberg then establishe the following technical lemma: Lemma 324 Selberg Let σρ enote the sum of the ivisors of ρ Let φρ be Euler s function an once again let Q be the prouct of all primes p q Then for z > 1, we write: 1 = σρ z ρ,q=1 2 = σρ z ρ,q=1 3 = σρ z ρ,q=1 µ 2 ρ σρ 1 σρ, φρ z µ 2 1 ρ φρ µ 2 1 ρ φρ 1 σρ z 1 σρ 2 z Then 1 = z 2 2 = p q p q 1 1 + O ze log z, p 1 1 p {log z + γ + κ1 + κ q 1 } + O e log z 3 = 1 1 { log z + γ + κ 1 + κ q 3 } + O e log z p 2 p q

38 Here γ is Euler s constant, while κ 1 κ q = p = p q log p log p + 1 p 1 p log p + 1 p For a proof of this lemma, the reaer is again referre to Section 22 of Selberg s Lectures on Sieves [4] In the next section, we will use these preliminaries to prove the Brun-Titchmarsh Theorem 322 Brun-Titchmarsh Theorem We now wish to apply the preliminaries establishe in the previous section to another problem Consier the interval I x = [b, b + x] We exclue one resiue class for each prime less than z We then look for an upper boun of the number of integers left We choose z just large enough to get the smallest possible upper boun that the metho allows Without loss of generality, we can assume that the resiue class is always represente by zero Then we look at the case where we are excluing all n which are ivisible by p for p z We assume that we have alreay exclue the n that are ivisible by p q We then apply a Λ 2 sieve with P, the prouct of all primes p with q + 1 p z an Λ = ρ We also recall that φq is the prouct of all primes p q Using Lemma 322, we fin 1 φq Q x 1, 2 2 ρ 1 ρ 2 + θ q ρ 31 1 2 1, 2 b n b+x n,p =1 We now introuce new variables y ρ accoring to the equation which implies 0moρ ρ 0mo ρ = µρ y ρ φρ, y ρ φρ = µρ The first term of 31 in terms of these new variables is then φq Q x 1, 2 ρ 1 ρ 2 = φq 1 2 Q x 1, 2 ρ,q=1 y 2 ρ φρ

39 In aition, we have the boun ρ ρ,q=1 σρ φρ y ρ From the normalization of ρ 1 = 1, we have the sie conition: ρ,q=1 y ρ = 1 32 φρ We lose nothing by assuming that the y ρ 0, so we may write the upper boun as a n a+x n,p =1 1 φq Q x ρ,q=1 yρ 2 φρ + θ σρ 2 q φρ y ρ ρ,q=1 y 2 33 ρ φρ ρ,q=1 an we rop the conition 32 We now wish to choose y ρ 0 to minimize the expression 33 with z chosen as large as possible We recall that P is the prouct of all primes p with q + 1 p z We follow Selberg s approach an choose 1 σρ if σρ < z, y ρ = z 0 if σρ z Then the upper boun 33 becomes a n a+x n,p =1 1 2 φq Q x 3 q +θ 1 2, 2 where 1, 2 an 3 are efine in Lemma 324 We then use the prouct expansions given in the same lemma an write z = e u x, where u will be chosen later, inepenent of x After some manipulation, we obtain the upper boun 2x log x + 2γ + 2κ 1 + 2κ q 1 + 2u θ q 2 e2u + O The optimal choice of u is given by e 1 2 log x e 2u = 2 θ q

40 With this choice of u, we have the upper boun 2x log x + 2γ + 2κ 1 + 2κ q 2 + log 2 + O θ q e 1 2 log x We can now choose q so that 2κ q + log 2 θ q is maximize We know there is an optimal choice because by the Lemma 322, this expression tens to negative infinity as q grows without boun By maximizing this expression, we maximize the constant enominator term fq = 2γ + 2κ 1 + 2κ q 2 + log 2 θ q We approximate κ 1 by restricting its efining sum to the primes less than 300 We note that κ 1 is a sum of positive terms, so we have κ 1 p 300 log p log p + 1 0368582 p 1 p We now calculate the values of fq for q 19 We have the following values: q fq 1 0584743117 2 1683355406 3 2319869574 5 2700101125 7 2810290556 11 275298288 13 2816814684 17 2729530369 19 < 2600232097 We see that the maximum of fq occurs for q = 13 In Selberg s lectures, he chose q = 7 because that was the highest value he calculate By calculating larger values of θ q, we were able to fin a larger value of fq

41 We fin that z = e 2u x = 2x 1001 = θ 13 3551 x < x Thus if we exclue one resiue class from I x for each prime p < x, there remain at most 2x log x + 28168 numbers if x > x 0 for some x 0 We know simply that such an x 0 exists Fining the value x 0 is a much more ifficult problem without much benefit Theorem 325 We have πx + y πx < 2y log y + 28168 34 for y > x 0 an all x > 0 Similarly for a, k = 1, if πx; a, k enotes the number of primes p x which are congruent to a moulo k, we have for y/k > x 0 πx + y; a, k πx; a, k < 2y φk, 35 log y/k + 28168 Proof Equation 34 follows easily For equation 35, we note that sifting the section of the arithmetic progression that lies in an interval of length y is equivalent to sifting an interval of length y/k We also note that the sifting range for the arithmetic progression oes not inclue the primes iviing k Therefore, we have an improvement of Selberg s version of the Brun-Titchmarsh theorem

