ON BOUNDED GAPS BETWEEN PRIMES AND MODERN SIEVE THEORY

Transcription

1 ON BOUNDED GAPS BETWEEN PRIMES AND MODERN SIEVE THEORY STIJN S.C. HANSON Contents. Introduction.. The History of Twin Primes.. Sieving.3. Oen Problems and Conditional Results 3. Notation and Identities 4 3. The Large Sieve Inequality and the Barban-Bombieri-Vinogradov Theorem 5 4. The Selberg Sieve General GPY-Selberg Sieve Maniulations Choosing y 7 5. Bounded Gas between Primes 6. Beyond Bounded Gas 4 A. Large Sieve Theorems 6 B. Coefficient Sequences 9 C. Assorted Lemmata 33 Acnowledgements 34 References 34 Abstract. We discuss the recent advances made towards the notorious twin rime conjecture and its generalisations. What follows is a lengthy exloration of the various techniques used in the field, both in their general setting and their secific alication to the roblem of gas between rime numbers, along with a new result by the author on a conjecture of Polignac s.. Introduction.. The History of Twin Primes. Euclid s roof of the infinitude of rimes [6, Boo IX, Proosition ] is well-nown and any rofessional mathematician can reroduce it from memory. It is an easy and intuitive result which raises the question as to how these numbers are distributed. Looing through the sequence of rimes we note that there seem to be rather a large number of rimes whose difference is recisely two 3 and 5; and 3; 347 and 349 but the earliest extant examle of the question as to whether there are infinitely many of these so-called twin rimes was first osed in a more general form by de Polignac [6]. Conjecture de Polignac, 849. Let be any ositive even integer and let n be the n th rime number. Then, for infinitely many n N, we have n+ n =. Any which satisfies Polignac s conjecture is called a Polignac number and the twin rime conjecture simly states that is a Polignac number. This conjecture was artially resolved by Maynard [3] and Zhang [7] with the following theorem

2 Theorem Maynard-Zhang, 3. There exist infinitely many airs of rime numbers whose difference is no more than 6. This bound was originally roved by Zhang [7] with 6 relaced by 7 7 which then got reduced to 4,68 by Polymath [7]. Maynard [3] then reduced it to 6 and we will give a further imrovement in this reort. This clearly shows that there exists at least one Polignac number 6 and this, together with the following theorem from Pintz[5], asserts that there are infinitely many Polignac numbers. Theorem 3 Pintz, 3. There is an ineffective constant C such that every interval of the form [m, m + C] contains a Polignac number. Therefore, if there is at least one Polignac number then there exist infinitely many Polignac numbers. Unfortunately this theorem is ineffective it offers no way to generate future Polignac numbers but it does give direction for future research, articularly some results by the author [] on the behaviour of Polignac numbers and arithmetic rogressions. We might also lie to as whether we have similar bounded behaviour for non-consecutive rimes and that question has also been resolved by Maynard [3]: Theorem 4 Maynard, 3. Let m and n be ositive integers. Then there are infinitely many n N such that n+m n Am 3 e 4m for all sufficiently large m and some ositive, real constant A. This shows that we have bounded behaviour for rime airs, whether consecutive or not. There are some revious results in this direction, dealing with the normalised rime gas n+ n log n where the log n reresents the exected ga. In 93, Westzynthius [6] roved that lim su n n+! n log n and, in 5, Goldston, Pintz and Yildirim [8] roved lim inf n n+ n log n = 3 =. 4 The Zhang result has the Golston-Pintz-Yildirim result as a corollary but, recently, Bans, Freiman and Maynard [] have announced an imrovement to these results which we will not have time to delve into but mention for the sae of comleteness. Theorem 5 Bans-Freiman-Maynard, 4. For at least.5% of all ositive real numbers x, the sequence of normalised rime gas has a subsequence which tends to x... Sieving. Sieving as a technique for counting and/or generating rimes has had a long and interesting history dating bac to the Grees and Eratosthenes. Unfortunately this is out of the scoe of this wor but much of the history can be found in []. The majority of the sieving in this reort uses the large sieve and the Goldston-Pintz- Yildirim GPY version of the Selberg sieve: the large sieve gives us the Barban-Bombieri- Vinogradov theorem [] and the GPY sieve [8] acts as a detector for multile rimes in an interval. Informally, the Barban-Bombieri-Vinogradov theorem tells us that the average number of rimes less than x on an arithmetic rogression does not grow much faster than

3 x / log x A for any ositive A. More secifically, if we define the level of distribution of the rimes by Definition 6. Let A be a ositive real number; m, n be the greatest common divisor of m and n; πx be the rime counting function; πx; a, q be the rime counting function on the arithmetic rogression n a mod q and ϕ be Euler s totient function. If, for all ɛ R +, su πx; a, q ϕq πx c Axlog x A 5 q x θ ɛ a,q= holds for some ositive real constant c A, deending only on A, and all sufficiently large x R +, then θ, ] is called the level of distribution of the rimes. The Barban-Bombieri-Vinogradov theorem is then equivalent to saying that the rimes have level of distribution /..3. Oen Problems and Conditional Results. A generalisation of the Barban- Bombieri-Vinogradov theorem is the following conjecture: Conjecture 7 Elliott-Halberstam, 97. [5] The rimes have level of distribution. Under the assumtion that the Elliott-Halberstam conjecture holds, Goldston, Pintz and Yildirim [8] roved the Maynard-Zhang theorem with a bound of 6. Maynard[3] himself has given an imrovement of this bound to for consecutive rimes and a bound of 6 for n+ n. There is also a natural generalisation of Polignac s conjecture and Dirichlet s theorem of rime numbers in arithmetic rogressions[4] which we would hoe to treat using the methods of Zhang, Pintz and Maynard. Conjecture 8 Dicson, 94. Let a,..., a, b,..., b be a finite collection of integers with b i for all i and let f i n = a i + b i n be a collection of linear forms. If there is no ositive integer m dividing all the roducts i= f in for all integers n then there exist infinitely many natural numbers n such that all of the linear forms are rime [3]. You can see that, by setting = and using the linear forms n, n +, this imlies the twin rime conjecture and roving Dicson with the linear forms n, n + for all N is equivalent to Polignac s conjecture. Finally, if we assume that there are infinitely many twin rimes, there is the question of just how frequently they show u. Hardy and Littlewood [9] gave the following conjecture as to how many there should be: Conjecture 9 Hardy-Littlewood, 93. Let P n be the number of rime airs less than n and let P be the set of all rime numbers. Then n P n C 6 log n where C = P\{} 3

