The Twin Prime Conjecture Ricardo Barca

Home , Sieve theory, Twin prime, Viggo Brun

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Ricardo Barca. The Twin Prime Conjecture. 2018. hal-01882333v2

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HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Public Domain THE TWIN PRIME CONJECTURE

Ricardo G. Barca

Abstract

Consecutive primes which diﬀer only by two are known as twin primes; the function π2(x) describes the distribution of the twin primes in the set of natural numbers. The twin prime conjecture states that there are inﬁnitely many primes p such that p + 2 is also prime. At the present time, the knowledge about twin primes comes mostly from sieve methods. Using a formulation of the sieve of Eratosthenes supported on a sequence of k-tuples of remainders we obtain a lower bound for the number of twin primes in [1, x]. Thus, we prove that π2(x) → ∞ as x → ∞.

1 Introduction

The twin prime conjecture is one of the problems in number theory that still remains without resolution. Twin prime conjecture: There exist infinitely many primes p such that p + 2 is a prime. In 1849, the french mathematician de Polignac made the statement (more general than twin prime conjecture) that for every positive even integer k there exist infinitely many primes p such that p + k is also a prime [6]. For k = 2 we have the twin prime conjecture. The most significant result for twin primes is due to Viggo Brun [3], who proves, in 1915, that the sum of the reciprocals of the odd twin primes,

B = (1/3 + 1/5) + (1/5 + 1/7) + (1/(11) + 1/(13)) + (1/(17) + 1/(19)) + ...

converges to a number known as the Brun’s constant. This outcome was the first use of the Brun sieve, and it was the beginning of modern sieve theory. Since the series of all prime reciprocals diverges to infinity, the result from Viggo Brun say us that twin primes are very scarce. On April 17, 2013, the american mathematician (born in Shangai) Yitang Zhang claims that there are infinitely many pairs of prime numbers that differ by 70 million or less [4]. For any m ≥ 1, let Hm denote the quantity 7 Hm = lim infn→∞ (pn+m − pn), where pn denotes the n-th prime. In his paper, Zhang proved that H1 ≤ 7 × 10 ; the twin prime conjecture is equivalent to the assertion that H1 is equal to two. After the appearance of Zhang’s paper, improvements to the upper bound on H1 were made. In addition, a Polymath project was made up to examine the arguments of Zhang [5]; one year after Zhang’s announcement, the bound has been reduced to 246. Let x > 0 be a positive real number; let π2(x) be the number of twin primes p and p + 2 such that p ≤ x. In this paper we prove (Main Theorem) the following: π2(x) → ∞ as x → ∞. One of the principal ways of addressing problems like the twin prime conjecture or the Goldbach’s conjecture have been through the use of sieve methods. The Sieve theory has developed along of the last century, starting from the Viggo Brun methods. For the sake of completeness and convenience of the reader, in this paper we shall repeat some explanations and arguments already given in [9]. Also, we use concepts and tools developed in [9], as well as the notations defined in this paper, unless we specifically state otherwise. We begin by defining the sieve of Eratosthenes, using the notation of the book by Cojocaru and√ Ram Murty [2]. Let A = {n ∈ Z+ : n ≤ x}, where x ∈ R, x > 1, and let P be the sequence of all primes. Let z = x, and

Y P (z) = p. p∈P p

Now, to each p ∈ P, p < z, we associate the subset Ap of A , deﬁned as follows: Ap = {n ∈ A : n ≡ 0 (mod p)}. Then, when we sift out from A all the elements of every set Ap, the unsifted members√ of A 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

\ Ad = Ap. p|d

Thus, from the inclusion–exclusion principle, we obtain

1 X S(A , P, z) = µ (d) |Ad| , (1) d|P (z) where µ(d) is the M¨obiusfunction. Moreover, from (1), the Legendre formula can be derived

X X jxk S(A , P, z) = µ (d) |Ad| = µ (d) . d d|P (z) d|P (z) √ The sieve of Eratosthenes is useful for finding√ the prime numbers between x and x; for our purposes, we need first to isolate the subset of twin primes between x and x and then to obtain a lower bound for the size of this subset, for every x sufficiently large. However, we can not hope to find a suitable lower bound for the size of this set using the usual sieve methods, due to a well-known phenomenon in sieve theory, called the ‘parity barrier’ or the ‘parity problem’ (see [9, Introduction, Subsection 1.1]). So far, all attempts to solve the twin prime 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.

n 2 3 5 7 n 2 3 5 7 1 1 1 1 1 31 1 1 1 3 2 0 2 2 2 32 0 2 2 4 3 1 0 3 3 33 1 0 3 5 4 0 1 4 4 34 0 1 4 6 5 1 2 0 5 35 1 2 0 0 6 0 0 1 6 36 0 0 1 1 7 1 1 2 0 37 1 1 2 2 8 0 2 3 1 38 0 2 3 3 9 1 0 4 2 39 1 0 4 4 10 0 1 0 3 40 0 1 0 5 11 1 2 1 4 41 1 2 1 6 12 0 0 2 5 42 0 0 2 0 13 1 1 3 6 43 1 1 3 1 14 0 2 4 0 44 0 2 4 2 15 1 0 0 1 45 1 0 0 3 16 0 1 1 2 46 0 1 1 4 17 1 2 2 3 47 1 2 2 5 18 0 0 3 4 48 0 0 3 6 19 1 1 4 5 49 1 1 4 0 20 0 2 0 6 50 0 2 0 1 21 1 0 1 0 51 1 0 1 2 22 0 1 2 1 52 0 1 2 3 23 1 2 3 2 24 0 0 4 3 25 1 1 0 4 26 0 2 1 5 27 1 0 2 6 28 0 1 3 0 29 1 2 4 1 30 0 0 0 2

Figure 1

Let P be the sequence of all primes; and given pk ∈ P, let mk = p1p2p3 ··· pk. From now on, and throughout 2 this paper, for convenience, we take x to be a real number greater than p4 = 49, unless we speciﬁcally state otherwise. √ 2 2 Note that if pk is the greatest prime less than x, every real number x > 49 satisﬁes pk < x ≤ pk+1 < mk.

