A new elementary proof of the Prime Number Theorem
Florian K. Richter Abstract We give a new elementary proof of the Prime Number Theorem by comparing averages of the Möbius function dilated by primes and almost primes.
1. Introduction One of the most fundamental results in mathematics is the Prime Number Theorem, which describes the asymptotic law of the distribution of prime numbers in the integers.
Prime Number Theorem. Let π(N) denote the number of primes smaller or equal to a positive integer N. Then π(N) lim = 1. (1.1) N→∞ N/log N The Prime Number Theorem was conjectured independently by Gauÿ and Legendre towards the end of the 18th century and was proved independently by Hadamard and de la Vallée Poussin in the year 1896. Their proofs are similar and rely on sophisticated analytic machinery involving complex analysis, which was developed throughout the 19th century by the combined eort of many great mathematicians of this era, including Euler, Chebyshev, and Riemann (to name a few). Historically, this method became known as analytic. We refer the reader to [Apo00; Gol73a; Gol73b] for more details on the history behind the analytic proof of the Prime Number Theorem, and to [New80] for an abridged version of it. Even though it was believed for a long time not to be possible, more than 50 years after the rst proof Prime Number Theorem, an elementary proof of it was found by Erd®s and Selberg [Erd49; Sel49]. In this context, elementary refers to methods that avoid using complex analysis and instead rely only on rudimentary facts from calculus and basic arithmetic identities and inequalities. Their approach was based on Selberg's famous fundamental formula: X X log2(p) + log(p) log(q) = 2x log(x) + O(x). (1.2) p6x pq6x We refer to [Gol04] and [SG09] for the history behind the Erd®s-Selberg proof, and to [Lev69] for a streamlined exposition thereof. See also [Sha50] for a short proof of (1.2), and [Dia82] and [Gra10] for more general surveys on this topic. A new way of deriving the Prime Number Theorem from (1.2), which bears similarities to the proof we give in this paper, was recently discovered by McNamara [McN]. A dierent elementary proof of the Prime Number Theorem was found by Daboussi [Dab84], using what he called the convolution method (cf. [Dab89, p. 1]). We refer the reader to the book of Tenenbaum and Mendés France [TM00, Chapter 4] for a friendly rendition of Daboussi's proof. Finally, a third elementary proof, which is dierent from the proofs of Erd®s-Selberg and Daboussi, was provided by Hildebrand [Hil86]. We remark that Hildebrand's proof is not self- contained, as it relies on a non-trivial corollary of the large sieve ([Mon71, Corollary 3.2]) as a starting point. In this paper we provide a new elementary proof of the Prime Number Theorem, using ideas inspired by recent developments surrounding Sarnak's and Chowla's conjecture. With the exception of Stirling's approximation formula, which is used in Section 3 without giving a proof, our proof is self-contained. 1 2. The proof
Let us write [N] to abbreviate the set {1,...,N}. Given a nite set A ⊂ N and a function f : A → C, de ne the Cesàro average of f over A and the logarithmic average of f over A respectively by P 1 X log f(n)/n f(n) := f(n) and f(n) := n∈A . En∈A |A| En∈A P 1/n n∈A n∈A The Prime Number Theorem possesses numerous well-known equivalent formulations. One of which asserts that lim µ(n) = 0, (2.1) N→∞ nE∈[N] where µ: N → {−1, 0, 1} is the classical Möbius function. Expression (2.1) was already known to von Mangoldt [Man97, pp. 849851] and the equivalence between (2.1) and (1.1) is classical and dates back to Landau (see [Lan99, 4] and [Lan11]).1 Many proofs of the Prime Number Theorem work by establishing (2.1) instead of (1.1); we pursue the same strategy here. Let P denote the set of prime numbers. As was already observed by Daboussi [Dab75, Lemma 1] and Kátai [Kát86, Eq. (3.1)], it follows from the Turán-Kubilius inequality that for every ε > 0 there exists such that for all one has s0 s > s0
log (2.2) lim sup E E 1 − p1p|n 6 ε, N→∞ n∈[N] p∈P∩[s] where 1p|n denotes the function that is 1 if p divides n and 0 otherwise. One way of interpreting (2.2) is to say that for large s and almost all n ∈ N the number of primes smaller or equal to s that divide n is approximately equal to P 1/p. Even though (2.2) is commonly viewed p∈P∩[s] as a corollary of the Turán-Kubilius inequality, we remark in passing that its proof is signi cantly shorter and easier (it follows by choosing B = P ∩ [s] in Proposition 2.1 below). An important role in our proof of the Prime Number Theorem will be played by a variant of (2.2), asserting that
log (2.3) lim sup E E 1 − m1m|n 6 ε, N→∞ n∈[N] m∈B for some special types of nite and non-empty subsets B ⊂ N. To clarify which choices of B work, besides B = P∩[s] for large s as in (2.2), we provide an easy to check criterion. In layman's terms, our criterion says that B is good for (2.3) if two integers n and m chosen at random from B have a high likelihood of being coprime. The precise statement is as follows.
