Ricardo Guillermo

Barca The

Goldbach Conjecture

Ricardo G Barca
email: rbarca@frba.utn.edu.ar

The Goldbach Conjecture

Introduction

1.1 The sieve method and the Goldbach's problem

In the year 1742 Goldbach wrote a letter to his friend Euler telling him about a conjecture involving prime numbers. Goldbach's conjecture: Every even number greater than 4 is the sum of two primes. The Goldbach Conjecture is one of the oldest unsolved problems in number theory [START_REF] Guy | Unsolved Problems in Number Theory[END_REF]. This conjecture was verified many times with powerful computers, but could not be proven. In May 26, 2013, T. Oliveira e Silva verified the conjecture for n ≤ 4 × 10 17 [START_REF] Oliveira E Silva | Goldbach conjecture verification[END_REF]. Mathematicians had achieved some partial results in their efforts to prove this conjecture. Vinogradov proved, in 1937, that every sufficiently large odd number is the sum of three primes [START_REF] Vinogradov | The Representation of an Odd Number as the Sum of Three Primes[END_REF]. Later, in 1973, J.R. Chen showed that every sufficiently large even number can be written as the sum of either two primes or a prime and the product of two primes [START_REF] Chen | On the representation of a large even integer as the sum of a prime and the product of at most two primes[END_REF]. In 1975, H. Montgomery and R.C. Vaughan showed that 'most' even numbers were expressible as the sum of two primes [START_REF] Montgomery | The exceptional set in Goldbach's problem[END_REF]. Recently, a proof of the related ternary Goldbach conjecture, that every odd integer greater than 5 is the sum of 3 primes, has been given by Harald Helfgott [START_REF] Helfgott | The ternary Goldbach conjecture is true[END_REF].

In this paper we prove (Main Theorem, Section 8) the following: There exists an integer K α > 5 such that every even integer greater than p 2 k (k > K α ) is the sum of two primes. We shall also derive an upper bound for the number K α .

It is well known that one of the principal ways of attacking the problem of the Goldbach's conjecture has been through the use of sieve methods. Viggo Brun [START_REF] Brun | Le crible d'Eratosthène et le théorème de Goldbach[END_REF] was the first to obtain a result, as an approximation to Goldbach's conjecture: Every sufficiently large even integer is a sum of two integers, each having at most nine prime factors. Later, other mathematicians in the area of sieve theory have improved this initial result.

In the context of sieve theory, the sieve method consist in removing elements of a list of integers, according to a set of rules; for instance, given a finite sequence A of integers, we could remove from A those members which lie in a given collection of arithmetic progressions. In the original sieve of Eratosthenes, we start with the integers in the interval [1, x], where x is a positive real number, and sift out all those which are divisible by the primes p < √ x. Therefore, any integer that remains unsifted is a prime in the interval [ √ x, x]. We begin by describing briefly the sieve problem; we use, as far as possible, the concepts and notation of the book by Cojocaru and Ram Murty [START_REF] Cojocaru | An Introduction to Sieve Methods an Their Applications[END_REF], chapters 2 and 5. Let A be a finite set of integers and let P be the sequence of all primes; let z ≥ 2 be a positive real number. Furthermore, to each p ∈ P, p < z we have associated a subset A p of A . The sieve problem is to estimate, from above and below, the size of the set A \ p∈P p<z A p , which consists of the elements of the set A after removing the elements of all the subsets A p . We call the procedure of removing the elements of the subsets A p from the set A the sifting process. The sifting function S(A , P, z) is defined by the equation S(A , P, z) = A \ p∈P p<z A p , and counts the elements of A that have survived the sifting process. Now, let P z be the set of primes p ∈ P, p < z; and for each subset I of P z , denote by

A I = p∈I A p .
Then, the inclusion-exclusion principle gives us

S(A , P, z) = I⊆Pz (-1) |I| |A I | ,
where for the empty set ∅ we have A ∅ = A . We often take A to be a finite set of positive integers, and A p to be the subset of A consisting of elements lying in some congruence classes modulo p.

Using this notation, we can now define formally the sieve of Eratosthenes. Let A = {n ∈ Z + : n ≤ x}, where x ∈ R, x > 1, and let P be the sequence of all primes. Let z = √ x, and

P (z) = p∈P p<z p.
Now, to each p ∈ P, p < z, we associate the subset A p of A , defined as follows: A p = {n ∈ A : n ≡ 0 (mod p)}. Then, when we sift out from A all those elements of every set A p , the unsifted members of A in the interval [ √ x, x] are the integers that are not divisible by primes of P less than z; that is to say, any integer remaining in [ √ x, x] is a prime. Furthermore, if d is a squarefree integer such that d|P (z), we define the set

A d = p|d A p .
So, from the inclusion-exclusion principle we obtain

S(A , P, z) = d|P (z) µ (d) |A d | , (1)
where µ(d) is the Möbius function; and from [START_REF] Halberstam | Sieve Methods[END_REF] it can be derived the well-known formula of Legendre

S(A , P, z) = d|P (z) µ (d) |A d | = d|P (z) µ (d) x d .
In a first instance, the sieve of Eratosthenes is very useful for finding the prime numbers between √ x and x. However, from a theoretical point of view, the experts in sieve theory are interested in estimating for every x the number of integers remaining after the sifting process has been performed.

The use of the Möbius function is a simple way to approach a sieve problem; however, satisfactory results are rather hard to achieve, unless z is very small. We shall illustrate this with the special case given in the book by Halberstam and Richert [START_REF] Halberstam | Sieve Methods[END_REF], Chapter 1, Section 5.

Let A = {n ∈ Z + : n ≤ x}, and let 2 ≤ z ≤ x. As usual in sieve theory, instead of |A | we can use a close approximation X to |A |. Furthermore, for each prime p we choose a multiplicative function w(p) so that (w(p)/p)X approximates to |A p |. Then, for each squarefree integer d we have that (w(d)/d)X approximates to |A d |, and we can write

|A d | = w(d) d X + R d ,
where R d is the remainder term. Then, substituting this into (1),

S(A , P, z) = d|P (z) µ (d) w(d) d X + R d = d|P (z) µ (d) w(d) d X + d|P (z) µ (d) R d . (2)
On the other hand, since w is a multiplicative function, Furthermore, if we impose the conditions |R d | ≤ w(d) and w(p) ≤ A 0 , for some constant A 0 ≥ 1, we get

S(A , P, z) = XW (z) + θ (1 + A 0 ) z .
See [1, Theorem 1.1] for details. Now, taking X = x, w(p) = 1, A 0 = 1, these conditions are satisfied, and we obtain S(A , P, z) ≤ x p∈P p<z

1 - 1 p + 2 z . (3)
From this illustrative case we can see that the error term will be very large provided that z is not sufficiently small compared with x. In spite of this, taking z = log x, the formula in (3) can be used to obtain an elementary upper bound for π(x). See [START_REF] Halberstam | Sieve Methods[END_REF]Ch. 1,(5.8)].

Now, suppose that we express the Goldbach's problem as a sieve problem; it is clear that in order to prove this conjecture what we require is a lower bound for the sifting function. However, there is a well-known phenomenon in sieve theory, called the 'parity barrier' or the 'parity problem', which was explained first by Selberg (see [START_REF] Selberg | The general sieve method and its place in prime number theory[END_REF]). It appears that sieve methods cannot distinguish between numbers with an even number of prime factors and an odd number of prime factors. The parity problem was described briefly by Terence Tao [START_REF] Tao | Open question: The parity problem in sieve theory[END_REF] as follows: 'If A is a set whose elements are all products of an odd number of primes (or are all products of an even number of primes), then (without injecting additional ingredients), sieve theory is unable to provide non-trivial lower bounds on the size of A.' This means that in order to solve the Goldbach's problem we need first to find a suitable sieve, and then to introduce new procedures for estimating the sifting function, very far from the usual methods in the current sieve theory.

A sieve for the Goldbach's problem

Let P be the sequence of all primes; and given p k ∈ P, let m k = p 1 p 2 p 3 • • • p k . From now on, and throughout this paper, for convenience, we take x to be an even integer greater than p 2 4 = 49. Note that if p k is the greatest prime less than √ x, every even number x > 49 satisfies p 2 k < x < p 2 k+1 < m k ; this fact is very important for our purposes, as we shall see later. Now, how can we construct a sieve to tackle the Goldbach's problem? Given a positive even integer x, as we have seen in the previous subsection, using the sieve of Eratosthenes we can get the primes between √ x and x. Assume that among the primes between √ x and x there is at least a prime q such that x -q is also a prime. Then, to attack the Goldbach's problem we need a sieve that sift out all the integers in the interval [1, x] which are divisible by the primes p < √ x, as the sieve of Eratosthenes does, and that additionally sift out, from the primes q remaining in [ √ x, x], all those such that x -q is not a prime.

Then, in order to construct such a sieve, we propose to modify the sieve of Eratosthenes as follows: First, we sift out all those integers n in the interval [1, x] such that n ≡ 0 (mod p), where p < √ x; thus, any integer that remains unsifted is a prime in the interval [ √ x, x]. Next, we sift out all those integers n that remains in [ √ x, x] such that n ≡ x (mod p). It is easy to see that any number that remains unsifted in [ √ x, x] is a prime q such that x -q is not divisible by the primes p < √ x; so, either x -q = 1 or x -q is a prime. Let us define formally this sieve, which we call the Sieve associated with x, or alternatively the Sieve I. Let A = {n ∈ Z + : n ≤ x}. Let P be the sequence of all primes; and let z = √ x. Let

P (z) = p∈P p<z p = m k .
Now, to each p ∈ P, p < z, we associate the subset A p of A , defined as follows: A p = {n ∈ A : n ≡ 0 (mod p) or n ≡ x (mod p)}. Furthermore, if d is a squarefree integer such that d|P (z), we define the set

A d = p|d A p .
In this case, the sifting function S(A , P, z) = A \ p∈P p<z A p counts the primes q in the interval [ √ x, x], such that x -q is also a prime in [ √ x, x]; and furthermore counts 1 and x -1 whenever x -1 is a prime. As in the case of the sieve of Eratosthenes-Legendre, the inclusion-exclusion principle gives us

S(A , P, z) = d|P (z) µ (d) |A d | .
Now, S(A , P, z) > 2 implies that x is the sum of two primes; and if this is proved for all x, the Goldbach's conjecture would be proved. Then, the solution of the Goldbach's problem depends on establishing a positive lower bound for the sifting function. However, we can not hope to find a suitable lower bound by using the usual sieve methods, due to the parity problem, which was already mentioned in this Introduction. So far, all attempts to solve the Goldbach's problem by the usual sieve techniques did not have the expected success. For these reasons, the strategy used in this paper differs quite a lot from the usual approach in sieve theory. In the next subsection we shall begin by introducing another way of formulating a sieve problem.

The sequence of k-tuples of remainders

In this paper we propose to use another formulation for this kind of sieves, which is able to show all the details of the sifting process, and will allow us to obtain a lower bound for the number of elements that remain unsifted. For this purpose, we begin by introducing the notion of sequence of k-tuples of remainders. Let {p 1 , p 2 , p 3 , . . . , p k } be the ordered set of the first k prime numbers. Suppose that for every natural number n we form a k-tuple, the elements of which are the remainders of dividing n by p 1 , p 2 , p 3 , . . . , p k ; so, we have a sequence of k-tuples of remainders. If we arrange these k-tuples from top to bottom, the sequence of k-tuples of remainders can be seen as a matrix formed by k columns and infinitely many rows, where each column is a periodic sequence of remainders modulo p h ∈ {p 1 , p 2 , p 3 , . . . , p k }. It is easy to prove that the sequence of k-tuples of remainders is periodic, and the period is

m k = p 1 p 2 p 3 • • • p k .
Suppose that within the periods of every sequence of remainders modulo p h (a given column of the matrix), we define some (not all) of the remainders as selected remainders, no matter the criterion for selecting the remainders. Consequently, some k-tuples have one or more selected remainders, and other k-tuples do not have any selected remainder. If a given k-tuple has one or more selected remainders, we say that it is a prohibited k-tuple; otherwise we say that it is a permitted k-tuple. We shall define more formally the sequence of k-tuples and related concepts in Section 2. Now, in a general context, a sieve is a tool or device that separates, for instance, coarser from finer particles. Then, given a sieve device we can define a 'sieve problem', for instance, to count the number of finer particles that pass through the sieve device. We can think of a sequence of k-tuples as a 'sieve device', in the sense that when a set of integers is 'fed' into the sieve device (the sequence of k-tuples), it separates the integers associated to permitted k-tuples from integers associated to prohibited k-tuples. The sieve problem, in this case, is to estimate the number of integers that 'pass through' the sieve device; that is, to estimate the number of permitted k-tuples attached to some of the integers in the input set.

Given an even integer x > 49, we formulate the Sieve I (the Sieve associated with x) by means of a sequence of k-tuples as follows. Let P be the sequence of all primes; let z = √ x, and let p k be the greatest prime less than z. With the index k corresponding to the prime p k , we construct the sequence of k-tuples of remainders, where the rules for selecting remainders are the following: If a given k-tuple of the sequence has 0 as an element, or has its hth element equal to the remainder of dividing x by p h ∈ {p 1 , p 2 , p 3 , . . . , p k }, these elements are defined as selected remainders. So, within the periods of every sequence of remainders modulo p h (a given column of the matrix), the remainder 0 is always a selected remainder, and besides, if p h does not divide x, the resulting remainder is a second selected remainder. Let A be the set consisting of the indices n of the sequence of k-tuples that lie in the interval [1, x]. For each p ∈ P, p < z, the set A p ⊂ A consists of the indices n for which the corresponding element in the sequence of remainders modulo p is a selected remainder. Then, the indices of the prohibited k-tuples lying in A are sifted out; and the indices of the permitted k-tuples lying in A remain unsifted. The sifting function is given by the the number of permitted k-tuples whose indices lie in the interval A . In Section 8 we shall define more formally the formulation of the Sieve I based on a sequence of k-tuples.

Remark 1.1. Note that given a k-tuple whose index is n < x, if n ≡ 0 (mod p) or n ≡ x (mod p) for at least one p < √ x, then it is a prohibited k-tuple; and if n ≡ 0 (mod p) and n ≡ x (mod p) for every p < √ x, then it is a permitted k-tuple.

Figure 1

Therefore, given an even integer x ≥ 49 (p 2 k < x < p 2 k+1 ), it is easy to see that the ordered set of k-tuples whose indices lie in the interval [1, x] of the sequence is only an alternative formulation of the Sieve associated with x (the Sieve I), which was described before by using the usual sieve theory notation. We shall prove (Theorem 8.1) that the indices (greater than 1) of the permitted k-tuples lying within [1, x] are primes q such that either x -q is a prime or x -q = 1. Note that this form of the sieve gives us a detailed picture of the sifting process; other reasons for using this formulation for sieves based on a sequence of k-tuples will be explained later.

Example 1.1. Figure 1 illustrates how the Sieve I can be used to find some Goldbach partitions for the even number x = 52. We proceed as follows:

1. We make a list of the primes less than √ 52. We obtain {2, 3, 5, 7}.

2. We compute the remainders of dividing x = 52 by the prime moduli of the list. We obtain {0, 1, 2, 3}.

3. In every k-tuple we select each 0, and also the elements {1, 2, 3}, corresponding to the moduli {3, 5, 7}, respectively. (The selected remainders are circled.)

4. Now, we colour gray the permitted k-tuples. The arrows show the corresponding Goldbach partitions. Note that there is no permitted k-tuple for the partition 47 + 5.

The auxiliary Sieve II

To prove the Main Theorem we need to find a lower bound for the sifting function of the Sieve associated to x, for every even number x > p 2 k , where k is sufficiently large. However, we can see that, no matter the formulation, the Sieve I is a 'static' sieve; that is, given an even number x, we can formulate a specific Sieve I for this even number x. For our purposes, we need a 'dinamic' sieve, which is able of working as x → ∞. Suppose that given x > 49 and using the Sieve I we have a way to compute the number of permitted k-tuples whose indices lie in [1, x]; then, we could prove the Main theorem by constructing a sequence of sieves associated with every even number x > 49. That is, we could construct a sequence where the elements are sequences of k-tuples, each one for every even number x > 49, and compute the number of permitted k-tuples whose indices lie in the interval [1, x] of each sequence of k-tuples. Now, using the Sieve I, the implementation of this idea finds some difficulties. For instance, if x = 50 the Sieve I can be described as follows: Since the greatest prime less than √ 50 is p 4 = 7, we have k = 4; so, we construct the sequence of 4-tuples of remainders. In every 4-tuple of the sequence, if the hth element is 0, or is equal to the remainder of dividing x by p h ∈ {p 1 , p 2 , p 3 , p 4 }, this element is a selected remainder. Let A be the set consisting of the indices of the sequence of 4-tuples that lie in the interval [START_REF] Halberstam | Sieve Methods[END_REF]50].

Suppose that we go to the next even integer x = 52. In this case, we have again k = 4, and the sequence of 4-tuples of remainders is the same as before, but now the set A consists of the indices that lie in [START_REF] Halberstam | Sieve Methods[END_REF]52], and the selected remainders take specific values for x = 52. In addition, as x runs through the even numbers, when x > 121 we have k > 4, because the greatest prime less than √ x will be p k > p 4 = 7. The difficulty resides in the handling of all these variables as x runs through all the even numbers. On the other hand, given k ≥ 4, when x is divisible by a prime p h ∈ {p 1 , p 2 , p 3 , . . . , p k }, the remainder is 0; so, in each sequence of remainders modulo p h (1 ≤ h ≤ k) that form the sequence of k-tuples, there could exist one or two selected remainders within the period of the sequence (if there is only one selected remainder, it is always 0). This is an additional serious difficulty in order to derive a formula for computing the sifting function.