4 . Notation and Identities.. Notation. We collect here a list of the symbols we will use in the rest of the reort for easy reference. x a ositive real number a rime number D the value log log log N W the rimorial D θ level of distribution of the rimes P the set of all rime numbers I an interval S I the set of all squarefree numbers in I L log x [x] the largest integer less than or equal to x m, n the greatest common divisor of m and n πx the number of rime numbers less than or equal to x φn the Euler totient function, giving the number of m < n such that m, n = Λn the von Mangoldt function ψx the Chebyshev function Λn n x ϑx the Chebyshev function log x en the exonential function e πin σn the number-of-divisors function σ n the number of ways of writing n as the roduct of natural numbers Γt the gamma function x t e x dx g the totally multilicative function defined on rimes by g = χ a Dirichlet character τχ the Gauss sum χaea/q a q A x the characteristic function of the set A, occasionally written without the argument if the meaning is clear f; a, q the discreancy a Z/qZ fn fn ϕq n x n x n a mod q n,q= f g Dirichlet convolution of two arithmetic functions f and g... Asymtotic Notation. We will write fx = Ogx or fx gx if and only if there is some ositive constant M such that, for all sufficiently large x, fx Mgx. If the imlied constant deends on some value A we will write fx = O A gx or fx A gx. We will also write fx = ogx if and only if fx as x. gx.. Identities. ϕmn ϕmϕn 8 µd = n= 9 d n d n µd ϕd = n ϕn 4

5 ϕz χ mod q N log Q ϕn n Q χaχb = a b mod q, if a, q = χnτχ = a q χaean/q, if χ is a rimitive character 3 τχ = q, if χ is a rimitive character 4 σ n log N 5 n<n [m, n] = mn d m,n ϕ[m, n] = ϕmϕn n log ϕd 6 gd 7 d mn log n 8 where q is always taen to be the modulus of the Dirichlet character 3. The Large Sieve Inequality and the Barban-Bombieri-Vinogradov Theorem In order to rove Maynard s theorem we will first need some information on rime numbers, secifically about their growth on arithmetic rogressions. We say that the rimes have level of distribution θ if P n P x A xl A 9 ϕq q x θ o a Z/qZ n a mod q n<x n,q= n<x for some sufficiently slowly decaying o and any A R +. The Barban-Bombieri- Vinogradov theorem will show that the rimes have level of distribution at least / and that is enough for our uroses. Theorem Large Sieve Inequality. For any comlex sequence a n n N ; any M, N N; and all real Q > M+N M+N a n χn Q + N a n q Q χ mod q n=m+ n=m+ where χ mod q reresents a sum of rimitive Dirichlet characters modulo q. Proof. Let χ be a rimitive Dirichlet character modulo q. If τχ is the Gauss sum of χ then, by 3, M+N n=m+ a n χn = τχ χa a q M+N n=m+ a n ean/q. 5

6 Define Sa/q = M+N n=m+ a nean/q. Then χ mod q M+N n=m+ a n χn So, by 4, this is no larger than q χas q χ mod q a= a q = χ mod q q = q τχ χ mod q a= q χas a= q χa S q a S q a= χ mod q a q. a q χa. As non-zero Dirichlet characters are roots of unity and non-zero only when a, q =, χ mod q χa equals ϕq whenever a, q = and is otherwise. Therefore is less than or equal to Hence we have shown that q ϕq q Q χ mod q M+N n=m+ ϕq q a q a,q= a n χn a S q q Q. 3 a q a,q= M+N n=m+ a n e an q. 4 We can view the double sum as a sum over all ositive roer fractions whose denominators do not exceed q. Moreover, as a, q =, the fractions are all distinct so we can aly Gallagher s large sieve inequality Theorem 3 to show that this is less than or equal to Finally note that πn + δ δ = min a q a q M+N n=m+ a n N + δ a a q min a q a q q M+N n=m+ a n. 5 q q Q 6 comleting the roof. We will immediately aly this to roving that rimes have level of distribution /. Theorem Barban-Bombieri-Vinogradov. Let ɛ, /. Then su P n P n q x / ɛ a Z/qZ ϕq xl A 7 n x for any fixed A R +. n a mod q n x n,q= 6

7 Setch of roof. Consider the Chebyshev ψ, ϑ functions and define πx; a, q = P n 8 ψx; a, q = ϑx; a, q = n x n a mod q n x n a mod q x a mod q Λn 9 log. 3 Using artial summation along with the facts that πx; a, = πx and ϑx; a, = ϑx we see π; a, q = ϑx; a, q x ϑt; a, q + log x tlog t dt ϑx; a, x ϑt; a, dt + ϕq log x tlog t ϑx; a, q ϑx; a, x log x ϕq log x + ϑt; a, q tlog t dt x ϑt; a, ϕq tlog t dt log x ϑ [,x]; a, q + max ϑ[,x] ; a, q [ ] t=x t x log t Furthermore ϑ [,x] ; a, q + max t x ϑ [,t]; a, q. 3 ψx; a, q = x a mod q log + x / a mod q log + x /3 a mod q t= log + = ϑx; a, q + O ϑx / + ϑx / Note that the sum in the O term is finite as x / log x = and so ϑx /n = for all n > log x. Also, by the rime number theorem, and so Id est Then so it is sufficient to show that ϑx /n x/n log x /n log x/n = x /n 33 ϑx / + ϑx /3 + x / + x /3 log x x /. 34 ψx; a, q = ϑx; a, q + O x /. 35 ϑ [,x] ; a, q ψ [,x] ; a, q + Ox / 36 su q x / ɛ a Z/qZ ψx; a, q ϕq ψx xl A. 37 This can be achieved by decomosing the von Mangoldt function via the Heath-Brown identity Lemma 3 and noting that all of the items we are convolving are coefficient sequences and that log obeys the Siegel-Walfisz condition, allowing us to aly the generalised Barban-Bombieri-Vinogradov theorem see Theorem 39 in Aendix B for details. 7

8 4. The Selberg Sieve 4.. General GPY-Selberg Sieve Maniulations. We will now loo at the Selberg lambda-squared sieve [] and its alications to the roblem of bounded gas. The Selberg sieve is any sieve of the form a n λ d 38 n d n for some arithmetic function λ n. This will let us weight a detector function for rimes in constellations. We will show the articular sieve to use later in this reort but we need to develo a few notions first. Let H = {h, h,..., h } be some finite set. We say that H is admissible if, for all rimes, there is some integer a such that h i a mod for all h i H. We choose to study these sets as these are the sets which ossess the otential to have infinitely many translates n + H, all of whose elements are rime. From here on we will consider H to be some fixed admissible set. If we tae some rime j then there must be some a j such that h a j mod j for all h H. By the Chinese remainder theorem there must be a unique solution to the system of congruence equations v a j mod j 39 for all j D, where D = log log log x. Then, as these j are recisely the rime divisors of the value W, we must have v + h h a j mod j. 4 Thus v + h is rime to W. These definitions allow us to define the articular detector function we will use. So let us define the GPY sieve sum S = S ρs 4 where S = 4 S = N n<n n v mod W N n<n n v mod W id i n+h i λ d,...,d P n + h i i= id i n+h i λ d,...,d. 43 From now on we will omit the i on the subscrit of the summation for ease of reading. Note that if S > then at least one of the terms in the outer sum must be ositive. As squared factors don t affect the sign of a term this imlies that P n + h i ρ > 44 i= for some n satisfying the revious conditions. Thus there must be at least ρ rimes in the translate n + H. Let F be some fixed iecewise-differentiable function suorted on { } R = x,..., x [, ] : x i. 45 i= 8

