BEST POSSIBLE DENSITIES OF DICKSON m-tuples, AS A CONSEQUENCE OF ZHANG-MAYNARD-TAO ANDREW GRANVILLE, DANIEL M. KANE, DIMITRIS KOUKOULOPOULOS, AND ROBERT J. LEMKE OLIVER Abstract. We determine for what roortion of integers h one now knows that there are infinitely many rime airs, + h as a consequence of the Zhang-Maynard-Tao theorem. We consider the natural generalization of this to k-tules of integers, and we determine the limit of what can be deduced assuming only the Zhang-Maynard-Tao theorem.. Introduction and statement of results The twin rime conjecture states that there are infinitely many airs of integers n, n + 2 which are simultaneously rimes. More generally, Hardy and Littlewood conjectured that each entry of the k-tule n + h,, n + h k should be rime infinitely often, unless there is a trivial reason why this cannot haen. This trivial reason is about divisibility by small rimes. For examle, the trilet n + 2, n + 4, n + 6 can never all be rime if n > because at least one of them must be a multile of 3. So we call a k-tule admissible if, for each rime, the reductions of the numbers h,..., h k modulo do not cover all of Z/Z. With this definition in hand, the Hardy-Littlewood conjecture states that if h,..., h k is an admissible k-tule, then there are infinitely many integers n for which the numbers n + h,..., n + h k are all rime. Even though we are still far from roving the full Hardy-Littlewood conjecture, there has been remarkable rogress made towards it recently. Firstly, in May 203, Yitang Zhang [8] made headlines by roving that there are bounded gas between rimes, and secifically that lim inf n n+ n < 70,000,000, where n denotes the n-th rime. Then, in November 203, James Maynard [3] and Terence Tao indeendently showed, using somewhat different techniques, that 70,000,000 can be relaced by 600, and, as it stands right now, the best bound known is 246, due to the Polymath roject [6]. Even more imressively, they roved that for any integer m there is aninteger k = k m such that if h,..., h k is an admissible k-tule, then there are infinitely many integers n for which at least m of the numbers n+h,..., n+h k are rime. Obviously, k m m, and in [6] it was shown that one can take k 2 = 50 and k m e 3.82m. We call a k-tule of integers h,..., h k a Dickson k-tule if there are infinitely many integers n for which n + h,..., n + h k are each rime. The Hardy-Littlewood conjecture is equivalent to the statement that all admissible k-tules of integers are Dickson k-tules, and the Maynard-Tao theorem imlies that every admissible k m -tule of integers contains Date: October 29, 204. The first and third authors are suorted by Discovery Grants from the Natural Sciences and Engineering Research Council of Canada. The second and fourth authors were suorted by NSF Mathematical Sciences Postdoctoral Research Fellowshis.
2 A. GRANVILLE, D. M. KANE, D. KOUKOULOPOULOS, AND R. J. LEMKE OLIVER a Dickson m-tule. In articular, the Maynard-Tao theorem imlies that Dickson m- tules exist, yet no exlicit examle of a Dickson m-tule is known! Nevertheless, a simle counting argument which also aeared in [2] yields the following attractive consequence of the Maynard-Tao theorem. Corollary.. A ositive roortion of m-tules of integers are Dickson m-tules. Proof. Let k = k m, so that m k, and define R = k and x = NR for some very large integer N. We let N = {n x, x] : n, R = }, so that N = 2xφR/R. Any subset of k elements of N is admissible, since it does not contain any integer 0 mod for any rime k. There are N k such k-tules. Each contains a Dickson m-tule by the Maynard-Tao theorem. Now suose that there are T x Dickson m-tules within N. Any such m-tule is a subset of exactly N m k m of the k-subsets of N, and hence N m N T x, k m k and therefore as desired. T x N m / N m k m N /k m = m φr 2x m, kr The main goal of this aer is to rovide better lower bounds on the roortion of m- tules that are Dickson m-tules. If m is the roortion of such m-tules, then Corollary. imlies that m > 0. We are interested in determining the best ossible lower bound on m assuming only the results of Zhang, Maynard, and Tao and treating them as black boxes : If D m is the set of Dickson m-tules, then there is an integer k = k m for which: D m is translation-invariant, that is to say, if h,..., h m D m and t Z, then h + t,..., h m + t D m ; D m is ermutation-invariant, that is to say, if h,..., h m D m and σ S m, then h σ,..., h σm D m ; for any admissible k-tule, x,..., x k, there exist distinct h,..., h m {x,..., x k } such that h,..., h m D m. We call any set A Z m that has the above roerties m, k-lausible. Then we define { } A [ N, N] m. δm, k = min lim inf : A Z m, A is m, k lausible ; N 2N m and, for any k m, we have that m δm, k by the Zhang-Maynard-Tao theorem. Our main result is the following. Theorem.2. For k m, we have, uniformly δm, k = log 2mOm k m log log 3k m.