For all these reasons it is preferable to work with a more general kind of sieve, for which the sequence of k-tuples is more 'homogeneous' than that corresponding to the Sieve I, in the sense that in each sequence of remainders modulo p h (1 < h ≤ k) that form the sequence of k-tuples of this new sieve there exist always two selected remainders in every period of the sequence. So, we introduce another sieve, which we call simply the Sieve II. We describe the Sieve II in the form proposed before, by means of a sequence of k-tuples, as follows. Let P be the sequence of all primes; and let p k (k ≥ 4) be a prime of the sequence. With the index k corresponding to the prime p k , we construct the sequence of k-tuples of remainders, where the rules for selecting remainders are the following: In every sequence of remainders modulo p h (1 < h ≤ k) that form the sequence of k-tuples there are always two selected remainders r and r modulo p h ; in the sequence of remainders modulo p 1 = 2 there is only one selected remainder r modulo p 1 . Let B be the set consisting of the indices of the sequence of k-tuples that lie in the interval [1, y], where y is an integer that satisfies y > p k . For each p ∈ P, p ≤ p k , the set B p ⊂ B consists of the indices n for which the corresponding element in the sequence of remainders modulo p is a selected remainder. The indices of the prohibited k-tuples lying in B are sifted out; and the indices of the permitted k-tuples lying in B remain unsifted. The sifting function is defined by the equation

T (B, P, p k ) = B \ p∈P p≤p k B p ,
and counts the number of permitted k-tuples whose indices lie in B. We shall define more formally the Sieve II in Section 2. Remark 1.2. In this case, given a k-tuple whose index is n, if n ≡ r (mod p) or n ≡ r (mod p) for at least one p ≤ p k , where r, r are the selected remainders modulo p, then it is a prohibited k-tuple; and if n ≡ r (mod p) and n ≡ r (mod p) for every p ≤ p k , then it is a permitted k-tuple.

Note that the unsifted elements in B may be or may be not prime numbers; indeed, the Sieve II is a collection of sieves, one for each particular choice of the selected remainders.

n 2 3 5 7 n 2 3 5 7
1 1 1 1 1 • • • • • 2 0 2 2 2 176 0 2 1 1 3 1 0 3 3
• • • • • 210 0 0 0 0 Figure 2
Now, suppose that in the Sieve II we take B = {n : 1 ≤ n ≤ p 2 k }. Given an even number x > 49 that satisfies p 2 k < x < p 2 k+1 , we can construct the sequence of k-tuples associated to the Sieve I; and using the same k, we can construct the sequence of k-tuples associated to the Sieve II. So, we can compare for every even number x > 49 the sifting function of the Sieve I with the sifting function of the attached Sieve II. That is, we can compare the number of permitted k-tuples whose indices lie in the interval [1, x] of the sequence of k-tuples corresponding to the Sieve I, with the number of permitted k-tuples whose indices lie in the interval [1, p 2 k ] of the sequence of k-tuples corresponding to the Sieve II. We shall prove later (Lemma 8.2) for every even number x > 49 that, under the given conditions, the value of the sifting function corresponding to the Sieve I is greater than or equal to the minimum value of the sifting function corresponding to the Sieve II.

Example 1.2. For k = 4 (p k = 7), the period of the sequence of k-tuples is equal to 210. The first 35, and the last 35 of the 4-tuples in the interval [START_REF] Halberstam | Sieve Methods[END_REF]210] (the first period of the sequence), are pictured in Figure 2, for a given choice of selected remainders. The Sieve II is given by the k-tuples whose indices lie in [1, 7 2 ].

We can now construct a sequence indexed by k where every element of this sequence is a sequence of k-tuples; that is, we have a sequence of sequences of k-tuples. In each of these sequences of k-tuples we have a Sieve II, which is given by the ordered set of k-tuples whose indices lie in the interval [1, p 2 k ] of the sequence of k-tuples. So, our problem now is, given the Sieve II, how to compute the number of permitted k-tuples whose indices lie within [1, p 2 k ] (the sifting function). We shall see how the study of the sequences of k-tuples reveals the way to derive a lower bound for this number.

Using the inclusion-exclusion principle for computing the number of permitted

k-tuples in a period of the sequence of k-tuples of the Sieve II Usually, the sieve method consist in operate on the formula given by the inclusion-exclusion principle to obtain bounds for the sifting function, as we have illustrated in the first subsection. In our approach, the starting point is also the inclusion-exclusion principle, but only as a first step towards obtaining a lower bound for the sifting function of the Sieve II. That is, from the formula given by this principle we shall compute the number of permitted k-tuples within a period of the corresponding sequence of k-tuples, as follows.

Let us consider again the Sieve II, but now taking B = {n : 1 ≤ n ≤ m k }; that is, B is now the set of the indices corresponding to the first period of the sequence of k-tuples. Given p ∈ P, 2 < p ≤ p k , we have |B p | = 2m k /p, since p|m k and there are two selected remainders r, r for each modulus p > 2, by definition. Furthermore, given a squarefree integer d such that d|m k , 2 d, the set B d is the intersection of the subsets B p such that p|d (p = 2). Hence,

|B d | = 2 ν(d) d m k (d|m k , 2 d),
where ν(d) is the number of distinct prime divisors of d. Furthermore, we have the identity

d|m k 2 d µ (d) 2 ν(d) d = 2<p≤p k p∈P 1 - 2 p . (4)
On the other hand, the subset B p1 consist of the integers n ∈ B such that n ≡ r (mod p 1 ), where r is the selected remainder for the modulus p 1 in the sequence of k-tuples of the Sieve II. Then |B p1 | = m k /p 1 , since p 1 |m k and there is only one selected remainder for the modulus p 1 , by definition. Furthermore, given a squarefree integer d such that d|m k , 2 | d, the set B d is now the intersection of the subsets B p such that p|d, and one subset B p is B p1 . Hence,

|B d | = 2 ν(d)-1 d m k (d|m k , 2 | d).
Now, by the inclusion-exclusion principle,

T ({n : 1 ≤ n ≤ m k }, P, p k ) = d|m k µ (d) |B d | = d|m k 2 d µ (d) 2 ν(d) d m k + d|m k 2|d µ (d) 2 ν(d)-1 d m k = = d|m k 2 d µ (d) 2 ν(d) d m k - 1 2 d|m k 2 d µ (d) 2 ν(d) d m k = 1 2 d|m k 2 d µ (d) 2 ν(d) d m k .
So, using (4) we can see that the number of permitted k-tuples whose indices lie in the interval [1, m k ] (the first period of the sequence of k-tuples associated to the Sieve II) is given by

T ({n : 1 ≤ n ≤ m k }, P, p k ) = 1 2 m k 2<p≤p k p∈P 1 - 2 p , (5)
whatever the selected remainders r, r (mod p), for every p ∈ P, p ≤ p k . Note that since m k is a multiple of every p ∈ {p 1 , p 2 , p 3 , . . . , p k }, the size of the sets B p , and so the size of the sets B d can be computed exactly. This is the reason why in this case we can obtain a precise result from the inclusion-exclusion principle.

1.6 The structure of the first period of the sequence of k-tuples of remainders Until now, we have arranged the elements of each k-tuple horizontally, from left to right; and we have arranged the k-tuples of the sequence vertically, from top to bottom. Hence, the first period of the sequence of k-tuples can be seen as a matrix, with columns from h = 1 to h = k, and

m k = p 1 p 2 p 3 • • • p k rows. Note that for each h (1 ≤ h ≤ k),
we also have a sequence of h-tuples with period m h = p 1 p 2 p 3 • • • p h , which fits into the period m k a whole number of times.

Suppose that we rotate (for convenience) the entire sequence 90 degrees counterclockwise. Then, the index n of the sequence of k-tuples increases from left to right, and the index k of the elements of each k-tuple increases from the bottom up. Consequently, we can think of the first period of the sequence of k-tuples as a matrix formed by k rows and m k columns. Each row of this matrix, from h = 1 to h = k, is formed by the remainders of dividing the integers from n = 1 to n = m k by the modulus p h . For every n (1 ≤ n ≤ m k ), the corresponding column matrix is the k-tuple of the remainders of dividing n by the moduli p 1 , p 2 , . . . , p k .

Note that if we let k → ∞, the period of the sequence and the size of the involved k-tuples grow simultaneously.

Example 1.3. Figure 3 illustrates the first period of the sequence of 4-tuples pictured in Figure 2, but now arranged horizontally from left to right.

The sequences of k-tuples in general shall be defined more formally in Section 2, but now we need the following definition: In particular, let us consider the sequence of k-tuples associated to the Sieve II. Since this sequence is periodic, it suffices to consider its first period, between n = 1 and n = m k (the interval I[1, m k ]). Note that for p k ≥ 7 (k ≥ 4), the interval I[1, p 2 k ] is completely included within the first period of the sequence of k-tuples. Although this is the interval that interests us, in order to understand the properties of the sequence of k-tuples, and its behaviour as k → ∞, it is necessary to study the whole fundamental period of the sequence, not just the interval I[1, p 2 k ]. The following step in our approach consists of dividing into two parts the first period of the sequence of k-tuples, as follows: the left interval I[1, p

I[p 2 k + 1, m k ] h .
If we think of the first period of the sequence of k-tuples as a matrix, we can see that this matrix has been now partitioned into two blocks: the left block, formed by the columns from n = 1 to n = p 2 k ; and the right block, formed by the columns from n = p 2 k + 1 to n = m k . Each row of the left block is formed by the remainders of dividing the integers from n = 1 to n = p 2 k by the modulus p h (1 ≤ h ≤ k); and each row of the right block is formed by the remainders of dividing the integers from n = p 2 k + 1 to n = m k by the modulus p h .

Recall that within the first period of the sequence of k-tuples (the interval I[1, m k ]), the exact number of permitted k-tuples is given by [START_REF] Chen | On the representation of a large even integer as the sum of a prime and the product of at most two primes[END_REF]. Furthermore, for every h such that 1 ≤ h < k, since the number of permitted h-tuples in a period of the sequence of h-tuples is given also by [START_REF] Chen | On the representation of a large even integer as the sum of a prime and the product of at most two primes[END_REF], and the period m h divides m k , we can compute precisely the number of permitted h-tuples in each interval I[1, m k ] h as well. We can see that for every h (1 ≤ h ≤ k), the number of permitted h-tuples in I[1, m k ] h is the same, whatever the choice of the selected remainders in the sequence of h-tuples. However, within both the left interval I[1, p 2 k ] h and the right interval I[p 2 k + 1, m k ] h , the number of permitted h-tuples could change when the selected remainders in the sequence of h-tuples are changed, because the positions of the permitted h-tuples along the interval I[1, m k ] h are modified.

Note that for every sequence of h-tuples of remainders (1

≤ h ≤ k), the intervals I[1, m k ] h , I[1, p 2 k ] h and I[p 2 k + 1, m k ] h are itself
sieve devices, that separate prohibited h-tuples from permitted h-tuples.

On the other hand, attached to the first period of the sequence of k-tuples there is a k × 2 matrix, where for every h (1 ≤ h ≤ k), the entry in the row h and first column is the number of permitted h-tuples in I[1, p 2 k ] h , and the entry in the row h and second column is the number of permitted h-tuples in

I[p 2 k + 1, m k ] h .
Of course, the entries in the matrix depends on the choice of the selected remainders in the sequence of k-tuples. Note that if we take y = p 2 k in the Sieve II, the sifting function is the entry in the first row and first column of this matrix; that is, to compute the sifting function for the Sieve II we ought to be able to compute this entry in the matrix. Note that this quantity is related to the entry in the first row and second column of the matrix, since the sum of both entries is given by [START_REF] Chen | On the representation of a large even integer as the sum of a prime and the product of at most two primes[END_REF].

A question may have already occurred to the reader at this point: What is the advantage of the formulation of the sieves based in a sequence of k-tuples of remainders? We shall explain the principal reason in what follows.

Let us consider the sequence of k-tuples of the Sieve II, in horizontal position, where k ≥ 4. For a given choice of selected remainders, the interval I[1, m k ] of this sequence is a sieve device, that sift out the prohibited k-tuples that lie in I[1, m k ], and allows to survive the permitted k-tuples in this interval. Furthermore, for every h (1 ≤ h < k) there is a sequence of h-tuples of remainders as well. And the interval I[1, m k ] h of every sequence of h-tuples is also a sieve device, that sift out the prohibited h-tuples and allows to survive the permitted h-tuples in I[1, m k ] h . So, we have decomposed the sifting process into several stages, from h = 1 to h = k, where each 'partial' sieve device contributes to the whole sifting process. Hence, we can study the behaviour of this partial sieve devices to determine the behaviour of the whole sieve; the advantage of this perspective will become apparent in the rest of this section. Note that as h goes from 1 to k, the number of permitted h-tuples decreases, as a result of the sifting process in each stage of the whole sifting process. Of course, there is also a similar structure in the left block and the right block of the first period of the sequence of k-tuples.

The density of permitted k-tuples

In the Sieve II we have taken first the set B = {n : 1 ≤ n ≤ p 2 k }, and so, the sifting function

T ({n : 1 ≤ n ≤ p 2 k }, P, p k ) is equal to the number of permitted k-tuples in the interval I[1, p 2
k ] of the sequence of k-tuples associated to the Sieve II. Note that the evaluation of this sifting function is what we need to solve the Goldbach problem. However, this sifting function depends on the choice of the selected remainders in the sequence of k-tuples associated to the Sieve II, and it can not be computed exactly. The obtaining of a lower bound for this sifting function is the main task that we must perform in this paper.

On the other hand, in the Sieve II we have next taken the set B = {n : 1 ≤ n ≤ m k }; here, the sifting function T ({n : 1 ≤ n ≤ m k }, P, p k ) is equal to the number of permitted k-tuples in the interval I[1, m k ] of the sequence of k-tuples associated to the Sieve II. In this case, the sifting function does not depend on the choice of the selected remainders in the sequence of k-tuples, and it can be computed precisely using [START_REF] Chen | On the representation of a large even integer as the sum of a prime and the product of at most two primes[END_REF].

Then, a natural question arises: How can we take advantage of the exact computation of T ({n : k ], since we know this quantity for the interval I[1, m k ], by [START_REF] Chen | On the representation of a large even integer as the sum of a prime and the product of at most two primes[END_REF]. Certainly, it is unlikely that our assumption on the proportion of permitted k-tuples in these intervals could be true; however, we could say that in some sense this assumption is 'aproximately' true. This suggests the possibility of working with this amount (the proportion of permitted k-tuples in a given interval) to obtain the expected results. Now, assume that for every k the proportion of permitted k-tuples in I[1, p 2 k ] were greater than some constant C > 0; in this case, the number of permitted k-tuples within this interval would be greater than Cp 2 k ; this would imply that the number of permitted k-tuples within I[1, p 2 k ] tends to infinity with k; and, indeed, this is the result that we wish to prove. However, it seems unlikely that this constant exists, since from (5) it follows that the proportion of permitted k-tuples in the interval I[1, m k ] is given by

1 ≤ n ≤ m k }, P, p k )
1 2 k h=2 1 - 2 p h , (6)
which tends slowly to 0, as k → ∞. This fact makes working with this amount (the proportion of permitted k-tuples) not very useful.

For this reason it is more convenient to work with a new quantity, that we call the density of permitted k-tuples, or simply the k-density, which is defined formally in Section 3. It is defined for a given interval as the quotient of the number of permitted k-tuples within this interval and the number of subintervals of size p k . That is, for a given interval, is the average number of permitted k-tuples within the subintervals of size p k . We denote by c k and δ k the number of permitted k-tuples, and the density of permitted k-tuples within the period I[1, m k ] of the sequence of k-tuples, respectively (see Definition 2.9 and Definition 3.3). Since c k does not depend on the choice of the selected remainders in the sequence of k-tuples, neither does δ k depend on that choice. We shall prove later (Lemma 3.2 and Theorem 3.4) that δ k increases and tends to infinity as k → ∞. For some values of k, Table 1 gives c k , the ratio c k /m k , and δ k . Note that given the proportion of permitted k-tuples within a given interval, multiplying by p k we obtain the density of permitted k-tuples within this interval.

Suppose that the minimum value of the density of permitted k-tuples within the left interval I[1, p 2 k ] were greater than some constant C > 0; this would imply that the number of permitted k-tuples within this interval is greater than Cp k , since the number of subintervals of size p k in I[1, p 2 k ] is equal to p k . And this, in turn, would imply that the number of permitted k-tuples in I[1, p 2 k ] tends to infinity, as k → ∞. In other words, the sifting function of the Sieve II would tend to infinity, as k → ∞; and this is the result that we need in order to prove the Main theorem.

Attached to the first period of the sequence of k-tuples there is now another k × 2 matrix, where for every h (1 ≤ h ≤ k), the entry in the row h and first column is the density of permitted h-tuples in I[1, p 2 k ] h , and the entry in the row h and second column is the density of permitted h-tuples in

I[p 2 k + 1, m k ] h .
The entries in this matrix also depends on the choice of the selected remainders in the sequence of k-tuples. Note that the entry in the first row and first column of this new k × 2 matrix, multiplied by p k , is equal to the entry in the first row and first column of the k × 2 matrix described in the preceding subsection. The relationships in the matrix of h-densities, between the elements of the rows, and between the elements of the columns are very important for our purposes, as we shall see later.

Simple explanation of the main ideas

Let us consider again the interval I[1, m k ] (the first period) of the sequence of k-tuples of the Sieve II, in horizontal position, for k sufficiently large. As we have seen before, we can consider the first period of the sequence of k-tuples as a matrix of k rows and m k columns. Recall that for every h (1 ≤ h ≤ k), the rows from 1 to h are part of a sequence of h-tuples; and we shall say that h is the level of this sequence. Note that for k large enough δ h increases between h = 4 and h = k (see Lemma 3.2 and Corollary 3.3); in this case δ k > δ 4 = 1/2 (see Table 1).