9 Also let R = N θ/ δ for some small, fixed, δ R +. Then we want our Selberg weight function to be of the form λ d,...,d = µd i d i i= r,...,r d i r i r i,w = µ i= r i log i= ϕr i F r log R,..., log r log R. 46 whenever i= d i, W =, when i= d i < R and when i= d i is square-free. Let λ d,...,d = otherwise. We roceed by finding asymtotic equations for S and S. Lemma. Let y r,...,r = µr i ϕr i i= d,...,d r i d i λ d,...,d i= d i 47 and y max = su r,...,r y r,...,r. Then S = N W r,...,r y r,...,r y i= ϕr i + O max ϕw Nlog R W + D 48 Proof. Exanding the square and switching the order of summation gives S = λ d,...,d λ e,...,e N n<n d i n+h i e i n+h i n v mod W = λ λ d,...,d e,...,d d,...,d e,...,e N n<n n v mod W [d i,e i ] n+h i. 49 We will suose that D max H and conjecture that W and the [d i, e i ] are corime. If not then there exists some d such that either d [d i, e i ], [d j, e j ] or d W, [d i, e i ] for all i j. In the first case d must divide n+h i and n+h j, both of which are corime to W so d also divides h i h j maxh, giving us that h i h j, W = h i h j. Thus d, W, contradicting our earlier statement. On the other hand λ d,...,d is only suorted when the d i and W are corime so we can suose that W, [d, e ],..., [d, e ] are airwise corime. This means that, by the Chinese remainder theorem, we can turn the inner sum in 49 into a sum over a single residue class q = W i= [d i, e i ] and this sum has a value of N/q + O. So S = N W d,...,d e,...,e λ d,...,d λ e,...,d i= [d i, e i ] + O λ λ d,...,d e,...,d d,...,d e,...,e 5 where reresents the restriction that W, [d, e ]..., [d, e ] are airwise corime. 9

10 Let λ max = su d,...,d λ d,...d. Then, as λ d,...,d is non-zero only if i= d i < R, λ λ λ d,...,d e,...,e max d,...,d e,...,e i= d i<r = λ max σ d Now we rewrite the main term using 6 as N λ W λ d,...,d e,...,e d d,...,d i= i e i e,...,e d<r λ maxr log R. 5 u i d i,e i ϕu i. 5 Recall that λ d,...,d is suorted only when the roduct d d is squarefree so the d i must be airwise corime and also recall that this roduct and W must also be corime. Therefore [d i, e i ] is corime to W. Then the above equals N λ d,...,d ϕu i λ e,...,e W 53 u,...,u i= d,...,d e,...,e u i d i,e i i= d i i= e i with the restriction on the sum simly now being that d i, e j = for all i j but this can be removed by multilying through by s i,j d i,e j µs i,j. This turns the main term into or N W N W u,...,u u,...,u ϕu i i= d,...,d e,...,e u i d i,e i ϕu i i= s i,j i j s i,j i j l,m, l m l m s l,m d l,e m µs l,m µs l,m d,...,d e,...,e u i d i,e i s i,j d i,e j, i j λ d,...,d λ e,...,e i= d i i= e i λ d,...,d λ e,...,e i= d i i= e i We can furthermore require that s i,j is corime to both u i and u j as, if s i,j, u i, we would have some common divisor between d i, e i and d i, e j. This would mean that e i and e j share some common factor but then λ e,...,e = and so these terms do not contribute to the sum. Similarly we can further restrict the sum so that s i,j is corime to s i,a and s b,j for all a j and b i. We denote this restricted sum by. Write y r,...,r = µr i ϕr i i= If we fix d,..., d with µ i= d i we find d,...,d r i d i λ d,...,d i= d. 56 i

11 r,...,r d i r i y r,...,r i= ϕr i = r,...,r d i r i = µr i i= λ e,...,e e,...,e i= e i = = r,...,r d i r i e i λ e,...,e e,...,e i= e i r,...,r r i e i /d i µd i i= e,...,e e,...,e r i e i λ e,...,e i= e i µr i i= µd i r i i= λ e,...,e j= e j r i e i /d i µr i. 57 If any of the e i /d i do not equal then one of the divisor sums of the Möbius function will be. Therefore we can write e i = d i for all i and so r,...,r d i r i y r,...,r i= ϕr i = λ d,...,d i= µd id i. 58 Thus any choice of y r,...,r which is suorted on r,..., r such that r = i= r i < R is square-free and r, W = will give us some λ d,...,d. Now let then we find that, by taing r = r i i= d i λ max = su λ d,...,d d,...,d = su µd i d i d,...,d i= = y max su d i d,...,d µ i= d i= i = y max su d,...,d µ i= d i= i y max = su r,...,r y r,...,r 59 r,...,r d i r i r<r µr d i ϕd i y r,...,r i= µr i i= ϕr i r <R/ i= d i µ i= d ir µ r <R/ i= d i µ i= d ir i= d ir σ r ϕ i= d ir µ i= d ir σ r ϕ i= d ir

12 = y max su d,...,d µ i= d i d i= d i µd ϕd r <R/ i= d i µ i= d ir i= d ir σ r ϕ i= d ir µ by equation. Setting u = dr and noting that σ dr σ r shows µu σ u λ max y max y max log R 6 ϕu u<r by standard estimates on the mean values of arithmetic functions. We can use this to bound our error term Oλ maxr log N by OymaxR logn 4. Substituting our exression for y r,...,r into equation 55 and using our new estimate for the error term we get S = N ϕu i W µs l,m µa l µb l ϕa l ϕb l y a,...,a y b,...,b u,...,u i= s i,j i j l,m l m + OymaxR log R 4 6 where we define a m = u m l m s m,l and b m = u m l m s l,m. As u m is corime to all s l,m we note that µa m, µb m and the main term becomes N µu i µs l,m W ϕu i ϕs l,m y a,...,a y b,...,b 63 u,...,u i= s i,j i j l,m l m Using the fact that y d,...,d is suorted only on i= d i, W = it must be that the s i,j are corime to W. Thus we only need to consider s i,j = and s i,j > D. The contribution when s i,j > D is y maxn µu µs W ϕu ϕs i,j ϕs s u<r u,w = s i,j >D µs i,j y maxϕw Nlog R W + 64 D by standard estimates on the mean values of multilicative functions. We can now just restrict our attention to the case where s i,j = for all i j. This gives S = N W u,...,u y u,...,u y i= ϕu i + O max ϕw Nlog R W + + y D maxr log R 4 but R = N θ ɛ N ɛ, W N ɛ and log R log N N ɛ for all ɛ R +, giving the required result. We derive an asymtotic bound for S in a similar way but we first need to brea it u. Define S m = N n<n n v mod W l= P n + h m λ d,...,d d,...,d d i n+h i

