Chapter 25 Primes in Short Intervals

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1 Chapter 25 Primes in Short Intervals October 10, Preliminaries to the modern theory A famous unsolved problem concerning prime numbers is the twin prime conjecture, that there are infinitely many pairs of primes which differ by 2. This suggests that there are considerable local oscillations in the primes, and this has motivated a large body of work concerned with investigating the possibility of gaps between primes which are significantly smaller than the average gap. Since 2004 a very powerful theory has been developed. This modern theory is motivated by the following observations. Consider a k tuple h 1, h 2,..., h k of distinct non negative integers for which it is believed that for infinitely many integers n the n + h 1,..., n + h k are simultaneously prime. Suppose we use a sieving technique to remove most n for which n+h 1,..., n+h k are not all prime. Whilst it may not be possible to establish that, for each of the remaining n, the members of the k tuple n+h 1,..., n+h k are all prime there is a better chance of finding several primes in many of the k tuples. Definition 1. Let h = h 1,..., h k be a k tuple of distinct non negative integers and let ν p h) denote the number of different residue classes modulo p among the h 1,..., h k. If ν p h) < p for every p, then h is called admissible. It is clear that if h is inadmissible, then there can only be a finite number of n for which the n + h 1,..., n + h k are simultaneously prime. 1

2 2 Conjecture 1 The prime k tuple conjecture). It is conjectured that if h is admissible, then there are infinitely many n such that n + h 1,..., n + h k are simultaneously prime. It is useful to establish that there are admissible sets with fairly small largest element. In Theorem 7.16 it is shown that there is a positive constant C such that ρn) = max M M+N n=m+1 p n = p>n πn) + CNlog N) 2. It is readily seen that if M is chosen to give the maximum, then the n contributing a positive amount to the sum cannot lie in the zero residue to any modulus not exceeding N and that for any p > N the number of n counted is smaller than the number of residue classes modulo p. Thus these n form an admissible set. Moreover taking n n 0 where n 0 is the smallest such n gives an admissible set which lies in [0, N]. One also has the following theorem. Theorem 1. Suppose that k 2 and the primes p 1,..., p k satisfy k < p 1 <... < p k. Then any translate of the k tuple p forms an admissible set. In particular h = {0, p 2 p 1,..., p k p 1 } is an admissible set and p k can be chosen so that p k < k log k + k log log k + Ok). Proof. The last part of the theorem follows from the prime number theorem. To prove the first part, suppose on the contrary that there is a q > 1 such that every residue class modulo q contains a p j. Then q k < p 1. On the other hand there is a j such that p j 0 mod q) and so p j = q k. One can consider applying the Hardy Littlewood method to this question. Suppose that h 1 < h 2 < < h k. Let Rx; h) = n x h k logn + h 1 ))... logn + h k )) where indicates that the sum is restricted to n for which n+h 1,..., n+h k are simultaneously prime, and let Sα) = p Nlog p)eαp) 1)

3 3 where N = x h k. Then Rx, h) = U k 1 Sα α k )S α 2 )... S α k )e α h α α k )h 1 ) dα. Suppose that we can replace each Sα) by its expected approximation when α is close to a rational number with a small denominator and the contribution from the remaining α is relatively small. We are deliberately rather imprecise as this is purely speculative. Thus if P = N δ for some small fixed positive number δ we would hope to obtain something of the form Rx, h) J q P a ) c q b) a h ϕq) c a2 + + a k )h 1 q a k 2 )... c q a k )e q where is over a modulo q with a 2,..., a k.q) = 1, b = a a k, and J = T β 2 + +β k )T β 2 )... T β k )e ) β h β 2 + +β k )h 1 dβ U k 1 and T β) = N eβn). n=1 It is believed generally that this should hold. The number J is the number of m 1,..., m k with 0 m j x and m j h j = n 1 h 1. Hence J = x + O1). Thus it is expected that Rx; h) xsh; P ) where Sh; P ) = q P fq; h)

4 4 and fq; h) = a c q b) ϕq) c q a k 2 )... c q a k )e a 2 h a k h k bh 1 ). 2) q It is readily verified that f is a multiplicative function of q. Moreover when q = p t with t 2, since a 2,..., a k, q) = 1 that for at least one j we have p a j, and so c p ta j ) = 0. Thus f has its support on the squarefree numbers. Now consider the case q = p. Then a 2,..., a k.p) = 1 holds for all a with 1 a j p except a 2 = = a k = p. If we sum over all a with 1 a j p we obtain p k 1 N where N is the number of solutions of r 1 r j h j h 1 mod p) with 1 r j p 1. Thus N = p ν p h). The term with a 2 =... = a k = p contributes p 1) k and so fp; h) = p ν ph))p k 1 p 1) k p 1) k = 1 ν p h)/p)1 1/p) k 1. 3) When p D = 1 i<j k h j h i we have ν p h) = k. Thus fp; h) p 2. Hence Sh; P ) converges absolutely to Sh) as P where and Sh) = fq; h) = 1 + fp; h)) = p p q=1 1 ν ) ph) 1 1 k 4) p p) Sh) k log log3d)) k k log log3 max h j )) k. 5) j It is clear from this that when the h j are distinct, if h is inadmissible, then Sh) = 0. If h is admissible, then we have ν p h) mink, p 1) and so 1 ν p h)/p 1/p when p k and is 1 k/p when p > k. Thus there is a positive number Ck) such that, when the h j are distinct, h is admissible if and only if Ck) < Sh). 6) This suggests a conjecture. Conjecture 2. Suppose that h is admissible. Then, as x, Rx; h) xsh).