Suppose that for every level h (1 ≤ h ≤ k), the permitted h-tuples were placed in positions that follow an approximately regular pattern along the interval I[1, m k ] of the corresponding sequence of h-tuples, whatever the choice of the selected remainders in the sequence of k-tuples. In this case, for each level h, the density of permitted h-tuples in both intervals I[1, p 2 k ] and I[p 2 k + 1, m k ] of the sequence of h-tuples should be close to the density within I[1, m k ]; that is, it should be close to δ h , since m k is a multiple of m h . Therefore, since δ h increases from level h = 4 to level h = k, the density of permitted h-tuples in the interval I[1, p 2 k ] of every sequence of h-tuples should also increase from h = 4 to h = k, whatever the choice of the selected remainders. So, the density of permitted k-tuples in the interval I[1, p 2 k ] of the sequence of k-tuples (no matter the choice of the selected remainders) should be greater than δ 4 . Now, in order to explain in a way as simple as possible how through these ideas can be achieved a favorable outcome in the resolution of the Goldbach's problem, we shall make the following assumption. We assume that there exists an ] of the sequence of k-tuples (for all the choices of the selected remainders) is greater than δ 4 = 1/2 for every k > K α . So, since the interval I[1, p 2 k ] is the union of p k subintervals of size p k , we can see that the quantity of permitted k-tuples in I[1, p 2 k ] (the value of the Sifting function of the Sieve II) must be greater than p k δ 4 = p k /2, for k > K α . Now, we have seen before that for all even number x that satisfies p 2 k < x < p 2 k+1 , the minimum value of the Sifting function of the Sieve II is a lower bound for the Sifting function of the Sieve I. From this it follows that the Sifting function of the Sieve I is also greater than p k /2 for all even number x such that p 2 k < x < p 2 k+1 (k > K α ). And it is easy to see that this fact proves the Goldbach's conjecture for all even number x > p 2 k , where k > K α . We shall prove the existence of K α and we shall give an upper bound for this number.

Periodic sequences of k-tuples

General Notation. We write (a, b) for the greatest common divisor of a and b, if no confusion will arise. In addition, lcm is used as an abbreviation for the least common multiple. Given a set A, we denote by |A| the cardinality of A.

For each a ∈ R, the symbol a denotes the floor function, and the symbol a denotes the ceiling function.

In the Introduction we began by describing a first kind of sieve to attack the Goldbach's problem, which we call the Sieve associated with x (or Sieve I); then, we have introduced the notion of sequence of k tuples of remainders as a new formulation for sieves in general, and for this sieve in particular. The Sieve associated with x (Sieve I) is directly related to the Goldbach's problem; we defer to Section 8 the formal definition of this sieve. On the other hand, we have also described in the Introduction a second sieve more general, which we call the Sieve II. As we have seen in the Introduction, the sequence of k-tuples corresponding to the Sieve II is more homogeneous than that corresponding to the Sieve I, in the sense that in every sequence of remainders modulo p h (1 < h ≤ k) there are always two selected remainders. This fact is very important in order to compute the minimum value of the sifting function of the Sieve II.

The Sieve II is not directly related to the Goldbach's problem, but, as we have seen in the Introduction, the minimum number of permitted k-tuples in the interval I[1, p 2 k ] of the sequence of k-tuples corresponding to the Sieve II (the minimum value of the sifting function of the Sieve II), is a lower bound for the number of permitted ktuples in the interval I [1, x] of the sequence corresponding to the Sieve I (the sifting function of the Sieve I), where p 2 k < x < p 2 k+1 ; we shall prove this fact in Section 8. In this section we define formally the Sieve II; and we shall deal with the properties of this sieve until Section 7.

We begin by defining the concepts of sequence of remainders and sequence of k-tuples of remainders, and other associated concepts.

Definition 2.1. Let P be the sequence of all primes, and consider the subset {p 1 , p 2 , p 3 , . . . , p k } of the first k primes.

(1) Given p h (1 ≤ h ≤ k), we define the periodic sequence {r n }, where r n denotes the remainder of dividing n by the modulus p h . We denote the sequence {r n } by the symbol s h . The period of the sequence is equal to p h . See Example 2.1.

(2) We define the sequence {(r 1 , r 2 , r 3 , . . . , r k ) n }, the elements of which are k-tuples of remainders obtained by dividing n by the moduli p 1 , p 2 , p 3 , . . . , p k . We arrange the sequence of k-tuples of remainders vertically; we usually omit the comma separators in the k-tuples. Then, the sequence of k-tuples can be seen as a matrix formed by k columns and infinitely many rows, where each column of the matrix is a periodic sequence s h (1 ≤ h ≤ k). We call the index k the level of the sequence of k-tuples of remainders. See Example 2.2.

Example 2.1. For the modulus p 3 = 5 we have s 3 = {1, 2, 3, 4, 0, 1, 2, 3, . . .}.

Example 2.2. Table 2 shows the first elements of the sequence of 5-tuples of the remainders of dividing n by {2, 3, 5, 7, 11}. Given a sequence {r n } with prime modulus p k we assign to the remainders r n one of the two following states: selected state or not selected state. Definition 2.3. Given a sequence of k-tuples of remainders, we define a k-tuple to be prohibited if it has one or more selected remainders, and we define it to be permitted if it contains no selected remainders. Definition 2.4. We denote by m k the product

p 1 p 2 p 3 • • • p k .
Proposition 2.1. The sequence of k-tuples of remainders is periodic, and its fundamental period is equal to

m k = p 1 p 2 p 3 • • • p k . Proof. Let s h (1 ≤ h ≤ k)
be the sequences of remainders that form a given sequence of k-tuples. Let m be a multiple of all the primes p 1 , p 2 , p 3 , . . . , p k . The period of every sequence s h is equal to p h ∈ {p 1 , p 2 , p 3 , . . . , p k }. Therefore, for every sequence s h , the remainders are repeated for all the integer intervals of size m , starting from the index n = 1 onward. Since p 1 , p 2 , p 3 , . . . , p k are primes, the product m k is the lcm. Consequently, the fundamental period of the sequence of k-tuples is equal to m k .

So far, we have defined the sequence of k-tuples of remainders without defining any rules for selecting remainders; note that without selected remainders, the sequence of k-tuples does not work as a sieve. Before defining these rules, we shall consider another important question concerning the sequence of k-tuples of remainders. As we have seen in the Introduction, in the case of the Sieve II it will be necessary to deal with the behaviour of the sequence of k-tuples as k increases indefinitely. Consequently, we need two more definitions before defining the rules for selecting remainders. Definition 2.5. Sum of sequences.

Let {p 1 , p 2 , p 3 , . . . , p k } be the set of the first k primes. Let {(r 1 r 2 r 3 . . . r k ) n } be the sequence of k-tuples of the remainders of dividing n by the k prime moduli {p 1 , p 2 , p 3 , . . . , p k }, and let {(r k+1 ) n } be the sequence of the remainders of dividing n by the prime modulus p k+1 . We define the sum {(r 1 r 2 r 3 . . . r k ) n } + {(r k+1 ) n }, of the sequence {(r 1 r 2 r 3 . . . r k ) n } and the sequence {(r k+1 ) n }, to be the sequence of (k + 1)-tuples given by the equation

{(r 1 r 2 r 3 . . . r k ) n } + {(r k+1 ) n } = {(r 1 r 2 r 3 . . . r k r k+1 ) n } ,
and formed by the ordered juxtaposition of each k-tuple of the first sequence with each element (index n modulo p k+1 ) of the second sequence. Definition 2.6. Let P be the sequence of all primes, and let p k ∈ P. Let s k be the sequence of the remainders of dividing n by the modulus p k . Let {s k } be the sequence of sequences s k . We define the series denoted by s k to be the sequence {S k }, where S k denotes the partial sum:

S 1 = s 1 , S 2 = s 1 + s 2 , S 3 = s 1 + s 2 + s 3 , . . . . . . . . . . . . S k = s 1 + s 2 + s 3 + s 4 + • • • + s k ,
and the symbol refers to the formal addition of sequences. In each partial sum S k , the greatest prime modulus p k will be called the characteristic prime modulus of the partial sum S k . The index k will be called the level, and we shall say that S k is the partial sum of level k.

Example 2.3. Table 3 shows the partial sum S 4 and the formal addition of the sequence of remainders s 5 to obtain the partial sum S 5 .

On the one hand, we can look at a given partial sum S k as a sequence indexed by n, of the k-tuples of remainders obtained by dividing n by the moduli p 1 , p 2 , p 3 , . . . , p k . On the other hand, the partial sum S k can be seen as a finite sequence indexed by the set {1, . . . , k} (k ∈ Z + ), of sequences of remainders modulo p h ∈ {p 1 , p 2 , p 3 , . . . , p k }, where the indices {1, . . . , k} increase from left to right. And the series s k is the sequence indexed by k, of the partial sums S k . Now we are ready to define the rules for selecting remainders in the sequences s h (1 ≤ h ≤ k) that make up every partial sum S k of the series s k .

Definition 2.7. Let s h (1 ≤ h ≤ k) be one of the sequences of remainders that form the partial sum S k .

Rule 1. If h = 1, in the sequence of remainders s 1 there will be selected one remainder, the same one in every period of the sequence.

Rule 2. If 1 < h ≤ k, in every sequence of remainders s h there will be selected two remainders, the same two in every period of the sequence.

Example 2.4. Table 4 shows the partial sum of level k = 4, where the selected remainders can be seen marked between the square brackets [ ]. Note that the 4-tuples 1 and 7 are permitted k-tuples.

Properly speaking, a given partial sum S k is a sequence of k-tuples of remainders. However, from now on, when we refer to a given partial sum S k , we mean S k together with the selected remainders, unless we specifically state otherwise. Now we are ready to define formally the Sieve II. Definition 2.8. Let P be the sequence of all primes; and let p k (k ≥ 4) be a prime of the sequence. Let B be the set consisting of the indices of the partial sum S k that lie in the interval [1, y], where y is an integer that satisfies y > p k . For each p = p h ∈ P (1 ≤ h ≤ k), the subset B p of B consists of the indices whose remainder modulo p = p h is one of the selected remainders r or r . The indices of the prohibited k-tuples lying in B are sifted out; and the indices of the permitted k-tuples lying in B remain unsifted. See Remark 1.2. The sifting function

) = B \ p∈P p≤p k B p ,
is given by the the number of permitted k-tuples whose indices lie in the interval B.

Hereafter until the end of the paper, we take B = {n : 1 ≤ n ≤ p 2 k }. In the following theorems we prove some other properties of the partial sums of the series s k , which will be used throughout this paper. Proposition 2.2. Let S k be a given partial sum. Let s k+1 be the sequence of remainders modulo p k+1 . Let r (0 ≤ r < p k+1 ) be one of the remainders modulo p k+1 of the sequence s k+1 . Let n ∈ Z + be the index of a given k-tuple of S k . Then, when we juxtapose the elements of the sequence s k+1 to the right of each k-tuple of S k , we have the following.

(1) If the k-tuple at position n is prohibited, then the (k + 1)-tuple of S k+1 at position n will be prohibited as well.

(2) If the k-tuple at position n is permitted and n ≡ r (mod p k+1 ), then:

(a) The (k + 1)-tuple of S k+1 at position n is prohibited if and only if r is a selected remainder;

(b) The (k + 1)-tuple of S k+1 at position n is permitted if and only if r is not a selected remainder.

Proof. By definition, a given k-tuple is prohibited if it has one or more selected remainders; if it has no selected remainder, the k-tuple is permitted. The proof is immediate. Definition 2.9. For a given partial sum S k , we denote by c k the number of permitted k-tuples within a period of S k . Proposition 2.3. Let S k be a given partial sum. We have:

c k = (p 1 -1)(p 2 -2)(p 3 -2) • • • (p k -2).
Proof. It follows from [START_REF] Chen | On the representation of a large even integer as the sum of a prime and the product of at most two primes[END_REF], by simplifying the expression.

n 2 3 5 7 1 1 1 1 1 2 [0] [2] 2 2 3 1 [0] [3] [3] 4 [0] 1 4 4 5 1 [2] [0] [5] 6 [0] [0] 1 6 7 1 1 2 0 8 [0] [2] [3] 1 9 1 [0] 4 2 10 [0] 1 [0] [3] 11 1 [2] 1 4 12 [0] [0] 2 [5] 13 1 1 [3] 6 14 [0] [2] 4 0 15 1 [0] [0] 1 16 [0] 1 1 2 . . . . .
Lemma 2.4. Let S k be a given partial sum. Let p k be the characteristic prime modulus of the partial sum S k . Let c k be the number of permitted k-tuples within the period of S k . We have

p 2 k = o(c k ).
Proof. Using Proposition 2.3, we have [START_REF] Hardy | An Introduction to the Theory of Numbers[END_REF], it is easy to see that lim k→∞ p 2 k /c k = 0.

p 2 k c k = p 2 k (p 1 -1) (p 2 -2) (p 3 -2) • • • (p k -2) = (7) = 1 (p 1 -1) (p 2 -2) (p 3 -2) • • • (p k-2 -2) p k p k-1 -2 p k p k -2 . Let g k-1 denote the gap p k -p k-1 ; so, p k /(p k-1 -2) = (p k-1 + g k-1 )/(p k-1 -2). By the Bertrand-Chebyshev theorem, we have g k-1 < p k-1 =⇒ (p k-1 +g k-1 )/(p k-1 -2) < 2p k-1 /(p k-1 -2). It follows that lim k→∞ p k /(p k-1 -2) < lim k→∞ 2p k-1 /(p k-1 -2) = 2. Since lim k→∞ p k /(p k -2) = 1, returning to
Lemma 2.5. Let S k be a given partial sum. Let p k be the characteristic prime modulus of the partial sum S k . Let m k be the period of S k . We have

p 2 k = o(m k ).
Proof. Since m k = p 1 p 2 p 3 • • • p k by definition, the proof follows at once from Proposition 2.3 and Lemma 2.4.

Proposition 2.6. The Construction Procedure Let S k and S k+1 be consecutive partial sums of the series s k . Let m k and m k+1 be the periods of S k and S k+1 , respectively. Consider the following procedure. First we take p k+1 periods of the partial sum S k . Next we juxtapose the remainders of the sequence s k+1 to the right of each k-tuple of S k (that is to say, we perform the operation S k + s k+1 ). This produces a whole period of the partial sum S k+1 .

Proof. By Proposition 2.1, the period m k of the partial sum S k is equal to m k = p 1 p 2 p 3 • • • p k . If we repeat p k+1 times the period of the partial sum S k , the total number of k-tuples will be m k p k+1 = p 1 p 2 p 3 • • • p k p k+1 = m k+1 . Thus, when we add the sequence s k+1 , the number of (k + 1)-tuples of S k+1 that we obtain is equal to m k+1 , that is to say, a period of S k+1 .

By the Construction Procedure, to get a period of the partial sum S k+1 , we first take p k+1 periods of the partial sum S k . The following proposition shows that the distribution of the permitted k-tuples that are within the p k+1 periods of the partial sum S k over the residue classes modulo p k+1 is uniform.

Proposition 2.7. The permitted k-tuples within the first p k+1 periods of the partial sum S k are uniformly distributed over the residue classes modulo p k+1 .

Proof. Let c k be the number of permitted k-tuples within a period of S k . Let [y] = [0], [START_REF] Halberstam | Sieve Methods[END_REF], [START_REF] Cojocaru | An Introduction to Sieve Methods an Their Applications[END_REF], . . . , [p k+1 -1] be the residue classes modulo p k+1 . Let n ∈ Z + be the index of a given permitted k-tuple within the first period of the partial sum S k . Thus, within p k+1 periods of the partial sum S k there are p k+1 permitted k-tuples with indices n = m k x + n, where x = 0, 1, 2, 3, . . . , p k+1 -1 represents each period. Because (m k , p k+1 ) = 1, for each residue class [y] the congruence m k x + n ≡ y (mod p k+1 ) has a unique solution x. Therefore, since there are c k permitted k-tuples within the period of S k , it follows that there are c k permitted k-tuples within each residue class modulo p k+1 , and the resulting distribution is uniform.

Corollary 2.8. If there are m consecutive periods of the partial sum S k (including the first), where m is a multiple of p k+1 , the permitted k-tuples within these m periods are also uniformly distributed over the residue classes modulo p k+1 .

Definition and properties of the density of permitted k-tuples

In this section, we define more formally the concept of the density of permitted k-tuples, and we prove that the density of permitted k-tuples within a period of the partial sum S k is increasing and tends to ∞ as k → ∞. The following lemma gives a formula for computing δ k .

Lemma 3.1. We have

δ k = p 1 -1 p 1 p 2 -2 p 2 p 3 -2 p 3 • • • p k-1 -2 p k-1 (p k -2) .
Proof. The size of the period of the partial sum S k is equal to

m k = p 1 p 2 p 3 p 4 • • • p k-1 p k (see Proposition 2.1).
Therefore, the number of subintervals of size p k is equal to (p

1 p 2 p 3 p 4 • • • p k-1 p k )/p k = p 1 p 2 p 3 p 4 • • • p k-1 .
On the other hand, the number of permitted k-tuples within a period of S k is equal to

c k = (p 1 -1)(p 2 -2)(p 3 -2) • • • (p k-1 -2)(p k -2)
, by Proposition 2.3. Consequently, by definition, we obtain

δ k = (p 1 -1) (p 2 -2) (p 3 -2) • • • (p k-1 -2) (p k -2) p 1 p 2 p 3 p 4 • • • p k-1 = = p 1 -1 p 1 p 2 -2 p 2 p 3 -2 p 3 • • • p k-1 -2 p k-1 (p k -2) .
The next lemma shows that δ k is increasing if k > 1.

Lemma 3.2. Let S k and S k+1 be consecutive partial sums of the series s k . If δ k denotes the density of permitted k-tuples within a period of S k , and δ k+1 denotes the density of permitted (k + 1)-tuples within a period of S k+1 , then

δ k+1 = δ k p k+1 -2 p k .
The proof is given in Section 6.