2 Now, the formulation of the sieve of Eratosthenes using the classic sieve theory notation does not allow us to discriminate between the twin primes and the other primes, for counting only the first ones. In this paper we propose to use another formulation for this kind of sieves, using a sequence of k-tuples, which is able to show all the details of the sifting process (see [9, Introduction, Subsection√ 1.3]). Let pk be the greatest prime less than x. In the formulation of the sieve of Eratosthenes by means of a sequence of k-tuples, the index k is the corresponding to pk, and A is the set of integers that lie in the interval [1, x]. The rules for selecting remainders are the following: In the k-tuples of the sequence, all the zeroes are defined as selected remainders. Therefore, within the periods of every sequence of remainders modulo ph (a given column of the matrix), the element 0 is always a selected remainder. We more formally define the formulation of this sieve using a sequence of k-tuples in Section 2. √ Remark 1.1. Note that given a k-tuple whose index is√n ≤ x, if n ≡ 0 (mod p) for at least one p < x, then it is a prohibited k-tuple, and if n 6≡ 0 (mod p) for every p < x, then it is a permitted k-tuple. √ Clearly, the indices of the permitted k-tuples lying in√ the interval [ x, x) are prime numbers. Therefore, the indices of a pair of permitted k-tuples that lie in the interval [ x, x) and differ by two form a pair of twin primes. In general, a pair of permitted k-tuples whose indices differ by two is called a k-doubleton (see Definition 2.3). 2 2 Therefore, given a real number x > 49 (pk < x ≤ pk+1), it is easy to see that the ordered set of k-tuples whose indices lie in the interval [1, x] is only an alternative formulation of the sieve of Eratosthenes, which was described before by using the usual sieve theory notation. Figure 1 illustrates the sequence of k-tuples for x = 52, where the selected remainders are circled, and the permitted k-tuples are colored gray. Note that this form of the sieve of Eratosthenes gives us a detailed picture of the sifting process. Since the sequence of k-tuples is periodic, it suffices to consider the first period of the sequence, between n = 1 and n = mk. Note that given k ≥ 4, the fundamental period of the sequence of k-tuples is itself a sieve for the integers in the interval [1, mk], that sift out the indices of the prohibited k-tuples that lie in this interval; the indices that have survived the sifting process are the corresponding to the permitted k-tuples. An important fact is that for an interval whose size is the period mk of this sequence of k-tuples, we can compute the exact number of permitted k-tuples which form k-doubletons within this interval. We shall prove that

k Y 2 (ph − 2) (2) h=3

counts the permitted k-tuples which form k-doubletons within the interval [1, mk], k ≥ 2 (Proposition 2.5).

7 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 ...

5 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 ...

3 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 ... k 2 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 ...

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 ...

... 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0

... 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0

... 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0

... 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

... 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210

Figure 2

Let us consider the sequence of k-tuples of remainders in horizontal position (see [9, Deﬁnition 3.1]). (Figure 2 shows the ﬁrst period of the sequence in the horizontal position, for k = 4.) Suppose that we subdivide this period 2 2 into two intervals, as follows: the left interval I[1, pk], and the right interval I[pk + 1, mk]. We have now that the

3 2 period has been partitioned into two blocks: the left block formed by the columns from n = 1 to n = pk; and the right 2 block formed by the columns from n = pk + 1 to n = mk. The following question may have occurred to the reader at this point: What is the advantage of the formulation of sieves based on a sequence of k-tuples of remainders? We shall explain the principal reason in what follows. 2 Let us consider again the sequence of k-tuples in horizontal position, where k ≥ 4. The interval I[1, pk] of this 2 sequence is a sieve device that sifts out the prohibited k-tuples that lie in I[1, pk] and allows the permitted k-tuples in this interval to remain. Furthermore, for every h (1 ≤ h < k), there is also a sequence of h-tuples of remainders 2 and the interval I[1, pk]h of every sequence is also a sieve device that sifts out the prohibited h-tuples and allows the 2 permitted h-tuples in I[1, pk]h to remain. Thus, 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 these partial sieve devices to determine the behaviour of the overall sieve; the advantage of this perspective will become apparent in the rest of this section. It is obvious 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 entire sifting process. In particular, in the case of the prime twins problem, there is another important advantage of the formulation of the sieve of Eratosthenes by means of a sequence of k-tuples of remainders. As we have seen, the sieve of Eratosthenes is given by√ the ordered set of k-tuples whose indices lie in the interval [1, x] and the indices of the permitted k√-tuples that lie in [ x, x) are all the primes in this interval; so, these indices can be twin primes, or other primes in [ x, x). However, given the sequence of k-tuples of remainders, the number of permitted k-tuples which form k-doubletons within the period of the sequence can be computed separately from the others permitted k-tuples (see (2)). Therefore, we can precisely compute the number of permitted h-tuples forming h-doubletons within the interval I[1, mk]h of every sequence of h-tuples from h = 1 to h = k. Thus, we can obtain the density of permitted h-tuples forming h-doubletons within each of those intervals.

Table 1: Values of ck, ck/mk and δk.

k pk mk ck ck/mk δk 4 7 210 30 0.142 1.000 5 11 2310 270 0.116 1.286 6 13 30030 2970 0.098 1.286 7 17 510510 44550 0.088 1.484 8 19 9699690 757350 0.078 1.484 9 23 223092870 15904350 0.072 1.640 10 29 - - 0.066 1.924 11 31 - - 0.062 1.924 12 37 - - 0.058 2.174 13 41 - - 0.056 2.290 14 43 - - 0.054 2.290 15 47 - - 0.050 2.398 16 53 - - 0.048 2.602 17 59 - - 0.048 2.798 18 61 - - 0.046 2.798 19 67 - - 0.044 2.980 20 71 - - 0.044 3.070 21 73 - - 0.042 3.070 22 79 - - 0.040 3.238 23 83 - - 0.040 3.320 24 89 - - 0.038 3.480 ......

