Extending Erdős-Kac and Selberg–Sathe to Beurling Primes with Controlled Integer Counting Functions by Malcolm Rupert B.S., Western Washington University, 2007 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in The Faculty of Graduate Studies (Mathematics) THE UNIVERSITY OF BRITISH COLUMBIA (Vancouver) April 2013 c©Malcolm Rupert 2013 Abstract In this thesis we extend two important theorems in analytic prime number theory to a the setting of Beurling primes, namely The Erdős–Kac theorem and a theorem of Saith and Selberg. The Erdős–Kac theorem asserts that the number of prime factors that divide an integer n is, in some sense, normally distributed with mean log log n and variance √ log logn. Saith proved and Selberg substantially refined a formula for the counting function of products of k primes with some uniformity on k. A set of Beurling primes is any infinite multiset {pi | i ∈ N} ⊂ R>1 such that pi ≤ pi+1 for all i and limi→∞ pi =∞. The set of Beurling primes has a corresponding multiset of Beurling integers formed by all finite products of Beurling primes. We assume that the Beurling integer counting function is approximately linear with varying conditions on the error term in order to prove the stated results. An interesting example of a set of Beurling primes is the set of norms of prime ideals of the ring of integers of a number field. Recently, Granville and Soundararajan have developed a particularly simple proof of the Erdős–Kac theorem which we follow in this thesis. For extending the theorem of Selberg and Sathe much more analytic machinery is needed. ii Table of Contents Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v 1 Introduction and Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 What is a Beurling Prime? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Examples of Sets of Beurling Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Summary of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2 Extending the Erdős–Kac Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2.2 The Main Technical Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3 A Set of Beurling Primes that Induces Large Gaps in its Set of Integers . . . . . . . . . . . . 18 4 Extending a Theorem of Selberg and Sathe . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 4.1 The Non-Uniform Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 4.2 Perron’s Formula for Beurling Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 4.3 The Bound on ζB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 4.4 The Integrations of ζB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.5 The Deduction of Theorem 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 iii Acknowledgements I would like to adamantly thank my advisor Dr. Greg Martin for all of his help and direction in completing this thesis and in making me a better mathematician. I would also like to thank Dr. Lior Silberman for his helpful suggestions and corrections. Finally I want to thank Robert Fraser and Michael Forbes for their assistance in formating. iv Dedication For my wife Haley and my parents Dave and Addie for all of their support throughout my education. v Chapter 1 Introduction and Summary 1.1 What is a Beurling Prime? When proving results in analytic number theory one does not always use the entire algebraic structure of the integers. Instead the analytic and asymptotic information is much more useful in proving classical results such as the prime number theorem or its generalizations. It turns out that many properties of the integers can be established in the more general context of a countably generated multiplicative semigroup of the positive reals satisfying appropriate analytic assumptions. Beurling noticed that one can prove many classical results in this generalized context. Fix a multiset P of real numbers {pi : i ∈ N} such that 1 < p1 ≤ p2 ≤ · · · and such that limi→∞ pi =∞. The elements of P will be called “Beurling primes”. The associated set of “Beurling integers” is the multiset B of all finite products of Beurling primes. Note that 1 ∈ B, as it is the empty product. Let the set of natural primes be denoted by P := {2, 3, 5, 7, 11, ...}. The fact that B can be a multiset introduces an interesting dynamic and can cause some confusion. For instance one can choose the set of Beurling primes to be all the natural primes with and additional 5 included which is, in some way, distinct from the other 5 already in P. In this case there will actually be n+ 1 ways to write 5n as an element of B. Beurling showed that if one assumes that the integer counting function is roughly like that of the natural numbers then the prime number theorem holds, where naturally the error term depends on the choice of B as it will for all results in this thesis. More precisely, Theorem 1.1 (Beurling, 1937). Given a a set of Beurling primes, P , and the corresponding set of Beurling integers, B, if the integer counting function, NB(x) := ∑ n≤x n∈B 1, has the asymptotic formula NB(x) = Ax+O(x(log x)λ) for some A > 0 and λ < −3/2 then one gets the following asymptotic formula for piB(x) := ∑ n≤x n∈P 1, the Beurling prime counting function: piB(x) ∼ x log x . (1.1.1) Furthermore this bound is optimal: exists a Beurling primes system with λ = −3/2 and where 1.1.1 does 1 not hold. There are several, possibly, surprising things about this result. First it is interesting that having twice the number of integers does not seem to effect the number of primes at all; therefore the additional integers introduced must overwhelmingly come from composite numbers. We will make this heuristic more exact in Chapter 4 when we discus the asymptotic formula for the counting function of Beurling integers with exactly k prime factors. Also see that adding a single very small element to P will greatly increase the number of integers. The next thing that one may find surprising is the way in which the results can be proved, since the methods are very similar, for the most part, to those in the natural prime case. In addition to this theorem one may also ask for what set of Beurling primes do we get the prime number theorem with a “good” error term. If a set of Beurling primes has a power savings in the error term then using essentially the same methods as in the natural prime case one can prove a prime number theorem with an error term of O(xe−cθ √ log x). In particular define the Beurling zeta function to be ζB(s) = ∑ n∈B n −s, then one can show that the Beurling zeta function extends analytically past σ = 1, one can then create a similarly shaped zero-free region and perform the same integration techniques as when P = P to get such a prime number theorem. 1.2 Examples of Sets of Beurling Primes A simple example of a set of Beurling primes is the natural primes with some small set of elements either added or removed. Note that we could add primes into our system with multiplicity. In the case where we remove a finite set of primes, T ⊂ P, the prime number theorem follows trivially from the classical prime number theorem and one can explicitly calculate the integer counting function NB(x) = ∑ n∈N p-n ∀ p∈T 1 = Ax+O(1) (1.2.1) where A = ∏ p∈T ( 1− 1p ) . To see this note that Ax is an exact formula when x is a multiple of N =∏ p∈T p and can only be off by a finite amount in [0, N ] and the error is periodic with period N . Note that in (1.2.1) the error term depends on B, as will all error terms in this thesis. The next example provides a way to prove the prime ideal theorem [10]. Let K be a number field such that n = [K,Q] and let OK be its ring of integers. Let P be the set of norms of prime ideals of OK . Then if one can show that the total number of ideals with norm less then x is Ax + O(xc), where A > 0 is called the ideal density and c = 1 − 1/n. By the unique factorization of prime ideals and the multiplicative property of the norm we then apply the Beurling prime number theorem 1.1 to get the prime ideal theorem. 