PRIME CHAINS AND PRATT TREES

Transcription

1 PRIME CHAINS AND PRATT TREES KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA ABSTRACT We study the distribution of prime chains, which are sequences p,, p k of primes for which p j+ mod p j ) for each j We first give conditional upper bounds on the length of Cunningham chains, chains with p j+ = 2p j + for each j We give estimates for P x), the number of chains with p k x k variable), and P x; p), the number of chains with p = p and p k px The majority of the paper concerns the distribution of Hp), the length of the longest chain with p k = p, which is also the height of the Pratt tree for p We show Hp) c log log p and Hp) log p) c for almost all p, with c, c explicit positive constants We can take, for any ε > 0, c = e ε assuming the Elliott-Halberstam conjecture A stochastic model of the Pratt tree is introduced and analyzed The model suggests that for most p x, Hp) stays very close to e log log x INTRODUCTION Impose on the set of positive integers a relation as follows: a b if there is a positive integer m with b = am + We are interested in properties of the chains with respect to this partial ordering, especially the chains consisting only of primes, the prime chains A simple example is 3, 7, 29, 59 In a chain n n 2 n k of length k, there are positive integers m,, m k with n i+ = m i n i + which we refer to as links) for i k Cunningham Chains When k 2 is fixed, the links m,, m k are fixed and there are no obvious congruential obstructions, it is unknown whether there are infinitely many such prime chains, an affirmative answer being implied by Dickson s prime k-tuples conjecture In the special case where m = = m k = 2 there are no congruential obstructions and the chains in question are known as Cunningham Chains these are sometimes called Cunningham Chains of the first kind, whereas a sequence p, p 2 = 2p,, p k = 2p k is called a Cunningham Chain of the second kind) A basic example is 2, 5,, 23, 47, while the longest one known is the chain with p = and k = 6, discovered by Carmody and Jobling in 2002 Date: August 3, Mathematics Subject Classification Primary N05, N36; Secondary 60J80 The research of K F was supported in part by National Science Foundation grants DMS and DMS , that of S K was supported in part by Grants from the Russian Foundation for Basic Research and NSh from the Program Supporting Leading Scientific Schools, and that of F L was supported in part by projects PAPIIT and SEP-CONACyT 79685

2 2 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA see [4]) In such a chain p,, p k we have ) p j = 2 j p + 2 j = 2 j p + ) Let kp) be the length of the longest Cunningham Chain starting from p By a standard upper bound for Sophie Germain primes coming from sieve theory [30], Theorem 24), kp) = for all primes p x except for Ox/ log 2 x) of them How large can kp) be in terms of p? By the first equality in ) and Fermat s little theorem, we have kp) ord p 2 and in particular kp) p for all p The distribution of ord p 2 is largely unknown, and it is conjectured Artin s Conjecture) that ord p 2 = p for infinitely many p; that is, 2 is a primitive root of infinitely many primes We quote a well-known result of Hooley [33] Theorem H Hooley) Assume the Riemann Hypothesis for the Dedekind zeta functions ζ Kr s) attached to the number fields K r = Q2 /r, e 2πi/r ), where r runs over the primes Then Ax), the number of primes p x for which 2 is a primitive root, satisfies Ax) Cπx), C = r where πx) is the number of primes x rr ) ) = Using the second equality in ) together with Hooley s result, we greatly improve conditionally) the bound on kp) Theorem Assume the conclusion of Theorem H, that is, Ax) Cπx) Then lim sup p kp) log p C We next show, unconditionally, that substantially larger values of kp) are very rare To state the results, let θ + be the supremum of real numbers θ so that there are x +o) primes p x such that p has a prime factor > x θ The best known result is due to Baker and Harman, who showed in [6] that θ It is conjectured that θ + = Theorem 2 For each fixed A > /θ +, there are Ox θ A +o) ) primes p x with kp) > log p) A 2 General prime chains We now turn to problems about counting prime chains with variable links, which will be the main focus of this paper Dirichlet s theorem on primes in arithmetic progressions implies that there are infinitely long prime chains We consider problems where the links are allowed to vary, but additional constraints are put on the chains, eg given beginning p, given end p k, p k /p x, etc The study of such restricted types of prime chains has arisen, for example, in investigations of iterates of Euler s totient function φn) and Carmichael s function λn) eg [7], [8], [9], [22], [38], [39], [40]), the value distribution of λn) [25], common values of φn) and the sum-of-divisors function σn) [26], and the complexity of the Pratt primality certificate [0], [44])

3 PRIME CHAINS AND PRATT TREES 3 Some basic counting functions of general prime chains are P k x), the number of prime chains with fixed length k and p k x, and P x), the total number of prime chains with p k x, so that P x) = k P kx) For fixed k we have 2) P k x) xlog 2 x) k k )! log x When k =, this is a restatement of the Prime Number Theorem The asymptotic 2) for k = 2 is implicit in the work of Erdős [2], for k = 3 it follows from Erdős and Pomerance [24], and the general case was shown by Bassily, Kátai and Wijsmuller [9], where 2) is proved uniformly for k log 3 x Here and throughout, log log 4 x k x is the k-th iterate of the natural logarithm The idea behind 2) is that for most primes p, p has about log 2 p prime factors, uniformly distributed on a log log-scale A related result is proved in [6] It is straightforward to prove by induction the uniform upper bound P k x) xlog 2 x) k x 3, k ) log x using the methods of [22] Theorem 35 therein, for example), but this is only non-trivial for k log 2 x in light of Theorem 3 below log 3 x If the asymptotic 2) is true uniformly for k c log 2 x with c >, this suggests that most prime chains with p k x have length about log 2 x, and that P x) x An upper bound of this order is easy to prove, but lower bounds are more difficult The asymptotic 2) provides a lower bound for P x), but we can do better by considering all k at once Theorem 3 We have xlog x) 036 P x) 2 + o))x as x log 2 Note that p q means q mod p) Thus, our study of prime chains is intimately connected with the distribution of primes in arithmetic progressions, fundamental results about which can be found in the monograph of Davenport [8] Let πx; q, a) be the number of primes p x with p a mod q), and let lix) = x dt/ log t For the lower bound 2 in Theorem 3, we require that πx; p, ) lix) for most p p xβ, where β > 0 is fixed Consider the statement 3) m Q max y x liy) πy; m, ) φm) x R The Bombieri-Vinogradov Theorem [8], Ch 28) implies that for every A > 0 there is a B > 0 so that 3) holds with Q = x /2 log x) B and R = log x) A If 3) holds with Q = x θ for some θ > and suitable R, then we can prove a stronger lower bound for 2 P x) Assuming the Elliott-Halberstam Conjecture, for any fixed ε > 0, 3) holds with Q = x ε and R = log x) 2, and we can deduce P x) x/log x) o) see the proof of Theorem 3 in Section 3) With a little stronger hypothesis but still one which is widely believed), we can prove an asymptotic for P x) Moreover, we deduce the normal order