Proposition 2.1. Let B ⊂ N be nite and non-empty. Then 1/2 log log log (2.4) lim sup E E 1 − m1m|n 6 E E Φ(n, m) , N→∞ n∈[N] m∈B m∈B n∈B where Φ: N×N → N ∪ {0} is the function Φ(m, n) := gcd(m, n) − 1. Proof. Using the Cauchy-Schwarz inequality and expanding the square gives 2 2 log log (2.5) 1 − m1m|n 6 1 − m1m|n = S1 − 2S2 + 1, nE∈[N] Em∈B nE∈[N] Em∈B where log and log . Since every -th number S1 := En∈[N]E (l1l|n)(m1m|n) S2 := En∈[N]E m1m|n m is divisible by , we havel,m∈B . Thism implies∈B . Similarly, m En∈[N]m1m|n = 1 + O (1/N) S2 = 1 + O(1/N) one can verify that , which gives En∈[N]lm1l|n1m|n = gcd(l, m) + O (1/N)
log log 1 log log 1 S1 = gcd(l, m) + O = 1 + Φ(l, m) + O . El∈B Em∈B N El∈B Em∈B N 1A proof of this equivalence can also be found in most textbooks on number theory, see for example [IK04, p. 32], [TM00, 4.4], or [Apo76, 4.9]. 2 Substituting and log log into (2.5) nishes the S2 = 1 + O(1/N) S1 = 1 + El∈BEm∈BΦ(l, m) + O(1/N) proof of (2.4).
This next proposition guarantees the existence of two nite sets B1 and B2 with a number of useful properties. With the help of this technical result we will be able to nish the proof of the Prime Number Theorem rather quickly.