13 All we have done is to brea u the set H as the arameter to the GPY sieve term S. Clearly we have S = m= Sm and our next lemma gives us a bound for these summands. Lemma 3. Let y m r,...,r m = µr i gr i i= d,...,d r r i d i d m= λ d,...,d i= ϕd i where g is the totally multilicative function defined on rimes by g =. Also let y max m = su r,...,r y r m,...,r. Then, for any fixed A R +, we have S m N y m r,...,r = ϕw log N r,...,r i= gr i + O y max m ϕw Nlog N W D y + O max N. 68 log N A Proof. As before we exand out the square and swa the order of summation in S m to give S m = λ λ d,...,d e!,...,d d,...,d e,...,d N n<n n v mod W [d i,e i ] n+h i 67 P n + h m. 69 As with the estimation of S we can use the Chinese remainder theorem to write the inner sum as a sum over a single rimitive residue class modulo q = W i= [d i, e i ] if W, [d, e ],..., [d, e ] are airwise corime. Also note that the inner sum equals unless [d m, e m ] =. Then the inner sum contributes P n + h m P n + O N n<n n a mod q N n<n n a mod q = ϕq + O ϕq + su a,q= N n<n N n<n P n + P n N n<n n a mod q N n<n n a mod q P n ϕq P n ϕq N n<n N n<n P n + O P n where = X N + OEN, q. 7 ϕq X N = N n<n P n; 7 EN, q = + su P [N,N ; a, q. 7 a,q= 3

14 Thus S m = X N ϕw d,...,d e,...,e d m=e m= λ d,...,d λ e,...,e i= ϕ[d i, e i ] + O λ λ d,...,d e,...,e d,...,d e,...,e EN, q. 73 Looing at the error term we see that, by the restricted suort of λ d,...,d we only need to worry about i= d i < R. I.E. when q i= d ie i < R W with q being square-free. Combinatorially, if r is square-free then there are no more than σ 3 r choices of d,..., d, e,..., e for which W i= [d i, e i ] = r. Then the error term contributes ymaxlog R µr σ 3 ren, r. 74 r<r W Trivially we see that EN, r + N N. Using Cauchy-Schwarz and the fact that ϕq ϕq the rimes have level of distribution θ this error term is ymaxlog R µr σ 3 r N / / µr EN, r ϕr r<r W y maxlog R N / logr W O N / y maxn log N A log N A r<r W for any fixed A R + with the enultimate ste coming from standard multilicative function means. We treat the main term similarly to in the revious lemma and rewrite the condition d i, e i = by multilying through by s i,j d i,e j µs i,j. Again we can restrict the s i,j to be corime to u i, u j, s i,a and s b,j for all a i and b j and denote the summation with resect to these restrictions by. Using the fact that di, e i are square-free we get a main term of X N gu i ϕw µs l,m 76 u,...,u i= s i,j i j i= l,m l m d,...,d r r i d i d m= d,...,d e,...,e u d i,e i s i,j d i,e j, i j d m=e m= where the roduct of the gu i comes from equation 7. Now let y r m λ d,...,d,...,r = µr i gr i i= ϕd i λ d,...,d λ e,...,e i= ϕd iϕe i and note that this will be unless r m = and unless the r i are all square-free. Substituting this into 76 we get X N µu i µs i,j ϕw gu i gs i,j ym a,...,a y m b,...,b 78 u,...,u i= s i,j i j l,m l m

15 where a j = u j i j s j,i and b j = u j i j s i,j for each j, as before. Arguing as in the receding lemma we find that the contribution from s i,j is of size Thus y m max N ϕw log N y m max S m = X N ϕw u<r u,w = µu gu ϕw Nlog R W D log N u,...,u y u m,...,u i= gu i + O µs i,j gs i,j s i,j >D s µs gs. 79 y m max y + O max N log N A ϕw Nlog R W D for all A R +. Finally, by the rime number theorem, X N = N + O N. This log N log N error term contributes y max m N µu y max m ϕw Nlog R 3 ϕw log N gu 8 W 3 u<r u,w = which is absorbed in the first error term above due to log R being significantly bigger than D, finishing the roof. Lemma 4. If r m = then y m r,...,r = a m y r,...,r m,a m,r m+,...,r ϕa m Proof. Substituting 58 into the definition of y r m,...,r y m r,...,r = i= µr i gr i a,...,a r i a i d,...,d r i d i d m= i= ymax ϕw log R + O W D µd i d i ϕd i we get a,...,a d i a i Now, by changing the order of summation, the sum over d,..., d equals y a,...,a µd i d i i= ϕa i ϕd i or, by relacing d i by d i r i a,...,a r i a i y a,...,a i= ϕa i d,...,d r i d i a i d m= d,...,d d i a i /r i d m= i= i= µd i r i d i r i ϕd i r i 8 8 y a,...,a i= ϕa i

16 Note that, due to multilicativity and the fundamental theorem of arithmetic µdd ϕd = µ ϕ d r d r d = + µ ϕ r = r = r. 86 Looing secifically at the sum over d we see that, by switching order of summation and roduct, 85 equals µd i r i d i r i = µr i r i ϕd i r i ϕr i i m d i a i /r i i m = i m µr i r i ϕr i a i /r i µa i /r i ϕa i /r i = i m µa i r i ϕa i 87 where the slitting and recombining of the multilicative functions holds due to the fact that the a i are squarefree else the contribution is. Substituting this bac into 85 and then 83 gives y m r,...,r = µr i gr i r i i= a,...,a r i a i y a,...,a µa i i= ϕa i ϕa i i m By the restricted suort of y a,...,a we can secify that a i, W =. This imlies that a i /r i, W = which, in turn, imlies that either a i = r i or a i > D r i. For j m the total contribution from a j r j is y max gr i r i µa j µa j µa i ϕa i= j ϕa j ϕa i a j >D r j ri ai y maxϕw log R W D y maxϕw log R W D i= gr i r i ϕr i a m<r a m,w = i i j,m by standard results on the mean values of multilicative functions and noting that r i = r. Thus the main contribution is when a i + j = r j for all j m. This gives y m r,...,r = i= gr i r i ϕr i a m y r,...,rm,am,rm+,...,r ϕa m ymax ϕw log R + O W D