4 4 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA of the function fp), the number of prime chains with a given end p k = p again, with k variable) Theorem 4 Assume that 3) holds with Q = x log 2 x) δ, R = log x) 2 for some fixed δ > 0 Then P x) cx for some constant c > 0 Moreover, fp) has normal order c log p We do not, however, have any guess as to the size of c, other than the upper bound provided by Theorem 3 Theorems 3 and 4 will be proved in section 3 3 Chains with a given starting prime Our next objective is to count prime chains with a given starting prime p As the number of such chains is infinite, we consider P x; p), the number of prime chains with p = p and p k /p x When p is very small compared with x, almost all primes q x lie above p in some chain The proof of Theorem 45 of [22] implies that if p log x) c for some small absolute constant c > 0, then P x; p) πpx) px/ log x Also, by the argument leading to Theorem 4, if p is fixed, then we expect P x; p) x For substantially larger p we expect that fewer primes q px will lie above p, especially if p is a fixed power of x Theorem 5 For 2 p x, we have the effective estimate { } log xlog3 x + O)) P x; p) x exp log 2 x x ) In particular, for every ε > 0 there is an effective constant Cε) so that P x; p) Cε)x +ε Theorem 5 will be proved in Section 4 using a novel sieve method based on matrices of Dirichlet series It is an important tool in the recent proof by Ford, Luca and Pomerance [26] that the equation φa) = σb) has infinitely many solutions, settling a well-known 50-year old problem of Erdős In [25], Theorem 5 is applied to the problem of determining if for every positive integer n, there is another positive integer m with λn) = λm) Remarks Theorem 5 is only nontrivial for large p, namely for { } log xlog3 x + O)) p exp log 2 x On the other hand, considering only chains of length 2 gives the lower bound P x; p) πxp, p, ) and we therefore expect that P x; p) x/ log x if x p δ for fixed δ > 0 We conjecture an even stronger upper bound Conjecture We have P x; p) x

5 PRIME CHAINS AND PRATT TREES FIGURE Pratt tree for p = Pratt trees The Pratt tree for a prime p is the tree with root node p, and below p are links to the Pratt trees for primes q which divide p This tree was introduced by V Pratt in 975 [44], who showed how to use it in conjunction with Lucas primality test [7], 4) to create an efficient certificate of primality for a given odd prime p The certificate consists of a primitive root g of p, which is proved to be such by verifying g p )/q mod p) for each prime q p ) To prove each of these q is prime, one iterates this procedure Pratt showed that this primality certificate provides a proof that p is prime in polynomial time the number of bit operations is Olog p) C ) for some constant C) For a general number n, this method is not a practical for determining if n is prime, since it requires factoring all the shifted primes in the tree Today, there are polynomial time algorithms for determining whether or not a given integer is prime [2] Pomerance [43] gave another method for producing primality certificates with a smaller upper bound on the complexity It is likely that the Pratt certificate is more complex for some primes, but it is an open problem whether the Pratt certificate has longer complexity for most primes see of [43]) All the leafs of the Pratt tree will be labeled with the prime 2 Figure shows one example One measure of the complexity of the Pratt tree is the total number of nodes, which is the quantity fp) introduced in 2 Another important statistic is the height of the tree, denoted Hp) We can also interpret Hp) as the length of the longest prime chain with p k = p For example, H05277) = 7 In the next graphs, we show histograms of Hp) for i) all primes p 0 9 and ii) for 000 randomly chosen primes near 0 40

6 6 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA 20 Pratt tree height, primes <,000,000,000 number of primes millions) Height 350 Pratt tree height, random primes near 0^ number of primes Height FIGURE 2 Pratt tree height

7 PRIME CHAINS AND PRATT TREES 7 Understanding the extreme values of Hp) is notoriously difficult At one extreme, a prime with Hp) = 2 is a Fermat prime, that is, p = 2 2m + for some integer m Only 5 such primes are known, corresponding to m {0,, 2, 3, 4}, and it is widely believed that there are only finitely many see [7], 32) On the other hand, the following conjecture seems plausible Conjecture 2 For every k 3, there are infinitely many primes with Hp) = k If, for example, there are infinitely many primes of the form 2 m 3 n 5 r 7 s 257 t u +, then Conjecture 2 holds for k = 3 At the other extreme, we have the trivial bound Hp) log p +, and pose the following log 2 question Problem Is it true that Hp) log p for infinitely many p? The answer is yes if 22) holds, but 22) seems very difficult to settle It is plausible that there is some C > 0 so that for every prime p, the least prime q mod p) satisfies q plog p) C If so, then the special chains p = 2 3 p k where for each j, p j+ is the least prime mod p j ), have the property that Hp k ) log p k log 2 p k Primes with Hp) very small or very large should be rare One main goal in this paper is to understand the behavior of Hp) for a typical prime p A seemingly natural candidate for the longest chain in the Pratt tree is the special chain p = p 0 p p k = 2, with p j the largest prime factor of p j for each j Let Lp) denote the length of this chain Recently, Banks and Shparlinski [8] proved that for almost all p, 4) Lp) o)) log 2 p log 3 p p ) We can do better, at least heuristically For a randomly chosen integer n, the largest prime factor of n is n /u with probability ρu), where ρ is the Dickman function, the unique continuous solution of the differential-delay equations see Section of [32]) ρu) = 0 u ), uρ u) = ρu ) u > ) It is natural to conjecture that the largest prime factor of the shifted primes p has the same distribution We can thus model the special chain by assuming that log 2 p j log 2 p j+ are independent random variables with distribution function ρe x ) for x 0 These random variables have mean µ = x d 0 dx ρu) ρex )) dx = u du By the law of large numbers, if k is large then we expect log 2 p log 2 p k kµ most of the time Hence, we should have Lp) /µ) log 2 p for most p Numerically, /µ = Better unconditional lower bounds can be obtained for Hp) than that provided by 4) In [35], Kátai showed that for some small unspecified positive constant c, Hp) c log 2 p