Proposition 2.2. For all there exist and such that for all and η > 0 ρ ∈ (1, 1 + η] k0 ∈ N k > k0 all there exist two nite, non-empty sets with the following properties: l > 1 B1,B2 ⊂ N (a) all elements in B1 are primes larger than l and all elements in B2 are a product of k distinct primes larger than l; (b) the sets B1 and B2 have the same cardinality when restricted to ρ-adic intervals, by which j j+1 j j+1 we mean |B1 ∩ (ρ , ρ ]| = |B2 ∩ (ρ , ρ ]| for all j ∈ N ∪ {0}; (c) log log Φ(m, n) η for i = 1, 2, where Φ is as in Proposition 2.1. Em∈Bi En∈Bi 6 Proof of the Prime Number Theorem assuming Proposition 2.2. Fix an arbitrary η > 0. Let ρ ∈ (1, 1 + η] and k0 ∈ N be as guaranteed by Proposition 2.2. Let k be an arbitrary even number bigger or equal to k0 and let l ∈ N be any number which the reader should think of as being much larger than k. By Proposition 2.2, we can then nd two nite and non-empty sets B1,B2 ⊂ N satisfying properties (a) through (c). For i ∈ {1, 2} we have
log log 1 E µ(n) − E E µ(mn) = E µ(n) − E E m1m|n µ(n) + O N n∈[N] m∈Bi n∈[N/m] n∈[N] m∈Bi n∈[N]
log 1 6 E E 1 − m1m|n + O N n∈[N] m∈Bi √ 1 6 η + O N , where the last inequality follows from property (c) combined with (2.4). This implies
log √ 1 (2.6) E µ(n) = E E µ(mn) + O η + N , i = 1, 2. n∈[N] m∈Bi n∈[N/m]
Since any element m ∈ Bi has at most k prime factors, each of which is bigger than l, the asymptotic density of numbers that are not coprime to m is very small; in fact it is smaller than k/l. Combining this observation with the multiplicativity of the Möbius function, i.e. µ(mn) = whenever and are coprime, we deduce that µ(n)µ(m) m n En∈[N/m]µ(mn) = En∈[N/m]µ(m)µ(n) + O(k/l). Note that µ(m) = −1 for all m ∈ B1 because B1 consists only of primes, whereas µ(m) = 1 for all m ∈ B2 because any element in B2 is a product of k distinct primes and k is even. Thus, (2.6) becomes log √ k 1 (2.7) E µ(n) = −E E µ(n) + O η + l + N n∈[N] m∈B1 n∈[N/m] in the case i = 1, and log √ k 1 (2.8) E µ(n) = E E µ(n) + O η + l + N n∈[N] m∈B2 n∈[N/m] in the case i = 2. Finally, using that B1 and B2 have the same size within ρ-adic intervals and that is very close to , we conclude that the expressions and ρ 1 Em∈B1 En∈[N/m]µ(n) Em∈B2 En∈[N/m]µ(n) are approximately the same; in fact, log log E E µ(n) = E E µ(n) + O(η). (2.9) m∈B1 n∈[N/m] m∈B2 n∈[N/m] From (2.7), (2.8), and (2.9) it follows that √ k 1 . Since and can be En∈[N]µ(n) = O( η + l + N ) η k/l made arbitrarily small, this completes the proof of (2.1).
3 3. Proof of Proposition 2.2 The starting point for our proof of Proposition 2.2 are Chebyshev-type estimates on the number of primes in intervals. In particular, we derive a rough lower bound on the number of primes in (8x, 8x+1], as well as a rough upper bound the number of primes in (8x, 8x+ε] for small ε.
Proposition 3.1. There are x0 1 and ε0 > 0 such that x > (i) | ∩ (8x, 8x+1]| 8 for all x x , and P > √x > 0 (ii) x x+ε ε 8x for all and . |P ∩ (8 , 8 ]| 6 x x > x0 ε ∈ (0, ε0] The ideas used in the proof of Proposition 3.1 are very much classical and date back to Cheby- shev. We begin with the following lemma.
Lemma 3.2. We have the asymptotic estimate
log(2) x ∩ [x] + O(1). (3.1) P > log x Proof. To obtain (3.1) for arbitrary positive reals x, it is enough to prove it for the special case when x is an even natural number, i.e., x = 2n. In this case, the key to proving (3.1) is to study the prime factorization of the binomial coe cient 2n. Observe that there are many numbers n bm/pc in the interval [m] that are divisible by p. Out of those, there are bm/p2c many divisible by p2, and out of those there are bm/p3c many divisible by p3, and so on. Therefore, if ν is the largest ν 2 ν exponent for which p 6 m, the power of p in m! is equal to bm/pc + bm/p c + ... + bm/p c. In light of this observation, we can calculate the multiplicity of a prime p in the prime factorization of 2n (2n)! using the formula n = (n!)2 X b2n/pic − 2bn/pic, (3.2) 16i6νp where is the largest exponent for which νp . Since i i , we can estimate νp p 6 2n b2n/p c−2bn/p c 6 1 Pνp i i . This yields i=1 b2n/p c − 2bn/p c 6 νp 2n Y pνp (2n)|P∩[2n]|, n 6 6 p∈P∩[2n] which, after taking logarithms, leaves us with
2n log ∩ [2n] log(2n). (3.3) n 6 P Stirling's approximation formula implies that log(m!) = m log(m) − m + O(log m). This can now be used to nish the proof by approximating 2n with in (3.3). log n 2 log(2)n + O(log n) Lemma 3.2 gives a reasonable lower bound on the asymptotic number of primes, which plays a crucial role in the proof for part (i) of Proposition 3.1. For the proof of part (ii) we need an upper bound.