5 This is highly speculative, of course, and establishing this is well beyond what can be done in the current state of knowledge. The likelihood of discovering primes in the k tuple n + h 1,..., n + h k depends on the avoidance of the zero residue class modulo p for all primes p, so in other words h needs to be admissible. A measure of this is the singular series Sh) and we can expect that this will arise naturally in the analysis. We can also deduce from 5) and the next theorem that there is a plentiful supply of admissible k tuples. Theorem 2 Gallagher). Suppose that k 2 and H is the set of k tuples h of distinct integers h 1,..., h k with 1 h j H, and let A be the subset of those h which are also admissible. Then Sh) = H k + OH k 1+ε ). h A In view of the observation above that if h H is inadmissible, then Sh) = 0, it suffices to prove the conclusion with A replaced by H. When ν p h) = k, fq) = fq; h) satisfies and otherwise fp; h) C k p 2 fp; h) C k p where C k is a suitable positive number. Let D = 1 i<j k h j h i, so that D H kk 1)/2. Then fq; h) q 2 C ωq) k D, q) ε q ε 2 D, q). For convenience we introduce the parameter Q 1 which is at our disposal. Then fq; h) r q ε 2 r ε 1 t ε 2 Q ε 1 dd). q>q r D r D t>q/r q>q D,q)=r 5 Hence fq; h) Q ε 1 H ε. 7) q>q

6 6 For convenience we write gq; h) = ϕq) k fq; h). 8) The case k = 2 is somewhat special so we treat that first. By 8) and 3), q gq; h) = µq) 2 eah 1 h 2 )/q) and so gq; h) = µq) 2 h H a=1 a,q)=1 q h 2 H a=1 h 1 H a,q)=1 h 1 h 2 eah 1 h 2 )/q). The innermost sum is a/q 1 and we have q 1 a=1 a/q 1 q log q. Thus fq; h) HQ ε. h H 1<q Q The case k = 2 of the theorem now follows from 4) and 7) with Q = H. For the rest of the proof we suppose that k 3. Crudely, by 3) for any h gq ; h) g q) where g q) = c q a 2 )... c q a k )c q a 2 a k ) a a,q)=1 and this is also a multiplicative function of q with its support on the square free numbers). Consider the k numbers a 2,..., a k, a 2 a k. When a, p) = 1 at least two of these numbers are not multiples of p. Moreover in g p) the terms with exactly j of the a 2,..., a k, a a k divisible by p contribute p 1) j and since the a 2,..., a k, a a k are linearly dependent the number of such terms is at most k j) p 1) k 1 j. Hence g p) 2 k p 1) k 1 and g q)ϕq) k q ε 1. Hence fq; h) q ε 1 and so h H 1<q Q h H 1<q Q fq; h) h [1,H] k 1<q Q fq; h) H k 1 q Q h [1,H] k 1<q Q fq; h) H k 1 Q ε. 9)

7 Returning to 2) when q > 1 at least two of a 2,..., a k, a 2 a k are non-zero modulo q. If there are at least two such of the a i, then we pick two and call them b 2, b 3. The remaining a i can be listed in the form b 4,..., b k so that a 2 a k = b 2 b 3 b k. If only one of the a i is non-zero modulo q, then call it b 2 and take b 3 = a 2 a k. In that case any one of the other a i can be rewritten in the form in the form b 2 b 3 s mod q) where s is the sum of the remaining a t. Thus h [1,H] k gq; h) q 1 H k 2 b 2 =1 c q b 1 ) b 3 /q q 1 b 2 =1 c q b 2 ) b 2 /q b [1,q] k 3 c q b 4 )... c q b k )c q b b k ) where b = b 4,..., b k and where the summand over b is taken to be c q b 2 +b 3 ) when k = 3. In general this multiple sum does not exceed q ) k 3 ϕq) c q b) b=1 Since c q b) q, b) the sum here is at most rϕq/r) dq)q. Similarly Therefore q 1 b=1 c q b) b/q r q r q q/r 1 r h [1,H] k 1<q Q a=1 a/q/r) 1 dq)q log q. fq, h) H k 2 Q 1+ε. Hence, by 7) and 9) the choice Q = H secures the theorem Exercises 1. Suppose that k 2 and the 1 < q 1 < q 2 < < q k. Prove that if for each j we have p q j p > k, then q 1,..., q k forms an admissible set. 7

8 8 2. As usual let ν p h) denote the number of different residue classes modulo p represented by the h. Let h j = 2j 1) 2 j = 1,..., k). Prove that h is an admissible set. 3. Call a set h of distinct non-negative integers h 1,..., h k sf admissible when there is no prime p such that every residue class modulo p 2 contains at least one of them. Let Sx; h) denote the number of n x such that n + h 1,..., n + h k are simultaneously squarefree. i) Let fn) denote the characteristic function of the squarefree numbers. Prove that Sx; h) = n x fn + h 1 )... fn + h k ) and fn) = d 2 n µd). ii) Suppose that 0 < δ < 1/3k) and let y = x δ and Prove that for j = 1,..., k where fn; y) = d y d 2 n µd). Sx; h) = T j x; y) + Ox 1+ε y 1 ) T j x; y) = n x fn + h 1 ; y)... fn + h j ; y)fn + h j+1 )... fn + h k ). iii) Given a k tuple of positive integers d = d 1,..., d k let d = d 1... d k and given another one r we use d r to mean d j r j j = 1,..., k)a and d 2 to mean d 2 1,..., d 2 k. Write n + h for the k tuple n + h 1,..., n + h k. Let ρd) denote the number of solutions of d 2 n + h in n modulo d 2. Prove that ρd) d 2 and µd 1 )... µd k ) T k x; y) = x ρd) + Oy 3k ). d 2 d 1 y,...,d k y iv) Let ν p h) denote the number of different residue classes modulo p 2 amongst the h 1,..., h k. Suppose that k = 2. Prove that Sx; h) = x 1 ν ) ph) + Ox 1 δ ). p 2 p

9 4. Given a k tuple of positive integers d = d 1,..., d k let d = d 1... d k and given another one r we use d r to mean d j r j j = 1,..., k) and d 2 to mean d 2 1,..., d 2 k. Write n + h for the k tuple n + h 1,..., n + h k. Let ρd) denote the number of solutions of d 2 n + h in n modulo d 2 and let ρ d) denote the number of solutions of d 2 n+h in n modulo lcm[d 1,..., d k ] 2. Let ν p h) denote the number of different residue classes modulo p 2 amongst the h 1,..., h k. i) Prove that ρd) = d 2 lcm[d 1,..., d k ] 2 ρ d) and ρ d) 1. ii) Prove that maxd j )>y and deduce that where µd 1 )... µd k ) d 2 ρd) T k x, y) = x gm) = m=1 d [d 1,...,d k ]=m maxd j )>y m>y gm) m 2 2 kωm) m 2 + O xy ε 1) µd 1 ) 2... µd k ) 2 [d 1,..., d k ] 2 y ε 1 µd 1 )... µd k )ρ d). iii) Prove that ρd) is multiplicative, i.e. given d, e, define de = d 1 e 1,..., d k e k and deduce that if d, e) = 1, then ρde) = ρd)ρe). iv) Prove that gm) is multiplicative and has its support on the squarefree numbers. v) Deduce that gm) = ) 1 + gp)p 2. m 2 p m=1 vi) Prove that 1 + gp)p 2 = 1 ν p h)p 2. vii) Pillai [1936].) Prove that Sx; h) = x 1 ν ) ph) + Ox 1 δ ) p 2 p 9