8 8 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA for almost all primes p In fact, his bound holds for a more general type of Pratt-like tree see the end of this introduction for more) We show a quantitatively stronger bound, using 3) First, however, we show that, over primes p x, the distribution of Hp) is tight on the left of its median That is, the width of the distribution, at least to the left of the median, remains bounded as x, in contrast to what might be expected We conjecture that the distribution is also tight to the right of the mean see Conjecture 4 below) Theorem 6 Suppose g and h are increasing functions, satisfying 0 gx) hx), hx 2 ) hx) K and gx 2 ) gx) K for x Suppose, for large x, that Hp) hp) for at least cπx) primes x Then 5) Hp) hp) gp) for all primes p x with at most Oπx) exp{ c log 2 gx)}) exceptions Consequently, if K gx) as x, then 5) holds for almost all p Theorem 7 Suppose that 3) holds with Q = x θ and R = log x) 2 For any c < e +log/θ), there is an ε > 0 so that Hp) > c log 2 p for all but Ox/log x) +ε ) primes p x Corollary Assume the Elliott-Halberstam Conjecture Then, for every c < e, almost all primes satisfy Hp) > c log 2 p In light of the heuristic for Lp), we see that Lp) should be much smaller than Hp) for most p We believe that the conclusion of Corollary is best possible; see Conjecture 3 below If p p k = p with k = Hp), we have hence k log p = log p 6) Hp) j= log p j+ log p j log 2 p min j k log min j k log p j+ log p j ) k log p j+, log p j + O) It is not known that there is an infinite set of primes p with Hp)/ log 2 p If this is the case, then 6) implies that for all ε > 0, there are an infinite number of primes p such that p has a prime factor > p ε cf section ) In the opposite direction, we can show, for infinitely many p, a lower bound for Hp) which is quantitatively better than the bound in Theorem 7 In particular, if the Elliott-Halberstam conjecture is true, then lim sup Hp)/ log 2 p = Theorem 8 Assume that 0 < θ < and for every A > 0, 3) holds with Q = x θ and R = log x) A Then, for every c <, there is a constant K so that for large x, there log/θ) are x/log x) K primes p x satisfying Hp) > c log 2 p

9 PRIME CHAINS AND PRATT TREES 9 Theorems 6, 7 and 8 will be proven in Section 5 What can be said about upper bounds on Hp), valid for almost all p? Before our work it was not known that Hp) = olog p) for infinitely many primes p A seemingly easy method for finding p with small Hp) is to take p with p having only small prime factors, and use log q 7) Hp) + max q p ) log 2 However, the best we know is that there are infinitely many p such that all the prime factors of p are < p 0293 [6] In order to show Hp) = olog p) using 7), we would need a prime p so that all prime factors of p are p o) Using a different method, we prove that Hp) is substantially smaller than log p for most primes p Theorem 9 For c = and some δ > 0, we have Hp) log p) c for all but Ox exp{ log x) δ }) primes p x The proof of Theorem 9 is given in Section 6 Later, in Section 7, we apply a different method to deduce weaker upper estimates, but ones which hold with a much smaller exceptional set of primes Theorem there) As δ 0 +, the probability that the largest prime factor of n is between n δ and n δ tends to The same is expected for the largest prime factor of shifted primes p It is reasonable to conjecture that Hp) c log 2 p for some constant c In section 8 we give a heuristic argument, based on a stochastic model of the Pratt trees, for the assertion that this is true with c = e More crudely, by the argument leading to 2), we expect that the number of primes at level k of a Pratt tree for p x has normal order log 2 x) k /k! By Stirling s formula, this quantity drops below when k e log 2 x Conjecture 3 Hp) has normal order e log 2 p That is, for every ε > 0, we have for almost all primes p Hp) e log 2 p < ε log 2 p Recall Corollary, which says that the Elliott-Halberstam conjecture implies the lower bound implicit in Conjecture 3 For the primes considered in Figure, we have e log and e log The peak of the distribution in these two cases is a bit smaller than e log 2 p, and the argument in Section 8 explains this phenomenon Moreover, our model predicts the tightness of the distribution of Hp) cf Theorem 6) and also a large asymmetry in the distribution with respect to its median The following conjecture makes this more precise Conjecture 4 Hp) = e log 2 p 3 2 log 3 p + Ep), where Ep) is an exponentially tight sequence More precisely, for some fixed c > 0 we have, for any z 0, Ep) z for all but Oe cz πx)) primes x, and Ep) z for all but Oexp{ e cz }πx)) primes p x

10 0 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA The asymmetry in the distribution is already evident in the graphs in Figure Numerically, e log log and e log log Due to the fact that Hp) is integer valued, it is unlikely that Ep) behaves like a particular random variable In particular, depending on the size of x, the second most popular value of Hp) for p x may be one less or one greater than the most popular value already evident in Figure 2) It is possible that Ep) behaves like a discrete approximation to a random variable The model in Section 8 can also be used to guess the distribution of other statistics of Pratt trees For example, the quantity fp) defined earlier is the total number of primes in the tree, and our model predicts that fp) log p for most primes p see Theorem 4) Other interesting statistics are the width maximum number of primes at some level of the tree) and mass product of all primes in the tree; see [0]) We leave these topics to the interested reader 5 Further problems One can define a more general type of prime chain, by fixing a non-zero integer a and putting p q if p q + a) Everything we have shown in the case a = holds in the general case as well, with one small caveat One must avoid the special prime q = a, if a is a prime, and avoid primes occurring in chain cycles when a > 0 For example, when a = 6 we have the cycle Problems concerning the structure of the cycles for various a are interesting in their own right, and are investigated in [35], [36] and [42] We conclude this section with a conjecture about prime chains, which follows from the prime k-tuples conjecture but should be easier It can be considered a multiplicative analog of the statement that the primes contain arbitrarily long arithmetic progressions, recently proved by Green and Tao [29] Conjecture 5 For each k 3, there are infinitely many prime k-tuples p,, p k ) where, for some m, p j+ = mp j + for j k Even the case k = 3 is not known 6 Notation Some notational conventions have already been mentioned, eg πx) for the number of primes x and log k x for the k-the iterate of the logarithm of x Other number theoretic functions we need are the Möbius function µn), Euler s totient function φn), the number of distint prime divisors ωn) of n, and Ωn), the number of prime power divisors of n The letter p, with or without subscripts, always denotes a prime Constants implied by the O and symbols do not depend on any parameter unless indicated In Section 8, we use P for probability and E for probabilistic expectation 2 CUNNINGHAM CHAINS Proof of Theorem Let q p be a prime such that ord q 2 = q and q p + ) The powers of 2 generate all reduced residues modulo q, so there is some j q for which

11 PRIME CHAINS AND PRATT TREES k p k/ log p discoverer Carmody and Jobling 2002) Carmody 2003) Jobling 999) Brennen 998) Löh 989) Löh 989) Löh 989) Nelson 980); Sumiyama 983) Nelson 980) Lehmer 965) Cunningham 907) TABLE Smallest prime p with kp) = k 2 j p + ) mod q) Hence, kp) q More generally, let Γ q be the orbit of 2 modulo q, ie, Γ q = {2 j mod q) : j Z} Then 2) kp) min{ord q 2 : q p, p + Γ q } Let 0 < ε 2 enough then +ε and put x = ) log p By Lemma and partial summation, if p is large C q x ord q2=q log q C ε ) x > logp + ) 2 Thus, there is some q x with ord q 2 = q and q p + ) By 2), kp) q x In light of Theorem, it is natural to ask if 22) lim sup p kp) log p > 0 Table gives values of kp)/ log p for the smallest p for which kp) = k the smallest p for k = 5 apparently isn t known) These were taken from a web site maintained by Dirk Augustin third column values truncated) The proof of Theorem 2 uses Montgomery s large sieve estimate [5], Théorème 6):