17 due to the µr i term getting absorbed into gr i. Note that g ϕ = + = + O, 9 imlying that the whole roduct is equal to +O. Furthermore, as the contribution is zero unless i= r i is corime to W we must have > D. This then gives a second error term of size y max D ϕa m comleting the roof. 4.. Choosing y. We will now define y r,...,r = F a m R y max log R D y maxw log R W D, 9 log r log R,..., log r log R whenever r = i= r i satisfies r, W = and µr = for some iecewise-differentiable function F suorted on the -simlex { } R = x,..., x [, ] : x i. 94 It is not hard to see that this choice of y satisfies all of our suort conditions and we can now develo some slightly more exlicit bounds for S and S which is what we send the rest of this chater doing. Lemma 5. Define Then we have where F max = su t,...,t [,] F t,..., t + S = ϕw Nlog R I W + F + O I = Proof. By our definition of y we clearly see that y max = i= 93 F t,..., t t i. 95 i= F max ϕw Nlog R W + D. 96 F t,..., t dt,..., dt 97 su F t,..., t F max 98 t,...,t [,] so, substituting our choice of y into the exression for S given by Lemma, we get S = N µu i log u F W ϕu i log R,..., log u log R u,...,u u i,u j = u i,w = i= F + O max ϕw Nlog R W + D. 99 7

18 Any integers a and b with a, W = b, W = but a, b must share some common rime factor which is greater than D. Therefore we can dro the condition u i, u j = by introducing an error of size F maxn W F maxn W F maxn W >D u,...,u <R u i,u j u l,w = >D l= µu l ϕu l u<r u,w = ϕw log R W >D µu ϕu F maxϕw Nlog R. W + D by standard estimates on the mean values of multilicative functions. We now deal with the sum u,...,u u,w = Let κ = and suose that i= γ = L = W µu i log u F ϕu i log R,..., log u log R {, W, W log + O log D. 3 Then h = γ = γ ϕ whenever, W = and otherwise so, when extended multilicatively, hn = ϕn whenever n, W = and otherwise. Also note that L > and, by Chebychev s inequality 8, w z W log log z w log z w log z + O = O. 4 w All the conditions of Lemma 4 are thus satisfied and we can aly it times to which then equals ϕw log R F I F + O max ϕw log D log R. 5 as we would have W S = = = ϕw W W γ γ ϕ 6 8

19 Substituting this into 99 and using the fact that log D log R D gives the result. We get a similar result for S Lemma 6. Using the notation from the revious lemma S m = ϕw Nlog R + J m F W + F + O max ϕw Nlog R log N W + D where J m F = u<r u,w i= r i= 7 F t,..., t dt m dt... dt m dt m+... dt. 8 Proof. By substituting our choice of y into the exression given by Lemma 4 we get y r m µu log,...,r = ϕu F r log R,, log r m log R, log u log R, log r m+ log R,, log r 9 log R We can use estimates of multilicative functions on this to bound y max m asymtotically above by F max ϕw log R/W. Now set κ = and {, W γ = i= r i, W L = O + W i= r i i= r i log <log R log + W i= r i >log R log log R log R log log N as the < log R sum will dominate over the second and R N. Also hn, defined in the same way as in the revious lemma as and extended multilicatively, is now γ γ equal to µn whenever n is corime to W ϕn i= r i.. As in the revious lemma we note that w z log / logz/w is bounded in magnitude by a constant and so all the conditions of Lemma 4 are satisfied and we aly it once to 9 to get y r m,...,r = log R ϕw ϕr i F m Fmax ϕw log R r W r,...,r + O i W D as, similarly, i= S = ϕw i= r i W i= r. 3 i Now, substituting this bac into the result from Lemma 3, remembering that gives S m ϕw Nlog R µr i ϕr i = F m W log N gr i ri r,...,r r,...,r r i,w = r i,r j = r m= i= F + O max ϕw Nlog R W + D. 4 9

20 Similarly to before, we can remove the condition u i, u j = with an error F maxn W F maxn W >D u,...,u <R u i,u j u l,w = l= 4 >D 4 µu l ϕu l gu l u l u<r u,w = µu ϕu gu u F maxϕw Nlog N W + D 5 again, by multilicative function estimates. We now loo at the sum r,...,r m,r m+,...,r r i,w = i i j µr i ϕr i gr i r i F m r,...,r 6 and estimate it using Lemma 4 times, on each of them r,..., r m, r m+,..., r in turn. We do this, as a slight correction to [3], with κ = and set { 3+, W γ = 3 + 7, W L = O + log log D. 8 W so h = n, W =. Also γ γ = ϕ g S = = W whenver W and is otherwise. So hn = ϕn gnn γ ϕw W whenever Then, as before w z W γ log ϕw W log z w = w z W log + n N = + ζ ɛ n +ɛ log z w w z W log

21 for all small enough ɛ R +. So we get where J = as required. S m = ϕw Nlog R + W + log N J m F + O max ϕw Nlog N W + D F t,..., t dt m dt,..., dt m, dt m+,..., dt Putting these results all together while noticing that /D yields Proosition 7. With the notation as before, there exists some choice of λ d,...,d such that S = + oϕw Nlog R I W + F 3 S = + oϕw Nlog R + J m W + F log N 4 rovided that neither I F and J m are zero for any m. m= 5. Bounded Gas between Primes We are almost in a osition to rove the main results of Maynard s aer but we first need a rearatory result Proosition 8. Let the rimes have level of distribution θ, and let S denote the set of iecewise-differentiable functions F : [, ] R suorted on the -simlex R with I F and J m for all m. Also let m= M = su J m F θm, r =. 5 F S I F Then there are infinitely many integers n such that at least r of the elements of n + H are rime. In articular, lim inf n n+r n max i,j h i h j 6 Proof. Let R = N θ/ δ for some δ R +, as before. Then, by definition, we can select some F S such that J m F > M δi F 7 m= Now let ρ be some ositive real number and recall that S = S ρs. Then, using Proosition 7, we can choose λ d,...,d such that S = ϕw Nlog R log R J m W + F ρi F + o log N m= ϕw Nlog R I F θ W + δ M δ ρ + o. 8 If we set ρ = θm ɛ, for some ɛ R, then θ δ M δ ρ = ɛ δ θ M + 9

22 which is ositive for small enough δ. Then, as the fraction in 8 is also strictly ositive we must have that S >. So, as noted reviously, there must exist an N n < N such that at least ρ + rimes in n + H. Also note that ρ + = θm as long as ɛ is sufficiently small. As this holds for any interval [N, N we obtain the lim inf result. The M are a ratio of -fold integration so, as one might exect, their calculation is rooted dee within the study of the calculus of variations. This is beyond the scoe of this wor so we will have to content ourselves with taing other eole s calculations [3] [4] at face value. Our best nown values are Then M 5 > ; M 59 > 4; M > log + log log for sufficiently large. Theorem 9 Maynard-Polymath. Unconditionally we have that lim inf n n+ n 3 3 and, on assuming that the rimes have level of distribution θ = ɛ for every ɛ R +, lim inf n+ n n 3 lim inf n+ n 3. n 3 Proof. By Theorem we can tae θ = / ɛ for all ɛ R +. Then θm 59 = M 59 4 ɛm can be made to exceed by choosing ɛ to be sufficiently small, as M 59 > 4. By Proosition 8 this gives lim inf n+ n max h i h j 34 n i,j 59 for some 59 element admissible set. There exists such an admissible set H whose diameter is 3, roving the first art. The results for the rimes having level of distribution θ = are almost identical excet that you get θm 5 = M 5 ɛm 5 >. 35 The required 5 element admissible set is given by {,, 6, 8, } In generality, we can loo for a result for general rime airs, consecutive or nonconsecutive. Unfortunately we are currently unable to obtain an effective inequality so we must content ourselves with an asymtotic one. Theorem Maynard. Let m N. Then lim inf n n+m n m 3 e 4m. 36 An examle is given by {, 4, 6, 6, 8, 8, 3, 34, 36, 46, 48, 58, 6, 64, 66, 7, 84, 88, 9, 94, 6, 8, 4, 8, 6, 3, 36, 44, 48, 5, 56, 6, 68, 74, 78, 84, 9, 96, 98, 4,,, 6, 8, 34, 38, 4, 44, 46, 56, 68, 7, 76, 8, 86, 88, 94, 98, 3}