BEST POSSIBLE DENSITIES 3 In the Maynard-Tao Theorem we know that one can obtain k m e cm for some constant c > 0. The best value known for c is a little smaller than 3.82, and Tao [7, 2] showed that the Maynard-Tao technique cannot be directly used to obtain a constant smaller than 2. So we deduce that m δm, k m e c+om2 e 4m2, and that this can, at most, be imroved to e 2m2 using the currently available methods. A de Polignac number is an integer h for which there are infinitely many airs of rimes and + h the twin rime conjecture is the secial case h = 2. A ositive roortion of integers are de Polignac numbers, as follows from both results above, but we wish to determine the best exlicit lower bounds ossible. In Section 2, using quite elementary arguments, we will show that.2 2 δ2, k 2 0.002830695767... > 49 354, 50 and that this cannot be imroved dramatically without imroving the value of k 2 = 50. We remark that this bound is surrisingly good in light of the fact that we can only deduce that there is some h 246 for which there are infinitely many rime airs, + h. Finally, we note that it would erhas be more natural to consider the roortion ad m of admissible m-tules that are Dickson m-tules, rather than the roortion of all m- tules. To this end, in Section 4, we determine the roortion ρ ad m of all m-tules that are admissible. Surrisingly, this question has seemingly not been addressed in the literature, and we rove that.3 ρ ad m = as m. e om e γ log m m 2. The density of de Polignac numbers To rove a lower bound on the density of de Polignac numbers, we consider admissible sets B of integers that do not contain a Dickson air. If the rime k-tulets conjecture is true, then necessarily B =, while Zhang s theorem imlies that B k 2. Because of this uer bound, there must be maximal such sets, in that, for any t B such that B {t} is admissible, B {t} contains a Dickson air. This condition imlies that t B contains a de Polignac number, and we will obtain a lower bound on δ2, k 2 by varying t. To this end, for an admissible set B, define ηb to be the minimal lower density of sets A with the roerty that t B contains an element of A whenever B {t} is admissible. Moreover, for any integer l, set ηl to be the infimum of ηb as B runs over admissible sets of size l. We thus have that 2 δ2, k 2 min ηl, l k 2 and we will rove the following. Proosition 2.. For any integer l, we have that ηl e γ /l log l, and exlicitly that ηl l l y l+ for any ositive integer y l. y
4 A. GRANVILLE, D. M. KANE, D. KOUKOULOPOULOS, AND R. J. LEMKE OLIVER For the articular alication to 2, we take l {,..., 49} in Proosition 2. to find that δ2, k 2 min ηl > 0.002830695767 > l 49 354, while taking y = 3 yields that η49 < 0.005328005328 <, so that, using our techniques, 87 we can hoe for an imrovement in our lower bound for δ2, k 2 by a factor of at most about two. However, we can do better: the exlicit uer bound given in Lemma 3. below shows that δ2, 50 < 0.0048. 20 Proof of Proosition 2.. We begin with the lower bound for ηl. Suose that B is admissible of size l, so that for each rime l + there exists a residue class n mod such that n + b for each b B. If t n mod for all l + then B {t} is admissible and so contains a Dickson air, and, as remarked above, t must be one of that air, else B would have contained a Dickson air. If x is large then the number of such integers t x is l+ and we know that, for some b B, t b is a de Polignac number. Given t Z, an integer can be written in at most l ways as t b for b B, since B = l. Hence, ηb e γ l l log l l+ as l. We now turn to the uer bound. We will construct sets A and B with the roerties that B is admissible and that if t is such that B {t} is admissible, then there exists b = b t B such that t b t A. Let y {,..., l } and set v = l y. Moreover, define r = y and m = l, and select h large so that q := hv + is a rime > l. We define A to be the set of all integers a for which a +, r = and a 0,, 2,..., or h mod q. The density δa of A is h q y l y x y Our set B will be constructed as the union of two sets, B and B 2. For B, we take B := {b 0, b,..., b v } where each b i is chosen to satisfy b i ih mod q and b i mod m, and we choose B 2 to be a set of y integers covering the residue classes 2,..., mod for each rime y. We note that B := B B 2 has l elements and is admissible: it occuies the congruence classes,..., mod for all y and covers at most y classes modulo for > y. If t is such that B {t} is admissible, then t 0 mod for all rimes y. We can write t ih + j mod q for some i and j satisfying 0 i v and 0 j h. Consider t b i. This is j mod q and mod for all rimes y. Hence t b i A, so that ηl δa, and the result follows. Remark. Pintz [5] showed that there is an ineffective constant C such that every interval of length C contains a de Polignac number. The roof of the lower bound above furnishes a different roof of this result: the values of t that aear are eriodic modulo r := k 2, so, given a maximal set B, we obtain Pintz s result with C = r + max b B b min b B b..
BEST POSSIBLE DENSITIES 5 3. Proof of Theorem.2 We begin with a strong version of the uer bound of Theorem.2. Lemma 3.. If k m 2 with k/m, then e γ + o δm, k k log log k. Proof. Pick q as large as ossible so that m φq < k. We claim that A = {x,..., x m : x x 2... x m mod q} is m, k-lausible. This is because any admissible set with k elements may take at most φq distinct values modulo q, so by the igeonhole rincile, at least m of these values must be congruent modulo q, giving an m-tule contained in A. On the other hand, q is of size e γ + o k log log k as k/m, and A has density /q. This comletes the roof. The roof of the lower bound is somewhat more involved. A key ingredient is the Lovász Local Lemma: Lovász Local Lemma. Suose that E,..., E n are events, each of which occurs with robability and deends on no more than d of the others. If d /4 then the robability that no E j occurs is at least e 2n. The inut for the Lovász Local Lemma will come from the following technical result. Lemma 3.2. Let k m. Consider a translation and ermutation invariant set A 2x, 2x] Z m, and set N = {n x, x] Z : n, k = }. If A N m N m and x is large enough in terms of m and k, then A > 8m, log 3mOm k log log k xm. m m Before we rove Lemma 3.2, we use it to deduce the lower bound in Theorem.2. Proof of the lower bound in Theorem.2. Fix a large ositive x and suose that A [ 2x, 2x] m is an m, k-lausible set, yet A log 2m cm x m/ k m log log k m for some c > 0. If c is large enough, then Lemma 3.2 imlies that 3. A N m N m 8m, where N := {n x, x] Z : n, k = }. Now select integers n,..., n k uniformly at random from N. We claim that there is a ositive robability that the n i are distinct and that there is no m-element subset of {n,..., n k } in A, which imlies that A is not m, k-lausible. We use the Lovász Local Lemma, in which we wish to avoid the events n i = n j for i < j k and x,..., x m A for any choice of m elements x,..., x m among the integers n,..., n k. There are k 2 + k m such events. Each event deends on no more than m of the n i, and so deends on no more than m + m 2m