Corollary 3.3. By Lemma 3.2,

1. p k+1 -p k < 2 =⇒ δ k+1 < δ k . 2. p k+1 -p k = 2 =⇒ δ k+1 = δ k . 3. p k+1 -p k > 2 =⇒ δ k+1 > δ k .
δ k = p 1 -1 p 1 p 2 -2 p 2 p 3 -2 p 3 p 4 -2 p 4 p 5 -2 p 5 • • • p k-1 -2 p k-1 (p k -2) .
If we shift denominators to the right, we obtain

δ k = (p 1 -1) p 2 -2 p 1 p 3 -2 p 2 p 4 -2 p 3 p 5 -2 p 4 • • • p k-1 -2 p k-2 p k -2 p k-1 .
By definition,

θ k = p k+1 -p k -2 =⇒ p k+1 -2 = p k + θ k .
Consequently, we can write the expression of δ k as

δ k = 1 2 p 2 + θ 2 p 2 p 3 + θ 3 p 3 p 4 + θ 4 p 4 • • • p k-2 + θ k-2 p k-2 p k-1 + θ k-1 p k-1 = = 1 2 1 + θ 2 p 2 1 + θ 3 p 3 1 + θ 4 p 4 • • • 1 + θ k-2 p k-2 1 + θ k-1 p k-1 = = 1 3 1 + 1 p 1 1 + θ 2 p 2 1 + θ 3 p 3 1 + θ 4 p 4 • • • 1 + θ k-1 p k-1 1 + θ k p k p k p k + θ k . Then lim k→∞ δ k = 1 3 1 + 1 p 1 ∞ k=2 1 + θ k p k lim k→∞ p k p k + θ k . (8)
The infinite product between square brackets diverges if the series

1 p 1 + ∞ k=2 θ k p k (9)
diverges. In the series [START_REF] Montgomery | The exceptional set in Goldbach's problem[END_REF], if p k is the first of a pair of twin primes, by definition we have θ k = 0; otherwise we have θ k ≥ 2. Let ∞ j=1 1/q j denote the series where every prime q j is the first of a pair of twin primes. Since the series of reciprocals of the twin primes converges [START_REF] Brun | où les dénominateurs sont nombres premieres jumeaux est convergente où finie[END_REF], the series ∞ j=1 1/q j also converges. Therefore, the series

∞ k=1 1/p k - ∞ j=1 1/q j diverges, because ∞ k=1 1/p k diverges. By comparison with the series ∞ k=1 1/p k - ∞ j=1 1/q j ,
it follows that the series (9) diverges, because θ k /p k > 1/p k for the terms where θ k > 0. Thus, the infinite product in (8) tends to ∞ as well. On the other hand, by the Bertrand-Chebyshev theorem,

p k < p k+1 < 2p k =⇒ θ k < p k =⇒ p k /(p k + θ k ) > 1/2. Consequently, δ k → ∞ as k → ∞.

4

The average density of permitted k-tuples within a given interval

I[m, n]
Let S k be a given partial sum of the series s k . In Section 3 we showed that, for the interval I[1, m k ] of the partial sum S k (the first period), the density of permitted k-tuples does not depend on the choice of the selected remainders in the sequences s h (1 ≤ h ≤ k) that form the partial sum S k (see Lemma 3.1). However, it is easy to see that this assertion does not hold for all the intervals I[m, n] of the partial sum S k . In this section we prove that, within a given interval I[m, n] of the partial sum S k , the average of the values of the k-density for all the possible choices of the selected remainders is equal to δ k . First, we make some definitions. Definition 4.1. Let s h (1 ≤ h ≤ k) be the sequences of remainders that form the partial sum S k . A given choice of the selected remainders within the period of one of the sequences s h , or within the periods of all the sequences s h (1 ≤ h ≤ k), will be called a combination of selected remainders. We denote by ν h the number of combinations of selected remainders within the period of a given sequence s h . Since, by definition, for the sequences s h (1 < h ≤ k) there are two selected remainders within the period p h ,

ν h = p h 2 . (10)
In the sequence s 1 there is only one selected remainder within the period; then, p 1 = 2 =⇒ ν 1 = 2. We denote by N k the number of combinations of selected remainders within the periods of all the sequences s h (1 ≤ h ≤ k). Then

N k = p 1 1 p 2 2 p 3 2 • • • p k 2 . ( 11
)
Convention. From now on, when we refer to the average density of permitted k-tuples within a given interval I[m, n] of the partial sum S k , we mean that this average is computed taking into account all the combinations of selected remainders in the sequences s h that form the partial sum S k . We use the same convention when we refer to the average number of permitted k-tuples.

Definition 4.2. The operation of Type A. Let s h (1 ≤ h ≤ k) be the sequences of remainders that form the partial sum S k . For h > 1, let r, r (mod p h ) be the selected remainders within the period p h of the sequence s h . We define the operation that changes the selected remainders r, r (mod p h ) to r + 1, r + 1 (mod p h ) to be the Type A operation.

For the sequence s 1 , we also define the operation of changing the selected remainder r (mod p 1 ) to r + 1 (mod p 1 ) to be the operation of Type A.

Example 4.1. Table 5 shows the first period of the sequence of remainders s 4 (p 4 = 7), where initially we select the remainders [START_REF] Halberstam | Sieve Methods[END_REF] and [START_REF] Brun | où les dénominateurs sont nombres premieres jumeaux est convergente où finie[END_REF] and then we apply successively the Type A operation.

Table 5: First period of the sequence of remainders s 4 . Let s h (1 < h ≤ k) be the sequences of remainders that form the partial sum S k . Let r, r (mod p h ) be the selected remainders (in that order), within the period p h of the sequence s h . We define the Type B operation as follows:

n 1 [1] 1 1 1 1 [1] 1 2 2 [2] 2 2 2 2 [2] 3 [3] 3 [3]
1) The remainder r holds selected.

2) We change the other selected remainder r (mod p h ) to r + 1 (mod p h ), r = r + 1.

Example 4.2. Table 6 shows the first period of the sequence of remainders s 4 (p 4 = 7), where initially we selected the remainders [START_REF] Halberstam | Sieve Methods[END_REF] and [START_REF] Cojocaru | An Introduction to Sieve Methods an Their Applications[END_REF], and then we applied successively the Type B operation.

Table 6: First period of the sequence of remainders s 4 . Remark 4.1. Suppose that we choose two consecutive selected remainders r, r within the period of the sequence s h (1 < h ≤ k). So, we have one out of ν h combinations of selected remainders. Repeating the Type A operation ν A h -1 times, we obtain ν A h = p h different combinations of selected remainders. Now, if for each one of these combinations we leave unchanged the selected remainder r, and then we repeat ν B h -1 times the Type B operation, we obtain all the ν h combinations of selected remainders within the period of the sequence s h . This is expressed by the equation We denote by f k+1 the average of f k+1 for all the combinations of selected remainders in the sequence s k+1 (k ≥ 1). For the partial sum S 1 , let f 1 denote the average of f 1 for the 2 combinations of selected remainders in the sequence s 1 .

n 1 [1] [1] [1] [1] [1] [1] 2 [2] 2 2 2 2 2 3 3 [3] 3 3
ν h = p h 2 = p h ! 2! (p h -2)! = p h p h -1 2 = ν A h ν B h .
The following lemma gives a formula for computing the average fraction f k+1 . Lemma 4.1. For k ≥ 1 we have f k+1 = 2/p k+1 . For S 1 we have f 1 = 1/p 1 .

Proof. Let [0], [START_REF] Halberstam | Sieve Methods[END_REF], [START_REF] Cojocaru | An Introduction to Sieve Methods an Their Applications[END_REF], . . . , [p k+1 -1] be the residue classes modulo p k+1 . Let c I k be the number of permitted k-tuples within I[m, n] k . We denote by η 0 , η 1 , η 2 , . . . , η p k+1 -1 the number of permitted k-tuples within I[m, n] k whose indices belong to the residue classes [0], [START_REF] Halberstam | Sieve Methods[END_REF], [START_REF] Cojocaru | An Introduction to Sieve Methods an Their Applications[END_REF], . . . , [p k+1 -1] respectively. Therefore, c I k = η 0 + η 1 + η 2 + • • • + η p k+1 -1 . We wish to compute the average fraction of the permitted k-tuples within the interval I[m, n] k that are converted to prohibited (k + 1)-tuples within the interval I[m, n] k+1 , for all the ν k+1 combinations of selected remainders in the sequence s k+1 (k ≥ 1). Now,

ν k+1 = ν A k+1 ν B k+1 = p k+1 (p k+1 -1) 2 ,
by Remark 4.1. Consequently, we begin by taking the average over the ν A k+1 combinations obtained by Type A operations, and then we take the average of the previous averages over the ν B k+1 combinations obtained by Type B operations.

Step 1. Suppose that we choose two selected remainders r, r within the period of the sequence s k+1 . By Proposition 2.

ν A k+1 i=1 η r + η r c I k ν A k+1 = p k+1 i=1 η r + η r c I k p k+1 = 1/c I k p k+1 -1 r=0 η r + p k+1 -1 r =0 η r p k+1 = 1/c I k c I k + c I k p k+1 = 2 p k+1 .
Step 2. Now, if we take the average over the ν B k+1 = (p k+1 -1)/2 combinations of selected remainders obtained by repeated Type B operations from each one of the combinations obtained before, we obtain

f k+1 = ν B k+1 j=1 2 p k+1 ν B k+1 = 2 p k+1
, because p k+1 does not depend on the index j (1 ≤ j ≤ ν B k+1 ). For the partial sum S 1 , there are two residue classes modulo p 1 = 2 and one selected remainder. Therefore, it is easy to see that f 1 = 1/p 1 . Definition 4.6. It follows from Proposition 2.2 that when we juxtapose the remainders of the sequence s k+1 to the right of each k-tuple of S k , the permitted k-tuples of S k whose indices are not congruent to any of the two selected remainders of s k+1 modulo p k+1 are, as (k + 1)-tuples of S k+1 , still permitted. We denote by f k+1 the fraction of permitted k-tuples within the interval I[m, n] k of S k that are transferred to the interval I[m, n] k+1 of S k+1 as permitted (k + 1)-tuples. For the partial sum S 1 , let f 1 denote the fraction of the permitted 1-tuples within the interval I[m, n] k=1 .

We denote by f k+1 the average of f k+1 for all the combinations of selected remainders in the sequence s k+1 . For the partial sum S 1 , let f 1 denote the average of f 1 for the 2 combinations of selected remainders in the sequence s 1 . Now, using the preceding lemma, we can calculate the average fraction f k+1 .

Lemma 4.2. We have f k+1 = (p k+1 -2)/p k+1 . For S 1 , we have f 1 = (p 1 -1)/p 1 .

Proof. By Proposition 2.2, a given permitted k-tuple within the interval I[m, n] k of S k can be transferred to the interval I[m, n] k+1 of S k+1 either as a permitted (k + 1)-tuple or as a prohibited (k + 1)-tuple. Consequently, f k+1 + f k+1 = 1, and so f k+1 + f k+1 = 1. Therefore, using Lemma 4.1, we obtain

f k+1 = 1 -f k+1 = 1 -2/p k+1 = (p k+1 -2)/p k+1 .
For the partial sum S 1 , we have Finally, using the preceding lemmas, we calculate the average k-density within a given interval I[m, n], and show that it is equal to the k-density within the period of S k .

f 1 = 1/p 1 =⇒ f 1 = (p 1 -1)/p 1 .
c I k = |I [m, n]| p 1 -1 p 1 p 2 -2 p 2 p 3 -2 p 3 • • • p k -2 p k . ( 12
)
Now, the number of intervals of size p k within the interval

I[m, n] is equal to |I[m, n]|/p k . Consequently, by definition, δ I k = c I k |I[m,n]| p k = p k |I [m, n]| c I k . (13)
Therefore, substituting (12) for c I k in (13), and using Lemma 3.1, we obtain

δ I k = p k |I [m, n]| |I [m, n]| p 1 -1 p 1 p 2 -2 p 2 p 3 -2 p 3 • • • p k -2 p k = = p 1 -1 p 1 p 2 -2 p 2 p 3 -2 p 3 • • • (p k -2) = δ k .
5 The density of permitted k-tuples within the interval I[1, n] as n → ∞

Let S k (k > 2) be a partial sum of the series s k . Let p k be its characteristic prime modulus, and let m k be its period. Let δ k be the density of permitted k-tuples within the period of S k . Let I[1, n] (n ≥ m k ) be a given interval of k-tuples of the partial sum S k . We denote by c I k the number of permitted k-tuples, and by δ I k the k-density in I [1, n]. In this section we shall show that δ I k converges to δ k as n → ∞. First, we make a definition. The following lemma gives us a formula for the k-density in the interval I[1, n].

Lemma 5.1. We have

δ I k = n m k m k n δ k + p k c n .
Proof. By definition,

δ I k = c I k n p k .
Since n/m k is the number of times that the period of S k fits in the interval I[1, n], the interval

I[1, n/m k m k ] is that part of I[1, n]
whose size is a multiple of m k , and so n/m k m k /p k is the number of subintervals of size p k within this part of the interval I[1, n]. Consequently, multiplying by the k-density in the period of S k , we obtain

c η = n m k m k p k δ k .
Since c I k is the number of permitted k-tuples within I[1, n], we have c I k = c η + c . Then

δ I k = c I k n p k = c η + c n p k = n m k m k p k δ k + c n p k = n m k m k n δ k + p k c n .
Remark 5.1. Let c k be the number of permitted k-tuples within the period of S k . By definition,

δ k = c k m k /p k =⇒ c k = δ k m k /p k .
Now, using the formula from the preceding lemma, we find lower and upper bounds for the k-density within the interval

I[1, n]. Lemma 5.2. Let I[1, n] (n ≥ m k ) be an interval of k-tuples of a given partial sum S k . For k > 2, n m k m k n m k m k + (m k -1) δ k < δ I k < n m k + 1 m k n m k m k + 1 δ k . (14)
Proof.

Step 1. We first consider the case where n is not a multiple of m k . By Lemma 5.1,

δ I k = n m k m k n δ k + p k c n . (15)
To obtain bounds for δ I k , we proceed as follows. We begin by obtaining bounds for c . By Remark 5.1, δ k m k /p k is the number of permitted k-tuples within the period of S k . Since by assumption n is not a multiple of m k , it is easy to see that

0 ≤ c ≤ δ k m k /p k . ( 16
)
Next, we obtain bounds for the denominator in (15); since n is not a multiple of m k ,

n/m k m k + 1 ≤ n ≤ n/m k m k + (m k -1) . ( 17
)
Step 2. We obtain a lower bound for δ I k . If we replace the denominator in ( 15) by the upper bound in (17),

n m k m k n m k m k + (m k -1) δ k + p k c n m k m k + (m k -1) ≤ δ I k . ( 18
)
Note that if n is equal to the upper bound in (17), the size of the incomplete period differs from the period m k by one. On the other hand, it is easy to check, using Proposition 2.3, that within the period of the partial sum S k (k > 2) there is more than one permitted k-tuple. It follows that if n is equal to the upper bound in (17), then there is at least one permitted k-tuple within the incomplete period of I[1, n], and so c > 0. Therefore, if we replace c in (18) by the lower bound in (16),

n m k m k n m k m k + (m k -1) δ k < δ I k . ( 19
)
Step 3. We now obtain an upper bound for δ I k . If we replace the denominator in ( 15) by the lower bound in (17),

δ I k ≤ n m k m k n m k m k + 1 δ k + p k c n m k m k + 1 . ( 20
)
Note that if n is equal to the lower bound in (17), the size of the incomplete period is equal to 1, and so there can not be more than one permitted k-tuple within the incomplete period of I [1, n]. On the other hand, we saw in Step 2 that for a level k > 2, there is more than one permitted k-tuple within the period of the partial sum S k . It follows that if n is equal to the lower bound in (17), then c ≤ 1 < δ k m k /p k . Therefore, if we replace c in (20) by the upper bound in (16),

δ I k < n m k m k n m k m k + 1 δ k + p k δ k m k p k n m k m k + 1 = n m k + 1 m k n m k m k + 1 δ k . ( 21
)
Step 4. Now we complete the proof. Suppose that n is a multiple of m k . Then, the density of permitted k-tuples within the interval I[1, n] will be equal to the density within the period of S k ; that is, δ I k = δ k . Since the lower bound in ( 19) is less than δ k , and the upper bound in ( 21) is greater than δ k , we conclude that for every interval I[1, n] of the partial sum S k (k > 2), the inequalities ( 19) and ( 21) are always satisfied, and the lemma is proved.

Remark 5.2. It is easy to check that in ( 14) the upper bound is decreasing, and the lower bound is increasing, as n → ∞.

Finally, we show that the k-density in the interval I[1, n] of a given partial sum S k tends to δ k as the size n of the interval increases. Proposition 5.3. Let S k (k > 2) be a given partial sum of the series s k . As n → ∞, the density δ I k converges to δ k , whatever the combination of selected remainders in the sequences s h that form the partial sum S k .

Proof. Using the inequalities of Lemma 5.2, if we take limits as n → ∞,

lim n→∞ n m k m k n m k m k + (m k -1) δ k < lim n→∞ δ I k < lim n→∞ n m k + 1 m k n m k m k + 1 δ k .
Now, dividing the numerator and denominator by n/m k , we obtain

lim n→∞ m k m k + (m k -1) n m k δ k < lim n→∞ δ I k < lim n→∞ 1 + 1 n m k m k m k + 1 n m k δ k .
Since for a given level k, the values m k and δ k are constants, as n → ∞ we have n/m k → ∞, and the lower and upper bounds tend to δ k . This implies that δ I k converges to δ k as n → ∞.

6 The k-density within the intervals I[1, p 2 k ] and

I[p 2 k + 1, m k ]
Let S k (k ≥ 4) be a given partial sum of the series s k . In this section, we shall subdivide the interval I[1, m k ] of S k into two parts, and shall establish the relationship between the density of permitted k-tuples within one part and the density of permitted k-tuples within the other part. We begin by introducing some terminology and notation. Definition 6.1. Let S k and S k+1 be consecutive partial sums of the series s k . We use the notation p k → p k+1 or alternatively k → k + 1 to denote the transition from level k to level k + 1. For the level transition p k → p k+1 , we call the difference p k+1 -p k the order of the transition. Definition 6.2. When we juxtapose the remainders of the sequence s k+1 to the right of each k-tuple of S k , by Proposition 2.2, a given permitted k-tuple of S k , whose index is congruent to a selected remainder of s k+1 modulo p k+1 , is converted to a prohibited (k + 1)-tuple of S k+1 . In that case, we say that at the level transition k → k + 1 one permitted k-tuple is removed.