2 1.3 Summary of Results In chapter 2 we extend a major result in probabilistic number theory, the Erdős–Kac theorem, to the case of Beurling primes whenever Mertens’s theorem (Corollary 2.1), Chebychev’s bound, and a condition on the analog to the Euler φ function hold for our set of Beurling primes. Chebychev’s bound refers to the property that pi(x) x/ log x, which Hall [7] and Diamond [2] showed holds precisely when NB(x) = Ax+O(x(log x)γ) where γ < −1. We also recount from the work of others when exactly a set of Beurling primes has these qualities. Define ω(n) to be the number of prime factors of n ∈ B. The Erdős–Kac theorem states that the quantity ω(n)−log logn√ log logn is, in some sense, normally distributed when n is a considered random variable, uniformly distributed among the integers. Formally we calculate all the moments of the distribution and find that they are, in the limit, the moments of the normal distribution. Probability theory tells us that the normal distribution is determined by its moments which is how we justify the above statement of the theorem. Theorem 1.2. Let P be a set of Beurling primes with corresponding set of integers B such that the integer counting function has the asymptotic NB(x) = Ax + O(xθ) where A > 0 and θ < 1. Set the notation Ck := Γ(k + 1)/2 k/2Γ(k/2 + 1). Assume that k ≤ log log log log x. When k is even∑ n≤x n∈B (ωB(n)− log log x)k = A · Ckx(log log z)k/2 ( 1 +OA(k 3/(log log z)) ) +O(8kpi(z)k). When k is odd ∑ n≤x n∈B (ωB(n)− log log x)k Ckx(log log z)k/2 k 3 log log z + 8kpi(z)k In particular, it follows simply that (ωB(n) − log log n)/ √ log log n is normally distributed among natural numbers in the following sense. The asymptotic density lim x→∞ ( 1 NB(x) ·# { n ≤ x : a ≤ ωB(n)− log logn√ log log n ≤ b }) = 1√ 2pi ∫ b a e−t 2/2dt. There are many corollaries to the Erdős–Kac theorem including the Hardy–Ramanujan theorem [8] and the Erdős multiplication table theorem [4] among others. These theorems come for free when we prove the Erdős–Kac theorem but are interesting and definitely worth mentioning for their simple interpretation and significance. In Chapter 3 we provide an interesting examples of a set of Beurling primes which produces extremely large gaps in the corresponding integers. We do this by removing select elements of Pwhich have a lot of leverage in removing integers approximately exponentially larger. 3 Next, in chapter 4, we extend a theorem proved by Sathe and greatly simplified by Selberg which gives a formula for counting integers with k prime factors with some uniformity in k. We first derive a formula for a fixed k using the prime number theorem and illustrate, drawing inspiration from Beurling primes, why this formula cannot be uniform, despite some conflicting evidence. Next we use Perron’s formula on the partial summation of dz(n), the generalized devisor function, and find bounds on ζ(s) to calculate the integral with integrand ζ(s)z which Perron’s formula gives us. The rest of the chapter relates dz back to the function which we care about, σk, using some arithmetic lemmas. In the end we prove the following theorem and give some corollaries and examples. Theorem 1.3. Let P be a set of Beurling primes with corresponding set of integers B such that the integer counting function has the asymptotic NB(x) = Ax+O(xθ) where A > 0 and θ < 1. Suppose that R < p1 and that F (s, z) := ∏ p∈P ( 1− z ps )−1( 1− 1 ps )z . Set G(z) = Az · F (1, z)/Γ(z + 1). Then σk(x) = G ( k − 1 log log x ) x(log log x)k−1 (k − 1)! log x ( 1 + OR ( k (log log x)2 )) uniformly for 1 ≤ k ≤ R log log x. 4 Chapter 2 Extending the Erdős–Kac Theorem The goal of this chapter is to prove Theorem 1.2. We start by proving a statement about the average number of prime factors of Beurling integers in Corollary 2.2. To do this we first need a generalization of Mertens’s theorem on an asymptotic formula for the sum of reciprocals of prime numbers up to x, given in Corollary 2.1, which follows easily from previously known results. To prove Theorem 1.2 we build upon the simple proof that was recently discovered by Soundararajan and Granville in [6]. To do this we need to first replicate some arithmetic properties of the integers to the case of Beurling primes with controlled integer counting functions. We do this in Lemma 2.3 and in Theorem 2.4. Finally we prove Theorem 1.2, which follows relatively simply from 2.4. 2.1 Preliminaries In a recent paper [14] by Paul Pollack he expands on a result of Olofsson, in [13], which states that if NB(x) ∼ Ax for some A > 0 then a generalization of Mertens’s theorem holds, namely ∏ p≤x p∈P ( 1− 1 p )−1 ∼ Aeγ log x. From this result one can derive a generalization of another theorem of Mertens. Corollary 2.1. Let P be a set of Beurling primes and let B be the corresponding set of Beurling integers. If NB(x) ∼ Ax for some positive A, then ∑ p≤x p∈P 1 p = log log x+ logA+ γ − ∑ p∈P ∞∑ k=2 1 kpk + o(1). Proof. Assume that NB(x) ∼ Ax for some A > 0. Then by theorem A in [14] we have that 5 ∏ p≤x p∈P ( 1− 1 p )−1 = Aeγ(log x)(1 + o(1)). Taking logs on both sides and adding zero on the left hand side yields ∑ p≤x p∈P 1 p + ∑ p≤x p∈P ( log [( 1− 1 p )−1] − 1 p ) = logA+ γ + log log x+ o(1). Since the Taylor expansion of log [( 1− 1x )−1]− 1/x is∑∞k=2 1kpk one can see that ∑ p≤x p∈P ( log [( 1− 1 p )−1] − 1 p ) = ∑ p∈P ∞∑ k=2 1 kpk +O ∑ p>x p∈P p−2 . The series in the error term is O(x−1) since piB(x) NB(x) A x and the above double sum converges since, by the integral comparison test, we see that ∑ p∈P ∞∑ k=2 1 kpk = ∞∑ k=2 1 k ∑ p∈P 1 pk ∞∑ k=2 1 k ∫ ∞ x=1 x−k = ∞∑ k=2 1 k(k + 1) 1. Thus, we conclude that ∑ p≤x p∈P 1 p = log log x+ logA+ γ − ∑ p∈P ∞∑ k=2 1 kpk + o(1). From Corollary 2.1 we easily get an average value for ωB, the function which counts the number of prime factors of n ∈ B. This is well defined since elements of B are defined by a product of elements of P . Consider the examples P = P ∪ {6}. Then 6 ∈ B has index 2 in the multi set B, that is there are two ways to write 6 is a product of elements of P . In this case, if we write all elements as their defining product over P , then we get ωB(2 · 3) = 2 and ωB(6) = 1. Corollary 2.2. If NB(x) = Ax+O(x(log x)γ), where γ < −1 then ∑ n≤x n∈B ωB(n) = Ax log log x+ x logA+ γ −∑ p∈P ∞∑ k=2 1 kpk + o(x). 6 Proof. By switching the order of summation we can write ∑ n≤x n∈B ωB(n) = ∑ n≤x n∈B ∑ p|n p∈P 1 = ∑ p≤x p∈P ∑ n≤x p|n n∈B 1 = ∑ p≤x p∈P NB ( x p ) . By our assumption on the growth of NB(x) we see that the above yields our desired main term with some additional error terms. If one uses the trivial error term NB(x) = Ax+O(1) for when x ≤ 2 we determine that the above is ∑ n≤x n∈B ωB(n) = Ax ∑ p≤x p∈P 1 p +O ∑ p≤x/2 p∈P x/p logγ(x/p) +O ∑ x/2≤p≤x p∈P 1 . By Chebychev’s bound for Beurling primes we see that the second error term is O(x/ log x). If x > p21, then for the first error term we make a further split in the summation at √ x to see that ∑ p≤√x d∈P x/p loga(x/p) x log x ∑ p≤√x d∈P 1 p x log log x log x and ∑ √ x<p≤x/2 d∈P x/p loga(x/p) √x ∑ 2≤t≤√x 1 log t √x li(√x) x log x . Corollary 2.1 yields the desired result. From these results we can see that the average value for ωB(n) for n ≤ x is log log x. We can say much more than this. In fact we can prove a generalization a theorem by Erdős and Kac [5], that ωB(n) is normally distributed with mean log log n and variance √ log log n. Branching off from there, using only the first two moments of the distribution of ωB, we also get a generalization of a result by Hardy and Ramanujan which say that almost all integers n have about log log n prime factors. Furthermore Erdős’ multiplication table theorem - that is, as n goes to infinity, almost no numbers up to n2 are in the n × n multiplication table - is also provable in the setting of Beurling primes. Now we prove the Erdős – Kac theorem following the techniques introduced by Granville and Soundararajan in [6]. To this end let us start with some technical lemmas. We define the GCD the using the multiplicative structure of B, that is the GCD of n and m denoted, (n,m) is just the product of all elements of P which are in the prime factorization of both n and m. One has to be 7 a bit careful here because there may be multiple ways to write any integer as a product of Beurling primes. For instance if a set of Beurling primes contains 2 and √ 2 then it would be the case that ( √ 2 2 , 2) = 1. We also define φB and τB, using the product formula, as follows φB(n) = n · ∏ p|n p∈P ( 1− 1 p ) , and τB(n) = ∏ pα||n p∈P 1 + α = ∑ d|n d∈B 1 Lemma 2.3. Let P be a set of Beurling primes such that NB(x) = Ax+O(xθ) for some A > 0 and θ < 1. For r and d in B ∑ n≤x (n,r)=d n∈B 1 = Ax d φB(r/d) r/d +O ((x r )θ gθ(r/d) ) where gs(n) = ∏ p|n(1 + p s). Much of the theory of Arithmetic functions is similar in the context of Beurling primes and evaluating the sum in Lemma 2.3 for the natural primes is a special case of this lemma. Proof. Let d ∈ B be given and let r be given such that d|r. Then∑ n≤x n∈B (n,r)=d 1 = ∑ n≤x n∈B d|n (n/d,r/d)=1 1 = ∑ k≤x/d k∈B (k,r/d)=1 1. Now we use the fact that (µ ? 