We shall prove that the density of permitted k-tuples forming k-doubletons within the period I[1, mk] of the sequence of k-tuples, denoted by δk, increases and tends to infinity as k → ∞ (Lemma 3.2 and Theorem 3.4). For some values of k, Table 1 gives the number of permitted k-tuples forming k-doubletons within I[1, mk], denoted by ck, the ratio ck/mk, and δk. Let us consider a given partial sum Sk. Clearly, the density of permitted h-tuples forming h-doubletons in the interval I[1, mk]h of every partial sum Sh (1 ≤ h ≤ k) is δh, which increases from h = 1 to h = k for sufficiently large k. On the other hand, we shall prove that the behaviour of the density of permitted h-tuples forming h-doubletons 2 in every interval I[1, pk]h from h = 1 to h = k is rather similar to the behavior of δh, for sufficiently large k. Hence, 2 we can compute a lower bound for the density of permitted k-tuples forming k-doubletons in the interval I[1, pk], for sufficiently large k. Thus, from this result, we can prove the Main theorem.

4 2 Periodic sequences of k-tuples

In this section we formally deﬁne the formulation of the sieve of Eratosthenes based in a sequence of k tuples of P remainders. Let Sk be a partial sum of the series sk. The rule for selecting remainders in the sequences sh (1 ≤ h ≤ k) that constitute every partial sum Sk is the following.

Deﬁnition 2.1. Let sh (1 ≤ h ≤ k) be one of the sequences of remainders that form the partial sum Sk. In every sequence of remainders sh, the remainder 0 will be the unique selected remainder. Now, we are ready to formally deﬁne the following sieve, which we call simply Sieve E.

Deﬁnition 2.2. Let P be the sequence of all primes, and let pk (k ≥ 4) be a prime of the sequence. Let A be the set consisting of the indices n of the partial sum Sk that lie in the interval [1, y], where y is a real number that satisﬁes y > pk. For each p = ph ∈ P (1 ≤ h ≤ k), the subset Ap of A consists of the indices n whose remainder modulo p = ph is the selected remainder 0. 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 1.1. The sifting function

[ S(A , P, pk) = A \ Ap ,

p∈P p≤pk is given by the number of permitted k-tuples whose indices belong to the set A . √ Let x > 49 be a real number, and let pk be the greatest prime less than x. Let us consider the Sieve E and the 2 2 associated partial sum Sk (k ≥ 4). Assume A = {n : 1 ≤ n ≤ x}, where pk < x ≤ pk+1. The set of consecutive k-tuples of the partial sum Sk, whose indices lie in the interval [1, x], is an alternative formulation of the sieve of Eratosthenes; that is, the unsifted elements in the sieve of Eratosthenes are the indices of the permitted k-tuples that lie in the interval [1, x]. P In the following theorems we prove some other properties of the partial sums of the series sk, which will be used throughout this paper.

Proposition 2.1. Let Sk be a given partial sum, where k > 1. The ﬁrst k-tuple and the penultimate k-tuple of the period of Sk are permitted k-tuples.

Proof. Let sh (1 ≤ h ≤ k) be the sequences of remainders that form the partial sum Sk. Since 1 6≡ 0 (mod ph) for every ph ∈ p1, p2, p3, . . . , pk, the first k-tuple of the period has no element equal to 0. Therefore, by definition, the first k-tuple of the period is a permitted k-tuple. On the other hand, it is easy to check that all the elements of the last k-tuple of the period are zeroes. So, the penultimate k-tuple of the period has no element equal to 0. Therefore, by definition, is a permitted k-tuple as well.

Deﬁnition 2.3. Given a partial sum Sk (k > 1), we deﬁne a k-doubleton to be a pair of permitted k-tuples whose indices are n and n+2 respectively. We call these permitted k-tuples the First k-tuple and the Last k-tuple respectively.

Note . In the partial sum S1 the indices of the permitted 1-tuples are the odd numbers; so, the concept of 1-doubleton does not make sense.

Remark 2.1. Note that given a period of the partial sum Sk (k ≥ 2), by Proposition 2.1, the penultimate k-tuple of this period and the ﬁrst k-tuple of the next period form a k-doubleton.

Deﬁnition 2.4. Given a period of the partial sum Sk, the ﬁrst k-tuple and the penultimate k-tuple of this period are counted as part of a k-doubleton.

Remark 2.2. Let S2 be the partial sum of level k = 2. The period of S2 is m2 = p1p2 = 6; and it is easy to check that there are two permitted 2-tuples, at positions n = 1 and n = 5.

Proposition 2.2. Let Sk be a given partial sum, where k > 1. Let sk+1 be the sequence of remainders modulo pk+1. Let n and n + 2 be the indices of the First k-tuple and the Last k-tuple of a given k-doubleton of Sk. Then, when we juxtapose the remainders of the sequence sk+1 to each k-tuple of Sk, if the First k-tuple is removed (see [9, Deﬁnition 6.2]), the Last k-tuple is not removed; and vice versa.

Proof. In the sequence sk+1, the only selected remainder is 0. Assume that both the First k-tuple and the Last k-tuple are removed; this means that n ≡ 0 (mod pk+1), and n + 2 ≡ 0 (mod pk+1), by [9, Proposition 2.2]. It follows that 2 ≡ 0 (mod pk+1). But this is not possible since k > 1; it follows that if the First k-tuple is removed, the Last k-tuple is not removed; and vice versa.

5 Deﬁnition 2.5. When we juxtapose the remainders of the sequence sk+1 to each k-tuple of Sk (k > 1), if either the First k-tuple or the Last k-tuple of a k-doubleton is removed, we say that at the level transition k → k + 1 one k-doubleton is removed.

To get a period of the partial sum Sk+1, we ﬁrst take pk+1 periods of the partial sum Sk. Let us consider the permitted k-tuples that form k-doubletons (the First or the Last k-tuples of a given k-doubleton) and are within the ﬁrst pk+1 periods of the partial sum Sk. The following proposition shows that the distribution of all these permitted k-tuples over the residue classes modulo pk+1 is uniform.

Proposition 2.3. The permitted k-tuples that form k-doubletons and are within the ﬁrst pk+1 periods of the partial sum Sk, are uniformly distributed over the residue classes modulo pk+1.

Proof. Step 1. Let [y] = [0], [1], [2],..., [pk+1 − 1] be the residue classes modulo pk+1. Let n be the index of a given permitted k-tuple within the ﬁrst period of the partial sum Sk. Thus, within the ﬁrst pk+1 periods of the 0 partial sum Sk there are pk+1 permitted k-tuples with indices n = mkx + n, where x = 0, 1, 2, 3, . . . , pk+1 − 1 represents each period. Because (mk, pk+1) = 1, for each residue class [y] the congruence mkx + n ≡ y (mod pk+1) has a unique solution x.