Let s h (1 ≤ h ≤ k) be the periodic sequences of remainders that form the partial sum S k . Let m h be the period of every partial sum S h from level h = 1 to level h = k. Let c h be the number of permitted h-tuples, and let δ h be the h-density within the period of every partial sum S h (1 ≤ h ≤ k). Definition 6.3. If we write the index n of the sequences s h from top to bottom, and the level h from left to right (see Table 2) we say that the partial sum S k is in vertical position. Now, suppose that the partial sum S k is in vertical position, and we rotate it 90 degrees counterclockwise. Then, the index n of the sequences s h increases from left to right, and the level h increases from the bottom up. In this case, we say that the partial sum S k is in horizontal position.

For every partial sum S h from level h = 1 to level h = k in horizontal position, let us consider the interval I[1, m k ] h , whose size is the period m k of S k . Remark 6.1. Using Proposition 2.1, it is easy to check that the period of the partial sum S 1 is equal to m 1 = p 1 = 2. On the other hand, by Proposition 2.3, within every period of S 1 we have only one permitted 1-tuple. Therefore, the interval I[1, m k ] 1 of the partial sum S 1 is divided into subintervals of size m 1 = 2, each one containing one permitted 1-tuple. The position of the permitted 1-tuple is the same within every subinterval, and is determined by the selected remainder in the sequence s 1 . Note that the positions of consecutive permitted 1-tuples in the partial sum S 1 differ by two. Remark 6.2. By the preceding remark, the positions of the permitted 1-tuples show a regular pattern along the interval I[1, m k ] 1 of the partial sum S 1 . However, when we add the sequences s h from level h = 2 to level h = k, the selected remainders in each sequence s h remove permitted (h -1)-tuples from the partial sum S h-1 . Consequently, we obtain an interval I[1, m k ] k where the permitted k-tuples are spread along the interval, in positions that show an irregular pattern. Note that if we change the combination of selected remainders in the sequences s h (1 ≤ h ≤ k), within the interval I[1, m k ] k some permitted k-tuples 'disappear', and other permitted k-tuples 'appear', although the number of permitted k-tuples within the interval I[1, m k ] k of the partial sum S k does not change (see Proposition 2.3).

The following lemma gives us the number of permitted h-tuples within the interval I[1, m k ] h of every partial sum S h where h < k. Lemma 6.1. Let S k (k ≥ 4) be a given partial sum of the series s k . Let us consider the interval I[1, m k ] h in every partial sum S h , from level h = 1 to level h = k.

For any given partial sum S h (h < k), the number of permitted h-tuples within the interval

I[1, m k ] h is equal to c h p h+1 p h+2 • • • p k .
Proof. Choose a level h < k. By definition, we have

m k = p 1 p 2 p 3 • • • p h p h+1 p h+2 • • • p k = m h p h+1 p h+2 • • • p k . That is, the size of the interval I[1, m k ] h of the partial sum S h is equal to p h+1 p h+2 • • • p k times
the period m h of the partial sum S h . Consequently, it is easy to see that the number of permitted h-tuples within the interval

I[1, m k ] h is equal to c h p h+1 p h+2 • • • p k .
Remark 6.3. By Proposition 2.3 and Lemma 6.1, the number of permitted h-tuples within the interval

I[1, m k ] h (1 ≤ h ≤ k)
does not depend on the combination of selected remainders in the sequences s h that form the partial sum S h ; therefore, neither does the density of permitted h-tuples within this interval. Furthermore, since the size of I[1, m k ] h is multiple of m h , the density of permitted h-tuples within the interval I[1, m k ] h is equal to δ h (the density of permitted h-tuples within the period m h of the partial sum S h ). Now, let us denote by c h the number of permitted h-tuples within the interval I[1, m k ] h of every partial sum S h (1 ≤ h ≤ k), which is computed using Proposition 2.3 and Lemma 6.1. We have a question at this point: What is the behaviour of c h as h goes from level 1 to level k? This behaviour can be described as follows.

Remark 6.4. For every partial sum S h (h < k), suppose that we juxtapose the remainders of the sequence s h+1 to each h-tuple of S h . By Proposition 2.2, the permitted h-tuples within the interval I[1, m k ] h whose indices are included in two residue classes modulo p h+1 are removed by the selected remainders within the sequence s h+1 ; and the permitted h-tuples whose indices are not included in these residue classes are transferred to level h + 1 as permitted (h + 1)tuples within the interval I[1, m k ] h+1 of the partial sum S h+1 , whatever the combination of selected remainders in the sequence s h+1 . Since for every level h < k the size of the interval I[1, m k ] h is a multiple of p h+1 , by Proposition 2.7 and Corollary 2.8, the permitted h-tuples within the interval I[1, m k ] h of S h are distributed uniformly over the residue classes modulo p h+1 . Therefore, for each level h < k, a fraction 2/p h+1 of the permitted h-tuples within the interval I[1, m k ] h of S h have been removed, and a fraction (p h+1 -2)/p h+1 have been transferred to level h + 1 as permitted (h + 1)-tuples within the interval I[1, m k ] h+1 of S h+1 , whatever the combination of selected remainders in the sequence s h+1 .

The next lemma follows at once from the preceding remark. Lemma 6.2. Let h and h + 1 be consecutive levels, where 1 ≤ h < k. We have

c h+1 = c h p h+1 -2 p h+1 .
Let us examine now the behaviour of δ h as h goes from level 1 to level k. Since the selected remainders of the sequences s h+1 remove permitted h-tuples within the interval I[1, m k ] h of the partial sum S h , at each level transition h → h + 1, the number of permitted h-tuples decreases as the level increases from h = 1 to h = k. However, by Lemma 3.2 and Corollary 3.3, the h-density within the interval I[1, m k ] h of the partial sum S h grows at each transition p h → p h+1 of order greater than 2, because to compute the h-density we count the permitted h-tuples within subintervals of size p h , which grow by more than 2, overcompensating for the permitted h-tuples removed. If p h → p h+1 is a level transition of order 2, the h-density does not change, because the increase in the size p h is compensated for by the permitted h-tuples removed. (Note that p 1 → p 2 is the only level transition where δ h decreases.) Therefore, the h-density increases between h = 1 and h = k if we choose k so large that there are a sufficient number of level transitions of order greater than 2 between h = 1 and h = k.

We have the following lemma: Lemma 6.3. Let h and h + 1 be consecutive levels, where 1 ≤ h < k. Then

δ h+1 = δ h p h+1 -2 p h . ( 22
)
Proof. By Lemma 6.2, the number of permitted (h+1)-tuples in the interval I[1, m k ] h+1 is given by c h (p h+1 -2)/p h+1 . On the other hand, in the interval I[1, m k ] h+1 there is m k /p h+1 subintervals of size p h+1 . Then, by definition

δ h+1 = c h (p h+1 -2) /p h+1 m k /p h+1 .
Multiplying numerator and denominator by p h , using Lemma 6.1, and simplifying, we obtain

δ h+1 = p h (c h (p h+1 -2) /p h+1 ) p h (m k /p h+1 ) = c h m k /p h p h+1 -2 p h = c h m h /p h p h+1 -2 p h .
So, by definition,

δ h+1 = δ h p h+1 -2 p h .
Now, the proof of Lemma 3.2 is immediate.

Proof of Lemma 3.2. It follows at once from Remark 6.3 and the preceding lemma. Now, if we 'cut' the first period of S k into two parts, between the indices p 2 k and p 2 k + 1, we obtain a left-hand subinterval and a right-hand subinterval. k + 1, m k ] h . See Figure 4. As we have seen in the Introduction, the first period of the sequence of k-tuples can be seen as a matrix, with m k columns and k rows. Each row of this matrix (from h = 1 to h = k), is formed by the remainders of dividing the integers from n = 1 to n = m k by the modulus p h . In addition, this matrix has been partitioned into two blocks: the Left block formed by the columns from n = 1 to n = p 2 k ; and the Right block formed by the columns from n = p 2 k + 1 to n = m k . Each row of the Left block is formed by the remainders of dividing the integers from n = 1 to n = p 2 k by the modulus p h ; and each row of the Right block is formed by the remainders of dividing the integers from n = p 2 k + 1 to n = m k by the modulus p h .

Level h = k I 1, p 2 k h=k ∪ I p 2 k + 1, m k h=k = I [1, m k ] h=k Level h = h I 1, p 2 k h=h ∪ I p 2 k + 1, m k h=h = I [1, m k ] h=h Level h = 1 I 1, p 2 k h=1 ∪ I p 2 k + 1, m k h=1 = I [1, m k ] h=1
I[p 2 k + 1, m k ] h .
Although the number of permitted h-tuples within the interval I[1, m k ] h of every partial sum S h (1 ≤ h ≤ k) does not change if we choose another set of selected remainders, the positions of the permitted h-tuples along the period of S h will be changed. Then, it seems reasonable to expect that some permitted h-tuples will be transferred from the

Left interval I[1, p 2 k ] h to the Right interval I[p 2 k + 1, m k ] h
, or vice versa, because the size of the Left interval and the size of the Right interval are not a multiple of m h . Hence, the number of permitted h-tuples within the Left interval I[1, p 2 k ] h and within the Right interval I[p 2 k + 1, m k ] h is determined by the combination of selected remainders in the sequences s h that form the partial sum S h . Definition 6.6. For a given partial sum S h (that is to say, a partial sum where we have a given combination of selected remainders in the sequences that form the partial sum S h ), we use the notation δ L k h to denote the density of permitted h-tuples within the Left interval I[1, p 2 k ] h , and we use the notation δ R k h to denote the density of permitted h-tuples within the Right interval

I[p 2 k + 1, m k ] h .
By Remark 6.3, the h-density within the interval I[1, m k ] h does not depend on the combination of selected remainders in the sequences s h that form the partial sum S h . However, the transfer of some permitted h-tuples from the Left interval I[1, p 2 k ] h to the Right interval I[p 2 k + 1, m k ] h , or in the opposite direction, when we change the combination of selected remainders, brings about changes in the h-density within both intervals. The crossing of some permitted h-tuples from I[1, p

δ R k h within I[p 2 k + 1, m k ] h is also equal to δ h . Hence δ L k h > δ h ⇐⇒ δ R k h < δ h , (23)
δ L k h < δ h ⇐⇒ δ R k h > δ h . Definition 6.7. We often call δ L k h (δ R k h ) the true h-density to distinguish it from the average δ h within the Left interval I[1, p 2 k ] h (the Right interval I[p 2 k + 1, m k ] h ).
The following lemma shows that for every partial sum

S h (1 ≤ h ≤ k), the h-density within the Left interval I[1, p 2 k ] h can not be equal to the h-density within the Right interval I[p 2 k + 1, m k ] h . Lemma 6.4. Let S k (k ≥ 4) be a given partial sum of the series s k . Let us consider the interval I[1, m k ] h (whose size is the period m k of S k ), the Left interval I[1, p 2 k ] h and the Right interval I[p 2 k + 1, m k ] h , in every partial sum S h from level h = 1 to level h = k.
Let us denote by m h the period of the partial sum S h , and by c h the number of permitted h-tuples within a period of the partial sum S h (1 ≤ h ≤ k).

For every partial sum S h we have

δ L k h = δ R k h . Proof.
Step 1. By Remark 6.1, the positions of consecutive permitted 1-tuples in the partial sum S 1 differ by 2. It follows that the number of permitted h-tuples in every Left interval

I[1, p 2 k ] h (1 ≤ h ≤ k) is less than the size of the interval. In symbols, c L k h < p 2 k (1 ≤ h ≤ k).
Step 2. Let us consider a given partial sum S h , where 1 ≤ h < k. Consider the number of permitted h-tuples in the Left interval I[1, p 2 k ] h , denoted by c L k h , and the number of permitted h-tuples in the Right interval

I[p 2 k + 1, m k ] h , denoted by c R k h . By Lemma 6.1, we have c L k h + c R k h = c h p h+1 p h+2 • • • p k ; so, if c L k h is a multiple of p k , then c R k h
is a multiple of p k as well. In this case, (c L k h /p k ) is a whole number; and so (c

L k h /p k )/p k , is a reduced fraction, since c L k h < p 2 k , by Step 1.
Hence, it is easy to check that this fraction is not equal to (c

R k h /p k )/(m k-1 -p k ), since (c R k h /p k
) is also a whole number, and m k-1 -p k is not a multiple of p k . On the other hand, if c L k h is not a multiple of p k , then c L k h /p 2 k is a reduced fraction; and it is easy to check that this fraction is not equal to

c R k h /(m k -p 2 k ), since m k -p 2 k is not a multiple of p 2 k . In either case, we proved that c L k h /p 2 k is not equal to c R k h /(m k -p 2 k ).
Step 3. Now, let us consider the partial sum S k . Consider the number of permitted k-tuples in the Left interval I[1, p 2 k ], denoted by c L k k , and the number of permitted k-tuples in the interval I[1, m k ] (complete period of S k ), denoted by c k . By Proposition 2.3 we have

c k = (p 1 -1)(p 2 -2)(p 3 -2) • • • (p k -2). Now, if c L k k is not a multiple of p k , we can see that c L k k /p 2
k is a reduced fraction; and it is easy to check that this fraction can not be equal to c k /m k , since m k is a squarefree integer.

On the other hand, if c L k k is a multiple of p k , then (c L k k /p k ) is a whole number; and (c L k k /p k )/p k , is a reduced fraction, since c L k k < p 2 k ,
c k m k = (p 1 -1) (p 2 -2) (p 3 -2) • • • (p k-1 -2) (p k -2) p 1 p 2 p 3 • • p k-1 p k = = p 1 -1 p 1 p 2 -2 p 2 p 3 -2 p 3 • • • p k-1 -2 p k-1 p k -2 p k ,
and shifting the denominators to the right, we obtain

c k m k = p 2 -2 p 1 p 3 -2 p 2 • • • p k-1 -2 p k-2 p k -2 p k-1 1 p k .
Note that p 1 can not be canceled with any numerator of the fractions in parentheses, since all these are odd integers. Thus, it is easy to check that this fraction is not equal to the reduced fraction (c L k k /p k )/p k . In either case, the proportion of permitted k-tuples in the interval

I[1, p 2 k ], given by c L k k /p 2 k , is not equal to the proportion of permitted k-tuples in the interval I[1, m k ], given by c k /m k . Thus, if c L k k /p 2 k > c k /m k , it must be c R k k /(m k -p 2 k ) < c k /m k ; and vice versa; this implies c L k k /p 2 k = c R k k /(m k -p 2 k ).
Step 4. We prove the lemma. From Steps 2 and 3, for every partial sum S h (1

≤ h ≤ k) it follows that c L k h /p 2 k = c R k h /(m k -p 2 k ); multiplying by p h we obtain p h c L k h /p 2 k = p h c R k h /(m k -p 2 k ); and so δ L k h = δ R k h .
Even though the increase of the number of permitted h-tuples within one interval is equal to the decrease of the number of permitted h-tuples within the other interval, the increase of the h-density within one interval is not equal to the decrease of the h-density within the other interval. This is due to their being more subintervals of size p h within

I[p 2 k + 1, m k ] h than within I[1, p 2 k ] h , for k > 3.
The following lemma gives the relationship between the h-density within I[1, p 2 k ] h and the h-density within

I[p 2 k + 1, m k ] h . First, a definition: Definition 6.8. Let S h be the partial sums from level h = 1 to level h = k (k ≥ 4). Let I[1, p 2 k ] h be the Left interval, and let I[p 2 k + 1, m k ] h be the Right interval, in every partial sum S h (1 ≤ h ≤ k). For a given partial sum S h (1 ≤ h ≤ k), let δ L k
h be the density of permitted h-tuples within the Left interval I[1, p 2 k ] h , and let δ R k h be the density of permitted h-tuples within the Right interval

I[p 2 k + 1, m k ] h . We use the notation {δ L k h } to denote the set of values of δ L k
h , and we use the notation {δ R k h } to denote the set of values of δ R k h , for all the combinations of selected remainders in the sequences that form the partial sum S h . Lemma 6.5. There is a bijective function

f h : {δ L k h } → {δ R k h } such that f h (x) = δ h -(x -δ h ) p 2 k m k -p 2 k , and
f -1 h (x) = δ h + (δ h -x) m k -p 2 k p 2 k .
Proof. For a given level h (1 ≤ h ≤ k), if we change the combination of selected remainders in the partial sum S h , some permitted h-tuples will be transferred from the Left interval I[1, p 2 k ] h to the Right interval I[p 2 k + 1, m k ] h , or vice versa, as we have seen before. So, it is easy to see that there is a set of values of the number of permitted h-tuples within the Left interval, and a set of values of the number of permitted h-tuples within the Right interval. However, there exists a one-to-one correspondence between both sets, since the number of permitted h-tuples within the interval I[1, m k ] h is the same, whatever the combination of selected remainders in the sequences s h that form the partial sum S h , by Proposition 2.3 and Lemma 6.1. It follows that there is also a one-to-one correspondence between the set of values of δ L k h , and the set of values of δ R k h . So, for a given level h (1 ≤ h ≤ k), we can define a bijective function

f h : {δ L k h } → {δ R k h }. Now,
(δ h -δ R k h )(m k -p 2 k )/p h .
Since the number of permitted h-tuples entering the Left interval must be equal to the number of permitted h-tuples coming out of the Right interval,

δ L k h -δ h p 2 k p h = δ h -δ R k h m k -p 2 k p h =⇒ δ R k h = δ h -δ L k h -δ h p 2 k m k -p 2 k .
Therefore, we have a bijective function

f h : {δ L k h } → {δ R k h }, such that f h (x) = δ h -(x -δ h )(p 2 k /(m k -p 2 k )
), and it is easy to check that f

-1 h (x) = δ h + (δ h -x)(m k -p 2 k )/p 2 k .
7 The sifting function of the Sieve II 7.1 The behaviour of the h-density within the Right interval as k → ∞

In this section we shall establish a lower bound for the minimum value of the sifting function of the Sieve II, for k sufficiently large. However, before achieving this result, we need to establish a lower bound for the k-density within the interval I[1, p 2 k ] of the partial sum S k , where k is large enough. Now, for reasons that will be clear later, we begin by studying the behaviour of the h-density (1 ≤ h ≤ k) within the Right block of the partition of the first period of S k ; the following example illustrates this behaviour for a small level k.