12 2 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA Lemma 2 Suppose a set of positive integers x avoids ξp) residue classes modulo p for each prime p Then the cardinality of the set is 2x G, G = µ 2 ξp) n) p ξp) n x /2 p n Proof of Theorem 2 Let ε > 0 be very small Suppose that x p 2x and kp) > log x) A =: Q If x is large enough, there are Q ε/2 primes q Q with q having a prime factor > Q θ+ ε Let Q be the set of primes q Q such that Γ q > Q θ+ ε For any prime q with q having a prime factor > Q θ+ ε, we have either Γ q > Q θ+ ε or Γ q < Q θ+ +ε =: Q 0 Any prime in the latter category divides M = j Q 0 2 j ) As M 2 Q2 0 and θ + >, the number of such q is 2 Q2 0 Q ε if ε is small enough Therefore, Q > Q ε for all sufficiently large x Let ω := log x)/2 log Q) and let N be the set of squarefree positive integers n having precisely ω prime factors all from Q Clearly, n x /2 for all n N Further, ) ) ω Q Q ω N x 2 2A ε 2 ω ω Finally, if n N, then taking ξq) = Γ q, we have ξq) ) ω q ξq) Q θ+ ε = Q ω θ + +ε) x θ+ 2ε 2 Thus, q n and the lemma follows G x θ A 2ε, Remarks The link 2 is special If we consider generalized Cunningham Chains p, p 2 = mp 2 +,, p k = mp k +, where m is an even integer, it is trivial that k is bounded in terms of m for m > 2 Indeed, if q is a prime factor of m, then p j p j + mod q), and thus k q if p > q and k q if p < q Problem 2 In light of the fact that kp) = for most p, can it be proved unconditionally that p x kp) πx)? 3 ESTIMATES OF P x) Throughout this section, variables q, q and q 2 denote primes Proof of Theorem 3 Since f2) = and 3) fp) = + q p ) fq),

13 an easy induction shows that 32) fp) 2 log p log 2, and hence P x) = p x PRIME CHAINS AND PRATT TREES 3 fp) 2 log 2 log p p x ) 2 log 2 + o) x x ) by the Prime Number Theorem We next show that if θ <, 3) holds with Q = x θ and R = log x) 2, and if α satisfies 0 < α < and θ α > α, then 33) P x) xlog x) α The lower bound in Theorem 3 follows by taking θ = 0499 admissible by the Bombieri- Vinogradov theorem) and α = 036 Define the numbers a j by a j = inf 2 y exp{/θ) j } P y)log y) α y We have a j > 0 and a j+ a j for all j To show that lim a j > 0, suppose that j is large and exp{/θ) j } < z exp{/θ) j+ } By 3) and 3), P z) = + ) fq) p z q p ) q z θ fq)πz; q, ) = liz) ) fq) z q + O log z q z θ Since fq) for all q and q diverges, for some q q 0 we have fq) P z) liz) q By partial summation and the definition of a j, P z) liz) z θ z a j log z q 0 t 2 z θ q 0 <q z θ P t) dt q 0 dt tlog t) α [ θ α z = a j α log z) + O α )] z log z

14 4 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA Since θ α > α, the right side is > a j zlog z) α for large j Hence, the sequence {a j } is eventually constant Proof of Theorem 4 By the proof of Theorem 2 in [22] more precisely, combine Lemma 24, Corollary 25, 28) and Theorem 2), we obtain 34) P x) = cx + Ox/log 2 x) δ ) for some c 0 Section 2 of [22] is devoted to the study of the normal behavior of the function ln), the number of times one must iterate the Euler function, starting from n, in order to reach For example, l9) = 4 since φ9) = 8, φ8) = 4, φ4) = 2 and φ2) = It turns out that F n) = ln) { 0 n n odd even is completely additive, and for odd primes p, F p) = F p ) = af q) q a p This is very similar to relation 3), the only difference being the behavior at proper prime powers, which play an insignificant role in the arguments in [22] In [22], F p) log p, however we do not have fp) log p from 3), exactly because of the influence of prime powers; from 32), fp) log q p ) q) Hence, we cannot immediately deduce that c > 0 in 34) To show this, we use the method of the proof of Theorem 3 For j let P y) log a j = inf 3 y 2 y e j y For e j < z e j+, let z 0 = z log 2 z) δ For large enough z, z 0 z/e For such z, we obtain for some fixed q 0 00 P z) liz) ) fq) z + O q log z q z 0 By integration by parts, hence, z0 q 0 liz) z log z a jz log z fq) q q 0 <q z 0 z0 q 0 z0 q 0 P t) t 2 dt t log 3 t = log t log 3 t + dt t log 3 t log z 0 + log 3 z 0 dt dt t log 3 t dt t log 2 tlog 3 t) 2 ; log 2 z 0 log 3 z 0 ) O) log z log 3 z

15 PRIME CHAINS AND PRATT TREES 5 for large z Hence, P z) a j z/ log 3 z and thus the sequence {a j } is eventually constant Thus, P z) z/ log 3 z and therefore c > 0 by 34) For the second part of the theorem, start with ) 2 )) fp) c log x = fp) 2 x 2c log x cx + O + c 2 πx) log 2 x log p x p x 2 x) δ = ) x log x fp) 2 c 2 x log x + O log p x 2 x) δ The sum on the right side is fq) 2 πx; q, ) + 2 q x q <q 2 x fq )fq 2 )πx; q q 2, ) Put x 0 = x log 2 x) δ We use 3) to handle the terms with q x 0 and q q 2 x 0 The terms with q > x 0 are dealt with using fq) log q and sieve methods By Theorem 24 of [30], we have πx; q, ) #{q x/k : q, qk + both prime} x 0 <q x k x/x 0 and x 0 <q q 2 x q <q 2 Hence, fp) 2 = p x πx; q q 2, ) k x/x 0 x φk) log 2 x x log x)log 2 x) +δ, #{q 2 x/kq ) : q 2, kq q 2 + both prime} q x k x/x 0 x φkq ) log 2 x q x k x/x 0 x log x)log 2 x) δ x fq)2 + x log x q log x q x 0 q <q 2 x q q 2 x 0 2fq )fq 2 ) q q 2 By the proof of Proposition 26 of [22], the two sums above are each 2 c2 log 2 x + O log x) 2 log 2 x) δ), and we conclude that p x fp) c log x ) 2 x log x log 2 x) δ ) ) x log x + O log 2 x) δ