1)(n) = e(n), where e is the identity in the ring of arithmetic functions under Dirichlet convolution. We use this identity to detect when k and rd are coprime to get ∑ n≤x n∈B (n,r)=d 1 = ∑ k≤x d k∈B ∑ c|(k, r d ) c∈B µB(c). Now we switch the order of summation which yields that the above is ∑ n≤x n∈B (n,r)=d 1 = ∑ c| r d c∈B ∑ k≤x d c|k k∈B µB(c) = ∑ c| r d c∈B ∑ k≤ x cd k∈B µB(c) = ∑ c| r d c∈B [ µ(c) Ax cd +O ( |µ(c)| · ( x cd )θ)] . Finally, note that φ(n) = µ(n) ? n as seen by examining the product formula for the functions involved. Therefore 8 Ax d 1 r/d ∑ c| r d c∈B µB(c) r/d c +O ((x r )θ ∑ c| r d c square free c∈B ( r cd )θ ) = Ax d φB(r/d) r/d +O ((x r )θ gθ(r/d) ) . 2.2 The Main Technical Work Toward the goal of proving Erdős–Kac theorem, in the style of Granville and Soundararajan we introduce a sort of “balanced prime counting function”. For a given p ∈ P define the function, f : B → Q, as fp(n) = 1− 1/p p|n−1/p otherwise. For example, a sort of balanced prime divisor counting function for n ∈ B is ∑ p≤x p∈P fp(n) = ωB(n)− ∑ n≤x n∈B 1 p . Furthermore one can generalize this function for any r ∈ B. If r = ∏si=1 pαii then set fr(n) = ∏si=1 fpi(n)αi . There is not an obvious reason why this “balancing” would be helpful, but it is. In other proofs of the Erdős– Kac theorem the expression ∑ n≤x (ω(n)− log log x)k is expanded and, carefully, one can cancel out the large terms and bound the terms that remain. The balanced prime divisor counting function takes care of all of the cancellation automatically. This makes the proof shorter and easier. Theorem 2.4. Let x be large, set z = x1/k and let k < log log log log(z). Define notation for the moments of the normal distribution to be Ck := Γ(k + 1)/[2k/2 · Γ(k/2 + 1)]. If k is even then ∑ n≤x n∈B (∑ p≤z p∈P fp(n) )k = A · Ckx(log log x)k/2 ( 1 +OA((k 3/ log log z)) ) +O(8kpi(z)k). If k is odd then 9 ∑ n≤x n∈B (∑ p≤z p∈P fp(n) )k Ckx(log log x)k/2(k3/ log log x)) +O(8kpi(z)k). Proof. By reordering the summation we get that ∑ n≤x n∈B (∑ p≤z p∈P fp(n) )k = ∑ p1,...pk≤z pi∈P ∑ n≤x n∈B fp1···pk(n). (2.2.1) If R = ∏s i=1 p αi i then set r = ∏s i=1 pi. Let n ∈ B be given. If d = (n, r) then fR(n) = ∏ pα||R fp(n) α = ∏ pα||R p|n ( 1− 1 p )α ∏ pα||R p-n (−1 p )α = ∏ pα||R p|d ( 1− 1 p )α ∏ pα||R p-d (−1 p )α = fR(d). From this fact we can sort the sum, ∑ n≤x fR(n), by what the value of (r, n) takes, which will bring us back to one of our lemmas. ∑ n≤x fR(n) = ∑ d|r fR(d) ∑ n≤x (n,r)=d n∈B 1 By Lemma 2.3, we know the formula for the inner sum. Therefore we get that the above is ∑ n≤x fR(n) = ∑ d|r ( Ax d φB(r/d) r/d +O ((x r )θ gθ(r/d) )) fR(d). Define Ids(n) := ns. To estimate the error term over the sum of divisors of r see that when r is square free gθ takes the simpler form gθ(r/d) = σθ(r/d) = 1 ? Idθ(n). Furthermore for d|r square free our |fR(d)| has the simple form |fR(d)| = 1r/d φ(d)d = φ(d)r . Therefore (x r )θ∑ d|r gθ(r)|fR(d)| = (x r )θ∑ d|r φ(d) r ∑ c|(r/d) (r/cd)θ = xθ r ∑ d|r φ(d) dθ ∑ c|(r/d) 1 cθ x θ r ∑ d|r d1−θτ(r/d) (2.2.2) 10 Since d|r is square free and ωB(r) ≤ k the quantity τ(d) ≤ 2k so ∑ d|r τ(r/d)d 1−θ 4kr1−θ. Therefore the quantity in the last line of (2.2.2) is (xr )θ 4k. Therefore we conclude that ∑ n≤x n∈B fR(x) = Ax r ∑ d|r fR(d)φB(r/d) +O ((x r )θ 4k ) . (2.2.3) We define the function G(R) := ∏ pα||R [ 1 p ( 1− 1p )α + ( −1p )α ( 1− 1p )] and claim that the following summation formula holds, G(R) = 1r ∑ d|r fR(d)φB(r/d). To see that this is true, observe that if one expands the product each term will simply be a product of some terms of the form 1p ( 1− 1p )α and the rest of the terms of the form ( −1p )α ( 1− 1p ) , which yields that G(R) = ∑ d|r ∏ p|d 1 p ( 1− 1 p )α · ∏ p-d (−1 p )α( 1− 1 p ) . To be pedantic and separate everything out neatly, we get that G(R) = ∑ d|r ∏ p|d 1 p ∏ p|d ( 1− 1 p )α · ∏ p|r/d (−1 p )α∏ p-d ( 1− 1 p ) . The first product yields 1/d. The second and third product combine to give fR(d) and the final product is the product formula for φB(r/d)r/d . We conclude that ∑ n≤x n∈B fR(n) = Ax ·G(R) +O ((x r )θ 4k ) . Applying the above to the right hand side of (2.2.1) gives us ∑ p1,...pk≤z pi∈P ∑ n≤x n∈B fp1···pk(n) = ∑ p1,...pk≤z pi∈P Ax ·G(p1 · · · pk) +O (( x p1 · · · pk )θ 4k ) Summing the error term over all k-tuples of primes less than z yields ∑ p1,...pk≤z pi∈P ∑ n≤x n∈B fp1···pk(n) = Ax · ∑ p1,...pk≤z pi∈P G(p1 · · · pk) +O(8kpi(z)k) (2.2.4) since ∑ p≤z p −θ z−θpi(z). Note that G(R) = 0 if and only if R has a prime divisor, p, such that p||R. 11 Thus we can eliminate many of the terms from the sum ∑ p1,...pk≤z pi∈P G(p1 · · · pk) = ∑ p1,...pk≤z pi∈P p1···pksquare-full G(p1 · · · pk). For p1 · · · pk = qα11 · · · qαss to be square-full we must have αi ≥ 2 for all i, thus we see that s ≤ k/2. We rewrite the above sum over square-full numbers according to how many primes appear in each numbers factorization. This shows us that the above is = ∑ s≤k/2 ∑ q1<...<qs≤z qi∈P ∑ α1,...,αs≥2∑ αi=k k! α1! · · ·αs!G(q α1 1 · · · qαss ). (2.2.5) Here the factorial term comes when we restrict the qi’s to an increasing sequence. When k is even then the terms when αi = 2 for all i are the main contributors. When s = k/2, we get the terms ∑ q1<···<qk2≤z qi∈P k! 2k/2 G(q21 · · · q2k/2). When R = ∏s i=1 q 2 i , then we evaluate G(R) to be G(R) = s∏ i=1 ( 1 qi ( 1− 1 qi )2 + (−1 qi )2( 1− 1 qi )) = s∏ i=1 1 qi ( 1− 1 qi ) . If we remove the increasing condition on the above sum and pull out all constants we get our Gaussian moments, Ck. Thus the terms in (2.2.5) when s = k/2 can be seen to be k! 2k/2(k/2)! ∑ q1,...,qs≤z qi∈P qi distinct k/2∏ i=1 1 qi ( 1− 1 qi ) . We can bound the above sum by above if we ignore distinctness and by below if we overcompensate for distinctness by ignoring the largest summands. Let pin be the Beurling prime which comes nth closest to maximize 1t ( 1− 1t ) . Note that t = 2 maximizes this function so when n is small pin should be relatively close to 2. To be more specific on how to acheive a lower bound, take q1, ..., qj as given, then the sum over just the qj+1 < z is clealy ≥ ∑ p≤z p∈P p 6=pii,1≤i≤j 1 p − 1 p2 12 Applying this argument for all 1 ≤ j ≤ k/2, and applying the trivial upper bound we get the pair of inequalities ( ∑ p≤z p∈P p6=pii,1≤i≤k/2 1 p − 1 p2 )k/2 ≤ ∑ q1,...,qs≤z qi∈P qi distinct k/2∏ i=1 1 qi ( 1− 1 qi ) ≤ (∑ p≤z p∈P 1 p − 1 p2 )k/2 . The inner sum of the lower bound can differ from the inner sum of the upper bound by at most k/8. So, by Corollary 2.1 we get that the term from (2.2.5) with s = k/2 contributes k! 2k/2(k/2)! (log log z +OA(k)) k/2 = k! 2k/2(k/2)! (log log z)k/2 ( 1 +O ( k3/(log log z) )) . (2.2.6) To deal with the terms when s < k/2 use the trivial estimate that 0 ≤ G(qα11 · · · qαss ) ≤ 1/(qα11 · · · qαss ). From this we get that these terms are bounded above by ∑ s<k/2 k! s! ∑ q≤z q∈P 1 q s ∑ α1,...,αs≥2∑ αi=k 1 α1! · · ·αs! . The number of ways of writing k = ∑ 1≤i≤s αi≥2 αi is the same as the number of ways of writing k − s =∑ 1≤i≤s αi≥1 αi. Picture lining k − s objects. Then we can separate them into s groups, of size at least 1, by placing s − 1 partitions between any combination of their k − s − 1 gaps. The number of ways to write k−s = ∑1≤i≤s αi≥1 αi is therefore ( k−s−1 s−1 ) . Now using Corollary 2.1 we bound the terms when s < k/2 above by < ∑ s<k/2 k! s!2s ( k − s− 1 s− 1 ) (log log z +OA(1)) s. (2.2.7) Now recall equations (2.2.4), (2.2.6) and (2.2.7). Stringing these together we get that 13 ∑ n≤x n∈B ∑ p≤z p∈P fp(n) k = ∑ p1,...pk≤z pi∈P ∑ n≤x n∈B fp1···pk(n) =A x ∑ pi≤z 1≤i≤k pi∈P G(p1 · · · pk) +O(8kpiB(z)k) =A Ckx(log log z +OA(1 + k 3/ log log z))k/2 +O x ∑ s<k/2 k! 2ss! ( k − s− 1 s− 1 ) (log log z +OA(1)) s + 8kpiB(z)k Note that we can bound the sum ∑ s<k/2 k! 2ss! ( k−s−1 s−1 ) (log log z)s by a convergent geometric sum since the ratio of consecutive terms are 2(s+1)s(k−s) log log z which is much smaller than 1. Hence ∑ s<k/2 k! 2ss! ( k − s− 1 s− 1 ) (log log z)s k! 2k/2(k/2)! (log log z)dk/2e−1. Now we want to say something about the sum ∑ n≤x(ωB(n)− log log x)k with some uniformity over k. Recall Theorem 1.2. LetP be a set of Beurling primes with corresponding set of integers B such that the integer counting function has the asymptotic NB(x) = Ax + O(xθ) where A > 0 and θ < 1. Set Ck := Γ(k + 1)/2k/2Γ(k/2 + 1). Assume that k ≤ log log log log x. When k is even∑ n≤x n∈B (ωB(n)− log log x)k = A · Ckx(log log z)k/2 ( 1 +OA(k 3/(log log z)) ) +O(8kpi(z)k). When k is odd ∑ n≤x n∈B (ωB(n)− log log x)k Ckx(log log z)k/2 k 3 log log z + 8kpi(z)k Proof of Theorem 1.2. Set z = x1/k to see that 14 ωB(n)− log log x = ∑ p≤z fp(n) + ∑ p|n p>z 1 + ∑ p≤z 1/p− log log x . If n ≤ x then we know that n can have, at most, k − 1 prime factors larger than z, thus ∑ p|n p>z 1 k. To handle the third sum recall Corollary 2.1 to see that ∑ p≤z 1/p− log log x = log log x1/k− log log x+O(1) = log log x 1/k log x +O(1) = log(1/k) +O(1) log k. Therefore, we conclude that ωB(n)− log log x = ∑ p≤z fp(n) +O(k). Using the above result and the binomial theorem, then for some positive constant c, say the explicit constant from the error term in the above equation, we get (ωB(n)− log log x)k = (∑ p≤z fp(n) )k +O ( k−1∑ `=0 (ck)k−` ( k ` )∣∣∣∣∑ p≤z fp(n) ∣∣∣∣` ) . This should look familiar. Now we will apply Theorem 2.4 to evaluate this sum. ∑ n≤x (ωB(n)− log log x)k = ∑ n≤x (∑ p≤z fp(n) )k +O (∑ n≤x k−1∑ `=0 (ck)k−` ( k ` )∣∣∣∣∑ p≤z fp(n) ∣∣∣∣` ) (2.2.8) Directly from the statement of Theorem 2.4 we can correctly conclude, if k < log log log log z and is even, then the main term is ∑ n≤x (∑ p≤z fp(n) )k = A · Ckx(log log z)k/2 ( 1 +OA(k 3/ log log z) ) +O(8kpi(z)k). We will also use Theorem 2.4 to bound the error tern in (2.2.8). If ` is even then, just as above, Theorem 2.4 bounds the sum in the error term. If ` is odd then the error term of 2.2.8 is bounded using the Cauchy–Schwarz inequality and Theorem 2.4 and is seen to be 15 ∑ n≤x n∈B ∣∣∣∣∑ p≤z p∈P fp(n) ∣∣∣∣` ≤ (∑ n≤x n∈B (∑ p≤z p∈P fp(n) )`−1)1/2(∑ n≤x n∈B (∑ p≤z p∈P fp(n) )`+1)1/2 √ C`−1C`+1x(log log z)`. In particular Theorem 1.2 proves that lim x→∞ 1 NB(x) ∑ n≤x ( ωB(n)− log log x√ log log x )k = Ck k is odd0 k is even which coincide with the moments of the normal distribution. Now we will show that the difference ∑ n≤x ( ωB(n)− log log x√ log log x )k − ∑ n≤x ( ωB(n)− log logn√ log log n )k = o(x). (2.2.9) This will be enough to prove that ωB(n)−log logn√ log logn is asymptotically normally distributed since the normal distribution is determined by its moments. In other words the asymptotic density lim x→∞ ( 1 NB(x) ·# { n ≤ x : a ≤ ωB(n)− log logn√ log log n ≤ b }) = 1√ 2pi ∫ b a e−t 2/2dt. To prove (2.2.9) see that the triangle inequality for the Lk norm gives ∣∣∣∣( ∑ 1<n≤x |ωB(n)− log log x|k )1/k − ( ∑ 1<n≤x |ωB(n)− log logn|k )1/k∣∣∣∣ ≤ ( ∑ 1<n≤x (log log x− log log n)k )1/k 16 To calculate a bound for the sum ∑ 1<n≤x(log log x − log logn)k we split the sum at x/ log x. For values of n between x/ log x and x we see that log log x− log logn log log x log x . Therefore ∑ x/ log x<n≤x (log log x− log logn)k x · ( log log x log x )k = o(x). For values of n between 1 and x/ log x we use the trivial bound ∑ 1<n≤x/ log x (log log x− log logn)k x(log log x) k (log x)k−1 = o(x). 17 Chapter 3 A Set of Beurling Primes that Induces Large Gaps in its Set of Integers Recall the standard set of primes of the natural numbers is being denoted by P. In this short chapter P will be an infinite subset of P and B will be the comprised of the elements of N that have a prime factorization containing primes only found in P . That is B is the smallest semigroup containing all elements of P . If we define piB as the counting function of P and NB as the counting function B then we show the existence of such a P as to make very large gaps between consecutive elements of B. To be precise, our constructed B has the property that for every γ < 1 there is a x ∈ R such that there is a xγ sized gap of integers inside of (x, 2x). The only result that the results in this chapter rely heavily on is the fact that the counting function for integers less than x with prime factors all less than y, denoted by ψ(x, y) has the property that if y = log x then ψ(x, y) = xo(1). As a source for counting integers with small prime factors take [12], Section 7.1. Define C[x, y] as the function which counts integers less than x with at least 1 prime factor larger than y. Clearly we see that if x > 1 and y ≤ 2 then C[x, y] = [x]. Again it is simple to see that if y ≥ x then we get that C[x, y] = 0. It is plain to see that C[x, y] + ψ(x, y) = [x]. Proposition 3.1. For any γ ∈ (0, 1) there exists a largeNγ ∈ N such that there exist xγ consecutive integers in [x, 2x] with a prime factor larger than log x whenever x > Nγ . Proof. Let γ ∈ (0, 1) be given and set κ = (1−γ)/2, so that γ < κ < 1. Without loss of generality log x is an integer since, if not, a larger value of xwe be sufficient. Therefore log(x+xγ) < log 2+log x < log x+1. Now, consider the difference C[x+ xκ, log x]− C[x, log x] = C[x+ xκ, log(x+ xκ)]− C[x, log x] = xκ − xo(1). Therefore in the interval [x, x + xκ] we have xκ integers with a prime factor at least as big as log x with at most xo(1) exceptions. By the pigeonhole principle, we must have at least xκ−o(1) consecutive integers with a prime factor larger than log x. For x sufficiently large (say, larger than Nγ) we have that κ − o(1) > γ. This means that for such an x we have at least xγ consecutive integers between x and 2x all with a prime factor larger than log x. 18 A corollary to this proposition is the main result of the chapter. Corollary 3.2. There exists a set of Beurling primesP ⊂ P such that piB(x) ∼ pi(x) and for every γ ∈ (0, 1) there exist an x such that there is a xγ size gap between two consecutive Beurling integers between x and 2x. Proof. Define the sequence γn = 1 − 1n for n ∈ N. Let Nγ be chosen to be as in Proposition 3.1. Choose x1 > Nγ1 and inductively Choose xn > ( e2xn−1 , Nγn ) . For each n ∈ N find, according to Proposition 3.1, the xγnn consecutive integers that all have a prime factor larger than log(xn) and call the set of such integers In. Now, define Q(n) to be the largest prime factor of n and define Rn = {Q(k) : k ∈ In}. Note that if i < j are integers, then for a ∈ Rj we have that a > log xj > log(e2xj−1) ≥ 2xi. Also, for b ∈ Ri we have b ≤ 2xi. Hence for i 6= j we have thatRi ∩Rj = ∅. Furthermore note that the size ofRn is less than or equal to xγnn , and all of the elements ofRn are within the interval [log xn, 2xn]. Now we choose the set of Beurling primes which will satisfy the statement of the corollary Pwow := P \ ∞⋃ m=1 Rm. By the above discussion we see that for all n ∈ N the set In ∩ Bwow = ∅. Therefore for all n ∈ N we have xγnn sized gaps at each xn. Furthermore it is easy to see that piPwow(x) ≤ pi(x). But also piPwow(x) ≥ pi(x)− ∑ xn≤x xγnn . Call m := max{i ∈ N | xm ≤ x}. Then m∑ i=1 xγii ≤ xγmm + m−1∑ i=1 xi xγmm + (m− 1)xm−1. Since m − 1 ≤ xm−1 ≤ log xm ≤ log x and also m log? x log log x. Therefore the above is∑m i=1 x γi i (log x)2. Hence we see that pi(x) ∼ piPwow(x). 