Step 2. Let c be the number of permitted k-tuples which form k-doubletons, within a period of Sk. So, there are pk+1c of these permitted k-tuples within the ﬁrst pk+1 periods of the partial sum Sk. Now, from Step 1 follows that there are c of these permitted k-tuples within each residue class modulo pk+1. Thus, the pk+1c permitted k-tuples which form k-doubletons within the ﬁrst pk+1 periods of the partial sum Sk, are uniformly distributed over the residue classes modulo pk+1.

0 0 Corollary 2.4. If there are m consecutive periods of the partial sum Sk, where m is a multiple of pk+1, the permitted k-tuples that form k-doubletons within these m0 periods are also uniformly distributed over the residue classes modulo pk+1. The following proposition gives us an important formula for the number of permitted k-tuples forming k-doubletons within a period of Sk. First, a deﬁnition.

Deﬁnition 2.6. For a given partial sum Sk (k > 1), we denote by ck the number of permitted k-tuples which form k-doubletons within a period of Sk.

Proposition 2.5. Let Sk be a given partial sum (k > 1). We have: ck = 2(p3 − 2)(p4 − 2) ··· (pk − 2).

Proof. Let sh (1 ≤ h ≤ k) be the sequences that form Sk. By definition, we have Sk = Sk−1 + sk. Step 1. By definition, the number of permitted (k − 1)-tuples that form (k − 1)-doubletons within a period of the partial sum Sk−1 is ck−1. To construct the period of the partial sum Sk, we first take pk periods of Sk−1; so, there are pkck−1 of these permitted (k − 1)-tuples within the first pk periods of Sk−1.

Step 2. Now, when we juxtapose the remainders of the sequence sk to each (k−1)-tuple of Sk−1, by [9, Proposition 2.2], the permitted (k−1)-tuples whose indices are included in the residue class 0 modulo pk are removed, since 0 is a selected remainder within the period of sk, by definition. Among the permitted (k−1)-tuples removed, only one of them can belong to a given (k −1)-doubleton (see Proposition 2.2). Since, by Proposition 2.3, the permitted (k − 1)-tuples that form (k − 1)-doubletons within the first pk periods of Sk−1 are distributed uniformly over the residue classes modulo pk, a fraction 1/pk of these permitted (k − 1)-tuples are removed. However, by definition, if either the First k-tuple or the Last k-tuple of a k-doubleton is removed, the complete k-doubleton is removed. Hence, using Step 1, the number of permitted (k − 1)-tuples that form (k − 1)-doubletons, and are removed at the level transition (k − 1) → k is (pkck−1)2/pk = 2ck−1. Since by Step 1 the original number of permitted (k − 1)-tuples that form (k − 1)-doubletons within the first pk periods of Sk−1 is pkck−1, it follows that ck = pkck−1 − 2ck−1 = ck−1(pk − 2) of the permitted (k − 1)-tuples that form (k − 1)-doubletons are transferred to level k as permitted k-tuples that form k-doubletons within the period of the partial sum Sk.

Step 3. Let S2 be the partial sum of level k = 2. In a period of S2 there are two permitted 2-tuples (see Remark 2.1); by deﬁnition, each one of these permitted 2-tuples is counted as part of a 2-doubleton. So, the number of permitted 2-tuples that form 2-doubletons within every period of S2 is c2 = 2. From this and Step 2 follows that the number of permitted k-tuples forming k-doubletons within a period of the partial sum Sk is given recursively by the formula

c2 = 2,

ck = ck−1 (pk − 2) .

It follows that ck = 2(p3 − 2)(p4 − 2) ··· (pk − 2).

6 3 The density of permitted k-tuples which form k-doubletons

In this section, we define the concept of the density of permitted k-tuples which form k-doubletons, and we prove some properties with regard to the behaviour of this quantity. P Definition 3.1. Let Sk (k > 1) be a given partial sum of the series sk; let I[m, n] be a given interval of k-tuples. I[m,n] We denote by ck the number of permitted k-tuples forming k-doubletons within I[m, n]. Remark 3.1. Given a k-doubleton, the First k-tuple could be inside the interval I[m, n], while the Last k-tuple is I[m,n] outside; so, the First k-tuple is counted in ck , and the Last k-tuple is not. On the other hand, the Last k-tuple I[m,n] could be inside the interval I[m, n], while the First k-tuple is outside; in this case the Last k-tuple is counted in ck , and the First k-tuple is not. P Definition 3.2. Let Sk (k > 1) be a partial sum of the series sk; let I[m, n] be a given interval of k-tuples. The number of subintervals of size pk in this interval is equal to |I[m, n]|/pk. We define the density of permitted k-tuples which form k-doubletons in the interval I[m, n] by

I[m,n] I[m,n] ck δk = . |I [m, n]| /pk

I[] For the empty interval we define δk = 0. Note . The density of permitted k-tuples which form k-doubletons is the average number of these permitted k-tuples inside subintervals of size pk. P Definition 3.3. Let Sk (k > 1) be a given partial sum of the series sk; let mk be the period of Sk. Recall that I[1,mk] we have used the notation ck = ck for the number of permitted k-tuples forming k-doubletons within the interval I[1,mk] I[1, mk] (the first period of Sk). We normally use the notation δk = δk for the density of permitted k-tuples which form k-doubletons within the interval I[1, mk]. Since mk/pk is the number of subintervals of size pk within a period of Sk, by definition, we have

ck δk = . mk/pk

The following lemma gives a formula for computing δk. P Lemma 3.1. Let Sk (k > 1) be a given partial sum of the series sk. We have

p1 − 1 p2 − 2 p3 − 2 pk−1 − 2 δk = 2 ··· (pk − 2) . p1 p2 p3 pk−1

Proof. The size of the period of the partial sum Sk is equal to mk = p1p2p3 ··· pk−1pk (see [9, Proposition 2.1]). Therefore, the number of subintervals of size pk is equal to (p1p2p3 ··· pk−1pk)/pk = p1p2p3 ··· pk−1. On the other hand, the number of permitted k-tuples forming k-doubletons within a period of Sk is equal to ck = 2(p3 − 2)(p4 − 2) ··· (pk−1 − 2)(pk − 2) (see Proposition 2.5). Consequently, by deﬁnition, we obtain