Lemma 3.3. De ne β(σ) := σ log(σ) − (σ − 1) log(σ − 1). Then for all 1 < σ 6 16, β(σ) x ∩ (x, σx] + O(1). (3.4) P 6 log x Proof. For convenience, let us write σx for the quantity bσxc. Observe that every prime number x bxc in the interval divides the number σx. This implies that the number σx is greater or equal (x, σx] x x than Q p. By bounding Q p from below by x|P∩(x,σx]| and taking logarithms, p∈P∩(x,σx] p∈P∩(x,σx] we obtain σx log | ∩ (x, σx]| log(x). (3.5) x > P 4 Similarly as in the proof of Lemma 3.2, we can now use Stirling's approximation formula, log(m!) = m log(m) − m + O(log m), to estimate that σx log = bσxc log(bσxc) − bxc log(bxc) − (bσxc − bxc) log(bσxc − bxc) + O(log x) x = σx log(σx) − x log(x) − (σ − 1)x log((σ − 1)x) + O(log x) = σx log(σ) − (σ − 1)x log(σ − 1) + O(log x) Together with (3.5), this proves (3.4).
Proof of Proposition 3.1. The proof of part (ii) simply follows from Lemma 3.3 (applied with √ σ = 8ε) and the fact that the order of magnitude of ε is much smaller than the order of magnitude of β(8ε) as ε tends to 0. For the proof of part (i), we start by rewriting the interval (8x, 8x+1] in the form (8x, 8x+1] = x x [8x+1]\ S ( 8 , 8 ]. Lemma 3.2 gives the estimate | ∩ [8x+1]| 8x+1/3(x + 1) + O(1), 06n63x 2n+1 2n P > x n+1 x n x n+1 whereas Lemma 3.3 (applied with σ = 2) gives the estimate |P ∩ (8 /2 , 8 /2 ]| 6 8 /2 x + O(1). Therefore x+1 x x+1 x x x+1 8 X 8 8 8 ∩ (8 , 8 ] − + O(x) − + O(x). P > 3(x + 1) 2n+1x > 3(x + 1) x 06n63x This implies that if is su ciently large then x x+1 8x for all . x0 |P ∩ (8 , 8 ]| > x x > x0 Proposition 3.1 is the only number-theoretic component in our proof of Proposition 2.2. The rest of our argument is more combinatorial in nature. The way Proposition 3.1 is used in our proof of Proposition 2.2 is through the following corollary.