16 6 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA The second statement in the theorem follows 4 SIFTED CHAINS The proof of Theorem 5 uses a novel method based on matrices of Dirichlet series The underlying idea is a simple sieve; relax the condition that the numbers in the chain are prime, and instead only require that they do not have small prime factors Let y 2 and let q be the product of the primes y For a, q) =, let N a x; y) be the number of chains with n = a, with n k /n x and consisting of numbers coprime to q Also let If p > y, we have Nx, y) = max a,q)= N ax, y) P x; p) N p x, y) Nx, y) For example, when y = q = 2, the links in the chains counted by Nx; 2) are even and have product x A variant of Kalmár s argument from [34] shows that Nx; 2) x ρ, where ρ is the unique positive solution of 2 s ζs) = For y 3, however, the restrictions on the links m i are more complex to deal with For positive integers a, b, q and real s >, let Sa, b) = Sa, b; q, s) = m s m am+ b mod q) This Dirichlet series encodes the possible links m from a number n i a mod q) to a number n i+ b mod q) Let U q = Z/qZ) be the multiplicative group of integers coprime to q Let A k a, a k ) be the sum of m m k ) s over all k )-tuples m,, m k ) which could serve as links in a chain starting from a number n a mod q), ending with a number n k a k mod q) and with all numbers in the chain coprime to q we suppress the dependence on q and s in this and future definitions) Then A 2 a, a 2 ) = Sa, a 2 ) and for k 3, A k a, a k ) = Sa, a 2 )Sa 2, a 3 ) Sa k, a k ) a 2,,a k U q Let V k a ) be the column vector A k a, a k ) : a k U q ) For consistency, let V a ) be a vector with all zero entries except for an entry of in the a position Since A k+ a, a k+ ) = a k U q A k a, a k )Sa k, a k+ ), we obtain where M = Mq, s) is the transition matrix V k+ a ) = MV k a ), M = Sa, b) ) b,a U q

17 PRIME CHAINS AND PRATT TREES 7 The rows of M are indexed by b and the columns are indexed by a Finally, let F k a ) = a k A k a, a k ), so that 4) F k a ) =,, )V k a ) =,, )M k V a ); ie, F k a ) is the sum of the entries of column a in M k Lemma 4 We have N a x; y) inf x s s> k log x log 2 + F k a) Proof If n k /n x, then m m k x and hence x/m m k ) s Observe that the sum on k in Lemma 4, if extended to k =, is convergent if and only if M is a contracting matrix, ie, all the eigenvalues of M have modulus < Since M has positive real entries, the Perron-Frobenius Theorem implies that the eigenvalue with largest modulus is positive, real and simple Call this eigenvalue λs; y) Problem 3 Obtain good estimates for λs; y) In particular, determine the smallest s, as a function of y, for which λs; y) < We show below that if y is large and s + log 2 y, then λs; y) < Accurate estimation log y of λs; y) is difficult for large y, but the largest row sum of M serves as an upper bound For a generic matrix A, let R b A) be the sum of the entries in the row indexed by b, and let RA) be the maximum row sum of A Lemma 42 With M defined above, we have R b M) = αq)βb, q)), where αq) = p>y p s ), βd) = p d Proof Write d = b, q) and b = b Then d R b M) = m s = d s a U q am b mod q) k,q/d)= p p s k s #{a U q : ak b mod q/d)} The congruence ak b mod q/d) has a unique solution modulo q/d, and hence has φd) solutions a U q Thus, R b M) = φd) k s = φd) p s ) = αq)βd) d s d s k,q/d)= p q/d)

18 8 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA By Lemma 42, 42) RM) = αq)β2) = 2 s p s ) Since RAB) RA)RB), RM k ) RM) k To bound Nx, y), however, we require a bound on the largest column sum of M k Lacking a better method, we ll just use the trivial bound φq)rm) k We then have 43) F k a) a U q F k a) φq)rm k ) φq)rm) k, p>y and, applying Lemma 4, 44) Nx, y) x s N a x; y) φq) inf s:rm)< RM) a U q By standard prime number estimates, if < s 2, then p>y log p s ) = O/y 2s ) + p>y y p s dt t s log t e s ) log y s ) log y Also, 2 s = + 2 log 2)s ) + Os ) 2 ) for s 2 By 42), for y, s log 2 y/ log y and s = + o), we have 45) RM) 2 log 2)s ) Take y = log x)/ log 2 x and s = + log 2 y Using 44) and the Prime Number Theorem log y bound φq) q = e +o))y, we obtain the following Theorem 0 If y = log x log 2 x, then Nx; y) { log xlog3 x + O)) N a x, y) x exp log a U 2 x q } 5 PRATT TREE HEIGHT: LOWER BOUNDS Proof of Theorem 6 The conclusion is trivial if gx /2 ) 3K, so we will assume that gx /2 ) > 3K Let gx /2 ) m = K

19 PRIME CHAINS AND PRATT TREES 9 so that m 3 Put Q = x 2 m and let T be the set of primes x /2 < p x such that there is a prime q p ) with Q < q x /4 and Hq) hq) For p T, Hp) + hq) hq) hx) mk hp) gp) By sieve methods Theorem 42 of [30]), for large x and the theorem follows {x /2 < p x : p T } x log x Q<q x /4 Hq) hq) x log x 2 mc ) q In the proof of Theorem 7, we need some further estimates from sieve theory The first is the Brun-Titchmarsh inequality x 5) πx, q, ) φq) logx/q), and the second is a way to handle primes in progressions to large moduli Lemma 5 Uniformly for z 2 and 4 A x /2, we have πx, p p 2, ) xlog2 x) log A log 4 z p,p 2 z p p 2 x/a p,p 2 z p p 2 x/a Proof Without loss of generality, assume that z x /4 If p x and p mod p p 2 ) with p p 2 x/a, then p = + kp p 2, with k A For each p, there are O log x log z )2 ) choices for the pair p, p 2 Hence, by sieve methods eg Theorem 22 of [30]), ) 2 log x πx, p p 2, ) {n x : n mod k), log z Proof of Theorem 7 Suppose that k A all prime factors of n n )) are > z} k ) 2 log x x log z φk) log 2 z k A xlog2 x) log A log 4 z c < h < c < e + log/θ)