23 Proof. Consider the case where is large. For this roof any constants imlied by asymtotic notation will be indeendent of. By the Barban-Bombieri-Vinogradov theorem, we can tae θ = / ɛ and so, by our exlicit exression for M we have, for sufficiently large : θm 4 ɛ Set ɛ = / and then, by Lemma 4, log log log. 37 θm > m 38 if Cm e 4m for some C, indeendent of and m. Then, for any admissible set H = h,..., h with Cm e 4m, at least m + of the n + h i must be rime for infinitely many integers n. We can choose our set H to be equal to { π+ + + π+ }. These are all rime and not divisible by rimes less than so, for rimes <, π+i mod. Also note that H only has elements so there must be an oen residue class modulo any rime greater than or equal to. Therefore this choice of H is indeed admissible and has diameter π+ π+ log 39 by the rime number theorem. Thus, if we tae = Cm e 4m, we get as required. lim inf n n+m n log m e 4m log m + 4m m 3 e 4m 4 With a very different tone combinatorial rather than sieve theoretic to the rest of his results, Maynard roved one final result concerning the density of these rime tules. We will denote by P m the set of all m-tules h, h,..., h m such that all of n + h, n + h,..., n + h m are simultaneously rime infinitely often. Then we have the following result. Theorem Maynard. Let m N and let r N be sufficiently large, deending on m. Also let A be the set of all tules of r distinct integers. Then #{h,..., h m A : h,..., h m P m } #{h,..., h m } m. 4 That is to say: a ositive roortion of m-tules are simultaneously rime infinitely often. Proof. Let m be fixed and define = Cm e 4m, as before. If {h,..., h } is an admissible set then there is a subset {h,..., h m} {h,..., h m } such that there are infinitely many integers n for which all of the n + h i are rime. Now let A be the set starting with A and then, for every, removing all elements of the residue class modulo with the fewest integers. Then, for each rime, at most / elements will be removed. Thus #A r m r. 4 as the number of rimes less than or equal to will deend only on m. Moreover, any subset of A of size will be admissible as it cannot cover all residue classes modulo for any rime. Write s = #A and, as r is sufficiently large, we may assume that s >. 3

24 Now there are s sets H A of size. Each of these is admissible due to the fact that A is admissible. Therefore these must contain at least one subset {h,..., h m} which are simultaneously rime infinitely often. Conversely, any admissible set B A of size m is contained in s m m sets H A of size. Thus there are at least s s m = m m s! m!s!!s!s m! = i= s i i m s m m r m 43 admissible sets of size m which are simultaneously rime infinitely often. As there are r m r m sets {h,..., h m } A we then get #A r r m m r m m r Beyond Bounded Gas So we have these beautiful results but what now? The Polymath roject has further imroved uon these bounds, getting 46 as the bound for consecutive rimes and lim inf n n+m n me 4 5/83m 45 unconditionally[8]. Furthermore, if we consider how we arrived at the Barban-Bombieri- Vinogradov theorem then there is the following natural generalisation to the Elliott- Halberstam conjecture: Conjecture Generalised Elliott-Halberstam. Let x ɛ M, N x ɛ be such that x MN x and let α, β be coefficient sequences at scale M and N resectively. Then su a, q = α β; a, q xlog x A 46 q x θ for any fixed A R + and for all θ,. Under the assumtion that this holds, Polymath have roven that we have bounded gas between rimes of size 6 [8]. Also, if we suose that we have roved that any admissible set of size has infinitely many translates that contain ρ rimes then we can verify Conjecture in a few secific cases. Lemma 3. Let 59 and d =. Then, for every N N, contains at least one Polignac number. Proof. Consider the set {dn, dn,..., dn} 47 H = {, dn, dn,..., dn}. 48 If we tae any rime then, for all h H, we have h. Id est h mod 49 for all. Furthermore, if > then H cannot occuy all residue classes modulo as there are only elements. Therefore H is admissible. 4

25 As is large enough we now that there are infinitely many translates of H which contain two rimes infinitely often. Therefore there is some air n dn, n dn which are simultaneously rime infinitely often, and the difference between these must be one of the elements of the set 47. Let q N and consider the sequence given by n mod q. This has a subsequence given by n mod qd where d = 59#. This, in turn, has the subsequence qd, qd,..., 58qd, 59qd, 59qd,..., 58 59qd,... 5 where each of the arenthesised terms contains a Polignac number by the above theorem so we have roven []: Corollary 4. There are infinitely many Polignac numbers on any arithmetic rogression of the form q, q,... 5 for all q N. As there is only one even rime number we cannot have odd Polignac numbers, and we will call the arithmetic rogressions containing only odd numbers trivial, but we then have the question as to whether the following statement is true. Conjecture 5. Every non-trivial arithmetic rogression contains infinitely many Polignac numbers. Theorem 3 imlies the conclusion of Pintz Theorem 3 but in a more effective form. We can reduce the value of on imrovement in our nowledge of the M values and, secifically, under the assumtion of the Elliott-Halberstam conjecture we now that one of 3, 6, 9 or is a Polignac number and, under the generalised Elliott-Halberstam conjecture, one of 6 or is a Polignac number. This very clearly shows. Theorem 6. Polignac s conjecture is true if and only if every admissible tule of size is rime infinitely often. By using Pintz result that every interval of the form [m, m + C] contains a Polignac number if C is large enough we can calculate a lower bound for the uer asymtotic density of the set of Polignac numbers P. σp = lim su n lim su n P [, n] n n/c i= P [i C, ic] n. 5 All of the intersections above are non-emty but they might count some Polignac numbers twice so this is n/c i= lim su n n lim su n n/c n = c. 53 So the uer asymtotic density is ositive and we can aly Szemerédi s theorem [9] [5] to show [] 5