10 10 and hence that if h is sf admissible, then there are infinitely many n such that n + h j are simultaneously square free for j = 1,..., k. 5. Find the minimal diameter of 20 tuples which are sf admissible, i.e max h j h i is minimal Maynard s Theorem The principal idea is to use artefacts from sieve theory, especially the Selberg sieve, not directly in the form of a sieve but as a means to enhance terms of special interest, particularly those terms with relatively few prime factors. As a preliminary observation consider the starting point for the Selberg upper bound sieve Chapter 3) in the form ) 2 λ q a A q R q a and recall that one is planning to minimise this under the assumptions that λ 1 = 1 and that A d = 1 a A d a can be approximated by an expression of the form Xgd) d where X is a good approximation to A 1 and g is multiplicative. The minimising choice of λ q is given by SR, q) ) p λ q = µq) SR, 1) p gp) where SR, q) = p q r R/q,r,q)=1 µq) 2 p q gp) p gp). Typically one applies this when the sieve is of dimension k, e.g. gp) log p = k log y + O1). p p y

11 11 Under this kind of condition one might expect that and so λ q could be replaced by SR, q) Clog R/q) k p q λ q = µq) logk R/q) log k R p gp) p = µq) 1 log q ) k log R Indeed this is correct, and whilst there is some loss in precision in the final conclusion there is one significant advantage, namely that this choice of λ q can be applied quite effectively to any sieving question where the dimension is k. Let 1 P denote the characteristic function of the set of primes P. Then the basic idea is to construct an expression of the form k ) ) 2 1 P n + h j ) ρ λ q N n 2N j=1 q R q Zn;h) where Z = k i=1 n + h i). A wrinkle introduced by Goldston, Pintz and Yıldırım is to use a more general λ q of the form ) log q λ q = µq)f log R where f is at our disposal. If one can show that this is positive, then it follows that there are n such that the number of primes amongst the n + h j is at least ρ + 1. Following Maynard [to appear] we will use a more sophisticated version of this. Let n + h denote the k tuple n + h 1,..., n + h k and let d denote the k tuple d 1,..., d k. We generally use the notation that given a k tuple d of positive integers d denotes d 1... d k and given another one r, then d r means that d j r j for each j. We also use [d, e] to denote the k tuple lcm[d 1, e 1 ],..., lcm[d k, e k ]. First of all we do some initial sieving for small primes so as to simplify some later expressions and s simple way to do this is to restrict our attention to a given residue class a modulo q where q = p Q p, Q = log log log N 10)

12 12 and N is a large integer parameter. When h is admissible we can suppose that there is an a modulo q such that for 1 j j we have a + h j, q) = 1. To see that this holds observe that it holds for each prime divisor of q and then apply the Chinese Remainder Theorem. The immediate effect of this can be seen via the heuristic argument of If one supposes in addition that n a modulo q, then the singular series takes the shape Sh) = p>q 1 k p ) 1 1 p) k 1 for large N. Now we consider N<n 2N n a mod q) k j=1 In the first instance we ought to consider ) 1 P n + h j ) ρ λd) = µd)gd) d R d n+h d,q)=1 λd)) 2. 11) for some suitable g. However we shall be carrying out diagonalisation of quadratic forms in the λ and this leads to a natural representation of the λd), when d is squarefree with d, q) = 1, in the form λd) = µd)d µr) 2 log r ϕr) f r1 log R,..., log r ) k. 12) log R d r r,q)=1 We further suppose that suppf = R = {x [0, 1] k : x 1 + x k 1}. 13) There are two major tasks to be undertaken. The first is to obtain a good approximation to 11) with 12) for a wide class F of f. In practice this means good approximations S f) and T f) to Sf) and T f) where Sf) = k S j f) j=1

13 with S j f) = S j = T f) = T = N<n 2N n a mod q) N n 2N n a mod q) The second is then to maximise the ratio 1 P n + h j ) S f) T f) d R d n+h d,q)=1 d R d n+h d,q)=1 13 λd)) 2. 14) λd)) 2. 15) over the class F. The optimal solution to this latter task is not known, although the former can be carried out for a very wide class, for example for f for which the partial derivatives are continuous on R, and even this requirement can be relaxed somewhat. A major input into the approximation for S j f) will be the Bombieri Vinogradov theorem 20.??) or a variant thereof. We define the level θ of distribution for the prime numbers to be the assumption that for every sufficiently small positive δ and every A > 0 we have max sup liy) a,q)=1 πy; q, a) ϕq) δ,a xlog x) A. q x θ δ y x The Bombieri Vinogradov theorem tells us that θ = 1 is permissible. However it is useful to be able to see at once the consequence of the Elliott 2 Halberstam conjecture θ = 1) or some intermediate improvement in the Bombieri Vinogradov theorem. Let R j denote the set of k tuples t 1,....t j 1, t j+1,..., t k with t R for some t j. We define F to be the class of functions f, not identically 0, defined on R such that for each j, given t = t 1,..., t j 1, t j+1,..., t k with t i 0 and t t j 1 + t j t k 1 the function fj t j ) = ft) is absolutely continuous on [0, 1 t 1 t j 1 t j+1 t k ]. Given an f F it is useful first to extend its definition to [0, 1] k by taking it to be 0 outside R and then to define k 1 F = sup ft) + sup f t) t R t R j t j dt j. j=1 0

14 14 Theorem 3 Maynard). Let k 2. Suppose that the primes have level of distribution θ where 0 < θ 1, let δ be a sufficiently small positive number and let N be a large positive integer. Put R = N θ 2 δ, define Q and q as in 10) and f and R as in 13), and assume f F. Let h be an admissible set and choose a modulo q so that for each j we have a + h j, q) = 1. Let and I j = [0,1] k 1 1 ) 2 ft)dt j dt 1... dt j 1 dt j+1... dt k 0 J = ft) 2 dt [0,1] k and let Sf) and T f) be as in 14) and 15). Then, as N, Sf) = 1 + o1))ϕq)k Nlog R) k+1 q k+1 log N k j=1 I j and In particular T f) = 1 + o1))ϕq)k Nlog R) k q k+1 J. Sf) T f) = 1 + o1) ) ) k θ 2 δ j=1 I j. J The proof of Theorem 3 is divided into several stages. Fortunately the treatments of Sf) and T f) are similar. Initially we do not assume 12) but suppose only that the λd) are general real valued functions with support satisfying d 1... d k = d R, d, q) = 1 and d squarefree. Thus it can be supposed that d i, d j ) = 1 when i j. We begin with the normal diagonalisation process. To this end it is useful to define the multiplicative function ϕ 2 n) by ϕ 2 p) = p 2 and ϕ 2 p t ) = 0 when t 2. Lemma 4. For j = 1,..., k let κ j r) = µr)ϕ 2 r) j d r d λd) ϕd),