20 20 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA For some constant c, described below, let We will show, for some δ > 0, that P = {p : Hp) c log 2 p c } 52) P x) := {p x : p P} δ x log x Consequently, a positive proportion of primes p satisfy Hp) > h log 2 p, and Theorem 7 follows from Theorem 6 Since lim k k k!)/k = e, there is an integer k 2 such that Let α, β satisfy and > k!)/k c k + log/θ) e k/c < β < θ k exp k!) /k) β exp k!) /k) < α < θ k Suppose that δ is sufficiently small, depending only on the choice of c, θ, k, α, β Let x 0 be sufficiently large, depending on c, θ, k, α, β, δ, and put c = c log 2 x 0 ) Observe that 52) holds trivially for 2 x x 0, provided δ is small enough Throughout this proof, constants implied by the O and symbols may depend on c, θ, k, α, β, but not on δ Next, suppose that Y x 0 and that inequality 52) holds for 2 x Y Let S be a subset of the primes in P [Y β, Y θk ] Let NS) be the number of primes p 0 Y, 2Y ] so that there is a prime chain 53) p k p k p 0 with p k S For such p 0, we have Hp 0 ) k + Hp k ) k + c log 2 p k c = c log 2 2Y ) c + c log β + k + O c log 2 p 0 c if x 0 is large enough We will show, for appropriate S, that 54) NS) δ Y log Y, ) log Y which implies P 2Y ) P Y ) + δ Y log Y δ 2Y log 2Y Therefore, by induction over dyadic intervals, 54) implies 52), and hence the theorem

21 PRIME CHAINS AND PRATT TREES 2 To prove 54), we will consider chains 53) satisfying not only p k S, but also 55) p j+ p θ j 0 j k ), p Y θ With 55), we can use 3) to accurately count such chains We have NS) N S) N 2 S), where N S) is the number of chains 53) satisfying p k S and 55), and N 2 S) is the number of pairs of distinct chains satisfying these conditions with the same p 0 For brevity, write Ex, p) = πx, p, ) lix) p We first have N S) = ) π2y, p, ) πy, p, ), p k S p k p where we assume 53) and 55) in the summations By induction on j k, we shall now prove ) j 56) N S) = Y log 2 Y θj log 2 p ) j Y + O log Y p p j j )! log 2 Y k First, π2y, p, ) πy, p, ) = p k p j li2y ) liy ) p ) Y/ log Y + E2Y, p ) EY, p ) + O p 2 For a given p, there are O) chains p k p with p k Y β Also, /p = O) By 3), the totality of the error terms is OY/ log 2 Y ), and we obtain 56) for j = Now suppose 56) is true for some particular j k By partial summation, for Y β p j+ Y θj+, p j mod p j+ ) p /θ j+ p j Y θj The first integral is log 2 Y θj log 2 p j ) j + O p j j )! EY θj, p j+ ) Y θj = Y θ j p /θ j+ log 2 Y θj+ log 2 p j+ ) j j!p j+ + Ep/θ j+, p j+) + p /θ j+ ) j log 2 Y θj log 2 t p j+ )j )!t log t dt Y θ j p /θ j+ ) + O p 2 j+ Et, p j+ ) t 2 dt )

22 22 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA To estimate the aggregate of the error terms, note that for each p j+, there are at most O) chains p k p j+ with p k Y β For fixed t [Y β, Y θj ], 3) gives ) t Et, p j+ ) = O log 2 t p j+ t θ By another application of 3), Y β p j+ Y θj+ Ep /θ j+, p j+) p /θ j+ log Y ) max Y β P Y θj+ P p j+ 2P log Y max Y β P Y θj+ P /θ p 2P Ep /θ j+, p j+) p /θ j+ max Ey, p) y 2P ) /θ log Y This completes the proof of 56) with j replaced by j + By 56) with j = k and partial summation, N P [Y β, Y α ]) δy Y α ) log 2 Y θk log 2 t) k Y dt + O log Y Y k )!t log t β log 2 Y > δy Y α log 2 Y α k ) log 2 t) Y dt + O log Y Y k )!t log t β log 2 Y = δy ) logα/β)) k Y + O log Y k! log 2 Y By hypothesis, logα/β) > k!) /k Also, note that the summands in 56) with j = k) are /p k for Y β p k Y α Hence, if δ is small enough, there is a set S P [Y β, Y α ] such that 57) N S) δ + δ 3/2) Y log Y and 58) We have p k S where N 2,j counts pairs of connected chains p k p j+ p k p j+ p k δ k N 2 = N 2,j, j=0 p j p 0

23 PRIME CHAINS AND PRATT TREES 23 with each of the two chains satisfying 55), p j+ p j+ and p k, p k S We further write N 2,j = N 2,j + N 2,j, where N 2,j counts pairs of such chains with p j p j+ p j+y δ2 As before, for each pair p j+, p j+), there are O) choices for p k, p k,, p j+2, p j+2 By Lemma 5, N 2,0 π2y, p p, ) δ 2 Y log Y p,p Y β p p Y δ2 When j, an argument similar to that leading to 56) yields N 2,j Y log Y p p j j p j+,p j+ Using 5) and Lemma 5, we get writing q = p j+ p j+) [ N 2,j Y πqy δ2, q, ) qy δ 2 Y δ2 + log Y q 2q p j+,p j+ δ 2 Y log Y + Y δ 2 log 2 Y Y δ2 log Y provided that log x 0 δ 4 Hence, 59) N 2,j δ 2 Y log Y j πt, q, ) t 2 For chains counted by N 2,j, the Brun-Titchmarsh inequality suffices for the estimations When j and p j is given, as before we have π2y, p, ) Y p p j p j log Y By partial summation and 5), given p j+ and p j+, p j p j For j + r k, 50) Finally, by 58), we arrive at p j+ p j+ δ2 log Y + log/δ) p j+ p j+ p r p r p r+, 5) N 2,j δ 2 Y log/δ) log Y p r p r log/δ) p j+ p j+ p r+ dt ]

24 24 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA In a similar way, when j = 0 we have by 5) and partial summation, p p 2Y δ2 π2y, p p, ) log/δ) p 2 p 2 A second application of 50) then gives 5) in this case Finally, combining 59) and 5), we obtain N 2 S) δ 2 Y log/δ) log Y Together with 57), if δ is small enough then 54) holds, and this completes the proof Proof of Theorem 8 We proceed by induction as in the proof of Theorem 7 However, the proof is much simpler Let c and θ satisfy θ > /3 and and define K by c < c < log/θ ) < log/θ), θ K = θ θ 8 Let x 0 be large, depending on K, c, c, θ, θ and put c = c log 2 x 3 0) Let P = {p : Hp) c log 2 p c } In particular, P contains all primes x 3 0 We shall prove that 52) Qx) := P x/2, x] x log x) K for x x 0, which implies the lemma since c > c) By the Prime Number Theorem and the fact that K >, if x 0 is large enough then 52) holds for x 0 x x 3 0 Suppose y x 3 0 and 52) holds for x 0 x y Assume y < x 2y and put I = P x θ, x θ ] Suppose that x/2 < p x, and that q p, where q I Then Hp) + Hq) + c log 2 q c + c log 2 x + c log θ c > c log 2 p c,