Example 7.1. Let S k be a partial sum of the series s k . Suppose that we take first k = 4, and then we let k → ∞. We can see that, as the level k increases, for every partial sum S h from h = 1 to h = k, the size of the Right interval I[p 2 k + 1, m k ] h grows very fast, since p 2 k = o(m k ), by Lemma 2.5. Note that for h = 1 there is one permitted 1-tuple within every period of size p 1 = 2 of the partial sum S 1 , by Remark 6.1. So, as the size of the Right interval I[p 2 k + 1, m k ] 1 increases, the number of permitted 1-tuples grows very fast as well, and the density of permitted 1-tuples becomes more and more closely to the average δ 1 . Therefore, we can reach a level k not too large (for example k = 8 (p k = 19)) such that the distribution of the permitted 1-tuples that are within I[p 2 k + 1, m k ] 1 over the residue classes modulo p 2 = 3 is not far from uniform. So, the fraction of permitted 1-tuples within the Right interval of S 1 that are transferred to the Right interval of S 2 as permitted 2-tuples is approximately (p 2 -2)/p 2 (see Remark 6.4).

However, as h goes from level 1 to level k (where k is this level not too large), the number of permitted h-tuples that are within the Right interval decreases (see Remark 6.2), and p h increases; in addition, the number of combinations of selected remainders within the sequences that form every partial sum S h increases as well. See [START_REF] Oliveira E Silva | Goldbach conjecture verification[END_REF] and [START_REF] Rey Pastor | Mathematical Analysis[END_REF]. Therefore, as h goes from level 1 to level k -1, for some combinations of selected remainders, the distribution of the permitted h-tuples that are within I[p 2 k + 1, m k ] h over the residue classes modulo p h+1 becomes far from uniform. So, for these combinations of selected remainders, the fraction of permitted h-tuples within the Right interval of S h that are transferred to the Right interval of S h+1 as permitted (h + 1)-tuples moves away from (p h+1 -2)/p h+1 (see Remark 6.4), and consequently, the values of δ R k h moves away from δ h , as h goes from level 1 to level k (see Lemma 6.3). Now, we consider in the following two remarks the case where k is larger than in the previous case. k + 1, m k ] h ; furthermore, from the second condition it follows that c k p 2 k > p k . Note that in this case, c h p h for every h < k, since p h < p k and c h > c k . Hence, it is easy to see that for each level from h = 1 to h = k -1, the distribution of the permitted h-tuples that are within the Right interval over the residue classes modulo p h+1 will be almost uniform, whatever the combination of selected remainders in the sequences s h that form the partial sum S k ; so, the fraction of the permitted h-tuples within the Right interval of S h that are transferred to the Right interval of S h+1 as permitted (h + 1)-tuples will be very close to the average fraction (p h+1 -2)/p h+1 (see Remark 6.4 and Lemma 6.2). Therefore, for every level transition h → h + 1 (1 ≤ h < k), the relationship between δ R k h and δ R k h+1 will be very close to the relationship between δ h and δ h+1 given by Lemma 6.3 (see the proof of this lemma).

By the preceding remark and what we have seen in this remark we conclude that, if k satisfies the conditions required in Remark 7.1, for each level from h = 1 to h = k the values of δ R k h will be very close to δ h , whatever the combination of selected remainders in the sequences s h that form the partial sum S k .

The following lemma shows that as k → ∞, the true h-density within the Right interval I[p

δ R k h = c R k h (m k -p 2 k ) /p h (1 ≤ h ≤ k). (24)
Step 2. Let us denote by m h the period of the partial sum S h , and by c h the number of permitted h-tuples within a period of the partial sum S h . For every level from h = 1 to h = k, let c h be the number of permitted h-tuples within the interval I[1, m k ] h of the partial sum S h . Using Lemma 6.1, we obtain

c 1 = c 1 p 2 p 3 • • • p k , (25)
c 2 = c 2 p 3 p 4 • • • p k , . . . c h = c h p h+1 p h+2 • • • p k , . . . c k = c k .
Note that c h increases as the level decreases from h = k to h = 1 (see Proposition 2.3). For every level from

h = 1 to h = k, since I[1, m k ] h = I[1, p 2 k ] h ∪ I[p 2 k + 1, m k ] h ,
c h -p 2 k (m k -p 2 k ) /p h ≤ δ R k h ≤ c h (m k -p 2 k ) /p h (1 ≤ h ≤ k).
Extracting the common factors c h and m k , we obtain

c h m k /p h 1 -p 2 k /c h 1 -p 2 k /m k ≤ δ R k h ≤ c h m k /p h 1 1 -p 2 k /m k . Now, by definition, m k = p 1 p 2 p 3 • • • p h p h+1 p h+2 • • • p k = m h p h+1 p h+2 • • • p k .
Then, using (25), we can simplify both sides:

c h m h /p h 1 -p 2 k /c h 1 -p 2 k /m k ≤ δ R k h ≤ c h m h /p h 1 1 -p 2 k /m k .
By definition,

δ h = c h m h /p h .
Therefore, for every partial sum S h from level h = 1 to level h = k, whatever the combination of selected remainders in the sequences s h that form the partial sum S k , we have the bounds

δ h 1 -p 2 k /c h 1 -p 2 k /m k ≤ δ R k h ≤ δ h 1 1 -p 2 k /m k . ( 26
)
Step 3. Now, let > 0 be a given small number, and let N ≥ 12. For level h = k, from (26), we obtain

δ k 1 -p 2 k /c k 1 -p 2 k /m k ≤ δ R k k ≤ δ k 1 1 -p 2 k /m k .
On the one hand, p 2 k = o(m k ), by Lemma 2.5. On the other hand, c k = c k by (25), and so, by Lemma 2.4,

p 2 k = o(c k ).
Besides, it follows from Proposition 2.3 that c k < m k . Therefore, we can take N large enough that for level k > N ,

δ k - 2 < δ k 1 -p 2 k /c k 1 -p 2 k /m k ≤ δ R k k ≤ δ k 1 1 -p 2 k /m k < δ k + 2 , (27)
at level h = k.

Step 4. Now, the rightmost inequality in (27) implies

δ k 1 1 -p 2 k /m k -1 < 2 .
For a given level h < k, since k > N ≥ 12 by assumption, it is easy to verify using Lemma 3.2 and Corollary 3.3 (see Table 1), that δ h < δ k . Hence,

δ h 1 1 -p 2 k /m k -1 < 2 =⇒ δ h 1 1 -p 2 k /m k < δ h + 2 . ( 28
)
Step 5. The leftmost inequality in (27) implies

δ k 1 - 1 -p 2 k /c k 1 -p 2 k /m k = δ k p 2 k /c k -p 2 k /m k 1 -p 2 k /m k < 2 .
For a given level h < k, since k > N ≥ 12, we have δ h < δ k (see Step 4). On the other hand, c k = c k < c h < m k , where 1 ≤ h < k (see (25) and Proposition 2.3). Hence, replacing δ k by δ h , and c k by c h , we obtain

δ h 1 - 1 -p 2 k /c h 1 -p 2 k /m k < 2 =⇒ δ h - 2 < δ h 1 -p 2 k /c h 1 -p 2 k /m k . ( 29
)
Step 6. We now prove the lemma. By (26), ( 27), (28), and (29), for k > N we can write

δ h - 2 < δ h 1 -p 2 k /c h 1 -p 2 k /m k ≤ δ R k h ≤ δ h 1 1 -p 2 k /m k < δ h + 2 ,
for every level from h = 1 to h = k. This implies |δ R k h -δ h | < for every level from h = 1 to h = k (k > N ), whatever the combination of selected remainders in the sequences s h that form every partial sum S k .

7.2 A lower bound for the sifting function of the Sieve II Now we are ready to establish a lower bound for the true density of permitted k-tuples within the Left interval I[1, p

for every k > N .

Step 2. Let us consider the Right interval

I[p 2 k + 1, m k ] 5 of S 5 ; and let R = L p 2 k m k -p 2 k . (31)
By Lemma 6.5 and Remark 7.4, we can write

δR k 5 = δ 5 -δL k 5 -δ 5 p 2 k m k -p 2 k = δ 5 + δ 5 -δL k 5 p 2 k m k -p 2 k ,
and in view of (30), for k > N we obtain

δR k 5 < δ 5 + L p 2 k m k -p 2 k .
So, by (31) we have

δR k 5 < δ 5 + R , (32)
for k > N .

Step 3. Note that the inequality (32) is only valid for h = 5, whenever k > N . However, by Lemma 7.1 there exists an integer N > N such that δR k h < δ h + R if k > N , for every partial sum S h from level h = 1 to level h = k; in particular, for h = k (k > N ) we have δR k k < δ k + R , and from this we obtain

R > δR k k -δ k . (33)
On the other hand, again by Lemma 6.5 and Remark 7.4, we can write

δL k k = δ k + δ k -δR k k m k -p 2 k p 2 k = δ k -δR k k -δ k m k -p 2 k p 2 k . (34)
In view of (33), if we replace ( δR k k -δ k ) by R in the last member of (34), we obtain

δL k k > δ k -R m k -p 2 k p 2 k (k > N ),
and using again the relationship (31), for k > N we have

δL k k > δ k -L .
Step 4. We prove the lemma. Since N > N ≥ 5, it is easy to check using Lemma 3.2 that δ k > δ 5 if k > N . So, taken K α = N , it follows that δL k k > δ k -L > δ 5 -L for every k > K α , by Step 3. Since L < (δ 5 -δ 4 ) by hypothesis, for every k > K α we obtain δL k k > δ 5 -(δ 5 -δ 4 ) = δ 4 .

Definition 7.2. Let S k be the partial sum associated to the Sieve II; recall that in Section 2 we have taken B = {n : 1 ≤ n ≤ p 2 k }; let T (B, P, p k ) be the sifting function of the Sieve II. We denote by {T (B, P, p k )} the set of the values of T (B, P, p k ) for all the combinations of selected remainders in the sequences that form the partial sum S k . Now, we can obtain a lower bound for the sifting function of the Sieve II (that is, a lower bound for the number of permitted k-tuples within the Left interval I[1, p

)} > p k /2 if k > K α .

Proof of the Main Theorem

In this section we prove the Main Theorem. We begin by defining the sequence of k-tuples of the Sieve associated with x (the Sieve I), where x > 49 is an even number. Definition 8.1. Let x > 49 be an even number, and let k be the index of the greatest prime less than √ x. Let {b 1 , b 2 , b 3 , . . . , b k } be the ordered set of the remainders of dividing x by p 1 , p 2 , p 3 , . . . , p k . We define the sequence of k-tuples of remainders of level k, where in the sequences of remainders modulo p h (1 ≤ h ≤ k) that form this sequence of k-tuples are applied the following rules for selecting remainders: Rule 1. Within every period of size p h of the sequence s h (1 ≤ h ≤ k), the remainder 0 is selected. Rule 2. Within every period of size p h of the sequence s h (1 ≤ h ≤ k), the remainder b h is selected. Now we can define formally the Sieve I, as follows.

Definition 8.2. Let P be the sequence of all primes; let z = √ x, and let p k be the greatest prime less than z. Let A be the set consisting of the indices of the sequence of k-tuples of the preceding definition that lie in the interval [1, x].

For each p = p h ∈ P (1 ≤ h ≤ k), the subset A p of A consists of the indices n of the sequence of k-tuples such that the remainder of dividing n by the modulus p h is a selected remainder. Then, the indices of the prohibited k-tuples lying in A are sifted out; and the indices of the permitted k-tuples lying in A remain unsifted. See Remark 1.2. The sifting function

S(A , P, z) = A \ p∈P p<z A p ,
is given by the number of permitted k-tuples whose indices lie in the interval A .

Remark 8.1. Every sequence s h (1 ≤ h ≤ k) that form the sequence of k-tuples associated to the Sieve I consists of the remainders of dividing n by p h . If a remainder is equal to 0, it is always a selected remainder. If a remainder is equal to b h , it is also a selected remainder. If x is divisible by p h , then b h = 0 and therefore, in every period p h of s h there is only one selected remainder.

The following theorem shows that if n is the index of a permitted k-tuple belonging to the set A and 1 < n < x, then n is a prime such that either x -n = 1 or x -n is also a prime. Theorem 8.1. Let us consider the Sieve I, and its associated sequence of k-tuples. If n (1 < n < x) is an unsifted element of the set A , then n is a prime such that either x -n = 1 or x -n is also a prime.

Proof. Step 1. By definition, the set A consists of the indices of the sequence of k-tuples associated to the Sieve I, which lie in the interval [1, x]. Since n is an unsifted element of the set A , by definition, n is the index of a permitted k-tuple. In the sequences of remainders modulo p h (1 ≤ h ≤ k) that form the sequence of k-tuples associated to the Sieve I, if a remainder is equal to 0 then it is a selected remainder. Then, by definition, a permitted k-tuple in this sequence has no element equal to 0 (see Remark 1.1). This means that n is not divisible by any of the primes p 1 , p 2 , p 3 , . . . , p k less than z = √ x. Since 1 < n < x, it follows at once that n is a prime. Step 3. By Step 1, n is a prime; furthermore n ≡ x (mod p h ), where p h < √ x, by Step 2. This last implies that x -n is not divisible by any prime p h < √ x. Since √ x -n < √ x, it follows that either x -n = 1 or x -n is also a prime.

Note that, given the level k, and given an even integer x (p 2 k < x < p 2 k+1 ), there is a sequence of k-tuples associated to the Sieve I, which has specific selected remainders for this particular x. On the other hand, given k, there is a partial sum S k associated to the Sieve II, where there are multiple choices for selecting remainders, allowed by the rules defined in Section 2. Both are sequences of k-tuples of remainders, but they differ in the rules for selecting remainders in each one of them. The following lemma gives the relationship between the number of permitted k-tuples within the interval I[1, p 2 k ] of the partial sum S k (the sifting function of the Sieve II), and the number of permitted k-tuples within the interval I[1, x] of the sequence of k-tuples associated to the Sieve I (the sifting function of the Sieve I).

Recall that we denote by {T (B, P, p k )} the set of the values of T (B, P, p k ) for all the combinations of selected remainders in the sequences that form the partial sum S k associated to the Sieve II. Lemma 8.2. Let P be the sequence of all primes. Let x > 49 be an even number, and let k be the index of the greatest prime less than z = √ x; that is, p 2 k < x < p 2 k+1 . Consider the Sieve I, the Sieve II, and their associated sequences of k-tuples. We have S(A , P, z) ≥ min{T (B, P, p k )}.

Proof. By definition, the sequences of remainders modulo p h (1 < h ≤ k) that form the sequence of k-tuples associated to the Sieve I can have one or two selected remainders in every period (see Remark 8.1). However, the sequences s h (1 < h ≤ k) that form the partial sum S k associated to the Sieve II, by definition, have always two selected remainders in every period. Suppose that we perform on the sequence of k-tuples associated to the Sieve I the following operation: in each sequence of remainders modulo p h (1 < h ≤ k) that have only one selected remainder in every period, we choose an arbitrary second selected remainder (the same element in every period of the sequence). We obtain a partial sum S k with a particular combination of selected remainders, where the number of permitted k-tuples within the interval I[1, p 2 k ] is greater than or equal to min{T (B, P, p k )}. It is obvious that in the interval I[1, p 2 k ] of the sequence of k-tuples associated to the Sieve I before performing the operation, the number of permitted k-tuples is also greater than or equal to min{T (B, P, p k )}. Since I[1, p 2 k ] ⊂ I [1, x], it follows that S(A , P, z) ≥ min{T (B, P, p k )}.

We need one more lemma before proving the Main theorem.

Lemma 8.3. In the sequence of k-tuples associated to the Sieve I, if n (1 < n < x) is the index of a permitted k-tuple, then n = x -n is the index of another permitted k-tuple.

Proof. Step 1. Let {p 1 , p 2 , p 3 , . . . , p k } be the ordered set of the primes less than z = √ x; and let {b 1 , b 2 , b 3 , . . . , b k } be the ordered set of the remainders of dividing x by p 1 , p 2 , p 3 , . . . , p k . Recall that in the sequences of remainders modulo p h (1 ≤ h ≤ k) that form the sequence of k-tuples associated to the Sieve I, we have that 0 is a selected remainder, and b h is also a selected remainder. Therefore, a given k-tuple whose elements are neither 0 nor b h (1 ≤ h ≤ k), by definition, is a permitted k-tuple.

Step 2. Let r h (1 ≤ h ≤ k) be the elements of the permitted k-tuple whose index is n. By definition, for the permitted k-tuple whose index is n we have r h = b h (1 ≤ h ≤ k), since every b h is a selected remainder; this implies n ≡ x (mod p h ), for every prime p h < √ x. Hence n = x -n ≡ 0 (mod p h ), for every prime p h < √ x. It follows that the k-tuple whose index is n has no element equal to 0.

Step 3. Let r h (1 ≤ h ≤ k) be the elements of the k-tuple whose index is n = x -n. By definition, the permitted k-tuple whose index is n has no element equal to 0, since it is a selected remainder. This means that n is not divisible by any of the primes p 1 , p 2 , p 3 , . . . , p k less than √ x. It follows that n ≡ 0 (mod p h ) =⇒ n + x ≡ x (mod p h ) =⇒ n = x -n ≡ x (mod p h ), for every prime p h ∈ {p 1 , p 2 , p 3 , . . . , p k }. So, for the k-tuple whose index is n = x -n we have r h = b h (1 ≤ h ≤ k). From Step 1, Step 2 and this step, the k-tuple whose index is n is a permitted k-tuple.