19 Chapter 4 Extending a Theorem of Selberg and Sathe 4.1 The Non-Uniform Case We would like to expand the prime number theorem for Beurling primes (Theorem 1.1) to counting products of k elements of P . Consider the counting function of Beurling integers with exactly k prime factors less than x, denoted by σB,k(x). Then we have the classical result due to Landau [9] – for the set of natural primes, P – that for a fixed k the following asymptotic relationship holds σk,P(x) ∼ x(log log x)k−1 (k − 1)! log x . (4.1.1) It is tempting to think that this result is uniform in k since counting the integers less then x according to how many prime in their factorization shows us that ∑ k∈N σk,P(x) = bxc, while Taylor series tell us that∑ k∈N x(log log x)k−1 (k−1)! log x = x. It turns out that this formula is actually not uniform. Now, using the inspiration of Beurling primes we can show that (4.1.1) is not uniform over all k. Consider the set of Beurling primes P = P \ {3}. Then for a fixed k Theorem 4.1 shows that that (4.1.1) will hold, yet clearly ∑ k≥1 σB,k(x) = ∑ n≤x n∈B 1 = 2x/3 +O(1). So it seams that there cannot be such a wide range of uniformity. In the natural prime case a theorem of Sathe [15] [16] [17] [18], which was greatly simplified by Selberg [19], gives a formula for σk(x) which is uniform for all k < R log log x where R < 2. In the setting of Beurling primes we will prove an analogous statement in theorem 1.3. First, we prove the non-uniform case. Theorem 4.1. For a fixed positive integer k and a fixed set of Beurling primes with integer counting functions NB(x) = Ax+ O(x/logγx) with A > 0 and γ > 3/2, then the counting function for integers with exactly k prime factors has asymptotic formula σB,k(x) ∼ x(log log x) k−1 log x . 20 Beurling proved in [1] that it is possible that the prime number theorem will not hold for a set of Beurling primes if we have an error term in the integer counting function which has γ = 3/2. Since the prime number theorem is a base case of this theorem it is also true that 3/2 is best possible for Theorem 4.1. Proof. To avoid cumbersome notation we drop the subscript B, but keep in mind hat we are always working with Beurling integers and primes. Consider a slightly different function, pik(x), which counts Beurling integers less than x which are products of k distinct Beurling primes. I claim that pik(x) has the asymptotic formula pik(x) ∼ x(log log x) k−1 (k−1)! log x . We see that 0 ≤ σk(x)− pik(x) ∑ 1≤i≤k−1 # { pα11 · · · pαii ≤ x | pj ∈ B | ∑ 1≤j≤i αj≥1 αj = k } ∑ 1≤i≤k−1 ( k − 1 i− 1 ) # { p1 · · · pi ≤ x | pj ∈ B } pik−1(x) ∑ 1≤i≤k−1 ( k − 1 i− 1 ) k pik−1(x) (4.1.2) because, as discussed in the proof of Theorem 2.4, the number of ways to sum i positive integers to k is( k−1 i−1 ) . The case for when k = 1 is taken care of by the prime number theorem for Beurling primes. Now assume the induction hypothesis. Every product of k + 1 distinct primes has exactly k + 1 options for a prime p0 to omit so ∑ p0≤x p0∈P pik(x/p) = ∑ p0≤x p0∈P ∑ p1···pk≤x/p pi∈P 1 = (k + 1)pik+1(x) +O(pik(x)) (4.1.3) since ∑ p0≤x p0∈P pik(x/p) also count products of k + 1 prime factors with at most 1 repeated factor. Writing∑ p0≤x p0∈P pik(x/p) as a Riemann–Stieltjes integral and applying partial summation we get that 21 ∑ p0≤x p0∈P pik(x/p) ∼ x (k − 1)! · ∑ p≤x p∈P (log log x/p)k−1 p log(x/p) ∼ x (k − 1)! · ∑ n≤x n∈B (log log x/n)k−1 n log(x/n) IP(n) = x (k − 1)! · pi(x) · (log log x)k−1 x log x − x (k − 1)! ∫ x p1 pi(y) d dy [ (log log x/y)k−1 y log x/y ] . We see that the first term in the above is k x(log log x)k−1/(log x)2 so we suspect that this should go into the error term. To calculate the above integral we apply the prime number theorem to get the rather complicated expression ∫ x p1 pi(y) d dy [ (log log x/y)k−1 y log x/y ] ∼ ∫ x p1 −(log log x/y)k−1 y log y log x/y + (log log x/y)k−1 y log y(log x/y)2 − (k − 1)(log log x/y) k−2 y log y(log x/y)2 dy. Now combining the smaller integrals into an error term shows us that (k − 1)!(k + 1)pik(x) x = (1 + o(1)) ∫ x p1 (log log x/y)k−1 y log y log x/y dy +O (∫ x p1 (log log x/y)k−1 y log y(log x/y)2 dy + (log log x)k−2 log x ) (4.1.4) Set I = ∫ x p1 −(log log x/y)k−1 y log y log x/y dy. To see that the error term is genuinely smaller than I split the integral up at√ x and get the bound ∫ √x p1 −(log log x/y)k−1 y log y(log x/y)2 dy I log x for the integral over small values of y. For the integral for large values of y use the substitution v = log log x/y and repeated integration by parts to get that 22 ∫ x √ x −(log log x/y)k−1 y log y(log x/y)2 dy 1 log x ∫ x p1 −(log log x/y)k−1 y(log x/y)2 dy =− ∫ log log x log log √ x vk−1 ev dv (log log x) k−1 (log x)2 . So the above calculations the error term in (4.1.4) yields (k − 1)!(k + 1)pik(x) x ∼ ∫ x p1 (log log x/y)k−1 y log y log x/y dy. (4.1.5) All that is required to prove Theorem 4.1 is to evaluate the above integral. Start by noting that ∫ x x/e pik(x/u) log u ∫ x x/e du u x log x so we can replace the upper bound of the integral in (4.1.5) with x/e. Now make the substitution t = log y and then use partial fraction decomposition to see that (k − 1)!(k + 1)pik(x) x ∼ ∫ log x−1 1 (log(log x− t))k−1 t(log x− t) dt = 1 log x ∫ log x−1 1 (log(log x− t))k−1 s− t dt+ 1 log x ∫ log x−1 1 (log(log x− t))k−1 t dt. A simple calculation shows that the first integral ∫ log x−1 1 (log(log x− t))k−1 log x− t dt = 1 k [ − log(log x− t)k ]log x−1 1 = 1 k log(log x− 1)k ∼ 1 k log log x. To get a grasp on the size of the second integral make the further split at log x2 and see that 23 ∫ log x−1 (log x)/2 (log(log x− t))k−1 t dt (log log x)k−1 · ∫ log x−1 (log x)/2 dt t (log log x)k−1 and using the binomial theorem and the Taylor expansion of log(1− x/c) we get that ∫ (log x)/2 1 (log(log x− t))k−1 t dt = ∫ (log x)/2 1 (log log x+ log(1− t/ log x))k−1 t dt = ∫ (log x)/2 1 (log log x)k−1 +O((log log x)k−2) t dt =(log log x)k +O((log log x)k−1). Therefore we combine the calculations of these integrals to conclude that (k − 1)!(k + 1)σk(x) x ∼ ∫ log x−1 1 (log(log x− t))k−1 t(log x− t) dt ∼ ( 1 + 1 k ) (log log x)k log x . Rearrange terms gives us pik(x) ∼ (x log log x) k k! log x . Because σ1(x) = pi1(x) induction and (4.1.2) proves that σk(x) ∼ (x log log x) k k! log x Similar to the case with natural primes we will be able to prove a result with more uniformity. In the case of natural primes, this is called the Selberg–Sathe formula. In fact, we will use similar methods of proof for our more general result, but first we must show that these methods are actually valid and just as powerful when we change to using Beurling primes. In order to use the same methods we require that our zeta function be analytic to the left of the line σ = 1. To achieve an extended range of analyticity we require that our integer counting function has the asymptotic approximation NB(x) = Ax+O(xθ) (4.1.6) where A > 0 and 0 ≤ θ < 1. We will show that this asymptotic property will imply analyticity to the left of σ = 1 in Section 4.3. 24 4.2 Perron’s Formula for Beurling Primes The first thing that is necessary to prove Theorem 1.3 is Perron’s formula and error bounds on the remainder upon subtraction by a finite integral. That is we want to bound R in the following equation: lim T→∞ 1 2pii ∫ σ0+iT σ0−iT α(s) xs s ds = 1 2pii ∫ σ0+iT0 σ0−iT0 α(s) xs s ds+R(T0). Perron’s formula, in this case, is just a special case of an inverse Mellin transform [11]. To see this, note that the integer counting function is locally continuous (in fact locally constant), and converges in right half plains. Lemma 4.2. Let {an}n∈B be an arithmetic sequence and let α(s) := ∑ n∈B an · n−s be its associated Dirichlet Series. Define σa := inf{σ ∈ R | ∑ n∈B |an|n−σ <∞}, the absolute abscissa of convergence for α(s). If σ0 > max(0, σa) and x > 0, then ∑′ n≤x an = 1 2pii ∫ σ0+iT σ0−iT α(s) xs s ds+R where R = 1 pi ∑ x/2<n<x n∈B an si ( T log x n ) − 1 pi ∑ x/2<n<x n∈B an si ( T log n x ) +O ( 4σ0+x σ0 T ∑ n |an| nσ0 ) (4.2.1) ∑ x/2<n<2x n6=x n∈B |an|min ( 1, x T |x− n| ) + 4σ0 + xσ0 T ∑ n∈B |an|n−σ0 (4.2.2) Proof. The series α(s) is absolutely convergent on the interval [σ0 − iT, σ0 + iT ] so 1 2pii ∫ σ0+iT0 σ0−iT0 α(s) xs s ds = ∑ n∈B an 1 2pii ∫ σ0+iT0 σ0−iT0 (x n )s ds s . Therefore it suffices to evaluate 12pii ∫ σ0+iT σ0−iT y s ds s for y in the four intervals (0, 1/2], [1/2, 1], [1, 2], [2,∞). This is taken care of for us already since it is the same as the integrals that need to be evaluated for the natural prime case. For a reference see [12][section 5.1]. This proves (4.2.1) Now for the less exact but more user friendly bound note that si(x) min(1, 1/x). Also, see that | log n/x| = | log(1 + (n− x)/x)| |x− n|. So, if x/2 ≤ n ≤ 2x, then 25 si(T | log n/x|) min ( 1, 1 T | log n/x| ) min ( 1, 1 T |x− n| ) . Applying this bound to 4.2.1 we get that R ∑ x/2<n<2x n6=x n∈B |an|min ( 1, x T |x− n| ) + 4σ0 + xσ0 T ∑ n∈B |an|n−σ0 . 