42 Chapter 4 Large Sieve Inequality 41 Introuction Large sieve inequalities are a very general problem in analytic number theory The topic was first introuce to aress the problem of least quaratic non-resiues by Linnik The problem has change shape quite a bit since then In general, we consier a finite set X of harmonics that will solve some interesting equation For each element x X, we associate a sequence xn In a way, these are Fourier coefficients The large sieve problem then is to fin a constant C = CX, N 0 such that the following large sieve inequality hols: a n xn x X n N 2 CX, N a 2 2 41 for any complex numbers a n, where a 2 is the L 2 norm of a n That is, a 2 2 = a n 2 There are two main forms of the large sieve inequality, the aitive an the multiplicative In the aitive large sieve inequality, the harmonics are aitive characters of Z with xn = eαn for some α R, where ex = e 2πix We will use the notation ex throughout this chapter In the multiplicative large sieve inequality, we take the harmonics to be Dirichlet characters We only aress the aitive case here For the aitive large sieve inequality, we are intereste in estimating the trigonometric polynomials Sα = n a n eαn 42 where the support of a n is M < n M + N for some M For y R, we efine y to be the istance to the nearest integer If there is some δ > 0 so that α r α s > δ, if r s

43 then the α r are well-space In this case, the number of istinct α r is less than or equal to 1 + δ 1 We say that {α r } is a set of δ-space points Selberg, Montgomery an Vaughn prove the following theorem inepenently Theorem 411 For any set of δ-space points α r R/Z an any complex numbers a n with M < n M + N, where 0 < δ 1 2 an N 1 an integer, we have 2 a n eα r n δ 1 + N 1 a 2 2 43 r M<n M+N We now look at a special case of the large sieve inequality We take α r to be rationals a/q with 1 q Q an a, q = 1 Then α r are space by δ = Q 2 Hence we have the following: Theorem 412 For any complex numbers a n with 1 n N, where N is a positive integer, we have where q Q amo q 1 n N an 2 a n e Q 2 + N 1 a 2 2 44 q a 2 2 = a n 2 1 n N The notation enotes the restriction to a, q = 1 throughout For general a n, 44 gives the best possible boun However, we can improve the boun in special cases We are intereste when the a n are sparsely supporte an Q is much smaller than N To make this improvement, we apply the techniques of Selberg s sieve, which is new in this context 411 Improvement of Large Sieve By utilizing Selberg s sieve, we will prove the following: Theorem 413 We let {λ } be any finite sequence of real numbers such that λ 1 for all, λ 1 = 1 an λ 0, for all n 1 n

44 We efine D = { : λ 0}, D = max D + 1, an Q = {q Q : q, = 1 for all D} Finally, we assume 3 Q N/D 2 Then, for any complex numbers a n C with 1 n N, we have q Q amo q an 2 a n e λ DQ, N a n 2 λ q n N n n N n where DQ, N = 2N D λ + 2Q N 3 D λ 2 We now look at a special case: Corollary 414 Let {a n } be a sequence of complex numbers supporte on n whose smallest prime factor is larger than some z We efine Q z = {q Q : smallest prime factor of q is > z} We let D = 1 + p z p Then if 3 Q 2 N/D 2, we have with q Q z amo q an 2 a n e DQ, N a n 2, a n C q n N n N DQ, N = 2aN + 2Q an 1/2 with a = az, 3 a2 = 1/2, a3 = 1/3, a5 = 4/15, a7 = 8/35, a11 = 16/77 In general, az = 1 p 1 p z

45 Proof We choose λ = µ whenever the largest prime factor of is p r Then D λ = ap r We note that if n oes not have any prime factors p r, then λ = λ 1 = 1 Applying this choice of λ to the theorem gives the corollary n 42 Proof of Theorem In orer to prove Theorem 413, we use the following lemma Lemma 421 Given a set Q boune by Q, consier the following three statements: 1 For all a n C, q Q amo q an 2 a n e λ DQ, N a n 2 λ q n N n N n n 2 For all b n C, q Q amo q n N b n X n, a 2 DQ, N b n 2 q n N where the operator Xn, a/q is efine by 3 For all ca, q C, X n, a 1/2 = λ e q n an q λ n N n q Q amo q an 2 ca, qe DQ, N c 2 2 q where c 2 2 = ca, q 2 q Q amo q Then statement 1 follows from 2 an 2 is equivalent to 3

46 Proof Statement 1 follows from statement 2 by letting 1/2 b n = λ a n By uality, 2 is equivalent to 3 n Now we wish to prove 3 in orer to prove Theorem 413 To continue our analysis, we introuce a continuous function g : R R such that gx 1 for x N an gx 0 for all x We will later pick this function to optimize the results Let ĝy be the Fourier transform of g, so that We have n N ĝy = q Q amo q gn λ n Z n R gxe 2πixy x 1/2 an ca, q λ e q n q Q amo q 2 an ca, qe q 2 The iagonal terms of this sum will be the main contribution, so we treat the iagonal an off-iagonal terms separately We let A be the contribution of the iagonal terms an B be the contribution of the off-iagonal terms n N q Q amo q = gn λ n Z n q Q + gn λ n Z = A + B n 1/2 an ca, q λ e q n amo q ca, q 2 2 a 1 q 1 a 2 q 2 ca 1, q 1 ca 2, q 2 e a1 q 1 a 2 q 2 n