2 (p3 − 2) (p4 − 2) ··· (pk−1 − 2) (pk − 2) 2 (p1 − 1) (p2 − 2) (p3 − 2) ··· (pk−1 − 2) (pk − 2) δk = = = p1p2p3 ··· pk−1 p1p2p3 ··· pk−1 p1 − 1 p2 − 2 p3 − 2 pk−1 − 2 = 2 ··· (pk − 2) . p1 p2 p3 pk−1

The next lemma proves that δk is increasing if k > 1. P Lemma 3.2. Let Sk and Sk+1 be consecutive partial sums of the series sk, where k > 1. If δk denotes the density of permitted k-tuples which form k-doubletons within a period of Sk, and δk+1 denotes the density of permitted (k + 1)-tuples which form (k + 1)-doubletons within a period of Sk+1, then

pk+1 − 2 δk+1 = δk . pk

7 Proof. Taking the quotient δk+1/δk and simplifying, the proof follows immediately.

Corollary 3.3. By Lemma 3.2,

1. pk+1 − pk = 2 =⇒ δk+1 = δk.

2. pk+1 − pk > 2 =⇒ δk+1 > δk.

Now we prove that δk → ∞ as k → ∞. First, we make a deﬁnition.

Deﬁnition 3.4. Let pk > 2 and pk+1 be consecutive primes. We denote by θk the diﬀerence pk+1 − pk − 2.

Theorem 3.4. Let Sk (k > 1) be a given partial sum. Let δk be the density of permitted k-tuples which form k-doubletons within a period of Sk. As k → ∞, we have δk → ∞. Proof. By Lemma 3.1,

p1 − 1 p2 − 2 p3 − 2 p4 − 2 p5 − 2 pk−1 − 2 δk = 2 ··· (pk − 2) . p1 p2 p3 p4 p5 pk−1 If we shift denominators to the right, we obtain

p2 − 2 p3 − 2 p4 − 2 p5 − 2 pk−1 − 2 pk − 2 δk = 2 (p1 − 1) ··· . p1 p2 p3 p4 pk−2 pk−1

By deﬁnition, θk = pk+1 − pk − 2 =⇒ pk+1 − 2 = pk + θk. Consequently, we can write the expression of δk as

p2 + θ2 p3 + θ3 p4 + θ4 pk−2 + θk−2 pk−1 + θk−1 δk = ··· = p2 p3 p4 pk−2 pk−1 θ θ θ θ θ = 1 + 2 1 + 3 1 + 4 ··· 1 + k−2 1 + k−1 = p2 p3 p4 pk−2 pk−1 2 1 θ θ θ θ θ p = 1 + 1 + 2 1 + 3 1 + 4 ··· 1 + k−1 1 + k k . 3 p1 p2 p3 p4 pk−1 pk pk + θk Then

" ∞ # 2 1 Y θk pk lim δk = 1 + 1 + lim . (3) k→∞ 3 p1 pk k→∞ pk + θk k=2 The inﬁnite product between square brackets diverges if the series

∞ 1 X θk + (4) p1 pk k=2

diverges. In the series (4), if pk is the first of a pair of twin primes, by definition we have θk = 0; otherwise P∞ we have θk ≥ 2. Let j=1 1/qj denote the series where every prime qj is the first of a pair of twin primes. Since P∞ the series of reciprocals of the twin primes converges [3], the series j=1 1/qj also converges. Therefore, the series P∞ P∞ P∞ P∞ P∞ k=1 1/pk − j=1 1/qj diverges, because k=1 1/pk diverges. By comparison with the series k=1 1/pk − j=1 1/qj, it follows that the series (4) diverges, because θk/pk > 1/pk for the terms where θk > 0. Thus, the infinite product in (3) tends to ∞ as well. On the other hand, by the Bertrand-Chebyshev theorem, pk < pk+1 < 2pk =⇒ θk < pk =⇒ pk/(pk + θk) > 1/2. Consequently, δk → ∞ as k → ∞.

P Let Sk (k > 1) be a partial sum of the series sk. We recall that pk denotes its characteristic prime modulus, and mk denotes its period. In addition, recall that ck denotes the number of permitted k-tuples which form k-doubletons, and δk denotes the density of permitted k-tuples forming k-doubletons, within the period of Sk. I[1,n] Let I[1, n](n ≥ mk) be a given interval of k-tuples of the partial sum Sk. We denote by ck the number of I[1,n] permitted k-tuples forming k-doubletons, and by δk the density of permitted k-tuples which form k-doubletons I[1,n] within I[1, n]. The following proposition shows that δk tends to δk as n → ∞.

8 I[1,n] Proposition 3.5. As n → ∞, the density δk converges to δk. Proof. By deﬁnition,

I[1,n] I[1,n] ck δk = n . pk

Since n ≥ mk, we have a ﬁrst part of the interval I[1, n] whose size is a multiple of mk. Let us denote by cη the number of permitted k-tuples which form k-doubletons within this part of I[1, n]. So, bn/mkcmk/pk is the number of subintervals of size pk within this ﬁrst part of the interval I[1, n]; multiplying by the density of permitted k-tuples which form k-doubletons within the period of Sk, we obtain

j n k mk mk cη = δk. pk

On the other hand, let us denote by c the number of permitted k-tuples forming k-doubletons within the remainder I[1,n] part of I[1, n]. Since ck is the number of permitted k-tuples which form k-doubletons within I[1, n], we have I[1,n] ck = cη + c. Then

j k n j k m mk n I[1,n] k mk c c + c δk + c mk p c δI[1,n] = k = η = pk = δ + k . k n n n n k n pk pk pk

Now, as n → ∞, the values mk, pk, ck and δk are constants; and c is bounded by ck. Therefore, in the expression I[1,n] I[1,n] of δk , the factor that multiplies δk tends to one, and the second term tends to zero. This implies that δk converges to δk as n → ∞.