6 A. GRANVILLE, D. M. KANE, D. KOUKOULOPOULOS, AND R. J. LEMKE OLIVER other events. If x,..., x m are m out of the k random variables, then the robability that x,..., x m A is /8m by 3.. We finally note that the robability that ni = n j is / N R/2xφR as x, which is certainly /8m for x sufficiently large. The Lovász Local Lemma now imlies that there exists a subset n,..., n k of distinct elements of N for which x,..., x m A for all subsets {x,..., x m } of {n,..., n k }. In fact, the Lovász Local Lemma imlies that this is true for a roortion e k/2m2 of the k-subsets of N. Hence, A is not an m, k-lausible set, a contradiction. To rove Lemma 3.2, we first need another result. Given an m-tule h = h,..., h m, we denote by n h the number of congruence classes mod covered by h,..., h m. Note that n h min{, m} and that both uer and lower bounds are easily obtained for some m-tule h. Lemma 3.3. Let k m. There is an absolute constant c > 0 such that if h is a randomly selected m-tule from N = {n x, x] Z : n, k = } and x is large enough in terms of m and k, then the robability that n h m > log 3m cm log log k is k 6m. Proof. The result is trivial if k m 2, so we will assume that k > m 2. Similarly, we may assume that k is large enough. Moreover, note that there are k N = o x N m m-tules h,..., h m with non-distinct elements. So it suffices to show that if we randomly select a subset B = {b,..., b m } of distinct elements of N, then the robability that is k n B m > log 3m cm log log k 7m. By Mertens Theorem, the contribution of the small rimes is m 2 n B m 2 < k m m 2 for some absolute constant c > 0. If c 2c and we set fb = n B m, then we are left to show that P fb > log 3m cm/2 log log k m log 3m c m 7m.
Taking logarithms, we have log fb m n m 2 + O 2 m 2 < k m m 2 < k BEST POSSIBLE DENSITIES 7 X B + O log m m 2 < k, m n + O log m where X B = if n B < m, and X B = 0 if n B = m. Therefore, if we set XB = m 2 < k X B/ and we assume that c is large enough, then it suffices to show that P XB > c 3 log log3m + log log log k 7m. In turn, the above inequality is a consequence of the weaker estimate 3.2 P XB > log logmax{m 2, m log k} + c 7m, where c is a sufficiently large constant. The X can be viewed as indeendent random variables as we run over all ossible sets B. As in the birthday aradox, the robability that X = 0 is 2... m = ex m2 + Om 2 m 2 = + O For any r, if r, then we trivially have that E[e rx / ] e r/, and otherwise E[e rx/ ] = PX = 0 + PX = e r/ = + PX = e r/ m 2 r ex O. Therefore, for any values of s and r, we have that e rs P X s E[e rx ] = k E[e rx / ] r e r/ r ex r log log r + Or + Om 2 / log r. 2 e Om2 r/ 2 Thus, if r m 2, then setting s = log log r + c for c sufficiently large, we find that P X log log r + c e r. Substituting r = max{m 2, m log k} establishes 3.2 for k large enough, thus comleting the roof of the lemma. Proof of Lemma 3.2. We artition the elements of [ 2x, 2x] m into translation classes, utting two elements in the same class if and only if they differ by l, l,..., l for some integer l. Each translation class T intersecting [ x, x] m contains at least 2x elements of [ 2x, 2x] m and at most 6x. The main idea of the roof is that if we can find at least U elements of A N m whose translation class has at most M elements inside N m, then A 2x U/M. Note that n h is fixed over all h T, so we denote it by n T. The number of integers l for which h + l, l,..., l N m is 3x k n T for x large enough. If we set.