26 Theorem 7 Hanson, 4. The set of Polignac numbers contains arbitrarily long arithmetic rogressions. This has the following natural generalisation: Conjecture 8. The Polignac numbers contain infinitely long arithmetic rogressions. One might also as whether the Selberg sieve techniques used in the roof of bounded gas have alications to other roblems related to the distribution of rimes such as the binary Goldbach conjecture. Analytic number theory is certainly advancing quicly and only time can tell us what the limits of these techniques are. A. Large Sieve Theorems Here we gather together a collection of results needed to rove the large sieve inequality and the Barban-Bombieri-Vinogradov theorem. Lemma 9 Sobolev-Gallagher Inequality. Let g have a continuous derivative on [, ]. Then g gt + g t dt. 54 Proof. Using integration by arts gtdt = [tgt] and, by the fundamental theorem of calculus gx + x tg tdt = gt tg tdt 55 g tdt = gx + g gx = g 56 for x [, ]. Substituting out g and rearranging to ut things in terms of gx we find that gx = Setting x = / gives g gtdt + gt dt + gt dt + x / / tg tdt + gt dt + gt + g t x t g t dt + g t dt + t g tdt. 57 / / t g t dt g t dt g t dt dt 58 due to the fact that t and t are less than or equal to / on [, /] and [/, ] resectively. Lemma 3 Parseval s Identity. Let f be a function with Fourier series fx = a n= c nenx where c n = a fxe nxdx. Then a a c n = fx dx 59 a whenever a = q/ for some q Z. n= a 6

27 Proof. Let S N x = a N n= N c nenx. Then, if is the norm over L [ a, a]: f S N = f, f S N, f S N, f + S N, S N a = f N c n e nxfxdx a + a a a a n= N N m= N c m emx a N n= N a a a N n= N c n enxfxdx c n e nx. 6 As all of our sums are finite they converge and we can exchange the order of summation and integration to give f N c n c n N c n c n + N N a c a a 4a m c n em nxdx. 6 n= N n= N If m n then m n = A Z and a a m= N n= N eaxdx = πia [eax]a x= a = 6 whenever a is of the form q/ for some integral q. However, if m = n, then the integrand is and so the integral is a giving a f S N = fx N c n. 63 a a n= N To finish the roof we simly note that S N f as n imlying that f S N. Theorem 3 Gallagher s Large Sieve Inequality. Let α, α,..., α J R and a, a,..., a N C be two finite sequences with δ := min j α j α 64 where a is the distance from a to the nearest integer. Then J M+N a n enα j πn + δ j= n=m+ M+N n=m+ a a n. 65 Proof. Let gt = fα + δt and suose that f has a continuous derivative. Then g has a continuous derivative and, by the Sobolev-Gallagher inequality: fα = g = α+δ/ f α + δ α δ/ t δ fu + f u + α δf + δ t dt du. 66 Let Sα = N n= N a neαn and set fα = Sα. This has a continuous derivative so J J α+δ/ Sα j δ Su + SuS u du n= n= α j δ/ δ Su + SuS u du 67 7

28 as the intervals [α j δ/, α j + δ/] are non-overlaing modulo and S is eriodic with eriod. By the Cauchy-Schwarz inequality / / SuS u du Su du S u du. 68 On changing variables we see that Calculating directly we get S u + N = and / n= N Su du = S u + = / / c n e u + n = N n= N u S + du. 69 N n= N Using Parseval s identity with a = / gives / u S + N = a ne n = and / / u S + = N n= N a n e n eun 7 πin a n e n eun. 7 n= N N n= N a n 7 πina ne n N = 4π na n. 73 n= N Putting these values bac into 67 and using Parseval s identity again we get J / / Sα j δ Su du + u S + / / u S + / j= / δ N n= N δ + πn a n + π N n= N N / n= N a n / N n= N / na n / a n. 74 Note that [N/] N so, defining any extra a n to be if necessary: J M+N [N/] J a n e α j [N/] M n = a n+[n/]+m+ enα j j= n=m+ comleting the roof. j= n= [N/] δ + πn δ + πn [N/] n= [N/] M+N n=m+ a n+[n/]+m+ a n 75 8

29 Lemma 3 Heath-Brown Identity. Let K be some natural number. Then K K Λ = j µ K j µ j j K log 76 j= on [, x], where µ is the Möbius function restricted to [, x /K ] and f n is the n-fold convolution of f with itself. Proof. Using the identities that Λ = µ log and δ = µ N, where δn = n= is the identity for Dirichlet convolutions, we deduce that Λ = µ K K N log. 77 Writing µ = µ + µ > where µ and µ > reresent the restriction of the Möbius function to [, x /K] and x /K, resectively, we notice that the K-fold convolution µ K > vanishes on [, x] and, in articular, µ K > K N log = 78 on [, x]. Now, as Dirichlet convolutions distribute over addition, we can generalise the binomial theorem to Dirichlet convolutions. Secifically: µ K > = µ µ K K K = j j j= µ K j µ j. 79 Putting this bac into equation 78 we get K K = j µ K j µ j j K N log 8 j= on [, x]. But the j = term is simly Λ so the lemma follows by using the fact that Dirichlet convolution is a symmetric oeration and remembering the cancellation identity µ N = δ. B. Coefficient Sequences Definition 33. A coefficient sequence is a finitely-suorted sequence α : N R such that αn σn O L O 8 Definition 34. If α is a coefficient sequence and a mod q is a rimitive residue class then the signed discreancy α; a, q of α in the sequence a mod q is given by α; a, q = αn αn 8 ϕq n a mod q n,q= Definition 35. A coefficient sequence α is said to be at scale N for some N if it is suorted on some interval of the form [ OL A N, + OL A N]. Definition 36. A coefficient sequence α at scale N is said to obey a Siegel-Walfisz theorem if α,q= ; a, r σqr O NL A 83 for any q, r, A R + and any rimitive residue class a mod r. 9

30 Lemma 37. Let α be a coefficient sequence and let C R +. Then d x C αd C x C L x. 84 Proof. Beyond the scoe of this wor but can be found in [3]. Proosition 38. Let α, β be coefficient sequences and let a be a rimitive residue class modulo q. Then α β; a, q = χa αχm βχn. 85 ϕq χ χ mod q Proof. By exanding out the sums and remembering that Dirichlet characters are totally multilicative: αχm βχn = αmβnχmn. 86 m= n= m= m= n= Changing the first sum to be over = mn this is equal to αdβ/d = α βχ. 87 = χ d = Thus the right-hand side of 85 becomes χa α βχ ϕq ϕq χ mod q = n= α βχ. 88 The rimitive Dirichlet character modulo q is equal to,q so the right-most term is α β 89 ϕq,q= and the left-most term is, uon switching the order of summation and recalling the orthogonality relations for Dirichlet characters α β χaχ = α β. 9 ϕq = χ mod q = a mod q Summing these last two equalities gives the required result. We will use these results to rove a generalised form of the Barban-Bombieri-Vinogradov theorem Theorem 39 Generalised BBV. Let M and N be such that x MN x and x ɛ M, N x ɛ for some fixed ɛ,. Let α, β be coefficient sequences at scale M, N resectively. Suose also that β obeys a Siegel-Walfisz theorem, then su α β; a, q xl A 9 a Z/qZ q x / o for some sufficiently slowly decaying o and any A R +. Proof. Set Q = x / ɛ then, by the oversill rincile Lemma??, it is sufficient to show that su α β; a, q xl A. 9 a Z/qQ q Q 3