4.3 The Bound on ζB In our calculations leading up to Theorem 1.3 we will need a bound on ζB(σ + it) in the region where σ > 1− cθ/(2 log t) and |t| > t0 and the region 2 > σ > 1− cθ/(2 log t) and |t| ≤ t0, where t0 is chosen to be slightly away from the pole at s = 1. We need these bounds to evaluate the integral given to us by Perron’s formula in Lemma 4.2. To calculate these bounds we first need to know that ζB is analytic to the left of σ = 1. In [3] there is a generalization of this for any zeta function of a continuous measure, but this simpler fact can be seen by applying the same methods that one would use to show this for the zeta function on natural primes. Start by separating the tail of the zeta function. For σ > 1 ζB(s) = ∑ n∈B n−s = ∑ n≤x n∈B n−s + ∑ n>x n∈B n−s. Now we deal with the tail. Define (u) := NB(u)−Au, where NB(x) is as in (4.1.6). Note that (u) uθ. Now write the tail of the zeta function as∑ n>x n∈B n−s = ∫ ∞ x u−sdNB(u) = ∫ ∞ x Au−sdu− ∫ ∞ x u−sd(u). The first integral can now be evaluated for σ > 1 and, using integration by parts, the second integral converges for σ > θ. Using partial summation on the finite sum ∑ n≤x n∈B n−s gives us that for s 6= 1 ζB(s) = Ax1−s s− 1 + (x) · x −s + ∫ ∞ x (u) · du−s = Ax 1−s s− 1 + (x) · x −s − s ∫ ∞ x (u) · u−s−1du. Since the integrand in the above is an analytic function of s we see that ζB(s) is analytic in any region not containing s = 1 contained in the half plane σ > θ, by the uniqueness of analytic continuation. In particular, 26 when x = 1 we have ζB(s) = A s− 1 − s ∫ ∞ 1 (u) · u−s−1du. (4.3.1) It is possible to determine more information about ζBints(s) to the left of σ = 1. Landau’s methods in [10] showed that ζB has a zero free region to the left of the curve σ = cθlog t for some small constant c. Using this zero free region we can determine a bound for ζ(s) which will be useful in applying Perron’s formula in a later calculation. Lemma 4.3. For s ∈ {σ + it | σ > 1− c(1−θ)2 log t and |t| ≥ t0} we get the bound ζB(s) A log t. For the region {σit | 2 ≥ σ > 1− c(1−θ)2 log t and |t| ≤ t0} we have that ζB(s)z s = Az (s− 1)z +O ( 1 |s− 1|<z−1 ) . One should note that one could make this lemma a corollary to a more general bound on ζ(s) but all we require for Theorem 1.3 are these specific bounds. In particular we calculate bounds for ζ ′ ζ (s) and for | log ζ(s)| in the same regions as above. Proof. Take s = 1 + r + it and assume that r ≥ 1−θlog t . By logarithmic differentiation of ζB(s) with respect to s and the fact that piB is a non-decreasing function we have ∣∣∣∣ζ ′BζB (s) ∣∣∣∣ = ∣∣∣∣∫ ∞ 1− x−s · log x dpiB(x) ∣∣∣∣ ≤ ∫ ∞ 1− x−1−r · log x dpiB(x). Furthermore, by (4.3.1) we see that ∣∣∣∣ζ ′BζB (s) ∣∣∣∣ ≤ ζ ′BζB (1 + r) 1r ≤ log t1− θ . In particular the above holds for s1 = 1 + log t1−θ + it. Now we need to prove the same inequality for smaller values of r. Using Jensen’s inequality and the Borel–Caratheodory lemma, or (50) in [3] for σ ≥ 1 − 1−θ4 we have 27 ζ ′B ζB (s) = ∑ ρ 1 s− ρ +O ( log t 1− θ ) where the sum is taken over zeros of ζB(s) in the disk of radius (1− θ)/2 centered at 1 + it. Looking at this equality for s1 as defined above we get that ∑ ρ 1 s1 − ρ log t 1− θ . Now suppose we have an s = σ + it such that 1− c(1−θ)2 log t ≤ σ ≤ 1 + 1−θlog t then ζ ′B ζB (s)− ζ ′ B ζB (s1) = ∑ ρ ( 1 s− ρ − 1 s1 − ρ ) +O ( log t 1− σ ) again, where the sum is over zeros, ρ, in the disk or radius (1 − θ)/2 centered at 1 + it. For all zeros of ζB(s) in this disk we have |s− ρ| |s1 − ρ|. So upon differencing we get the bound 1 s− ρ − 1 s1 − ρ = s1 − s (s− ρ)(s1 − ρ) 1 |s1 − ρ|2 log t < 1 s1 − ρ. Therefore we have that ζ ′ B ζB (s) log t 1−θ for σ > 1− c(1−θ)2 log t and |t| ≥ t0. Moving along we turn the above work into a bound for log ζB(s), which in turn will provide us with a bound on ζB(s). By the analytic continuation of ζ in (4.3.1) we have that for σ > θ A σ − 1 < ζB(s) < Aσ 1− σ . Therefore if σ > 1 + 1−θlog t then ζB(s) < A ( 1 + log t1−θ ) and furthermore | log ζB(s)| ≤ log log t+O(1/(1− θ)). In particular the same s1 = 1 + 1−θlog t + it satisfies the above bound. As before we take a difference and compute log ζB(s)− log ζB(s1) = ∫ s s1 ζ ′B ζB (w)dw. Now consider an s such that 1− c(1−θ)2 log t ≤ σ ≤ 1 + 1−θlog t . Since we have a big-oh bound on the integrand, in the strip, of log t1−θ we get that | log ζB(s)| ≤ log log t+ O (1/(1− θ)) in the strip as well. Complex analysis tells us that log |ζB(s)| = < log ζB(s). Therefore ζB(s) A log t in whenever σ > 1− c(1−θ)2 log t and |t| > t0. 28 A bound for ζB(s)/s on the region where 1− c(1−θ)2 log t < σ < 2 and |t| < 2 is seen by examining the Laurent series expansion about s = 1. Since ζB(s) has a simple pole at s = 1 and is elsewhere analytic we have that on this region ζB(s)z s = Az (s− 1)z +O ( 1 |s− 1|<z−1 ) . 4.4 The Integrations of ζB Now we are ready to start evaluating some integrals of interest. By our version of Perron’s formula and the subsequent bounds on the error term – Theorem 4.2 – we can use the information gained about ζB through Lemma 4.3 to evaluate the partial summation of dz(n), defined for any z ∈ C. This arithmetic function is defined by ζB(s)z = ∑ n∈B dz(n)n −s. Call Dz(x) = ∑ n≤x dz(n). When k ∈ N the function dk(n) has the simple interpretation that it is the number of all ordered k-tuples (m1, ...,mk) ∈ Bk such that m1 · · ·mk = n. We start by evaluating Dz(x) this when z = ` is an integer. We will extend this to the general case by using the bound, |dz(n)| ≤ d|z|(n) ≤ dR(n) for any integer R ≥ |z|, along with Perron’s formula (4.2.2) and then evaluating a contour integral using Lemma 4.3. First we need some simple calculations. Lemma 4.4. Let P be a set of Beurling primes such that the associated integer counting function is as in (4.1.6). Furthermore define (x) := NB(x)− Ax, that is the error term of the integer summation formula. Then note that (x) xθ. Then we can evaluate∑n≤x n∈B 1 nv for v ∈ (0, 1] as follows: ∑ n≤x n∈B 1 nv = A log x+ CB +O ( xθ−1 ) v = 1 A · θ · x1−θ1−θ +O(θ · log x). v = θ A · v · x1−v1−v + ∫∞ 1− (u) u1+v du+Oθ,v(x θ−v) otherwise where CB = ∫∞ 1− (u) u2 du is the analog for the Euler constant, usually referred to as C0 in the natural prime case. Note that the integral in the case when v 6= 1, θ is a constant if v > θ and would fit into the error term if v < θ. Furthermore the sum ∑ n≤x n∈B loga n n = A loga+1 a+ 1 x+O(xθ−1 loga x) Proof. First consider the case when v = θ and evaluate by converting the sum to a Riemann–Stieltjes integral and applying integration by parts to see that 29 ∑ n≤x n∈B 1 nθ = ∫ x 1− 1 uθ dNB(u) = ∫ x 1− [Au+O(uθ)] θ uθ+1 du = A · θ · x 1−θ 1− θ +Oθ(log x). The case where v ∈ (0, 1) \ θ is calculated very similarly. We see that in this case ∑ n≤x n∈B 1 nv = ∫ x 1− 1 uv dNB(u) = ∫ x 1− [Au+O(uθ)] v uv+1 du = A · v · x 1−v 1− v + ∫ ∞ 1− (u) u1+v +Oθ,v(x θ−v). We calculate the sum of reciprocals or Beurling integers much in the same way as we would for the natural numbers. Now see that ∑ n≤x n∈B 1 n = ∫ x 1− 1 u dNB(u) = ∫ x 1− [Au+ (u)] 1 u2 du = A log u+ ∫ ∞ 1− (u) u2 du+O(xθ−1). Next calculate ∑ n≤x n∈B loga(n) n = ∫ x 1 loga(u) u dNB(u) =A ∫ x 1 loga(u) u du+ ∫ x 1 loga(u) u d(u) =A loga+1 a+ 1 x+ ∫ x 1 loga(u) u d(u) Where (u) := NB(u)−Au. Then repeated integration by parts gives us that 30 ∫ x 1 loga(u) u d(u) xθ−1 loga x+ ∫ x 1 loga(u) u2−θ du xθ−1 loga x. Lemma 4.5. Let ` be a natural number. Given a set of Beurling primes, P , such that the associated set of Beurling integers, B, is as in (4.1.6), then D`(x) = A `x · P`(log x) +O ( x 1− 1−θ 2`−1 ) where P` is a polynomial of degree `− 1, which is independent of the choice of P . Proof. We proceed by induction. First note that D1(x) = NB(x) = Ax + O(xθ). Now we have the following induction hypothesis: for all r < ` Dr(x) = A