8 A. GRANVILLE, D. M. KANE, D. KOUKOULOPOULOS, AND R. J. LEMKE OLIVER R = k, then Lemma 3.3 imlies that the roortion of elements of N m for which n m B φr 3.3 log 3m cm log log k R k is / 6m, rovided that x is large enough, with c begin an absolute constant. Since A contains at least N m / 8m elements in N m by assumtion, we find that A contains at least N m / 6m for which 3.3 holds. Therefore, we conclude that A N /{ m m } φr 24m log 3m cm log log k > xm log 3m Om R k log log k m m for x large enough, and the result follows. 4. The number of admissible k-tules Our goal in this section is to show relation.3. Given a rime, we say that an m-tule is admissible mod if its elements do not occuy all of the residue classes mod, so an m- tule is admissible if and only if it is admissible mod for every rime. By the igeonhole rincile, any set of m integers is admissible mod if > m, so to test for admissibility we need only work with the rimes m. This imlies, using the Chinese Remainder Theorem, that ρ ad m = m ρ ad m,, where ρ ad m, denotes the roortion of m-tules that are admissible mod. If m/ log m < m, then we note the trivial bounds / m ρ ad m,, with the lower bound coming from counting m-tules whose elements are not 0 mod. Therefore m/ log m< m ρ ad m, m/ log m< m m m log log m O = e log m. It remains to comute the contribution of rimes m/ log m. It is not difficult to determine an exact exression for ρ ad m, using an inclusion-exclusion argument: the robability that the elements of an m-tule h belong to a given subset of residue classes is m. There are choices of the residue classes. If the elements of h belong to exactly 2 residue classes mod then h was just counted twice and so we need to subtract the robability of this haening. That robability is 2 k, and there are 2 choices of the 2 residue classes. Continuing in the way, we find that ρ ad m, = j j m = j j+ j= m We note that ratio of two consecutive summands in absolute value is m j j m j + m = + j+ j j + 2 2 m +... 2 m { } m 2 ex log 2 log m
for all m/ log m. Therefore, we deduce that ρ ad m, = which imlies that m log log m O ρ ad m = e log m BEST POSSIBLE DENSITIES 9 m/ log m m + O which roves a quantitative version of relation.3. m =, log m eo m log log m log m e γ log m m, 5. A better density in the continuous case Analogous to the discrete question considered here, one can also ask about the continuous version, i.e. the set of limit oints L of the set of values of n+ n / log n, where n is the nth rime. One can deduce from a uniform version of Zhang s Theorem that for any 0 β β 2... β k with k = k 2 there exists i < j k such that β j β i L. By a small modification of the argument used to rove Corollary., Banks, Freiberg and Maynard [] showed that L [0, T ] has Lebesgue measure T/. It is not difficult to see that this is more-or-less best ossible, given the result used, since the set L 0 := {t : t /k } contains a β j β i for all ossible choices of the β s. References [] W. D. Banks, T. Freiberg, J. Maynard, On limit oints of the sequence of normalized rime gas, rerint. [2] A. Granville, Primes in intervals of bounded length, Bull. Amer. Math. Soc., to aear [3] J. Maynard, Small gas between rimes, Annals of Mathematics, to aear. [4] J. Maynard, Dense clusters of rimes in subsets, rerint. [5] J. Pintz, Polignac Numbers, Conjectures of Erdős on Gas between Primes, Arithmetic Progressions in Primes, and the Bounded Ga Conjecture, rerint. [6] D.H.J. Polymath, Variants of the Selberg sieve, and bounded intervals containing many rimes, Research in the Mathematical Sciences, to aear. [7] T. Tao, Public communication, htt://terrytao.wordress.com/ [8] Y. Zhang, Bounded gas between rimes, Annals of Mathematics, 79 no. 3,. 2-74. AG: Déartement de mathématiques et de statistique, Université de Montréal, CP 628 succ. Centre-Ville, Montréal, QC H3C 3J7, Canada E-mail address: andrew@dms.umontreal.ca DMK: Deartment of Mathematics, University of California-San Diego, 9500 Gilman Drive #0404, La Jolla, CA 92093, USA E-mail address: dakane@math.ucsd.edu DK: Déartement de mathématiques et de statistique, Université de Montréal, CP 628 succ. Centre-Ville, Montréal, QC H3C 3J7, Canada E-mail address: koukoulo@dms.umontreal.ca RJLO: Deartment of Mathematics, Stanford University, Building 380, Stanford, CA 94305, USA E-mail address: rjlo@stanford.edu