31 Using Proosition 38, the left-hand side is simly su q Q a Z/qZ ϕq χ χ mod q βχn χa αχm. 93 m= n= By the definition of imrimitivity we can rewrite every character in the above sum in the form χn = χ n n,e= where q = de and χ is a rimitive Dirichlet character with conductor d >. Then the above exression becomes su n χa αχ,e= m βχ,e= e Q a Z/qZ <d Q/e ϕde χ χ mod d m= n=. 94 Now, using ϕde ϕdϕe we can slit u the totient function term. Furthermore, if we define S e to be equal to su a Z/qZ <d Q/e then 94 is equal to ϕd χ χ mod d e Q n χa αχ,e= m βχ,e= m= n= 95 ϕe S e L su S e. 96 e Q Woring with S e itself shows that, by the triangle inequality and the fact that the absolute value of any Dirichlet character is at most, S e is no more than <d Q/e ϕd χ χ mod d αχ,e= m βχ,e= n. 97 m= We now brea aart the sum over d by dyadic decomosition. I.E. using the observation that <d Q/e a d [log Q/e] = L su n= <d + a d D Q D<d D a d 98 Putting these all together shows that 94 is asymtotically less than L O su e,d Q D<d D ϕd χ χ mod d αχ,e= m βχ,e= n. 99 m= n= 3

32 Fix some C R + and suose that D L C. Then, slitting the sum over the rimitive residue classes βnχn n,e= βnχn n,e= βnχn n ϕd n,e= a Z/dZ n a mod d n,d= + βnχn n,e= n,d= ϕd max βχ,e= ; a, d + a Z/dZ βnχn n,e=. n,d= We treat the first term using the Siegel-Walfisz condition 83 along with the fact that ϕd d L C and the second using Lemma 37 to obtain βnχn n,e= σde O NL A +OC. n for any A R +. Note that, because A can tae any ositive value, we can ignore any bounded contribution of L. So, multilying this through by αnχn n,e= ML O, and using the fact that MN x we get βnχn n,e= xl A n n xl A su e Q D L C su e Q D L C D<d D D<d D ϕd χ χ mod d τde O τde O xl A 3 for all A R +, by multilicative function means. This is accetable so we deal with the case where D > L C. Then, again by Lemma 37 αm m,e= ML O 4 m βn n,e= NL O. 5 n Using the large sieve inequality and the Cauchy-Schwarz inequality gives us T := αmχm ϕd m,e= βnχn n,e= D<d D χ mod d m n / αmχm ϕd m,e= βnχn n,e= D<d D D<d D ϕd χ mod d χ mod d m αmχm m,e= m χ mod d / D<d D M + D / N + D / / αm D D m,e= m n n ϕd χ mod d / βnχn n,e= n βn n,e= / / 3

33 LO MNM + D N + D /. 6 D Again, using the fact that MN x we get But L C D Q = x / ɛ so T x/ L O MN + MD + ND + D 4 / D x/ L O MN / + M / D + N / D + D D xl O D + N / + M / + x / D. 7 T xl O L C + x ɛ + x ɛ + x ɛ xl O C 8 which, for large enough C, is xl A for all A. Substituting both these values bac into 99 will only multily by a bounded ower of L which we can safely ignore. This roves the result. C. Assorted Lemmata Lemma 4. Let κ, A, A, L R + and let γ be a multilicative function satisfying and L γ w z γ log A 9 κ log z w A for any w z. Also let h be the totally multilicative function defined on rimes by h = γ γ and, finally, let G : [, ] R be a iecewise-differentiable function with Then log d µd log zκ hdg = S log z Γκ d<z G max = su Gt G t. t [,] Gxx κ dx + O A,A,κSLG max log z κ 3 where S = γ κ 4 is the singular series and the O term is indeendent of G and L. Proof. Found in [7] with slight notational changes. Lemma 4. Let, m N and suose there exists some constant C R +, indeendent of m and, such that Cm e 4m. Then 4 log log log > m. 5 33

34 Proof. Suose that the lemma does not hold. Then there is some Cm e 4m with log m log 4 + loglog log log = log 4 log log + log log + log. 6 log log Now log log C + log m + 4m so log C + log 4 + log log + log + log + log log log = log C + log 4 + log log log log log log log log log log + log log. 7 This is continuous and tends to a finite limit and so is bounded. Thus we can increase C sufficiently to violate this inequality for all, yielding a contradiction. This imlies that the lemma must hold. Acnowledgements The author would lie to than Martina Balagovic, Samuel Bourne, James Maynard, Terence Tao, Robert Vaughan and Charles Wing for many useful conversations and correction. Further thans must go to Christoher Hughes for suervising this roject and to Victor Beresnevich for suggesting the toic of sieves many moons ago. References [] Bans, W. D., Freiberg, T., Maynard, J. 4. On limit oints of the sequence of normalized rime gas. erint arxiv: [] Bombieri, E., Friedlander, J. B., Iwaniec, H Primes in arithmetic rogressions to large moduli. Mathematische Annalen, 773; [3] Dicson, L. E. 94. A new extension of Dirichlet s theorem on rime numbers. Messenger of Mathematics, 33; 55-6 [4] Dirichlet, P. G. L Beweis des Satzes, dass jede unbegrenzte arithmetische Progression, deren erstes Glied und Differenz ganze Zahlen ohne gemeinschaftlichen Factor sind, unendlich viele Primzahlen enthlt, Abhand. A. Wiss. Berlin 48. [5] Elliott, P. D. T. A., Halberstam, H. 97. A conjecture in rime number theory. Symosia Mathematica, 4; [6] Euclid c. 3 B.C.. Elements. [7] Goldston, D. A., Graham, S. W., Pintz, J., Yildirim, C. Y. 9 Small gas between roducts of two rimes. Proceedings of the London Mathematical Society, 983; [8] Goldston, D. A., Pintz, J., Yildirim, C. Y. 9 Primes in tules I. Annals of Mathematics, 7; [9] Hardy, G. H., Littlewood, J. E. 93. Some roblems of artitio numerorum ; III: on the exression of a number as a sum of rimes. Acta Mathematica, 44; -7. [] Hanson, S. S. C. 4. On Polignac s conjecture and arithmetic rogressions. erint arxiv: [] Harman, G. 7. Prime-Detecting Sieves, nd edition. Princeton University Press. [] Maynard, J. 3. Bounded length intervals containing two rimes and an almost rime II. erint arxiv: [3] Maynard, J. 3. Small gas between rimes. erint arxiv:3.46. [4] Montgomery, H., Vaughan, R. 7. Multilicative Number Theory I. Classical Theory, st edition. Cambridge University Press. 34