Corollary 3.4. There are and such that for all and all there x0 > 1 ε0 > 0 ε ∈ (0, ε0) δ ∈ (0, 1) exists with the following property: For all there exist D = D(ε, δ) ∈ (0, 1) n > x0 x, y ∈ [n, n + 1) with ε4 < y − x < ε such that x y x x+δ 8 y y+δ 8 ∩ (8 , 8 ] D , and ∩ (8 , 8 ] D . P > x P > y Proof. As guaranteed by Proposition 3.1, the number of primes in n n+1 is at least 8n . There- (8 , 8 ] n fore, by the Pigeonhole Principle, for some t ∈ [n, n + 1) the number of primes in (8t, 8t+ε] is at least ε8n/2n. We can then cover the interval (8t, 8t+ε] by K := dε−3e many smaller intervals in the following way:
4 4 4 4 4 (8t, 8t+ε] = 8t, 8t+ε ∪ 8t+ε , 8t+2ε ∪ ... ∪ 8t+(K−1)ε , 8t+Kε . By Lemma 3.3, each of the intervals (8t+iε4 , 8t+(i+1)ε4 ] contains at most O(ε28n/n) many primes, whereas the whole interval (8t, 8t+ε] contains at least O(ε8n/n) many primes. It follows that if ε is chosen su ciently small, say smaller than some threshold ε0, then we can nd two non-consecutive a, b ∈ {0, 1,...,K − 1} such that the intervals (8t+aε4 , 8t+(a+1)ε4 ] and (8t+bε4 , 8t+(b+1)ε4 ] contain at least ε48n many primes. Using the Pigeonhole Principle once more we can then nd for O( n ) every δ ∈ (0, 1) some x ∈ [t + aε4, t + (a + 1)ε4) and some y ∈ [t + bε4, t + (b + 1)ε4) such that the intervals x x+δ and y y+δ contain at least δε48n many primes. Since and are (8 , 8 ] (8 , 8 ] O( n ) a b non-consecutive, we have y − x > ε4, and since x, y ∈ [t, t + ε) we have y − x < ε. Lemma 3.5. Let and be as in Corollary 3.4. Let be arbitrary, let 4 be x0 ε0 ε ∈ (0, ε0) k > d4/ε e arbitrary, and let D = D(ε, δ) be as in Corollary 3.4, where δ := ε/k. Then for all n1, . . . , nk ∈ , and all with there exist such that {n ∈ N : n > x0} x > x0 n1 + ... + nk = bxc x1, . . . , xk ∈ R (I) xi xi+δ xi for all ; P ∩ (8 , 8 ] > D 8 /xi 1 6 i 6 k (II) for all ; |xi − ni| 6 1 1 6 i 6 k (III) x1 + ... + xk ∈ [x, x + ε).
5 Proof. Suppose n1, . . . , nk are natural numbers bigger than x0 with n1 + ... + nk = bxc for some x. In view of Corollary 3.4 there are xi,1, yi,1 ∈ [ni − 1, n) and xi,2, yi,2 ∈ [ni, n + 1) with 4 and 4 , such that xi,j xi,j +δ xi,j and ε < yi,1 − xi,1 < ε ε < yi,2 − xi,2 < ε |P ∩ (8 , 8 ]| > D8 /xi,j yi,j yi,j +δ yi,j for . Therefore, if then (I) |P ∩ (8 , 8 ]| > D8 /yi,j j = 1, 2 xi ∈ {xi,1, yi,1, xi,2, yi,2} and (II) are already satis ed. It remains to show that one can choose for every i and element xi ∈ {xi,1, yi,1, xi,2, yi,2} in a way such that the sequence x1, . . . , xk satis es (III) too. Set (1) and then inductively de ne if and x1 := x1 xi+1 := xi,2 x1 + ... + xi 6 n1 + ... + ni xi+1 := xi,1 otherwise. This results in a sequence x1, . . . , xk whose sum x1 + ... + xk lies between n − 1 and n. Note that for every choice of xi,1 we could have also chosen yi,1 instead, which would 4 increase the sum x1 + ... + xk by a threshold between ε and ε. Similarly, instead of xi,2 we could 4 have chosen yi,2 which would also increase the sum by a number between ε and ε. Therefore, by replacing a certain number of xi this way, we can change the sum x1 + ... + xk and guarantee that it lies between x and x + ε.