Finally, we prove the Main Theorem. Remark 8.2. Let x > 49 be an even number. Assume that in the sequence of k-tuples associated to the Sieve I there is a permitted k-tuple at position n = x -1. Then, by Lemma 8.3, there is another permitted k-tuple at position 1; and furthermore, n = x -1 is a prime, by Lemma 8.1, Step 1. So, 1 and x -1 will appear among the unsifted members of the set A . Note that in this case x is an even number of the form p + 1, where p is a prime. Theorem 8.4. The Main Theorem Let x be an even number, and let k be the index of the greatest prime less than z = √ x. Furthermore, let K α be the number whose existence is established in Lemma 7.2. Every even integer x > p 2 k (k > K α ) is the sum of two primes.

Proof.

Step 1. Recall that S(A , P, z) denotes the sifting function of the Sieve I; assume that S(A , P, z) ≥ 3. By Remark 8.2, among the unsifted members of the set A might appear 1 and x -1. So, we can see that there are at least S(A , P, z) -2 integers n in A such that n is a prime and x -n is also a prime, by Theorem 8.1.

Step 2. By Lemma 7.3, for every level k > K α we have min{T (B, P, p k )} > p k /2. On the other hand, S(A , P, z) ≥ min{T (B, P, p k )} for every even number x such that p 2 k < x < p 2 k+1 , by Lemma 8.2. It follows that S(A , P, z) > p k /2 for every even number x > p 2 k , where k > K α > 5, by definition (see Lemma 7.2). Then, by Step 1, if x > p 2 k (k > K α ) there must be at least one unsifted member n < x of A , which is a prime such that x -n is also a prime. The theorem is proved. 9 Estimation of the number K α In this section we shall give an upper bound for K α , whose existence is guaranteed by Lemma 7.2. In order to obtain this upper bound we need to advance our understanding of the behaviour of the maximum density of permitted k-tuples within the Right interval, and the minimum density of permitted k-tuples within the Left interval. k + 1, m k ] 4 ? These are the questions in which we are interested. By now, we know that the h-density within each interval I[1, m k ] h is equal to δ h (1 ≤ h ≤ k). Consequently, for a level k sufficiently large, when we subdivide every interval I[1, m k ] h into a Left interval and Right interval, it seems reasonable to expect that the behaviour of δ L k h and δ R k h , between level h = 1 and level h = k, is analogous to the behaviour of δ h . In particular, with regard to the Right block, the values of δ R k h (1 ≤ h ≤ k) approximate the respective average δ h more and more closely as the level k becomes large, by Lemma 7.1. Thus, it is easy to see that in the Right block of the partition, between h = 1 and h = k, the maximum values of δ R k h within the Right interval I[p 2 k + 1, m k ] h tend to be proportional to the values of δ h as k → ∞. This will allow us to derive a formula for δR k h , in the Right block of the partition. Now, given to levels h = i and h = j (1 ≤ i < j ≤ k), the following lemma shows that δR k j is greater than the value δR k i δ j /δ i , computed by assuming δR k h ∝ δ h , between level h = 1 and level h = k. Lemma 9.1. Let S h be the partial sums from level h = 1 to level h = k (k ≥ 4). Let us consider the Right interval in every partial sum S h (1 ≤ h ≤ k). Let i, j be two levels such that 1 ≤ i < j ≤ k. Then

δ j < δR k i δ j δ i < δR k j .
In other words, the value of δR k j exceeds the value calculated by assuming δR k h ∝ δ h , between level h = 1 and level h = k.

Proof. By Remark 7.3, for each level between h = 1 and h = k we have δR k h < δ h < δR k h . Now, for level h = i, there exists one combination of selected remainders in the sequences s h that form the partial sum S i , such that the density of permitted i-tuples within the Right interval I[p 2 k + 1, m k ] i is equal to the maximum value δR k i . Since by definition the size of the Right interval I[p 2 k + 1, m k ] i of the partial sum S i is equal to m k -p 2 k , the number of subintervals of size p i within this interval is (m k -p 2 k )/p i . Consequently, for this particular combination of selected remainders, the number of permitted i-tuples within the Right interval I[p 2 k + 1, m k ] i is equal to δR k i (m k -p 2 k )/p i . Therefore, using Lemma 4.2 at each level transition h → h + 1, up to level h = j, we obtain

δR k i m k -p 2 k p i p i+1 -2 p i+1 p i+2 -2 p i+2 • • • p j -2 p j = = δR k i p i+1 -2 p i p i+2 -2 p i+1 • • • p j -2 p j-1 m k -p 2 k p j ,
which is the average number of permitted j-tuples within the Right interval I[p 2 k + 1, m k ] j , for all the combinations of selected remainders in the sequences s h from level i + 1 to level j, starting with the combination corresponding to the maximum value δR k i . Now, dividing by (m k -p 2 k )/p j (the number of subintervals of size p j within the Right interval I[p 2 k + 1, m k ] j , for level h = j), we get

δR k i p i+1 -2 p i p i+2 -2 p i+1 • • • p j -2 p j-1 ,
which is the corresponding average j-density. We can write this expression as

δR k i p i+1 -2 p i p i+2 -2 p i+1 • • • p j -2 p j-1 = δR k i δ i pi+1-2 pi pi+2-2 pi+1
δ h -(k) < δ R k h < δ h + (k)
as k → ∞. Dividing by δ h we obtain

1 - (k) δ h < δ R k h δ h < 1 + (k) δ h .
Since (k)/δ h → 0 as k → ∞, the lemma follows. Proof. The proof uses exactly the same argument as given in the preceding lemma, replacing h by k throughout.

Definition 9.1. Suppose given the partial sum S k , and a particular combination of selected remainders in the sequences s h (1 ≤ h ≤ k) that form S k . Let I[p 2 k + 1, m k ] h be the Right interval for every partial sum S h from h = 1 to h = k, and let δ R k h be the true density of permitted h-tuples within every interval I[p 2 k + 1, m k ] h . We denote by φ h the 'true' factor by which we must multiply the h-density within the interval I[p 2 k + 1, m k ] h (denoted by δ R k h ) to obtain the (h + 1)-density within the interval I[p 2 k + 1, m k ] h+1 (denoted by δ R k h+1 ), for every level transition h → h + 1 from h = 1 to h = k -1. In symbols as k → ∞, whatever the combination of selected remainders in the sequences s h (1 ≤ h ≤ k) that form S k .

δ R k h+1 = δ R k h φ h (1 ≤ h < k).

Conclusion

By the Main theorem and the upper bound in (50) we can say that every even integer x > p 2 k , where k ≥ 89 (p 2 k ≥ 212521), is the sum of two primes. Now, it is a known fact that the strong Goldbach conjecture has already been verified for all n ≤ 4 × 10 17 [START_REF] Oliveira E Silva | Goldbach conjecture verification[END_REF]. Therefore, we conclude that every even number x > 4 can be expressed as the sum of two odd primes; and then, the binary Goldbach conjecture is proved.

  this into (2),S(A , P, z) = XW (z) + d|P (z) µ (d) R d .Hence, we can writeS(A , P, z) = XW (z) + θ d|P (z) |R d | (|θ| ≤ 1).

Definition 1 . 1 .

 11 Given a sequence of k-tuples, and using the order relation given by the index n, we define an interval of k-tuples, denoted by I[m, n] k , to be the set of consecutive k-tuples associated with an integer interval [m, n] ∩ Z + , where m is the index of the first k-tuple, and n is the index of the last k-tuple. We also use the notation I[m, n] = I[m, n] k for this interval. We define the size of I[m, n] by the equation |I[m, n]| = n -m + 1; and we use the notation I[] k , or alternatively I[], to denote the empty interval.

  for obtaining an estimate of T ({n : 1 ≤ n ≤ p 2 k }, P, p k )? Let us consider the interval I[1, m k ] (first period of the sequence of k-tuples of the Sieve II); furthermore, consider the intervals I[1, p 2 k ] and I[p 2 k + 1, m k ]. We can see that, for a given choice of selected remainders in the sequence of k-tuples, if the proportion of permitted k-tuples in I[1, p 2 k ] is less than the proportion in I[1, m k ], the proportion of permitted k-tuples in I[p 2 k + 1, m k ] must be greater than the proportion in I[1, m k ]; and vice versa. Suppose that the proportion of permitted k-tuples in the interval I[1, p 2 k ] were equal to the proportion of permitted k-tuples in the interval I[1, m k ]. In this case, we could compute at once the exact number of permitted k-tuples in the interval I[1, p 2

Definition 3 . 1 .Definition 3 . 2 ..Remark 3 . 1 .

 313231 Let S k be a given partial sum of the series s k ; let I[m, n] be a given interval of k-tuples. We denote by c I[m,n] k the number of permitted k-tuples within I[m, n]. By abuse of notation, we normally omit specific mention of the integer interval [m, n] ∩ Z + and write c I k instead of c I[m,n] k if no confusion will arise. Let S k be a partial sum of the series s k ; let I[m, n] be a given interval of k-tuples. The number of subintervals of size p k in this interval is equal to |I[m, n]|/p k . We define the density of permitted k-tuples in the interval I[m, n] (or simply the k-density) by δ I[m,n] k = c I k |I [m, n]| /p k . For the empty interval we define δ I[] k = 0. By abuse of notation, we often omit specific mention of the integer interval [m, n] ∩ Z + and write δ I k instead of δ I[m,n] k The density of permitted k-tuples is the average number of permitted k-tuples inside subintervals of size p k . Definition 3.3. Let S k be a given partial sum of the series s k ; let m k be the period of S k . Recall that we have used the notation c k = c I[1,m k ] k for the number of permitted k-tuples within the interval I[1, m k ] (the first period of S k ). We normally use the notation δ k = δ I[1,m k ] k for the density of permitted k-tuples within the interval I[1, m k ]. Since m k /p k is the number of subintervals of size p k within a period of S k , by definition, we have δ k = c k m k /p k . By Proposition 2.3, the number of permitted k-tuples within the interval I[1, m k ] (the first period of S k ), does not depend on which are the selected remainders in the sequences of remainders that form S k . Therefore, we think of I[1, m k ] as being a special interval, and this explains why we use the special notation c k for the number of permitted k-tuples within I[1, m k ], and δ k for the density of permitted k-tuples within I[1, m k ]. Example 3.1. The period of the partial sum S 4 is equal to m 4 = 2 × 3 × 5 × 7 = 30 × 7 = 210, and the number of permitted 4-tuples within the period is equal to c 4 = (2 -1)(3 -2)(5 -2)(7 -2) = 15. Then δ 4 = c 4 m 4 /p 4

Example 3 . 2 .

 32 The characteristic prime moduli of the partial sums S 4 and S 5 are p 4 = 7 and p 5 = 11. The period of the partial sum S 4 is m 4 = 2 × 3 × 5 × 7 = 30 × 7 = 210, and the number of permitted 4-tuples is c 4 = (2 -1) (3 -2) (5 -2) (7 -2) = 15. Then δ 4 = 15/30 = 0.500. On the other hand, the period of the partial sum S 5 is m 5 = 2 × 3 × 5 × 7 × 11 = 210 × 11 = 2310, and the number of permitted 5-tuples is c 5 = (2 -1) (3 -2) (5 -2) (7 -2) (11 -2) = 135. Then δ 5 = 135/210 ≈ 0.643. Note that since 7 and 11 are not twin primes, δ 5 > δ 4 (see Corollary 3.3). Now we prove that δ k → ∞ as k → ∞. First, we make a definition. Definition 3.4. Let p k > 2 and p k+1 be consecutive primes. We denote by θ k the difference p k+1 -p k -2. Theorem 3.4. Let S k be a given partial sum. Let δ k be the density of permitted k-tuples within a period of S k . As k → ∞, we have δ k → ∞. Proof. Lemma 3.1 implies

Definition 4 . 3 .

 43 The operation of Type B.

Definition 4 . 5 .

 45 Let S k and S k+1 be the partial sums of level k and k + 1. Let s k+1 be the sequence of remainders of level k + 1. Let I[m, n] k be an interval of k-tuples of S k , and let I[m, n] k+1 be an interval of (k + 1)-tuples of S k+1 , where the indices m, n are the same for both intervals. When we juxtapose the remainders of the sequence s k+1 to the right of each k-tuple of S k , then, by Proposition 2.2, the permitted k-tuples of S k , whose indices are congruent to a given selected remainder of s k+1 modulo p k+1 , are converted to prohibited (k + 1)-tuples of S k+1 . We denote by f k+1 the fraction of the permitted k-tuples within the interval I[m, n] k that are converted to prohibited (k + 1)-tuples within the interval I[m, n] k+1 . For the partial sum S 1 , let f 1 denote the fraction of the prohibited 1-tuples within the interval I[m, n] k=1 .

2 ,

 2 the indices of the permitted k-tuples within the interval I[m, n] k of S k that are converted to prohibited (k + 1)-tuples within the interval I[m, n] k+1 of S k+1 belong to one of the residue classes [r] or [r ]. It follows that the fraction of the c I k permitted k-tuples within the interval I[m, n] k of S k that are converted to prohibited (k + 1)-tuples within the interval I[m, n] k+1 of S k+1 is equal to (η r + η r )/c I k . Taking the average over the ν A k+1 combinations of selected remainders obtained by repeated Type A operations, we obtain

Definition 4 . 7 .

 47 Let S k be the partial sum of level k. Let I[m, n] be an interval of k-tuples of S k . We denote by c I k the average number of permitted k-tuples within the interval I[m, n]. We denote by δ I k the average density of permitted k-tuples within the interval I[m, n].

Theorem 4 . 3 .

 43 Let δ k be the density of permitted k-tuples within a period of the partial sum S k . Then δ I k = δ k . Proof. Let s h (1 ≤ h ≤ k) be the sequences of remainders that form S k . If there were no selected remainders within the sequences s h , all the k-tuples within the interval I[m, n] would be permitted k-tuples, and then c I k = |I[m, n]|, where |I[m, n]| is the size of the interval I[m, n]. However, since we have selected remainders in every sequence s h (1 ≤ h ≤ k), using Lemma 4.2 at each level transition from h = 1 to h = k, we can write

Definition 5 . 1 .

 51 Let n/m k denote the integer part of n/m k (n ≥ m k ). We denote by c η the number of permitted k-tuples within the intervalI[1, n/m k m k ] ⊆ I[1, n]. If n is not a multiple of m k ,we denote by c the number of permitted k-tuples within the interval I[ n/m k m k + 1, n] ⊂ I[1, n]; otherwise c = 0. We call the interval I[ n/m k m k + 1, n] the incomplete period of the interval I[1, n].

Definition 6 . 4 .

 64 Let S k (k ≥ 4) be a given partial sum, in horizontal position. We subdivide the interval I[1, m k ] (its first period) into two intervals: I[1, p 2 k ], which we call the Left interval, and I[p 2 k + 1, m k ], which we call the Right interval. We often denote the Left interval I[1, p 2 k ] by the symbol L k , and the Right interval I[p 2 k + 1, m k ] by the symbol R k . For every partial sum S h from level h = 1 to level h = k -1 there is also a Left interval I[1, p 2 k ] h , and a Right interval I[p 2

Figure 4 : 6 . 5 .

 465 Figure 4: Left and Right intervals

Remark 7 . 1 .

 71 For every partial sum S h from level h = 1 to level h = k (k ≥ 4), let us consider the interval I[1, m k ] h , the Left interval I[1, p 2 k ] h and the Right interval I[p 2 k + 1, m k ] h . We denote by c k the number of permitted k-tuples within the interval I[1, m k ] k , and we denote by c h the number of permitted h-tuples within every interval I[1, m k ] h where 1 ≤ h < k. Since p 2 k = o(m k ), by Lemma 2.5, and p 2 k = o(c k ), by Lemma 2.4, we can take k large enough that the size of the Left interval I[1, p 2 k ] is negligible compared to the size of the interval I[1, m k ], and furthermore the size of I[1, p 2 k ] is negligible compared to c k . Therefore, since c h > c k (see Proposition 2.3 and Lemma 6.1), almost all the permitted h-tuples in every interval I[1, m k ] h (1 ≤ h ≤ k) belong to the Right interval I[p 2 k + 1, m k ] h . Remark 7.2. Assume that k satisfies the conditions required in the preceding remark. Then, as we have remarked, almost all the permitted h-tuples in every interval I[1, m k ] h (1 ≤ h ≤ k) belong to the Right interval I[p 2

Lemma 7 . 2 .

 72 Let S k (k ≥ 5) be a partial sum of the series s k . There exists an integer K α > 5 such that δL k k > δ 4 for every k > K α . Proof. Step 1. Let L be a number such that 0 < L < (δ 5 -δ 4 ); let us consider the Left interval I[1, p 2 k ] 5 of the partial sum S 5 . Since the size of the interval I[1, p 2 k ] 5 tends to infinity as k → ∞, by Proposition 5.3 it must exist an integer N ≥ 5 such that δ 5 -δL k 5 < L

Step 2 .

 2 Let {b 1 , b 2 , b 3 , . . . , b k } be the ordered set of the remainders of dividing x by p 1 , p 2 , p 3 , . . . , p k . Let r h (1 ≤ h ≤ k) be the elements of the permitted k-tuple whose index is n. In the sequences of remainders modulo p h (1 ≤ h ≤ k) that form the sequence of k-tuples associated to the Sieve I, by definition, if a given remainder of the sequence is equal to b h ∈ {b 1 , b 2 , b 3 , . . . , b k }, then it is a selected remainder. Consequently, by definition, for the permitted k-tuple whose index is n we have r h = b h (1 ≤ h ≤ k); this implies n ≡ x (mod p h ), for every prime p h < √ x (see Remark 1.1).

9. 1 A

 1 formula for the maximum density of permitted k-tuples within the Right interval Let S k (k > 4) be a given partial sum of the series s k , where the first period is partitioned as we have seen before, in a Left block and a Right block. In particular, consider for level h = k the Left interval I[1, p 2 k ] k and the Right interval I[p 2 k + 1, m k ] k ; and for level h = 4 the Left interval I[1, p 2 k ] 4 and the Right interval I[p 2 k + 1, m k ] 4 . By Lemma 6.5 there is a bijection between the values of the k-density in I[1, p 2 k ] k , and the values of the k-density in I[p 2 k + 1, m k ] k , and there is also a bijection between the values of the 4-density in I[1, p 2 k ] 4 , and the values of the 4-density in I[p 2 k +1, m k ] 4 . However, what is the relationship between the k-density within the Left interval I[1, p 2 k ] k , and the 4-density within the Left interval I[1, p 2 k ] 4 ? And what is the relationship between the k-density within the Right interval I[p 2 k + 1, m k ] k , and the 4-density within the Right interval I[p 2

Lemma 9 . 4 .Remark 9 . 2 .