4 The behaviour of the density of permitted h-tuples forming h-doubletons between h = 2 and h = k P Suppose given a partial sum Sk (k > 1) of the series sk. Let sh (1 ≤ h ≤ k) be the periodic sequences of remainders that form Sk. Let mh be the period of every partial sum Sh from level h = 1 to level h = k. Let ch be the number of permitted h-tuples that form h-doubletons, and let δh be the density of permitted h-tuples that form h-doubletons, within the period of every partial sum Sh (2 ≤ h ≤ k). We need the following result, which is a corollary of Proposition 2.5. P Lemma 4.1. Let Sk (k ≥ 4) be the partial sum of the series sk associated to the Sieve E. Let us consider the interval I[1, mk]h in every partial sum Sh, from level h = 2 to level h = k. Let us denote by mh the period of the partial sum Sh, and by ch the number of permitted h-tuples forming h-doubletons within a period of the partial sum Sh (2 ≤ h ≤ k). For any given partial sum Sh (2 ≤ h < k), the number of permitted h-tuples forming h-doubletons within the interval I[1, mk]h is equal to chph+1ph+2 ··· pk.

Proof. Choose a level h (2 ≤ h < k). By deﬁnition, we have mk = p1p2p3 ··· phph+1ph+2 ··· pk = mhph+1ph+2 ··· pk. That is, the size of the interval I[1, mk]h of the partial sum Sh is equal to ph+1ph+2 ··· pk times the period mh of the partial sum Sh. Consequently, it is easy to see that the number of permitted h-tuples forming h-doubletons within the interval I[1, mk]h is equal to chph+1ph+2 ··· pk.

0 Now, let us denote by ch the number of permitted h-tuples forming h-doubletons within the interval I[1, mk]h of 0 every partial sum Sh (1 ≤ h ≤ k), which is computed using Proposition 2.5 and Lemma 4.1. The behaviour of ch as h goes from level 1 to level k is given by the following lemma. P Lemma 4.2. Let Sk (k ≥ 4) be the partial sum of the series sk associated to the Sieve E. Let us consider the interval I[1, mk]h in every partial sum Sh, from level h = 2 to level h = k. Let h and h + 1 be consecutive levels, where 1 ≤ h < k. We have

0 0 ph+1 − 2 ch+1 = ch . ph+1

9 Proof. Given the partial sum Sh (1 ≤ h < k), suppose that we juxtapose the remainders of the sequence sh+1 to each h-tuple of Sh. By [9, Proposition 2.2]), the permitted h-tuples forming h-doubletons within the interval I[1, mk]h whose indices are included in the residue class 0 modulo ph+1 are removed. Furthermore, for every permitted h-tuple forming a given h-doubleton that is removed, the other permitted h-tuple forming this h-doubleton is also ‘removed’ in the sense that it is transferred to level h + 1 as permitted (h + 1)-tuple within the interval I[1, mk]h+1 of the partial sum Sh+1 but is not part of any (h + 1)-doubleton in I[1, mk]h+1. The other permitted h-tuples forming h-doubletons whose indices are not included in the residue class 0 modulo ph+1 are transferred to level h + 1 as permitted (h + 1)- tuples forming (h + 1)-doubletons within the interval I[1, mk]h+1 of the partial sum Sh+1. Since for every level h < k, the size of the interval I[1, mk]h is a multiple of ph+1, by Proposition 2.3 and Corollary 2.4, the permitted h-tuples forming h-doubletons within the interval I[1, mk]h of Sh are distributed uniformly over the residue classes modulo ph+1. Therefore, a 2/ph+1 fraction of the permitted h-tuples forming h-doubletons within the interval I[1, mk]h of Sh have been removed, and a (ph+1 − 2)/ph+1 fraction have been transferred to level h + 1 as permitted (h + 1)-tuples forming (h + 1)-doubletons within the interval I[1, mk]h+1 of Sh+1. From this result, the lemma follows.

Let us now examine the behaviour of δh as h goes from level 1 to level k. Since the selected remainder of the sequence sh+1 remove permitted h-tuples forming h-doubletons within the interval I[1, mk]h of the partial sum Sh, at each level transition h → h+1, the number of permitted h-tuples forming h-doubletons decreases as the level increases 0 0 from h = 1 to h = k (the factor by which we must multiply ch to obtain ch+1 is given by Lemma 4.2). However, by Lemma 3.2 and Corollary 3.3, the density of permitted h-tuples forming h-doubletons within the interval I[1, mk]h of the partial sum Sh grows at each transition ph → ph+1 of order greater than 2; and if ph → ph+1 is a level transition of order 2, the density of permitted h-tuples forming h-doubletons within I[1, mk]h does not change.

Remark 4.1. Note that if ph → ph+1 is a level transition of order greater than 2, δh increases because to compute the density of permitted h-tuples forming h-doubletons we count this permitted h-tuples within subintervals of size ph, which grow by more than 2, overcompensating for the removed permitted h-tuples forming h-doubletons. On the other hand, if ph → ph+1 is a level transition of order 2, δh does not change because the increase in the size ph is compensated for by the removed permitted h-tuples forming h-doubletons. Note that the result from Lemma 3.2 can be written in the form

ph+1 − 2 ph+1 δh+1 = δh , ph+1 ph

where the factor (ph+1 −2)/ph+1 is related to the permitted h-tuples forming h-doubletons removed from I[1, mk]h (see Lemma 4.2), and the factor ph+1/ph is related to the increase of the size of the subintervals.

Lk Deﬁnition 4.1. For a given partial sum Sh (2 ≤ h ≤ k), we use the notation ch to denote the number of permitted 2 Rk h-tuples that form h-doubletons within the Left interval I[1, pk]h, and we use the notation ch to denote the number 2 of permitted h-tuples that form h-doubletons within the Right interval I[pk + 1, mk]h (see [9, Deﬁnition 6.3]). We use Lk 2 the notation δh to denote the density of permitted h-tuples that form h-doubletons within the Left interval I[1, pk]h, Rk and we use the notation δh to denote the density of permitted h-tuples that form h-doubletons within the Right 2 interval I[pk + 1, mk]h.

Recall that we use the notation δk for the density of permitted k-tuples that form k-doubletons within the interval I[1, mk]. Now, choose a level h such that 2 ≤ h < k. As we have seen in the proof of Lemma 4.1, we have mk = mhph+1ph+2 ··· pk; in other words, the size of the interval I[1, mk]h is equal to ph+1ph+2 ··· pk times the period mh of the partial sum Sh. Consequently, it is easy to see that the density of permitted h-tuples that form h-doubletons within the interval I[1, mk]h is equal to δh. Hence, for every level h (2 ≤ h ≤ k), we have

Lk Rk δh > δh ⇐⇒ δh < δh, (5)

Lk Rk δh < δh ⇐⇒ δh > δh.