25 PRIME CHAINS AND PRATT TREES 25 so that p P For x/2 < p x, p is divisible by at most two primes from I, and hence by our hypothesis 3), Qx) ) πx; q, ) πx/2; q, ) 2 q I lix) lix/2) ) 2 q + O x log x) K+ q I x ) 4 log x q + O x log x) K+ q I Since 52) holds for x θ < y x θ, we have q I q j 2 j x θ θ Q2 j x θ ) 2 j x θ θ θ ) log x log 2 θ θ θ K log x) K 8 = log x) K ) log x θ ) K Therefore, ) 2x Qx) log x) + O x x K log x) K+ log x) K if x 0 is large enough By induction on dyadic intervals, 52) holds for all x x 0 Remark If 3) holds with Q = x εx) for a function εx) 0 as x, and where Rx) = log x) fx), where fx) as x sufficiently fast depending on the decay of εx)), we can deduce that Hp) gp) log 2 p for infinitely many p, where gp) as p depending on the decay of εx)) 6 PRATT TREE HEIGHT: UPPER BOUNDS, I The proof of Theorem 9 is more complex than the proofs of the lower bounds in Theorems 7 and 8 Rather than trying to construct friable shifted primes, we utilize sieve methods, which tell us that the largest prime factor of p cannot be too large too often At the core is a sieve upper bound for k-tuples of primes which is uniform in k These bounds involve a factor, the singular series, for which average estimates are required There is another factor, c k) k, in the sieve estimate, which has the potential to derail any attempt to bound Hp) non-trivially We get around this problem by observing that if Hp) is

26 26 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA large, then there must be a prime chain in the Pratt tree for p which is very condensed in a multiplicative sense Lemma 6 There is a δ > 0 so that the following holds Let a,, a k be positive integers, let b,, b k be integers and let ξp) be the number of incongruent solutions of k i= a in + b i ) 0 mod p) If x 0, k δ log x log 2 x and B := p k ξp) p log p δ log x, then the number of integers n x for which a n + b,, a k n + b k are all prime and > k is )) 2k k! kb + k 2 log xs exp O 2 x, S = ξp) ) k log x) k log x p p) p Remarks When a j = for all j and k is of order log x, better upper bounds are possible using methods of Elsholtz [20] Proof Since ξp) = k for large p, S converges to a nonzero number if and only if ξp) < p for all p Also, ξp) k for all p Hence, if S = 0, the number of n in question is zero and there is nothing to prove Now assume S > 0 By Lemma 2, the number of n in question is x/g x), where Gz) = n z gn), gn) = µ 2 n) p n ξp) p ξp) When k is fixed, the argument in 53 of [30] implies that Gz) log z) k /k!s) We modify the argument to make the estimate uniform in k in the stated range By the argument on p of [30], we have gd) log d = gd) d z d z = d z On the right side, we have p z/d ξp) log p p p z/d gd) p z/d = p z/d ξp) log p p ξp) log p p k log p p + p z + h z gp)ξp) p gh) p h p>z/h log p z/p 2 <d z/p p d ξp) log p p + OB) = k logz/d) + OB + k) gd)

27 and p h p>z/h ξp) log p p k p h PRIME CHAINS AND PRATT TREES 27 log p p k p log z log p p + k log 2 z log z k log 2 z Thus, gd) log d = k gd) log z d + O Gz)B + k log 2 z)) d z d z Adding the sum on the right side to both sides yields 6) Gz) log z = k + ) gd) log z + rz)gz) log z d d z z Gt) = k + ) dt + rz)gz) log z, t where rz) B+k log 2 z By assumption, if δ is small enough and z x, then rz) log z 2 As on p 50 of [30], if we define k + Ey) = log log k+ y then we obtain from 6) the estimate It follows that the integral E y) = k + y log y y Gt) t ) dt, p h ry) ry) kb + k log 2 y) y log 2 y x E y) dy is absolutely convergent Hence, for some constant D and for z x, z rz)) Gz) log k z = k + Gt) dt log k+ z t { = e Ez) = D exp z { kb + k 2 log = D exp O 2 z log z } E y) dy )} By the argument on p 5 52 of [30], D = k!s Therefore, G { )} x) logk x kb + k 2 k!s exp log O 2 x, log x and the proof is complete

28 28 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA For given positive integers m,, m k, we will apply Lemma 6 with the forms Write f j n) = a j n + b j, so that f n) = n, f j+ n) = m j f j n) + j k ) j 62) a j = m m j j ), b = 0, b j = + m i m j j 2) Let Sm) = S be the associated singular series and let ξp, m) = ξp) We first give a global upper bound on Sm), and then show that Sm) is bounded in a suitable average sense Our result can be compared to the result of Gallagher [28], where an average of the singular series for prime k-tuples with fixed k is estimated Lemma 62 There is a constant c 2 > 0 so that a) Sm) c 2 log 2 4m m k )) k Furthermore, b) p k ξp, m) p i=2 log p k log 2 4m m k ) + O)) Proof Let x = m m k Assume that Sm) 0, so that all m i are even and 2 k x Then ξp, m) = k if p N, where N = m m k a i b j a j b i By 62), a j x and b j + k 2 j= x/2j x for each j Thus, N x kk )+ exp{olog 3 x)} Then, since k/p /p) k for p > k, Sm) p N i<j p) k = N φn) ) k Part a) follows from the elementary bound N/φN) log 2 N + 2) For part b), the Mertens estimates give k ξp, m) log p k log p k log p + k log 2 N p p p log N p p N p log N p N,p>log N k log 2 N + O)) Lemma 63 If n and k are positive integers, then there exists d n with d n /k and ωn) kωd) + )

29 PRIME CHAINS AND PRATT TREES 29 Proof If ωn) k, then take d = Otherwise, write the prime factorization of n as n = p e p er r, where r > k and p e < < p er r Define j by p e p e j j n /k < p e p e j+ j+ and put d = p e p e j Then r kj + ), for otherwise n > j kj+) i= k p e i i = l+)j+) l=0 i=lj+)+ p e i i p e p e ) j+ k j+ > n Lemma 64 Let k 2 and I {,, k } If the variables m i are fixed i k, i I), then for any prime p, ξp, m,, m k )) p I + p ) I + 0 m i <p i I) Proof Put m l = m,, m l ) for 0 l k and let 63) N l p) := ξp, m l ) 0 m i <p i I,i l) We have N 0 p) = If a+ I, then ξp, m a+ ) ξp, m a ) and hence N a+ p) N a p) If a + I, we have N a+ p) = {n mod p : p f n) f a n)m a f a n) + )} = 0 m i <p i I,i a+) 0 m i <p i I,i a) [ p {n mod p : p f n) f a n)} ] + {n mod p : p f n) f a n)} = p +ga) + p )N a p), where ga) = {i I : i a} If follows, by induction, that N a p) p +ga) p ) +ga) 0 a k ), and the lemma follows from the case a = k Lemma 65 Let k 4, and suppose that M i, N i are integers satisfying M i 2 N i 2kM i for i k We have { Sm) M M k 3e γ log k + O)) b exp O N i <m i N i +M i i k ) where b is the number of variables M i which are 2 k3 )} k log2 k, log k 2 and