 9492 Let S k (k > 4) be a partial sum of the series s k . Let us consider the Right intervalI[p 2 k + 1, m k ] k for the level k. We have δ R k k δ k → 1 as k → ∞.Note that in this case, δ k → ∞ as k → ∞, by Theorem 3.4.

Lemma 9 . 5 .

 95 Let S k (k > 4) be a partial sum of the series s k . The partial product

  2 k ], and the right interval I[p 2 k + 1, m k ]. So, since for every h (1 ≤ h ≤ k) there is a sequence of h-tuples of remainders, the interval I[1, m k ] h of each sequence turns out subdivided into two intervals: the left interval I[1, p 2 k ] h , and the right interval

Table 1 :

 1 Quotient c k /m k and density δ k .

	k	p k	m k	c k	c k /m k	δ k
	4	7	210	15	0.071 0.500
	5	11	2310	135	0.058 0.643
	6	13	30030	1485	0.049 0.643
	7	17	510510	22275	0.044 0.742
	8	19	9699690	378675	0.039 0.742
	9	23 223092870 7952175 0.036 0.820
	10 29	-	-	0.033 0.962
	11 31	-	-	0.031 0.962
	12 37	-	-	0.029 1.087
	13 41	-	-	0.028 1.145
	14 43	-	-	0.027 1.145
	15 47	-	-	0.025 1.199
	16 53	-	-	0.024 1.301
	17 59	-	-	0.024 1.399
	18 61	-	-	0.023 1.399
	19 67	-	-	0.022 1.490
	20 71	-	-	0.022 1.535
	21 73	-	-	0.021 1.535
	22 79	-	-	0.020 1.619
	23 83	-	-	0.020 1.660
	24 89	-	-	0.019 1.740
	.	.	.	.	.	.

integer K α > 5 such that the density of permitted k-tuples in the interval I[1, p 2 k

Table 2 :

 2 Sequence of 5-tuples of remainders.

Table 3 :

 3 Partial sums S 4 and S 5 .

			S 4		s 5		S 5
	n	2 3 5 7	11	2 3 5 7 11
	1	1 1 1 1	1	1 1 1 1 1
	2	0 2 2 2	2	0 2 2 2 2
	3	1 0 3 3	3	1 0 3 3 3
	4	0 1 4 4	4	0 1 4 4 4
	5	1 2 0 5	5	1 2 0 5 5
	6	0 0 1 6	6	0 0 1 6 6
	7	1 1 2 0	7	1 1 2 0 7
	8	0 2 3 1	8	0 2 3 1 8
	9	1 0 4 2 + 9 = 1 0 4 2 9
	10	0 1 0 3	10	0 1 0 3 10
	11	1 2 1 4	0	1 2 1 4 0
	12	0 0 2 5	1	0 0 2 5 1
	13	1 1 3 6	2	1 1 3 6 2
	14	0 2 4 0	3	0 2 4 0 3
	15	1 0 0 1	4	1 0 0 1 4
	16	0 1 1 2	5	0 1 1 2 5
	17	1 2 2 3	6	1 2 2 3 6
	18	0 0 3 4	7	0 0 3 4 7
	.	.	.	. .	.	. .	.	.	.

T (B, P, p k

Table 4 :

 4 Partial sum S 4 with selected remainders.

  2 k ] h to I[p 2 k + 1, m k ] h decreases δ L k h and increases the δ R k h , and vice versa. By Theorem 4.3, the average of δ L k

h within I[1, p 2 k ] h is equal to δ h , and the average of

  by Step 1. From Proposition 2.1 and Proposition 2.3 follows

  for the partial sum S h , assume that the density of permitted h-tuples within I[1, p 2 k ] h , and within I[p 2 k + 1, m k ] h , is equal to the average δ h . Then, suppose that some permitted h-tuples are transferred from the Right interval to the Left interval. We have an increase (δL k h -δ h ) ofthe h-density within the Left interval, and a decrease (δ h -δ R k h ) of the h-density within the Right interval. See (23). Because within the Left interval I[1, p 2 k ] h we have p 2 k /p h subintervals of size p h , by definition, the number of permitted h-tuples entering the Left interval is equal to (δ L k h -δ h )p 2 k /p h . In the same way, within the Right interval I[p 2 k + 1, m k ] h we have (m k -p 2 k )/p h subintervals of size p h , and then, by definition, the number of permitted h-tuples coming out of the Right interval is equal to

  2 k + 1, m k ] h of every partial sum S h (1 ≤ h ≤ k) converges uniformly to the average δ h . Lemma 7.1. Let S k (k ≥ 4) be a partial sum of the series s k . Let us consider the Right interval I[p 2 k + 1, m k ] h in every partial sum S h , from level h = 1 to level h = k. For every > 0, there exists N (depending only on ) such that k > N implies |δ R k h -δ h | < , for every partial sum S h from level h = 1 to level h = k, whatever the combination of selected remainders in the sequences s h that form every partial sum S k . Proof. Step 1. The size of the Right interval I[p 2 k + 1, m k ] h of the partial sum S h , by definition, is equal to m k -p 2 k , and so the number of subintervals of size p h within the Right interval is equal to (m k -p 2 k )/p h (1 ≤ h ≤ k). Denoting by c R k h the number of permitted h-tuples within I[p 2 k + 1, m k ] h , by definition, we have

  the number of permitted h-tuples within the Right interval I[p 2 k + 1, m k ] h can not be greater than c h , and so we have c R k h ≤ c h . On the other hand, the number of permitted h-tuples within the Left interval I[1, p 2 k ] h of the partial sum S h can not be greater than the size p 2 k of the Left interval. Therefore, c h -p 2 k ≤ c R k h . Consequently, replacing the numerator in (24) by c h -p 2

	k
	and by c h , we obtain

  2 k ] of the partial sum S k , denoted by δ L k k , for k sufficiently large. By Theorem 4.3, the average density of permitted k-tuples within I[1, p 2 k ] is equal to δ k , that is to say, is equal to the k-density within the period of S k . By Lemma 3.2 and Corollary 3.3, the density δ k increases at each level transition p k → p k+1 of order greater than 2, and this implies that for level k > 4 we have δ k > δ 4 . However, what happens to the true density δ L k k ? The next lemma shows that there existsK α ∈ Z + , K α > 5 such that δ L k k > δ 4for every level k > K α , whatever the combination of selected remainders in the sequences s h that form every partial sum S k .Recall the notation {δ L k h } to denote the set of values of δ L k h , and the notation {δ R k h } to denote the set of values of δ R k h , for all the combinations of selected remainders in the sequences that form the partial sum S h (1 ≤ h ≤ k). Furthermore, recall that δ h denotes the density of permitted h-tuples within the period of the partial sum S h . Definition 7.1. We use the notation δL k h and δL k h to denote, respectively, min{δ L k h } and max{δ L k h }, and the notation δR k h and δR k h to denote, respectively, min{δ R k h } and max{δ R k R k h . On the other hand, for any given level h, the average density of permitted h-tuples within the Left interval I[1, p 2 k ] h (the Right interval I[p 2 Remark 7.4. Note that for a given level h (1 ≤ h ≤ k), the image of δL k h under the function f h of Lemma 6.5 is δR k h , and the image of δL k h under f h is δR k

h }.

Remark 7.3. By Lemma 6.4, for every partial sum

S h (1 ≤ h ≤ k, k ≥ 4), we have δ L k h = δ k + 1, m k ] h ) is equal to δ h ,

by Theorem 4.3. Then, for each level between h = 1 and h = k, we have δL k h < δ h < δL k h ( δR k h < δ h < δR k h ). See (23). h . See (23).

  2 k ] of S k ), for k sufficiently large. Lemma 7.3. Let K α be the number whose existence is established in Lemma 7.2. For level k > K α , we have min{T (B, P, p k )} > p k /2. Proof. Step 1. Consider a given partial sum S k of the series s k . Recall the notation {δ L k k } to denote the set of values of δ L k k , for all the combinations of selected remainders in the sequences that form the partial sum S k ; and recall the notation δL k k to denote min{δ L k k }. Note that within the Left interval I[1, p 2 k ] of S k we have p k subintervals of size p k . So, the minimum number of permitted k-tuples within the Left interval I[1, p 2 Now, by Lemma 7.2, if k > K α then δL k k > δ 4 . From this and Step 1 it follows that min{T (B, P, p k )} > p k δ 4 , whenever k > K α . Using Lemma 3.1, it is easy to check that δ 4 = 1/2 (see Table 1), and so min{T (B, P, p k

	is p k	δL k k . Then, by definition, min{T (B, P, p k )} = p k	δL k k .	k ] of S k
	Step 2.

  Lemma 9.3. Let S k (k > 4) be a partial sum of the series s k . Let us consider the Right interval I[p 2 k + 1, m k ] h for a fixed level h < k. We have Remark 9.1. Note that since h is a fixed level, δ h does not change as k → ∞. Proof. By Lemma 7.1 there exist a function (k) such that (k) → 0 and

	and then
	k-1 h=4	p h+1 -2 p h	=	δ k δ 4	.
	δ R k h δ h	→ 1 as k → ∞.
					δ i	• • •	pj -2 pj-1	,

The numerical computation were carried out on a desktop computer, using a program written in language C.

Acknowledgements

The author wants to thank Dra. Patricia Quattrini (Departamento de Matematicas, FCEyN, Universidad de Buenos Aires) for helpful conversations and Dr. Hendrik W. Lenstra (Universiteit Leiden, The Netherlands) for his extremely useful suggestions.

and, using Lemma 3.2, it is easy to see that

Therefore, we can see that δR k i δ j /δ i is the average density of permitted j-tuples within the Right interval I[p 2 k + 1, m k ] j , for all the combinations of selected remainders in the sequences s h from level h = i + 1 to level h = j such that the combination of selected remainders in the sequences s h from level h = 1 to level h = i is the one corresponding to the maximum value δR k i . That is, δR k i δ j /δ i is an average, not a maximum value. Consequently, it is easy to see that δR k j must be greater than δR k i δ j /δ i , and then we can write

In particular for h = 4 and h = k (k > 4) it follows from Lemma 9.1 that

We can see that δR k k tends asymptotically to (δ k /δ 4 ) δR k 4 as k → ∞, since the difference ( δR k k -δ k ) tends to 0 as k → ∞, by Lemma 7.1. In other words, as k → ∞ the ratio δR k k / δR k 4 tends to be proportional to the ratio δ k /δ 4 . On the other hand, we have the elementary identity

which can be easily verified. (Note that this identity is not valid in the limit as k → ∞, since in this limit δR k 4 = δ 4 , by Lemma 7.1.) So, by (35) and the preceding identity, we also have

Therefore, by the rightmost inequality in (37) we can write

where β k 4 is some real number greater than 1. The number β k 4 measures how far the ratio ( δR k k -δ k )/( δR k 4 -δ 4 ) 'deviates' from being exactly proportional to the ratio δ k /δ 4 . Now, an obvious question occurs at this point: What is the behaviour of β k 4 as k → ∞? The answer to this question is given in the next subsection.

The behaviour of

We need some more lemmas.

Lemma 9.2. Let S k (k ≥ 4) be a given partial sum of the series s k ; consider the interval I[1, m k ] h in every partial sum S h , from level h = 1 to level h = k. We have the identity

Proof. Using the formula (22) at each level transition from h = 4 to h = k -1, we can write

Proof. Let us consider the levels h = 4 and h = k. Using the factors φ h at each level transition from h = 4 to h = k -1, we can write

(Note that each set of factors φ 4 , φ 5 , . . . , φ k-1 is attached to every particular combination of selected remainders in the sequences s h (1 ≤ h ≤ k) that form S k .) On the other hand, by (39) we have

Dividing side by side (42) and (43), we obtain

, and so Proof. Let S k (k > 4) be a given partial sum of the series s k ; consider the Right interval

Step 1. For level h = 4, there exists one combination of selected remainders in the sequences s h that form the partial sum S 4 , such that the density of permitted 4-tuples within the Right interval

Suppose that the factor by which we must multiply the h-density within the interval

whatever the combination of selected remainders in the sequences s h from h = 5 to h = k; that is, we are assuming that

(see identity (39)); that is, δR k k would be equal to

(see identity (36)).

Step 2. Clearly, the assumption made in the preceding step is not true (see ( 35) and ( 37)). We know certainly that for every combination of selected remainders in the sequences s h (5 ≤ h ≤ k) that form S k the factors φ h (4 ≤ h < k) are different from the corresponding average factors (p h+1 -2)/p h . And there exists one of these combinations of selected remainders (and its attached factors φ h (4 ≤ h < k)) such that the density of permitted k-tuples within the Right interval I[p 2 k + 1, m k ] k is the maximum value δR k k . Now, for a level k not too large, the partial product

corresponding to this particular combination of selected remainders could be quite larger than the average product (41), and so, δR k k could be quite far from the proportional value given by (45). So, since the factor β k 4 in the formula (38) stands for taking into account the excess over the proportional value, we conjecture that, in this case, β k 4 will be far above 1. However, as k → ∞, the partial product (46) approaches the average product (41), whatever the combination of selected remainders in S k , by Lemma 9.5; and furthermore, we know that δR k k tends asymptotically to (45) as k → ∞, by (37) and Lemma 7.1. Therefore, it is reasonable to think that β k 4 approaches 1 as k → ∞.

Step 3. On the contrary, suppose that β k 4 > C holds as k → ∞, for some constant C > 1. This means that as k → ∞, for every partial sum S k there exists a combination of selected remainders in the sequences s h that form S k for which the partial product (46) holds too far from the average partial product (41). That is, there exists a combination of selected remainders in every S k such that the inequality

holds as k → ∞, for some constant B > 1; and this contradicts Lemma 9.5. Therefore, we conclude that β k 4 → 1 by the right as k → ∞.

A formula for the maximum density of permitted k-tuples within the Left interval

We recall that the first period of the partial sum S k (k ≥ 4), in horizontal position, can be seen as a matrix, with m k columns and k rows; and we recall also that this matrix was partitioned into two blocks: the Left block formed by the columns from n = 1 to n = p 2 k ; and the Right block formed by the columns from n = p 2 k + 1 to n = m k . From (38) we have a formula for δR k k within the Right block of the partition. However, we need a similar formula for δL k k within the Left block of the partition; in order to derive this formula, we proceed as follows. Suppose that for the level [START_REF] Brun | Le crible d'Eratosthène et le théorème de Goldbach[END_REF]

where β k 4 is the number which appears in (38).

Proof.

Step 1. We compute the value of f 4 (see Lemma 6.5) at x = δL k 4 (the minimum density of permitted 4-tuples within the Left interval I[1, p 2 k ] 4 ). We obtain

Step 2. Next, we take the maximum density of permitted 4-tuples within the Right interval I[p 2 k + 1, m k ] 4 obtained in the previous step, and using the formula in (38), we get

Step 3. Finally, we compute the value of f -1 k (see Lemma 6.5) at x = δR k k (the maximum density of permitted k-tuples within the Right interval I[p 2 k + 1, m k ] k , obtained in the preceding step). See Remark 7.4. We obtain

An upper bound for K α

Before computing an upper bound for K α , we need to obtain an upper bound for β k 4 . Since β k 4 > 1 by definition, it is convenient to compute the upper bound for β k 4 in a range of values of k where this number is close to 1. By Lemma 9.6, we know that β k 4 → 1 as k → ∞. Now, how large must k be for β k 4 to be close to 1? In the case examined in Remark 7.1 and Remark 7.2, we have seen that, if k satisfies the conditions required in Remark 7.1, for each level from h = 1 to h = k the values of δ R k h will be very close to δ h , whatever the combination of selected remainders in the sequences s h that form the partial sum S k , and furthermore, for every level from h = 4 to h = k -1, the factor φ h will be very close to the average factor (p h+1 -2)/p h (see Lemma 6.3). It follows that for every combination of selected remainders in the sequences s h that form the partial sum S k , the partial product in (40) will be very close to the average partial product in (41). In particular, this is true for the combination of selected remainders corresponding to the maximum value δR k k . Therefore, δR k k will be very close to (45), and β k 4 must be close to 1 (see the proof of Lemma 9.6). So, we formulate the following criterion.

for any given level k > 4; and since δL k k can not be less than zero, we can write In order to compute the upper bound in (48), we need the value of δL k 4 . Note that by [START_REF] Rey Pastor | Mathematical Analysis[END_REF] there are 1260 combinations of selected remainders in the sequences s h (1 ≤ h ≤ 4) that form the partial sum S 4 . The minimum number of permitted 4-tuples within the Left interval I[1, p 2 k ] 4 of the partial sum S 4 (for k = 35) can be obtained by explicit computation of the number of permitted 4-tuples for every combination of selected remainders in S 4 , and then taking the minimum among these values 1 . Then, with the minimum number of permitted 4-tuples within I[1, p

Now we are ready to compute an upper bound for K α , using the formula (47); we proceed in the following way. For every k starting from level k = 35 we perform the following procedure: Step 2. Compute as the difference between δ 4 and the lower bound of the preceding step.

Step 3. Compute δ k .

Step 4. Compute a lower bound for the density of permitted k-tuples within the Left interval I[1, p 2 k ] k .

Step 5. Increase k by 1.

We use Lemma 5.2 and Lemma 3.1 in the steps 1 and 3 respectively; and we take δ 4 = 1/2 in Step 2. We use the formula (47) in the Step 4, replacing (δ 4 -δL k 4 ) by , β k 4 by its upper bound in (49), and δ k by the value obtained in the Step 3. The procedure is carried out until the lower bound for the density of permitted k-tuples within I[1, p 2 k ] k , computed in Step 4, overcome the value δ 4 = 1/2; at this point, the value of k is the upper bound for K α . We obtain

For some values of k, Table 7 gives p k , , δ k and the lower bound for δ L k k computed by using the preceding procedure.