15 where j indicates that the summation variable is a k tuple, say d, which is restricted by d j = 1, and let κr) = µr)ϕr) d r d λd) d. 15 Then and λd) = µd)ϕd) j r d r λd) = µd)d r d r κ j r) ϕ 2 r) 16) κr) ϕr). 17) Proof. Consider j r d r κ j r) ϕ 2 r) On substituting the definition of κ j this becomes j r d r µr) j s r s λs) ϕs) = j s λs) ϕs) µr). r d r s The innermost sum is a sum over r 1,..., r j 1, r j+1,..., r k with d i r i s i, and the general term is µr) = µr 1 )... µr j 1 )µr j+1 )... µr k ). Thus the sum over r i is µd i ) t i s i /d i ) µt i) = 0 unless s i = d i in which case it is µd i ). Thus d j = 1 and j r d r κ j r) ϕ 2 r) = µd)λd) ϕd), which is equivalent to 16). The inversion formula 17) follows in the same way. Lemma 5. Let K j = max r κ j r), K = max κr). r

16 16 Then S j = N j κ j r) 2 ϕq) log N ϕ r 2 r) + O ) K 2 j ϕq) k 2 Nlog R) k 2 q k 1 Q and T = N q r κr) 2 ϕr) + O K 2 Nlog R) k qq ). Proof. We set the pattern with S j. Not only do we need to substitute κ j for λ in the main term but we need suitable bounds for the λd) in any error terms which arise. Moreover, we need to do so in terms of κ and κ j rather than λ. We square out and invert the order of summation. Thus S j = d,e d j =e j =1 λd)λe) N<n 2N [d,e] n+h n a mod q 1 P n + h j ). We recall that for λd) 0 we have d squarefree and d, q) = 1. Therefore d h, d v ) = 1 when u v. Likewise for e. Also if p n + h u and p n + h v, then p h v h u and this is impossible since p > log log log N > max h v h u. Thus, when u v, [d u, e u ], [d v, e v ]) = 1, whence d u, e v ) = 1. Since d j = e j = 1 we have [d j, e j ] = 1. Hence in the inner sum we are left with the system of congruences n h i mod [d i, e i ]) i j and n a mod q). Then the innermost sum can be rewritten as 1. N+h j <p 2N+h j p h j h i mod [d i,e i ] i j) p a+h j mod q By construction a + h j, q) = 1 and h j h i, de) = 1 when i j. Let m = q X = k [d i, e i ], i=1 2N+hj N+h j dt log t

17 and E = λd)λe) max d,e sup b,m)=1 x 2N+H lix) πx; m, b) ϕm) where indicates the restrictions d j = e j = 1 and d u, e v ) = 1 when u v, and H = max j h j. Then S j = X d,e λd)λe) ϕm) + OE). By 16), on taking the maximum over d with d j = 1 we have Thus, max λd) max ϕd) j K j µr) 2 d,d j =1 d,d j =1 r ϕ 2 r) d r d,q)=1 max λd) K j max d,d j =1 d ϕd) ϕ 2 d) Q<p R p d A concomitant argument shows that = K j max d ϕd) ϕ 2 d) j s s,dq)=1 µs) 2 ϕ 2 s) ) k 1 K j log R) k 1. p 2 17 max d λd) Klog R)k. 18) Now consider the number of ways that the modulus m/q can arise in E. This is squarefree. Let p be a prime factor of m/q. Then p divides exactly one of the [d i, e i ]. There are k 1 choices of i and for any one choice there are 3 possibilities, p d i, e i ); p d i and p e i ; and p d i and p e i. Thus there are 3k 1)) ωm/q) 3k) ωm) possible d, e which give rise to m. Thus E Kj 2 log R) 2k µm) 2 3k) ωm) max sup lix) b,m)=1 πx; m, b) m qr 2 x 2N ϕm). We have µm) 2 3k) 2ωm) max sup lix) b,m)=1 πx; m, b) m qr 2 x 2N ϕm) µm) 2 3k) 2ωm) Nm 1 Nlog N) 3k)2. m qr 2

18 18 Hence, by Bombieri s theorem E K 2 j Nlog N) A. It remains to deal with the main term for S j and it is desirable to rid ourselves of the condition that d u, e v ) = 1 when u v. That this is possible without undue effect on the main term is due to the prior sieving resulting from the choice of the residue class a modulo q. Thus any primes p which can potentially divide d u, e v ) satisfy p > Q. We have ϕm) = ϕq) i j ϕ[d 1 i, e i ]), = ϕd i,e i )) and ϕd ϕ[d i,e i ]) ϕd i )ϕe i ) i, e i )) = ϕ 2 n i ). Hence n i d i,n i e i 1 ϕm) = 1 ϕq)ϕd)ϕe) n d,n e ϕ 2 n). We substitute this in the main term and invert the order of summation to obtain X j ϕ 2 n) λd)λe) ϕq) ϕd)ϕe). n d,e n d,n e We now take the first step in dealing with the condition d u e v ) = 1 for u v. We replace it by s uv d u,s uv e v µs uv ). There are various observations we can make with regard to the s uv. We have n u d u. Thus d v, n u ) = 1. Hence s uv, n u ) = 1. Likewise s uv, n v ) = 1. Also, when w v, s uw e w and e v, e w ) = 1. Hences uv, s uw ) = 1. Likewise, when w u, s uv, s wv ) = 1 and so in summary Thus j n s uv, n u ) = 1, s uv, n v ) = 1, s uv, s uw ) = 1, s uv, s wv ) = 1. 19) ϕ 2 n) d,e n d,n e λd)λe) ϕd)ϕe) = j ϕ 2 n) µs uv ) n s uv u v u v j d n d s uv d u ) λd) j ϕd) e n e s uv e v ) λe) ϕe)