5 Proof of the Main Theorem

For every partial sum Sh from level h = 1 to level h = k (k ≥ 4), let us consider the interval I[1, mk]h, the Left interval 2 2 I[1, pk]h and the Right interval I[pk + 1, mk]h. Recall the notation δh for the density of permitted h-tuples forming Lk Rk h-doubletons within the period of Sh, and recall the notation δh and δh for the density of permitted h-tuples forming 2 2 h-doubletons within the intervals I[1, pk]h and I[pk + 1, mk]h, respectively, where 2 ≤ h ≤ k. In this section, we shall prove that there exists a suﬃciently large integer Kα > 4 such that the number of permitted 2 k-tuples forming k-doubletons within the Left interval I[1, pk] of the partial sum Sk is greater than pk for every k > Kα.

10 P Definition 5.1. Let Sk be a given partial sum of the series sk. Given the level transition ph → ph+1 (2 ≤ h < k), Lk 2 we denote by Fh,h+1 the fraction of the permitted h-tuples forming h-doubletons within the Left interval I[1, pk]h of 2 Sh that are transferred to the Left interval I[1, pk]h+1 of Sh+1 as permitted (h + 1)-tuples forming (h + 1)-doubletons. P Remark 5.1. Let Sk be a given partial sum of the series sk. Suppose that in every level transition ph → ph+1 (2 ≤ Lk h < k), the value of Fh,h+1 is very close to the fraction (ph+1 −2)/ph+1, given by Lemma 4.2 for the intervals I[1, mk]h. Lk In this case, if k is sufficiently large, δh must increase between h = 2 and h = k, by the arguments given in Remark 4.1 for δh. Now, suppose that there exists a level h = h0 (1 < h0 < k), where the difference k −h0 is sufficiently large, such that Lk 0 δh does not increase between h = h and h = k. Clearly, this can only happen if there are level transitions ph → ph+1 0 Lk between h = h and h = k for which Fh,h+1 is too far below the fraction (ph+1 − 2)/ph+1 given by Lemma 4.2. P 0 0 Remark 5.2. Let Sk be a partial sum of the series sk, and consider a given level h = h (1 < h < k). Given consecutive primes ph and ph+1, we denote gh = ph+1 − ph. Assume that in each period (of size ph) of every sequence 0 sh from h = h + 1 to h = k of Sk there are gh selected remainders (the same in every period of the sequence). 0 As we have seen in [9, Remark 6.2], the selected remainders in every sequence sh (h < h ≤ k) remove permitted (h−1)-tuples from the partial sum Sh−1. However, under the preceding assumption, the fraction of permitted h-tuples forming h-doubletons within the interval I[1, mk]h of Sh that are transferred to the interval I[1, mk]h+1 of Sh+1 as 0 permitted (h + 1)-tuples forming (h + 1)-doubletons is equal to (ph+1 − gh)/ph+1, where h ≤ h < k (see the proof of Lemma 4.2). So, for sufficiently large k, the whole number of permitted h-tuples forming h-doubletons removed between h = h0 and h = k is rather greater than in the case where there is only one selected remainder in every period 0 0 of the sequences sh (h < h ≤ k). Furthermore, in every level transition ph → ph+1 (h ≤ h < k), the increase of δh under this assumption would be given by

ph+1 − gh ph+1 δh+1 = δh = δh, ph+1 ph (see Remark 4.1). This means that under the preceding assumption, the density of permitted h-tuples forming 0 h-doubletons within the interval I[1, mk]h of Sh does not increase between h = h and h = k. 2 Lemma 5.1. Let Sk (k > 4) be the partial sum associated to the Sieve E. Let I[1, pk]h be the Left interval for every Lk partial sum Sh from h = 1 to h = k. Furthermore, let δh be the density of permitted h-tuples forming h-doubletons 2 Lk within every Left interval I[1, pk]h (2 ≤ h ≤ k). There exists an integer Kα > 4 such that δk > δ4 for every k > Kα.

Proof. Step 1. Let us consider the density of permitted h-tuples forming h-doubletons within every interval I[1, mk]h from h = 2 to h = k, denoted by δh. Recall that, as the level increases from h = 2 to h = k, the value of δh increases at each level transition ph → ph+1 of order greater than 2, and, if ph → ph+1 is a level transition of order 2, δh does not change, as we have seen in the paragraph below Lemma 4.2 and in Remark 4.1. 2 Lk The fact that I[1, pk]h ⊂ I[1, mk]h (1 ≤ h ≤ k) suggests that for every h (2 ≤ h ≤ k) the value of δh is relatively close to δh, if k is sufficiently large. Thus, it seems reasonable to assume that for k sufficiently large Lk it must be δk > δ4. In the following steps we shall prove that, indeed, the preceding assumption is true. 0 0 Step 2. Consider Sk (k > 5) and choose a fixed level h = h (4 < h < k). (Note that δh0 > δ4, by Corollary 3.3.) Let ε > 0 be a very small real number. Let Kα be a positive integer sufficiently large that for every k > Kα the two following conditions are satisfied:

2 (a) The size of the Left interval I[1, pk]h0 is suﬃciently large that

Lk δh0 − ε < δh0 < δh0 + ε. (See Proposition 3.5.) (b) The number and proportion of level transitions h → h + 1 of order greater than 2 between h = h0 and h = k are very large.

0 Step 3. Let us consider the interval I[1, mk]h in every partial sum Sh from h = h to h = k. Under the condition assumed in Remark 5.2 (with regard to the selected remainders in the sequences sh), for every level transition 0 ph → ph+1 between h = h and h = k, the fraction of permitted h-tuples forming h-doubletons within the interval I[1, mk]h of Sh that are transferred to the interval I[1, mk]h+1 of Sh+1 as permitted (h + 1)-tuples 0 forming (h + 1)-doubletons is equal to (ph+1 − gh)/ph+1, where h ≤ h < k. Hence, in this case, for every 0 level transition ph → ph+1 of order greater than 2 (that is gh > 2) between h = h and h = k, the number of permitted h-tuples forming h-doubletons removed from I[1, mk]h, is rather greater than in the case where 0 there is only one selected remainder in every period of the sequences sh (h < h ≤ k), so, the density of 0 permitted h-tuples forming h-doubletons within I[1, mk]h does not increase as h goes from level h to level k (see Remark 5.2).