1/K Proof of Proposition 2.2. Let η > 0 be given. Choose K ∈ N su ciently large such that 8 < 1+ 1/K η and de ne ρ := 8 . Let x0 and ε0 be as in Corollary 3.4, pick any ε > 0 with ε < min{ε0, η/2}, and set 4 . We claim that and are as desired, meaning that for all and all k0 := d4/ε e ρ k0 k > k0 there exist two nite, non-empty sets satisfying properties (a), (b), and (c). l > 1 B1,B2 ⊂ N To verify this claim, let be arbitrary. By Proposition 3.1 there exists a constant k > k0 such that n n−1 n+1 n for all . Set and de ne C > 1 8 /n 6 |P ∩ (8 , 8 ]| 6 C8 /n n > x0 δ := ε/k k k k k N := 384 C K /D η, where D = D(ε, δ) is as in Corollary 3.4. Choose any s1 ∈ N with . s1 > max{x0, log(l)/ log(8)} First we will construct the set B2. Let A1 be a subset of s1N = {s1n : n ∈ N} with P 1/n N. Then, assuming A has already been found, de ne s := max(A + ... + A ) n∈A1 > i i+1 1 i and let A be any subset of s with P 1/n N. Following this procedure until i+1 i+1N n∈Ai+1 > i = k, we end up with a sequence of nite sets A1,...,Ak with the convenient property that any element in is unique in the sense that if 0 0 with A1 + ... + Ak n1, n1 ∈ A1, . . . , nk, nk ∈ Ak 0 0 then necessarily 0 0 . n1 + ... + nk = n1 + ... + nk n1 = n1, . . . , nk = nk n n+1 According to Proposition 3.1, for all n ∈ A1 + ... + Ak, the interval (8 , 8 ] contains at least 8n/n many primes. Since (8n, 8n+1] splits into exactly K many ρ-adic intervals (ρi, ρi+1], namely (8n, 8n+1] = (ρKn, ρKn+1] ∪ ... ∪ (ρKn+K−1, ρKn+K ], the Pigeonhole Principle implies that for at in in+1 n n+1 n least one in ∈ {Kn, . . . , Kn + K − 1} the interval (ρ , ρ ] ⊂ (8 , 8 ] contains at least 8 /Kn many primes. Note that bin/Kc = n = n1 + ... + nk for some n1 ∈ A1, . . . , nk ∈ Ak. In view of Lemma 3.5 we can therefore nd x1, . . . , xk ∈ R such that (I), (II), and (III) hold with x = in/K. From Property (I) it follows that for every there is a subset of xi xi+δ of i = 1, . . . , k Pni P∩(8 , 8 ] size xi . This allows us to de ne , which has the useful property that bD8 /Kxic Bn,2 := Pn1 ·...·Pnk x1+...+xk x1+...+xk+kδ in/K in/K+2ε in in+1 , because . Let Bn,2 ⊂ (8 , 8 ] ⊂ (8 , 8 ] ⊂ (ρ , ρ ] kδ 6 ε 6 η/2 be any subset of in in+1 of size , which exists because Qk xi Bn,1 P∩(ρ , ρ ] |Bn,2| |Bn,2| 6 i=1 D8 /Kxi 6 n i i +1 n 8 /Kn and |P ∩ (ρ n , ρ n ]| > 8 /Kn. Now de ne [ [ B1 := Bn,1 and B2 := Bn,2.