19 where indicates that 19) holds. We now substitute the κ j for the λ. Thus the above becomes ) j 1 µs uv ) κ ϕ 2 n) ϕ 2 s uv ) 2 j a)κ j b) n s uv u v u v where a = a 1,..., a k, b = b 1,..., b k and a u = n u s uv, b v = n v s uv. v v u u u v In particular a = b = ns where s = u v s uv. Thus the main term is X j ϕq) n 1 ϕ 2 n) s uv u v Since n j = 1 the terms with s > 1 contribute µs) ϕ 2 s) 2 κ ja)κ j b). 19 K2 j N ϕq) log N n R n,q)=1 d k 1 n)µn) 2 ϕ 2 n) s>1 s,q)=1 d kk 1) s)µs) 2 ϕ 2 s) 2. The inner sum is 1 + p>q 1 + ) kk 1) 1 p 2) 2 Q log Q and the sum over n is 1 + k 1)/p 2)) ϕq)log R)/q) k 1. Q<p R Thus the total contribution from the terms with s > 1 is Kj 2 ϕq) k 2 Nlog R) k 2. q k 1 Q For the remaining terms we have a = b = n. Thus they give X j κ j n) 2 ϕq) ϕ n 2 n).

20 20 We recall that X = 2N+h j N+h j 1 j κ j n) 2 ϕq) ϕ n 2 n) K2 j ϕq) dt = log t Q<p R N + O N ). Moreover log N log N) ) k 1 K2 j ϕq)) k 2 log R) k 1. p 2 q k 1 This completes the proof of the approximation for S j. The proof of the approximation for T is essentially the same, except that we do not use Bombieri s theorem and we do no have the restriction that d j = 1 to contend with. Thus on the initial application of the Chinese Remainder Theorem the main term is N m and the error term is O1). By 18) we see that the total contribution arising from this error is K 2 R 2 log R) 4k 2 which is acceptable. Then just as the function ϕ now plays the rôle that ϕ 2 played earlier, so the κ j is replaced by its understudy κ. Then the process of replacing λ by κ is identical, as is the elimination of the restriction d u, e v ) = 1. The functions κ j and κ introduced in Lemma 4 are clearly related to each other, as can be seen explicitly by 16) and 17). Thus when we insert the 16) into the definition of κ j and invert the order of summation we obtain when r j = 1) κ j r) = µr)ϕ 2 r) s r s κs) j µd)d ϕs) ϕd). d r d s Write e i = d i /r i and t i = s i /r i. Then the inner sum is µr)r ϕr) e e t e j =1 µe)e ϕe) = µr)rµt/t j) ϕr)ϕt/t j ) = rµs/s j) ϕs/s j ). On using the notation rt for r 1 t 1,..., r k t k we find that κ j r) = rϕ 2r) ϕr) 2 t κrt) µt)ϕt j)µt j ) ϕt) 2.

21 21 The terms with t > t j contribute K t j R t j,q)=1 µt j ) 2 ϕt j ) n>1 n,q)=1 k 1) ωn) µn) 2 ϕn) 2 and we have t j R t j,q)=1 µt j ) 2 ϕt j ) Q<p R ) p ϕq) log R p 1 q and 1 + Q>p 1 + k 1 ) ) Q 1. p 1) 2 Since also it follows that, when r j = 1, 2 rϕ 2 r) = 1 + O1/Q) ϕr) κ j r) = t j κr ) ϕt j ) + O ) Kϕq) log R qq 20) where r = r 1,..., r j 1, t j, r j+1,..., r k. Having come this far, we should take stock. maximise the ratio j r κ j r) 2 ϕ 2 r) κr) 2 r ϕr) We now make the reasonable assumption that log r1 κr) = f log R,..., log r ) k log R The ultimate aim is to 21) where f F which through 17) gives 12). The final step of the proof of Theorem 3 is to obtain smooth approximations to the main terms in Lemma 4. We have standard methods of carrying this out when k = 1, i.e r = r 1. We adopt the simple expedient of establishing a suitable one dimensional approximation and then applying it k times.

22 22 Suppose that g : [0, 1] R. Then we call g l piecewise absolutely continuous on [0, 1] when associated with g there is a partition a 0 = 0 < a 1 <... < a l = 1 of [0, 1] such that for 1 j l 1. g + a j 1 ) = lim x aj 1 + gx) and g a j ) = lim x aj gx) both exist, and 2. g is absolutely continuous on [a j 1, a j ] when we replace ga j 1 ) and ga j ) by g + a j 1 ) and g a j ) respectively. We define Gl, G) to be the class of l piecewise absolutely continuous functions g on [0, 1] such that sup gv) + v [0,1] 1 0 g v) dv G. We observe in passing that in practice it suffices for our application that g is continuous except for at most one x in [0, 1] where g and g have jump discontinuities. Lemma 6. Suppose that η : N R is multiplicative with its support on the squarefree numbers, and that 0 ηp) 2 and there is a constant C such that whenever p > C we have ηp) 1 p C p. 2 Suppose also that g Gl, G) and m N. Then n x n,m)=1 ηn)g ) log n log x 1 = A m gv)dv log x + O lg p m log p p ) p m p) where We also have A m = ϕm) m p m 1 + ηp) ) 1 1 ). p p A m ϕm) m.

23 23 In order to make a comparison with the main term, which is of size ϕm) m log x 1 it is useful to observe that the error term is 0 gv)dv, G ϕm) m log log 3m)3. Proof. We begin with the case when g is identically 1. Also we may suppose that ηp) = 0 when p m. Let ρ be the multiplicative function with ρp) = ηp) 1/p, ρp 2 ) = ηp)/p, ρp t ) = 0 t 3). Then u n ρn/u) u = ηn). Let l = rst 2 with r m and st, m) = 1, so that ρl) = 0 unless µrst) 0. Moreover ρr) = 1/r, ρs) C ωs) s 2, ρt 2 ) t 2 u t Cωu) /u. Thus, for ν = 0 or 1, log 2l) ν ρl) l y rst 2 y r m,st,m)=1 1 + p m log 2rst) ν µrst)2 rs 2 t 2 Cωs) u t log p p ) p m ). p C ωu) u Also Therefore x<l y n x ηn) = u,v uv x ρl) 1 log x log l) ρl). ρv) u = ρv) log x ) v + O1) v x = A m log x + O 1 + ) log p p p m p m l p).