r · xPr(log x) +O(x1− 1−θ 2r−1 ) where Pr is a polynomial of degree r − 1. Now we use the hyperbola method to write D`(x) = ∑ km≤x k,m∈B D`−1(m) = ∑ k≤y k∈B d`−1 (x k ) + ∑ m≤x/y m∈B d`−1(m)D1 ( x m ) −D1(y)D`−1 ( x y ) = A`x ∑ k≤y k∈B 1 k · P`−1(log x/k) +Ax ∑ m≤x/y m∈B 1 m · d`−1(m) − ( Ay +O ( yθ )) · ( A`−1 x y · P`−1(log x/y) +O ( (x/y) 1− 1−θ 2`−2 )) + O ∑ k≤y k∈B (x k )1− 1−θ 2`−2 + ∑ m≤x/y m∈B d`−1(m) ( x m )1− 1−θ 2`−2 . (4.4.1) As in the case for ` = 2 we can use partial summation (and the induction hypothesis) to evaluate all the sums that appear, expand and cancel terms in the error term and see that when we choose y = √ x we get an essentially minimal error term to get the desired expression for D`(x). To follow this plan we use the previos lemma, but in addition we need to evaluate one more sum. Namely note that 31 ∑ n≤x/y n∈B d`−1(m) m = ∫ x/y u=1 1 u d[D`−1(u)] = D`−1(x/y) x/y + ∫ x/y 1 D`−1(u) u2 du. Use the induction hypothesis and partial summation to see that the above is ∑ n≤x/y n∈B d`−1(m) m = D`−1(x/y) x/y +A ∫ x/y 1 P`−1(u) u du+O (∫ x/y 1 u −1− 1−θ 2`−2 du ) . Repeated integration by parts then shows us that if y = √ x then ∑ n≤x/y n∈B d`−1(m) m = P`(log x) +O ( (x/y) 1−θ 2`−2 ) Where P` is a polynomial of degree `−1. Therefore we see that the main term of (4.4.1) is, what we expect, namely A` · xP`(log x) for P` some `− 1 degree polynomial. To take care of the error term expand, apply Lemma 4.4 and the induction hypothesis to show that the error term of (4.4.1) is xy θ−12`−2 + x1− 1−θ2`−2 y 1−θ2`−1 x1+ 1−θ2`−1 because we choose y = √ x. Corollary 4.6. For dz defined above we have the following bounds on these two sums for any R > |z|: ∑ x 2 <n<2x |dz(n)| ·min ( 1, x T |x− n| ) ARx log(x)−R−2 (4.4.2) and xa T ∑ n∈B |dz(n)|n−a ARx log(x)−R−2 (4.4.3) where T is chosen to be T = exp( √ log x). Proof. Consider the first sum and split it up by summing over the subset, S ⊂ (x/2, x)∩B which are close to x and those far away from x. Specifically set S := {n ∈ B||n− x| ≤ x/(log x)2R−1}. Fist note that 32 ∑ n∈S |dz(n)|min ( 1, x T |x− n| ) ∑ n∈S |dz(n)|. Now we use the fact that |dz(n)| ≤ dR(n) whenever R > |z| is an integer and recall (4.5), our formula for DR(x), set y = (log x)−2R+1 to get that ∑ n∈S |dR(n)| = DR (x+ xy)−DR (x− xy) = AR [x · (PR (log x+ log(1 + y))− PR (log x+ log(1− y)))] +AR · xy (PR(log x+ log(1 + y)) + PR(log x+ log(1− y))) . To calculate a bound on the above we need to examine cancellation of the log part in the above expression. So using a Taylor approximation or two we get that (log x+ log(1 + y))p − (log x+ log(1− y))p = (log x+O(y))p − (log x+O(y))p =(log x)p [(1 +O(y/ log x))p − (1 +O(y/ log x))p] =(log x)p [O(p · y/ log x)] p · (log x)p−2R−2 and if p < R then the above is R · (log x)R−3. Therefore we must have ∑ n∈S dR(n) ARx(log x)−R−2. When we evaluate the sum of the remaining terms of (4.4.2) we are summing over n ∈ B that are far away from x so we use the other possible value of the minimum. ∑ n/∈S |dz(n)|min ( 1, x T |x− n| ) ∑ n/∈S |dz(n)| · x T |x− n| T −1 · (log x)2R+1 ∑ n/∈S |dz(n)| Now using our calculated bound when z is an integer in lemma 4.5 and the fact that |dz(n)| ≤ dR(n) so the above is ARx · (log x)−R−2. 33 Next we turn our attention to the sum in (4.4.3). We have previously seen, in Theorem 4.3, that ζB(a) A log x so this sum can be seen to be xa T ∑ n∈B |dz(n)|n−a x a T (ζB(a))R x exp( √ log x) (A log x)R ARx(log)−R−2. Theorem 4.7. Let R be any positive real number. If x ≥ p1, then uniformly for |z| ≤ R Dz(x) = Az · x(log x)z−1 Γ(z) +O ( x(log x)<z−2 ) . Proof. If a = 1 + 1/ log x then by Lemma 4.2.2 Dz(x)− 1 2pii ∫ a+iT a−iT ζB(s)z · x s s ∑ x 2 <n<2x |dz(n)| ·min ( 1, x T |x− n| ) + xa T ∑ n∈B |dz(n)|n−a. (4.4.4) We must evaluate the integral to get the main term. From Lemma 4.3 we have a bound on ζB(s) which, modulo a constant, is the same as in the natural prime case. Therefore the method of integration will not be any different than in the case for the natural primes. For a reference take [12] Theorem 7.17. Now to deal with the error term in (4.4.4) we simply apply Corollary 4.6. 4.5 The Deduction of Theorem 1.3 The following lemmas give us a way to turn our calculated values for Dz , in Theorem 4.4.4, into bounds on σk(x), the number of elements of B less than x with exactly k prime factors. Lemma 4.8. For a function bz(n) define F (s, z) = ∑ m∈B bz(m)m −s and suppose that the sum ∑ m∈B |bz(m)|(logm)2R+1/m is uniformly bounded for |z| ≤ R. For σ ≥ 1 let cz(n) = bz ? dz(n). Let Cz(x) := ∑ n≤x cz(n) be the summation function of cz . Then Cz(x) = A zF (1, z) Γ(z) x(log x)z−1 +O ( x(log x)<z−2 ) . 34 Proof. First we simply re-write the defining summation of Cz to see that Cz(x) = ∑ n≤x cz(n) = ∑ n≤x ∑ m|n bz(m)dz(n/m) = ∑ m≤x bz(m) ∑ n≤x/m dz(n) = ∑ m≤x/p1 bz(m)Dz(x/m) + ∑ x/p1<n≤x bz(m) where p1 is the smallest prime in P . Note that if a < p1 then Dz(a) = 1. Then using our formula for Dz(x) in Theorem 4.7 we can calculate that Cz(x) = A z z Γ(z) ∑ m≤x/p1 bz(m) m (log(x/m))z−1 +O ∑ m≤x |bz(m)| m (log(2x/m))<z−2 . By splitting the sum at √ x and using our uniform bound on ∑ m∈B |bz(m)|(logm)2R+1/m we see that the error term is in the above expression is = x(log x)<z−2 ∑ m≤√x |bz(m)| m + x(log x)<z−2 ∑ √ x<m≤x |bz(m)| m (log x)2<z−2 x(log x)<z−2. To handle the main term see that when m ≤ √x the binomial theorem gives us that (log(x/m))z−1 = (log x− logm)z−1 = (log x)z−1 +O ( logm · (log x)<z−2 ) . Therefore we can write ∑ m≤x/p1 bz(m) m log(x/m)z−1 =(log x)z−1 ∑ m≤x/p1 bz(m) m +O (log x)<z−2 ∑ m≤√x bz(m) m logm+ (log x)Rz−1 ∑ m> √ x bz(m) m 35 Furthermore, since we assumed that the sum ∑ m∈B |bz(n)|(logm)2R+1/m is uniformly bounded on the disk |z| < R we get that the above is =(log x)z−1F (1, z) +O ( (log x)<z−2 ∑ m∈B bz(m) m (logm)2R+1 ) giving the result. Now we want to use this arithmetic lemma in the case where the arithmetic function bz is chosen to satisfy the following ∑ n∈B bz(n)n −s = F (s, z) = ∏ p∈P ( 1− z ps )−1( 1− 1 ps )z . If we choose R < p1 then for |z| < R we have that ∑ m∈B bz(m)/m = ∏ p∈P ( 1− z p )−1( 1− 1 p )z is uniformly bounded in this range. Therefore, Lemma 4.8 tells us information about a function cz defined by the property ∑ n∈B cz(n)n −s = ζB(n)zF (s, z) = ∏ p∈P ( 1− z ps )−1 = ∑ n∈B zΩ(n)n−s. The information that 4.8 tells us is that the partial summation function of cz has the formula Cz(x) = ∑ k∈N σk(x)z k = Az F (1, z) Γ(z) x(log x)z−1 +O ( x(log x)<z−2 ) . (4.5.1) If k > logp1(x) then there are no Beurling integers, b, with exactly k prime factors such that b < x. So Cz(x) is a polynomial in z and therefore Cauchy’s theorem asserts that σk(x) = 1 2pii ∫ |z|=r Cz(x) zk+1 dz (4.5.2) for r < p1. Now we combine all of the results of this section in order to prove the Selberg–Sathe formula for Beurling primes. Recall the statement of Theorem 1.3 Let P be a set of Beurling primes with corresponding set of integers B such that the integer counting 36 function has the asymptotic NB(x) = Ax+O(xθ) where A > 0 and θ < 1. Suppose that R < p1 and that F (s, z) := ∏ p∈P ( 1− z ps )−1( 1− 1 ps )z . Set G(z) = Az · F (1, z)/Γ(z + 1). Then σk(x) = G ( k − 1 log log x ) x(log log x)k−1 (k − 1)! log x ( 1 + OR ( k (log log x)2 )) uniformly for 1 ≤ k ≤ R log log x. It is perhaps a bit surprising that doubling the number of integers in a set of Beurling integers does not seem to affect the number of primes by much. The next corollary shows us that all the integers are hiding plain sight. The generalization of the Hardy–Ramanujan theorem tells us that almost all, that is 100% of, integers less than x have about log log x prime factors. Alternatively, and perhaps more naturally, consider the system P = P ∪ {a}. For any A > 1 we can chose to add a prime a that is small enough to make NB(x) ∼ Ax, yet adding a single prime will not change the asymptotic formula for piB. Evaluate G(1) = A · F (1, 1)/Γ(2) = A∏p∈P (1− 1p)−1 (1− 1p) = A and see that this is enough to give us the following corollary. Corollary 4.9. If k ∼ log log x then σk(x) ∼ Ax(log log x) k−1 (k − 1)! log x . Before we dive into the proof let’s look at a simple example. Let P be the set