11 2 0 Step 4. Now, let us consider the Left intervals I[1, pk]h from h = h to h = k, of the partial sum Sk (k > Kα) associated to the Sieve E. On the one hand, by Step 2 (a) the density of permitted h0-tuples forming h0- 2 Lk doubletons within the Left interval I[1, pk]h0 , denoted by δh0 , is very close to δh0 . On the other hand, in the case of the Sieve E the condition assumed in Remark 5.2 is not present (see Deﬁnition 2.1). Therefore, for 0 Lk every level transition ph → ph+1 between h = h and h = k, clearly, the fraction Fh,h+1 must be much closer to the fraction (ph+1 − 2)/ph+1 given by Lemma 4.2 than to (ph+1 − gh)/ph+1, the fraction obtained under the condition assumed in Remark 5.2 (see the preceding step). Furthermore, by Step 2 (b), the number and proportion of level transitions h → h + 1 of order greater than 2 between h = h0 and h = k is very large (see Lk 0 Step 1). Therefore, δh must increase as h goes from level h to level k > Kα, by the same arguments given in Lk Remark 4.1 that explain the behaviour of δh (see Remark 5.1). Thus, δh will take some value greater than δh0 + ε for every k > Kα, if Kα is suﬁciently large.

Lk Step 5. Clearly, by the preceding step, δk > δh0 > δ4 for every k > Kα, for suﬃciently large Kα. The lemma is proved.

The next lemma shows that for suﬃciently large k, the number of permitted k-tuples that form k-doubletons within 2 the Left interval I[1, pk] of the partial sum Sk is greater than pk.

P Lk Lemma 5.2. Let Sk (k ≥ 5) be a partial sum of the series sk. Let ck denote the number of permitted k-tuples 2 forming k-doubletons within the Left interval I[1, pk] of the partial sum Sk. Let Kα be the number whose existence is Lk established in Lemma 5.1. For every k > Kα we have ck > pk. 2 2 Proof. Since the size of the Left interval I[1, pk] of the partial sum Sk is equal to pk, there are pk subintervals of size Lk Lk Lk pk in this interval; it follows that ck = δk pk. Then, ck > δ4pk for every k > Kα, by Lemma 5.1. Using Lemma 3.1 Lk we can check that δ4 = 1, and so ck > pk for every k > Kα.

Deﬁnition 5.2. Consider the interval I[1, mk] of the partial sum Sk. Given a permitted k-tuple in I[1, mk] which is not part of any k-doubleton, we call it a single permitted k-tuple. Let B be the set consisting of all the indices n 2 of the single permitted k-tuples within the Left interval I[1, pk]. Finally, we prove the Main Theorem.

Theorem 5.3. Main Theorem √ Let x > 49 be a real number, and let pk be the greatest prime less than z = x. Let us consider the Sieve E and the associated partial sum Sk (k ≥ 4). Let Kα be the number whose existence is established in Lemma 5.1. Recall the notation π2(x) for the number of twin primes in [1, x]. We have π2(x) → ∞ as x → ∞.

2 Proof. In the Sieve E, let A = {n : 1 ≤ n ≤ pk}. Clearly, the indices n > 1 of the permitted k-tuples within the Left 2 2 interval I[1, pk] are all the primes that lie in (pk, pk]. On the other hand, the quantity S(A , P, pk) − |B| − 1 is the 2 number of the permitted k-tuples within I[1, pk] whose indices are primes and are forming k-doubletons, that is, is 2 the number of twin primes in (pk, pk]. Now, S(A , P, pk) − |B| > pk for every k > Kα, by Lemma 5.2, so, S(A , P, pk) − |B| − 1 > pk − 1 (k > Kα). Therefore,

π2(x) ≥ S(A , P, pk) − |B| − 1 > pk − 1 (k > Kα), (6)

2 since [1, pk] ⊂ [1, x]. Since pk → ∞ as x → ∞, it follows from (6) that π2(x) → ∞ as x → ∞, and the theorem is proved.

We conclude that the Twin prime conjecture is true.

Acknowledgements

The author wants to thank Olga Vinogradova (Mathematics student, FCEyN, Universidad de Buenos Aires), Dra. Patricia Quattrini (Departamento de Matematicas, FCEyN, Universidad de Buenos Aires), Dr. Ricardo Duran (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.

12 References

[1] H. Halberstam and H.-E. Richert, Sieve Methods, Academic Press, London, 1974. [2] A. C. Cojocaru and M. Ram Murty, An Introduction to Sieve Methods an Their Applications, Cambridge University Press, London, 2005. [3] V. Brun, La série1/5+1/7+1/11+1/13+1/17+1/19+1/29+1/31+1/41+1/43+1/59+1/61+..., oùles dénominateurs sont nombres premieres jumeaux est convergente oùfinie, Bull. Sci. Math., 43, (1919), 124–128. [4] Y. Zhang, Bounded gaps between primes, Ann. of Math., 179, (2014), 1121–1174. [5] Polymath8, Bounded gaps between primes, http://michaelnielsen.org/polymath1/index.php [6] E. W. Weisstein, de Polignac’s Conjecture, MathWorld, http://mathworld.wolfram.com/dePolignacsConjecture.html [7] Terence Tao, Open question: The parity problem in sieve theory, https://terrytao.wordpress.com/2007/06/05/open-question-the-parity-problem-in-sieve-theory/. [8] A.Selberg, The general sieve method and its place in prime number theory, Proc. Int. Congress of Mathematicians Cam- bridge, MA (Providence, RI. American Mathematical Society), 1 (1950), 262–289. [9] R. Barca, Every sufficiently large even number is the sum of two primes, preprint, https://hal.archives-ouvertes.fr/hal-02075531v4. Ricardo G. Barca Universidad Tecnologica Nacional Buenos Aires (Argentina) E-mail address: [email protected]