24 24 Now we apply this to general g Gl, G). Let Ex) = n x ηn) A m log x and choose a j as provided by the definition of Gl, G). When x a j 1 < n x a j we have ) log n aj g = g a j ) g v)dv log x log n log x except possibly when n = x a j in which case the two sides differ by G. Now we multiply by ηn), sum over the n x a j 1, x a j ], interchange the order of summation and integration and apply the formula for E to obtain Am log x)a j a j 1 ) + Ex a j ) Ex a j 1 ) ) g a j ) aj a j 1 Am log x)v a j 1 ) + Ex v ) Ex a j 1 ) ) g v)dv + OG). We integrate the main term by parts to obtain aj a j 1 A m log x)gv)dv which on summing over j gives the desired main term. We insert the bound for E given by the first part of the proof and sum over j. This completes the proof of the lemma. We are now in a position to complete the proof of Theorem 3. We now make the choice 21) for some f in F. To simplify some of the formulae we then extend the definition of f to [0, 1] k by taking f to be 0 outside R. Again we concentrate on S j rather than T. We recall that κ j r) = 0 unless r j = 1, r, q) = 1 and r is squarefree, in which case, by 20) and 21), we have κ j r) = µt j )2 ϕt t j ) f j log r1 log R,..., log r j 1 log R, log t j log R, log r j 1 log R,..., log r k log R ) ) F ϕq) log R + O qq

25 25 where r = r 1,..., r j 1 t j, r j+1,..., r k. Thus K j F ϕq) q log R. Moreover, by Lemma 6, with ηp) = 1/p and m = qr we have κ j r) = log R) ϕqr) ) F ϕq) log R qr f jr) + O qq where f j r) = 1 0 f log r1 log R,..., log r j 1 log R, u j, log r j 1 log R,..., log r ) k du j. log R Recall that this holds when r j = 1, r, q) = 1 and r is squarefree and that otherwise κ j r) = 0. Thus, by Lemma 5, ϕq)nlog R)2 ) j µr) 2 ϕr) F S j = q 2 log N ϕ 2 r)r f jr) 2 2 ϕq) k Nlog R) k + O. q k+1 Q r r,q)=1 The general arithmetical factor in the main term in the sum can be rewritten as k µr i )ϕr i ) ϕ 2 r i )r i i=1 provided that the sum over r is restricted to r with r u, r v )=1 when u v. However if r u, r v ) > 1, then there is a prime p > Q such that p r u and p r v. Therefore when we remove the condition r u, r v ) = 1 the total error in so doing is F 2 ϕq)nlog R) 2 q 2 log N p>q F 2 ϕq) k Nlog R) k. q k+1 Q ϕp) 2 ϕ 2 p) 2 p 2 n<r n,q)=1 Thus the sum in the main term can be replaced by j r r,q)=1 f j r) 2 k i=1 µr i ) 2 ϕr i ) ϕ 2 r i )r i. ) k 1 µn) 2 ϕn) ϕ 2 n)n

26 26 Now we apply Lemma 6 to each variable r i in turn, i.e k 1 times, with and m = q. Thus ηp) = p 1 p 2)p 1 p = 1 pp 2) S j = ϕq)k Nlog R) k+1 I q k+1 j + O log N ) F 2 ϕq) k Nlog R) k q k+1 Q where I j is as in Theorem 3. This gives the first part of that theorem. The second part follows in the same way. Theorem 7 Maynard). Suppose that when k 2, we take f F and then I j = I j f) and J = Jf) are as in Theorem 3. Let k j=1 I jf) Then, for k sufficiently large, ρ = sup f F Jf). ρ > log k log log k 1. Corollary 8 Zhang). There are bounded gaps in the sequence of primes. This is immediate from Theorems 3, 7 and the fact that there are admissible sets with k elements as provided, for example, by Theorem 1. Corollary 9 Maynard, Tao). For each m N we have lim inf n p n+m p n ) m 2 e 4m. Corollary 10 Maynard). Let m N and let G = {g 1,..., g l } be a set of l distinct non negative integers. Let Mm, l, G) be the number of admissible m tuples contained in G and let Nm, l, G) be the number of admissible m tuples h contained in G such that there are infinitely many n for which each member of the m tuple n + h is prime. Then for l > l 0 m) l m Mm, l, G) m l m and Nm, l, G) Mm, l, G) m 1.

27 de Polignac s conjecture [1849] asserts that every even integer is the the difference of infinitely many pairs of primes. That the conjecture holds for a positive proportion of all even integers follows on taking m = 2 and g j = 2j 2 in the previous corollary, for then number of solutions of g j2 g j1 = 2d is at most l and so there must be l 2 /l = 1 different differences g j2 g j1 arising from the admissible pairs counted by N2, l, G). Corollary 11. There is an infinite subset D of N with positive lower asymptotic density such that for each d D there are infinitely many pairs of primes p 1, p 2 such that p 2 p 1 = d. Proof of Theorem 7. Let x = and let ξ be the positive solution to k/ log k logk/ log k) 1 + ξx = e ξ. 27 Then and so, for k sufficiently large e ξ ξ > x log k log log k < ξ < log k. Let g : [0, ) R be defined by { 1 0 y x, 1+ξy gy) = 0 x < y. We need to compute various integrals which we denote by α, β, γ, τ as follows. α = β = γ = τ = gy)dy = 1, gy) 2 dy = 1 ξ 1 ξe ξ, ygy) 2 dy = 1 ξ 1 ξ ξ 2 e ξ, y 2 gy) 2 dy = x ξ 2 2 ξ ξ 3 1 ξ 3 e ξ.

28 28 We now take ft) = { k i=1 gkt i) t R, 0 t / R. Since f is symmetric we have I j f) = I k f) for every j k. Thus ρ ki kf) Jf) and we now proceed to estimate I k f) and Jf). It is clear that with this choice most of the mass of f is close to the axes and so the boundary condition t t k 1 on R is relatively unimportant. Since we are concerned with only a lower bound for ρ, lower and upper bounds for I k f) and J respectively will suffice. An upper bound for Jf) is easy. We have Jf) [0, ) k i=1 k gkt i ) 2 dt = k k β k. Thus we can concentrate on I k f). Let S denote the set of k 1 tuples y 1,..., y k 1 with y i 0 and y y k 1 k x. Then we have k 1 ) 1 t1 t k 1 ki k f) = k gkt i ) 2 R k 1 i=1 0 k 1 k k α 2 gy i ) 2 dy S i=1 = k k α 2 β k 1 E gkt k )dt k ) 2 dt 1... dt k 1 where and Let E = k k α 2 k 1 gy i ) 2 dy S i=1 S = [0, ) k 1 \ S. σ = γ/β = 1 ξ 1 + ξ 1 e ξ 1 e ξ = 1 1 ξ + 1 e ξ 1.