of all natural primes excluding a finite set of primes T . Then in this case it is easy to see that our integer counting function counts all natural numbers which don’t contain any t ∈ T in their prime factorizations. Hence, by (1.2.1), we get the formula NB(x) = ∏ t∈T ( 1− 1t ) · x + O(1). Then σk is the counting function of natural numbers less than x with exactly k prime factors, none of which are in T . Set z = (k − 1)/ log log x, and set G(z) = ∏ p∈P ( 1− 1 p )z · ∏ p∈P ( 1− z p )−1 · 1 Γ(1 + z) Theorem 1.3 tells us that for R < p1 we get that the counting function of Beurling integers comprised of exactly k primes, in this case, has the formula σk(x) = G(z) x(log log x)k−1 (k − 1)! log x ( 1 + OR ( k (log log x)2 )) 37 uniformly for k ≤ R log log x. Also Corollary 4.9 tells us that, when k ∼ log log x the number of integers less than x with log log k prime factors, none of which are in T is approximately ∏ p∈T ( 1− 1 p ) x(log log x)k−1 (k − 1)! log x . Proof of Theorem 1.3. When k = 1 we have the classical result proved by Beurling in [1] so we may assume that k > 1. By Lemma 4.8, (4.5.1), and (4.5.2) we have that σk(x) = 1 2pii · x log x · ∫ |z|=r G(z)(log x)zz−kdz +O ( x log x ∣∣∣∣∣ ∫ |z|=r z−k−1(log x)zdz ∣∣∣∣∣ ) . (4.5.3) Now we choose r = (k − 1)/ log log x and see that the error term in the above is x log x · 2pir · r−k−1 = x(log x)r−1 · r−k = x (log x)2 ek−1 (log log x)k (k − 1)k . By Sterling’s formula [20] we get the above to be x(log log x) k (k − 1)!(log x) x(log log x) k−3 (k − 1)! log x . To tackle the integral in the main term just requires some technical maneuvering. See that integration by parts yields E := r 2pii ∫ |z|=r (log x)zz−kdz = 1 2pii ∫ |z|=r (log x)zz1−kdz. So by adding and subtracting the term G(r)2pii ∫ |z|=r(log x) zz−kdz and the termG′(r) ·E in two different ways we see that the integral in (4.5.3) of can be seen to be rewritten 38 12pii ∫ |z|=r G(z)(log x)zz−kdz = G(r) 2pii ∫ |z|=r (log x)zz−kdz + 1 2pii ∫ |z|=r (G(z)−G(r)) (log x)zz−kdz = G(r) 2pii ∫ |z|=r (log x)zz−kdz + 1 2pii ∫ |z|=r ( G(z)−G(r)−G′(r)(z − r)) (log x)zz−kdz. Cauchy’s theorem can be used to show that the first integral is (log log x)k−1/(k − 1)! which gives us the main term for our formula for σk(x). The second integral will now be shown to bounded by our error term in (4.5.3). Consider that we can rewrite the first term in the product of the integrand as G(z)−G(r)−G′(z − r) = ∫ z r (z − w)G′′(w)dw |z − r|2 since G′′(z) is bounded on the region of integration. Following [12][pg. 233] yields we write z = re2piiθ so that the second integral can be seen to be ∫ 1/2 −1/2 (sinpiθ)2e(k−1) cos 2piθdθ. Note the bounds | sinpiθ| ≤ |piθ| and cos 2piθ ≤ 1 − 8θ2 for θ ∈ [−1/2, 1/2]. Therefore, using integration by parts, we can bound the above as r3−kek−1 ∫ ∞ 0 θ2e−8(k−1)θ 2 dθ r3−kek−1(k − 1)−3/2 k(log log x)k−3/(k − 1)! since we chose r = (k − 1)/ log log x. This completes the proof of Theorem 1.3. 39 Bibliography [1] Arne Beurling. Analyse de la loi asymptotique de la distribution des nombres premiers généralisés. i. Acta Mathematica, 68:255–291, 1937. [2] Harold G. Diamond. Chebyshev estimates for Beurling generalized prime numbers. Proc. Amer. Math. Soc., 39:503–508, 1973. [3] Harold G. Diamond, Hugh L. Montgomery, and Ulrike M.A. Vorhauer. Beurling primes with large oscillation. Mathematische Annalen, 334:1–36, 2006. [4] P. Èrdeš. An asymptotic inequality in the theory of numbers. Vestnik Leningrad. Univ., 15(13):41–49, 1960. [5] P. Erdös and M. Kac. The Gaussian law of errors in the theory of additive number theoretic functions. Amer. J. Math., 62:738–742, 1940. [6] Andrew Granville and K. Soundararajan. Equidistribution in Number Theory, volume 237 of NATO Sci. Ser. II Math. Phys. Chem., chapter Sieving and the Erdős-Kac Theorem, pages 15–27. Springer, 2007. [7] R. S. Hall. Beurling generalized prime number systems in which the Chebyshev inequalities fail. Proc. Amer. Math. Soc., 40:79–82, 1973. [8] G. H. Hardy and S. Ramanujan. Proof that almost all numbers n are composed of about log logn prime factors [Proc. London Math. Soc. (2) 16 (1917), Records for 14 Dec. 1916]. In Collected papers of Srinivasa Ramanujan, pages 242–243. AMS Chelsea Publ., Providence, RI, 2000. [9] E. Landau. Sur quelques problèmes relatifs à la distribution des nombres premiers. Bull. Soc. Math. France, 28:25–38, 1900. [10] Edmund Landau. Neuer Beweis des Primzahlsatzes und Beweis des Primidealsatzes. Math. Ann., 56(4):645–670, 1903. [11] HJ. Mellin. Über den Zusammenhang Zwischen den Linearen Differential- und Differenzengleichun- gen. Acta Math., 25(1):139–164, 1902. [12] Hugh L. Montgomery and Robert C. Vaughan. Multiplicative Number Theory. I. Classical Theory, volume 97 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, 2007. [13] Rikard Olofsson. Properties of the Beurling generalized primes. J. Number Theory, 131(1):45–58, 2011. [14] Paul Pollack. On Mertens’ theorem for Beurling primes. To appear in Canadian Math Journal. 40 [15] L. G. Sathe. On a problem of Hardy on the distribution of integers having a given number of prime factors. I. J. Indian Math. Soc. (N.S.), 17:63–82, 1953. [16] L. G. Sathe. On a problem of Hardy on the distribution of integers having a given number of prime factors. II. J. Indian Math. Soc. (N.S.), 17:83–141, 1953. [17] L. G. Sathe. On a problem of Hardy on the distribution of integers having a given number of prime factors. III. J. Indian Math. Soc. (N.S.), 18:27–42, 1954. [18] L. G. Sathe. On a problem of Hardy on the distribution of integers having a given number of prime factors. IV. J. Indian Math. Soc. (N.S.), 18:43–81, 1954. [19] Atle Selberg. Note on a paper by L. G. Sathe. J. Indian Math. Soc. (N.S.), 18:83–87, 1954. [20] J. Sterling. Methodus differentialis: sive, tractatus de sommationes et interpolationes serium infinito- rum,. n.p., 1730. 41
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Extending Erdős-Kac and Selberg-Sathe to Beurling primes with controlled integer counting functions Rupert, Malcolm 2013
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Title | Extending Erdős-Kac and Selberg-Sathe to Beurling primes with controlled integer counting functions |
Creator |
Rupert, Malcolm |
Publisher | University of British Columbia |
Date Issued | 2013 |
Description | In this thesis we extend two important theorems in analytic prime number theory to a the setting of Beurling primes, namely The Erdős–Kac theorem and a theorem of Sathe and Selberg. The Erdős–Kac theorem asserts that the number of prime factors that divide an integer n is, in some sense, normally distributed with mean log log n and variance log log n. Sathe proved and Selberg substantially refined a formula for the counting function of products of k primes with some uniformity on k. A set of Beurling primes is any countable multiset of the reals with elements that tend towards infinity. The set of Beurling primes has a corresponding multiset of Beurling integers formed by all finite products of Beurling primes. We assume that the Beurling integer counting function is approximately linear with varying conditions on the error term in order to prove the stated results. An interesting example of a set of Beurling primes is the set of norms of prime ideals of the ring of integers of a number field. Recently, Granville and Soundararajan have developed a particularly simple proof of the Erdős–Kac theorem which we follow in this thesis. For extending the theorem of Selberg and Sathe much more analytic machinery is needed. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2013-04-19 |
Provider | Vancouver : University of British Columbia Library |
Rights | Attribution-NonCommercial-NoDerivatives 4.0 International |
DOI | 10.14288/1.0071950 |
URI | http://hdl.handle.net/2429/44300 |
Degree |
Master of Science - MSc |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
GraduationDate | 2013-05 |
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UBCV |
Scholarly Level | Graduate |
Rights URI | http://creativecommons.org/licenses/by-nc-nd/4.0/ |
AggregatedSourceRepository | DSpace |
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