n∈A1+...+Ak n∈A1+...+Ak
By construction, B1 consists only of primes larger than l and B2 only of numbers that are a product of exactly k distinct primes larger than l. Also by construction, they have the same cardinality when restricted to ρ-adic intervals. This means that properties (a) and (b) are satis ed. It remains to show that and also satisfy property (c). De ne ni−1 ni+1 for all B1 B2 Qni := P ∩ (8 , 8 ] i ∈ {1, . . . , k} and all n ∈ A and set Q := S Q . Since |Q | C8ni /n , we have i i i ni∈Ai ni ni 6 i
k k k ni−1 ni+1 X 1 Y X 1 Y X |P ∩ (8 , 8 ]| k k Y X 1 = 6 n −1 6 8 C . m m 8 i ni m∈Q1·...·Qk i=1 m∈Qi i=1 ni∈Ai i=1 ni∈Ai
6 Similarly, using xi ni , we see that |Pni | > bD8 /Kxic > D8 /2Kni k k X 1 X X 1 D Y X 1 (3.6) = > k k , m p1 · ... · pk 16 K ni m∈B2 n1∈A1,...,nk∈Ak p1∈Pn1 ,...,pk∈Pnk i=1 ni∈Ai which proves P 1/m E P 1/m for E := Dk/128kKkCk. Since B consists only m∈B2 > m∈Q1···Q 1 of primes, a straightforward calculation showsk that log log 1 E E Φ(n, m) 6 1 . m∈B1 n∈B1 P m∈B1 m Moreover, since B and B satisfy property (b), we have P 1/m 1 P 1/m, which 1 2 m∈B1 > ρ m∈B2 together with (3.6), the fact that P 1/n N, and our choice of N proves that B satis es ni∈Ai i > 1 property (c), that is, log log Φ(m, n) η. Finally, to prove that B also satis es property (c), Em∈B1 En∈B1 6 2 note that P is a subset of Q and hence B is a subset of Q · ... · Q . Since also P 1/m ni ni 2 1 k m∈B2 > E P 1/m, we conclude that log log Φ(m, n) 1 log log Φ(m, n). Now m∈Q1···Qk Em∈B2 En∈B2 6 E Em∈Q1···Qk En∈Q1···Qk the key idea is to use the following observation: Whenever U and V are coprime sets (meaning that for all and ) then log log is equal to gcd(m, n) = 1 m ∈ U n ∈ V Em∈UV En∈UV Φ(n, m) log log log log log log log log Φ(n, m) Φ(n, m) + Φ(n, m) + Φ(n, m). Em∈U En∈U Em∈V En∈V Em∈U En∈U Em∈V En∈V From this observation it follows that if log log and log log are smaller Em∈U En∈U Φ(m, n) Em∈V En∈V Φ(m, n) than 1 then log log log log log log Φ(n, m) 3 max Φ(n, m), Φ(n, m) . (3.7) mE∈UV nE∈UV 6 Em∈U En∈U Em∈V En∈V
Since Q1,...,Qk are pairwise coprime sets, we can iterating (3.7) to obtain ( ) log log k 1 E E Φ(n, m) 6 3 max P 1 : 1 6 i 6 k . m∈Q1···Qk n∈Q1···Qk m∈Qi m Since |Q | 8ni /n we get P 1/m 1 P 1/n N/8, which due to our choice of N ni > i m∈Qi > 8 ni∈Ai i > implies log log Φ(m, n) η. Em∈B2 En∈B2 6 References [Apo00] T. M. Apostol. A centennial history of the prime number theorem. In: Number theory. Trends Math. Birkhäuser, Basel, 2000, pp. 114. http://dx.doi.org/10.1007/978-3-0348-7023-8_1. [Apo76] T. M. Apostol. Introduction to analytic number theory. Undergraduate Texts in Mathematics. Springer-Verlag, New York-Heidelberg, 1976, pp. xii+338. http://dx.doi.org/10.1007/978-1-4757-5579-4. [Dab75] H. Daboussi. Fonctions multiplicatives presque périodiques B. In: Journées Arith- métiques de Bordeaux (Conf., Univ. Bordeaux, Bordeaux, 1974). Vol. tome 2425. Soc. Math. France, Paris, 1975, pp. 321324. http://www.numdam.org/item/AST_1975__24-25__321_0. [Dab84] H. Daboussi. Sur le théorème des nombres premiers. In: C. R. Acad. Sci. Paris Sér. I Math. 298.8 (1984), pp. 161164. [Dab89] H. Daboussi. On the prime number theorem for arithmetic progressions. In: J. Number Theory 31.3 (1989), pp. 243254. http://dx.doi.org/10.1016/0022-314X(89)90071-1.
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Florian K. Richter Northwestern University [email protected]
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