30 30 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA Proof Let L = log k + and r = k 3 L + We will perform a precise averaging of the factors in Sm) for primes p r, and we will use crude estimates for larger p If p > r and p m m k, we ll use the trivial bound ξp, m) For other primes p, each congruence f j n) 0 mod p) has exactly one solution For h > j, f j n) 0 mod p) and f h n) 0 mod n) have a common solution if and only if p a j b h a h b j ) Write where It follows that where a j b h a h b j = m m j g j,h m), g j,j+ m) =, g j,h m) = + h i=j+ ξp, m) k wp, m), m i m h h j + 2) wp, m) = {3 h k : j h 2 with p g j,h m)} Hence, 64) Let and ) ξp, m) k p p) p) ξp,m) k p) wp,m) ψ r n) = G j,h m) = p n, p>r p p, p p>r p g j,h m) p g i,h m) j+ i h 2) p m m k We then have 65) p m m k p>r p) wp,m) = j<h k h j+2 ψ r G j,h m))

31 PRIME CHAINS AND PRATT TREES 3 The number of pairs j, h) in the product on the right is J = k 2)k 3) 2 By 64), 65) and Hölder s inequality, 66) Sm) m m p r [ If we write m p r ) ξp, m) ) k k ψ r m i ) k ψ r G j,h m)) p p i= j,h ) ξp, m) p p ) k ψ r m i ) 2Lk )2 i= m ψ r m) s = d m ) k ] L L ) L 2Lk ) β s d), j,h m ψ r G j,h m)) 2JL ) 2JL then β s is multiplicative and supported on square-free integers composed of primes > r Furthermore, if p > r s +, then ) s p β s p) = e s/p ) 4s p p We then have for each i ψ r m i ) 2Lk )2 β 2Lk ) 2d) N i + M i d N i <m i N i +M i d N i +M i 2k + )M i + β 2Lk ) 2p) ) p p>r ) 2k + )M i + 8Lk2 km i Hence, 67) i= m p>r ) k 2Lk ) ψ r m i ) 2Lk )2 ) 2L km M k For fixed j, h), let M l = maxm j+,, M h ) and write p 2 M M k ) 2L 68) g j,h m) = m l m l+ m h )g j,l m) + g l,h m)

32 32 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA We ll use G j,h m) G j,hm) := p>r p g j,h m) p m m k g l,h m) p, and note that crudely G j,h m) g j,hm) km l + N l ) k 6kM l ) k Fix all m i satisfying j + i h and i l If M l 2 k3, then we have, by 68), N l <m l N l +M l ψ r G j,hm)) 2JL = d 6kM l ) k d, Q i l m i)= p>r β 2JL d) {N l < m l N l + M l : d G j,hm)} ) Ml d + β 2JL d) d 6kM l ) k M l + 8JL ) + p 2 r<p 6kM l ) k + 8JL ) p M l + exp [8JLlog 2 6kM l ) + log k log 2 r + O))] M l If M l < 2 k3, then Lemma 63 gives N l <m l N l +M l ψ r G j,hm)) 2JL m l 2k+)M l exp m l 3kM l e d G j,h m) d 6kM l ) 2JL r ωg j,hm)) k 3 L exp ωd) + ) r e ωd) {m l 3kM l : d G j,hm)} d 6kM l km eωd) l d d 6kM l km l + e ) p p 6kM l km l log6km l )) e M l k 3e+ )

33 j,h m PRIME CHAINS AND PRATT TREES 33 We conclude that ) 2JL ψ r G j,h m)) 2JL ) k 3e+ 2L M M k 69) Note that trivially [ p r Put r = 4 k3 We have similarly r <p r ) ξp, m) ) k ] L p p ξp, m) p M M k ) 2L ) k L p p r { )} k log2 k = exp O log k L ) p) k { = exp O Therefore, by 66), 67) and 69), Sm) M M k ) L S L exp {O where m S = m p r ) ξp, m) k p p) )} k log2 k log k )} k log2 k, log k Let I = {i : M i 2 k3 } and M = p r p By an elementary estimate, M e 2r Thus, for i I, M i km Hence, the number of N i m i < N i + M i in a given residue class modulo M is M i /M + +O/k))M i /M Thus, by Lemma 64 and the Chinese Remainder Theorem, S N i <m i N i +M i i I) i I M i i I N i <m i N i +M i i I) M M k p r M i + O M p I p r )) k p p ) k I p p m i mod M i I) p r ) k [ p I p 0 m i <p i I) ) ξp, m) ) k p p ] ξp, m) Noting that b = k I, the lemma follows from Mertens estimate

34 34 KEVIN FORD, SERGEI V KONYAGIN AND FLORIAN LUCA Proof of Theorem 9 Suppose that x is large, and let r, l 2 be integers to be chosen later, and depending on x Put 60) k j = jl 0 j r) and, for 0 < η fixed, let 2 6) x j = x j+) η 0 j r) If p x and Hp) log xr log 2 + k r, then there are even integers h,, h kr so that 62) p = p 0 = h p +, p = h 2 p 2 +,, p kr = h kr p kr +, with p 0,, p kr prime and p kr x r We further suppose that 63) k r log x r log 2 x) 2, The vector h,, h kr ) may not be unique, but we associate to each such p a single such vector For j r, let Q j be the set of primes p so that 64) p ki < x i i < j), p kj x j Then, 65) { p < x : Hp) log x } r log 2 + k r r Q j Fix j r and even integers h,, h kj satisfying h h kj x/x j By Lemma 62 b) and 63), p k j ξp; h kj,, h )) p j= log p k j log 2 x log x r log 2 x log x j log 2 x Fix c > 2e By Lemma 6, 63) and Stirling s formula, the number of p = p 0 x satisfying 62) is 66) x h h kj c k j ) kj+ Sh kj,, h ) log x j ) k j+ if x is large enough Now we estimate Q j Let b j k j, a parameter to be chosen later Let Q j, be the set of p Q j for which at least b j of the variables h,, h kj are 2 2k3 j, and Qj,2 = Q j \Q j, To estimate Q j,, fix a set B {,, k j } of size b j so that h i 2 2k3 j for i B Let I = { i k j : i B} and put a i = B {k i +,, k i+ } 0 i j ), I i = I {k i +,, k i+ } 0 i j )