29 The condition y S is equivalent to y y k 1 k x and this in turn is equivalent to 29 y y k 1 k 1 σ k x σk 1) k 1 = 1 σ)k 1) x + 1. k 1 For k sufficiently large we have and In particular if y S, then 1 σ)k 1) x + 1 > 0 1 σ x 1 k 1 = ξ 1 + O ξ 2). y1 + + y k 1 k 1 σ) 2 ζ 2 1 where Hence ζ = E k k α 2 ζ 2 1 σ x 1 ) 1 = ξ + O1). k 1 [0, ) k 1 y1 + + y k 1 k 1 ) 2 k 1 σ gy i ) 2 dy. i=1 We now square out the expression y1 + + y k 1 k 1 = ) 2 σ 2 k 1) 2 1 i<j k 1 y i y j + 1 k 1) 2 k 1 yi 2 i=1 2σ k 1 k 1 i=1 y i + σ 2 and evaluate the various integrals with reference to the integrals evaluated above. Thus k 2 E k k α 2 ζ 2 k 1 γ2 β k ) k 1 τβk 2 2σγβ k 2 + σ 2 β k 1.

30 30 By the definition of σ this becomes k k α 2 ζ 2 k 3 τβ γ2 β k 1 < k k α 2 ζ 2 β k 2 τ k 1. Thus We have Thus ) ρ > β 1 1 ζ2 τ. βk 1) log k log log k < ξ = log k log log k + O1), β 1 = ξ + Oξk 1 log k), ζ 2 = ξ 2 + Oξ), τ = xξ 2 + Oξ 2 ), 1 k 1 = 1 k + Ok 2 ). ζ 2 τ βk 1) = ξ + O1) ) xk 1 = Hence ρ > ξ 1 + O k 1 log k )) 1 1 log k + O log k) 2) ) if k is sufficiently large. ξ + O1) log k) logk/ log k) = 1 log k + O log k) 2). > log k log log k 1 Proof of Corollary 9. Let C be a constant chosen so that for every m N we have Cme 4m 4m + log m + log C > e2+4m. Hence for k max3, Cme 4m ) we have k log k e2+4m and so log k log log k 1 > 4m + 1.

31 31 Thus if k is large enough ) log k log log k 1) > m. k Taking the level of distribution θ to be 1 and choosing δ = 2 we see that 2 k every admissible k tuple h has the property that there are infinitely many n such that the k-tuple n + h contains at least m primes. By Theorem 1 there is a an admissible k tuple of diameter k log k m 2 e 4m. Proof of Corollary 10. Let k = max3, Cme 4m ) be as in the proof of Corollary 9 and let h be an admissible k tuple. By considering all possible m tuples h = h 1,..., h m which are subsets of h we see that at least one has the property that there are infinitely many n such that n + h 1,..., n + h m are simultaneously prime, i.e. the prime m tuple conjecture holds for this m-tuple. Starting from G we construct a subset G by successively removing elements from G. Given a prime p and a finite set L of integers we can construct a subset as follows. Let Lp; h) = {n L : n h mod p)} and Lp; h) = cardlp; h). Choose an h for which Lp; h) is minimal and take L = L \ Lp; h). Then cardl 1 1/p)cardL. We apply this operation successively to G for p k giving a subset G which satisfies cardg cardg p k 1 1 ) m l. p Thus on taking l to be sufficiently large we have s = cardg > k. Every subset h of G of cardinality k is an admissible set since it omits a residue class modulo p for every p k. There are s k) such h and, from above, each one contains at least one m tuple h for which the prime m tuples conjecture holds. Subsets b of G of cardinality k which contain h are exactly those in which the k m remaining elements of b are chosen at random from the the s m remaining elements of G. Thus there are precisely s m k m) such b. Hence there are at least ) s k / s m ) k m = s m+1)...s 1)s k m+1)...k 1)k m s m m l m admissible subsets of G of cardinality m which satisfy the prime m tuples conjecture. On the other hand there are l m) l m subsets h of G of cardinality m, and this completes the proof of Corollary 10.

32 Exercises 1. Suppose that k 2. Let R k [0, 1] k be defined by R k = {t : t i 0, t t k 1}, and let m N and ft) = 1 t 1 t k ) m. Given t 1,..., t j 1, t j+1,..., t k [0, 1] k 1 with t t j 1 + t j t k 1 let A j denote the interval [0, 1 t 1 t j 1 t j+1 t k ] and take it to be the empty set otherwise) and define and I j f) = i) Prove that k I j f) = j=1 A j ft)dt j Jf) = ft) 2 dt. R k k2m + 2)! 2m k)!m + 1) 2. ) 2 dt 1... dt j 1dt j+1... dt k ii) Prove that Jf) = 2m)! 2m + k)!. k j=1 iii) Prove that I ) jf) 1 = m + 1 ). Jf) 2m + 2 2m k iv) Goldston, Pintz, Yıldırım) Prove that if the level θ of distribution of the primes satisfies θ > 1, then there are infinitely many bounded gaps in 2 the sequence of primes. 2. Let R k be as in question 1. For t R k let α k t) = t t k and β k t) = t t 2 k. i) Suppose that a and a j are non negative integers. Prove, by induction on k or otherwise, that R k 1 α k t)) a k j=1 t a j j dt = a! k j=1 a j! k + a + k j=1 a j)!. ii) Suppose that a and b are non negative integers. Prove that R k 1 α k t)) a β k t) b dt = a!b! k + a + 2b)! k b j=1 b 1 + +b k =b 2b j )!. b j!

33 The multinomial theorem applied to β b k is useful here.) 3. Maynard) i) Let k = 5. In the notation of the previous question, when t R 5, let ft) = 1 α 5 t))β 5 t) α 5t)) β 5t) α 5t)). 33 Prove that 5 j=1 I jf) Jf) = ii) Prove that if the level of distribution θ is 1, then 25.3 Notes lim inf n p n+1 p n The study of the distribution of k tuples of primes is an area which is, as we write October 10, 2018), very active and indeed has a considerable air of excitement about it with substantial progress and widespread consequences. There is a long history of investigation into the existence of k tuple of primes, but until the last decade or so, in spite of the use of quite heavy machinery, there has been more speculation than progress. The twin prime conjecture that there are infinitely many prime pairs which differ by 2 was already well known when de Polignac conjectured 1849) the every even number is the difference of infinitely many pair of primes. The general prime k tuples conjecture, Conjecture 1, and a quantitative form equivalent to Conjecture 2, were made by Hardy and Littlewood [1922], and their justification was, as here, by a process which ignores any minor arcs and assumes good approximations to the generating function on the major arcs. Theorem 2 can be found in Gallager [1976]. 2. In an unpublished manuscript, originally intended as part of the seminal series of papers On some problems of partitio numerorum, in which they assumed the Generalised Riemann Hypothesis GRH), Hardy and Littlewood applied their eponymous method to show that lim inf n p n+1 p n log p n 2 3.

34 34 By a different method based on Brun s sieve Erdős [1940] showed that there is a constant c < 1 such that unconditionally lim inf n p n+1 p n log p n c. Rankin [1940] refined the Hardy Littlewood argument to obtain c = 3/5 on GRH, and in [1950] obtained c = 42/43)3/5) = on GRH by combining his method with that of Erdős. Ricci [1954] improved the sieve method so as to obtain c = 15/16 unconditionally. More substantial progress was made by Bombieri and Davenport. It is of some methodological and historic interest. In a similar manner to Hardy and Littlewood, and Rankin, they consider where Sα) is as in 1) and I = 1 Kα) = H eαh) 2 = h=1 0 Sα) 2 Kα)dα H H h )eαh) h= H is the Fejér kernel. They then evaluate I in two different ways. First of all, by the orthogonality of the additive characters enα), where The first sum is I = H p xlog p) Rx; h) = p,p x p p=h H H h)rx; h) h=1 log p)log p ). Hx log x + OHx). The positivity of the integrand means that a lower bound can be obtained by considering major arcs only. Moreover here the Bombieri Vinogradov theorem can be used to make the major arcs almost as wide and numerous as would pertain if one assumed the Generalised Riemann Hypothesis. This was the first use of the Bombieri Vinogradov theorem, and indeed it is this

35 work which stimulated Bombieri s discovery. Thus it can be shown that the integral is asymptotically at least and so the asymptotic lower bound H 2 x + 1 Hx log x 2 H H h)rx; h) 1 2 H2 x 1 Hx log x 4 h=1 p implies that lim inf n+1 p n n log p n 1. They also apply the Selberg sieve to 2 obtain an upper bound for the terms on the left with h near H, and here the Bombieri Vinogradov theorem is also applied in the sieve. This gives 35 c = = More refined kernels Kα) were introduced by Pil tai [1972] and Huxley [1973,1977] to obtain small improvements, c = , c = , c = respectively. A larger improvement was made by Maier [1988] by the adaptation of his matrix method to the argument so as to take advantage of oscillations in the primes over short intervals and this lead to the known bound being reduced by a factor of e γ where γ is Euler s constant. c = In the last decade or so there have been a series of major advances. In a seminal paper Goldston, Pintz, Yıldırım [2009] proved that lim inf n p n+1 p n log p n = 0 In fact they were able to prove rather more than this. For example they obtain [2010] an explicit upper bound p n+ p n log p n ) 1/2 log log p n ) 2 22)

36 36 which is satisfied for infinitely many n. and they showed that if the level of distribution exceeds 1, then there are infinitely many bounded gaps between primes. Indeed, if the level of distribution can be taken to be 1 2 the Elliott Halberstam conjecture [1970]), they were able to show that infinitely often p n+1 p n 16. All subsequent work is based on their method. There have been two sensational developments. Zhang [2014] proved a version of the Bombieri Vinogradov theorem in which the moduli of the arithmetic progressions are restricted to being numbers with only relatively small prime factors but, crucially, the level of distribution exceeds 1 by a small 2 amount. Then, although the moduli are restricted, nevertheless the modified Bombieri Vinogradov theorem contains enough information to enable an adaptation of the Goldston, Pintz, Yıldırım machinery to work. Thus Zhang obtains lim inf p n+1 p n 70, 000, ) n Then Maynard [to appear], returning to a version of the Goldston, Pintz, Yıldırım methods which predates their [2009] paper and which had been aborted as unsuccessful, was able to adapt their methods to establish infinitely many bounded gaps between the primes even if one only has a positive level of distribution for the primes. In particular, using the Bombieri Vinogradov theorem Maynard obtains lim inf n p n+1 p n ) These most recent methods involve quite heavy computations to obtain the sharpest bounds. For example, in the notation of exercise , Maynard considers Theorem 3 with ft) = d a i 1 α k t)) b i β k t) c i i=1 and finds that c.f. Exercise ) k j=1 I jf) Jf) = at Ma a T N a where the d d positive definite matrices M, N depend on the exponents b i, c i. He shows that this ratio is maximised when a is an eigenvector of MN 1 corresponding to the largest eigenvalue. He then takes k = 105 and considers

37 all choices of b i, c i with b i + 2c i 11, so that d = 42. It transpires that the largest eigenvalue is and so an appeal to Theorem 3 establishes that for any admissible 105 tuple h there are infinitely many n such that n + h contains at least two primes. He then displays a known admissible 105 tuple of diameter 600 discovered by Englesma to establish 24). Maynard s method also shows that when the level of distribution is 1 we have for which see exercise There is an ongoing Polymath project at 37 lim inf n p n+1 p n 12, 25) =Bounded_gaps_between_primes lead by Tao to combine all the methods, especially those of Maynard and Zhang, and the record as of 11th April 2014 is lim inf n p n+1 p n ) The methods described here have very great flexibility and have many potential applications. One is to Dickson s conjecture [1904] which states that if the g i, h i are integers and k i=1 g in + h i ) has no fixed prime divisor, then there are infinitely many n such that the g i n + h i are simultaneously prime. Another undertaken by Pintz [to appear] is to deal with questions involving consecutive primes and arithmetic progressions. In yet another Goldston, Graham, Pintz, Yıldırım [2011] have considered n for which dn) = dn + 1), ωn) = ωn + 1) and Ωn) = Ωn + 1) simultaneously. There are also applications to cognate problems in algebraic number fields References Bombieri, E., Davenport, H. [1966], Small differences between prime numbers, Proc. Roy. Soc. Ser. A ), Dickson, L. E. 1904), A new extension of Dirichlet s theorem on prime numbers, Messenger of Math ),

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