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# The Twin Prime Conjecture

Ricardo Guillermo Barca

► To cite this version:

Ricardo Guillermo Barca. The Twin Prime Conjecture. 2018. hal-01882333v3

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https://hal.science/hal-01882333v3

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# THE TWIN PRIME CONJECTURE

Ricardo G. Barca

## Abstract

Consecutive primes which differ only by two are known as twin primes; the function $\pi_2(x)$ describes the distribution of the twin primes in the set of positive integers. 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 $\pi_2(x) \to \infty$ as $x \to \infty$.

# 1 Introduction

## 1.1 The Twin prime problem and sieve methods

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 Alphonse 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. For $k = 2$ we have the twin prime conjecture.

One of the most significant results for twin primes is due to V. Brun [BRU], who showed that the sum of the reciprocals of the odd twin primes,

$$ (1/3 + 1/5) + (1/5 + 1/7) + (1/(11) + 1/(13)) + (1/(17) + 1/(19)) + \dots $$

converges to a number known as the Brun's constant; this major result constitutes the first use of Brun's sieve. Since the series of all prime reciprocals diverges to infinity, the result from Viggo Brun say us that twin primes are very scarce.

In 2013, the mathematician Yitang Zhang asserted that there are infinitely many pairs of prime numbers that differ by 70 million or less [ZHA]. For $m \ge 1$, let us denote $H_m = \liminf_{n \to \infty} (p_{n+m} - p_n)$, where $p_n$ denotes the $n$-th prime. In his paper, Zhang proved that $H_1 \le 7 \times 10^7$; the twin prime conjecture is equivalent to the assertion that $H_1$ is equal to two. After the appearance of Zhang's paper, improvements to the upper bound on $H_1$ were made. Furthermore, a collaborative work called the Polymath project was started to examine Zhang [POL]'s arguments; in 2014 the bound has been reduced to 246.

Let $x$ be a positive real number; let $\pi_2(x)$ be the number of twin primes $p$ and $p+2$ such that $p \le x$. In this paper we prove (Main theorem) the following: $\pi_2(x) \to \infty$ as $x \to \infty$.

In the present paper some theorems and the corresponding proofs are similar or almost a copy of other provided in the author's work [BAR]; we repeat them here for convenience of the reader.

One of the ways of dealing with problems like the twin prime conjecture or Goldbach's conjecture is by means of sieve methods; the sieve theory has developed along of the last century, starting from the Viggo Brun methods. Let us begin by considering once more the classical sieve of Eratosthenes; we repeat here some parts of the description given in [BAR, Introduction, subsection 1.1], for convenience of the reader. Let $\mathcal{A} = \{n \in \mathbb{Z}_+ : n \le x\}$, where $x \in \mathbb{R}$, $x > 1$, and let $\mathcal{P}$ be the sequence of all primes. Let $z = \sqrt{x}$ and

$$ P(z) = \prod_{\substack{p \in \mathcal{P} \\ p < z}} p. $$

Now, for each $p \in \mathcal{P}$, $p < z$, we define $\mathcal{A}_p = \{n \in \mathcal{A} : n \equiv 0 \pmod{p}\}$. Then, when we sift from $\mathcal{A}$ all the elements of every set $\mathcal{A}_p$, the unsifted members of $\mathcal{A}$ in $[\sqrt{x}, x)$ are the primes in this interval.

The sifting function $S(\mathcal{A}, \mathcal{P}, z)$ is defined by the equation

$$ S(\mathcal{A}, \mathcal{P}, z) = \left| \mathcal{A} \setminus \bigcup_{\substack{p \in \mathcal{P} \\ p < z}} \mathcal{A}_p \right| $$
and counts the elements of $\mathcal{A}$ that have survived the sifting process. In addition, we define

$$ \mathcal{A}_d = \bigcap_{p|d} \mathcal{A}_p, $$

where $d$ is a squarefree integer such that $d|P(z)$. Then from the inclusion-exclusion principle, we obtain

$$ S(\mathcal{A}, \mathcal{P}, z) = \sum_{d|P(z)} \mu(d) |\mathcal{A}_d|, \qquad (1) $$

where $\mu(d)$ is the Möbius function.

As usual in sieve theory, in place of $|\mathcal{A}|$ we can use a close approximation $X$ to $|\mathcal{A}|$. Furthermore, for each prime $p$ we can use a multiplicative function $w(p)$ such that $(w(p)/p)X$ approximates to $|\mathcal{A}_p|$. Thus, for each squarefree integer $d$, we have that $(w(d)/d)X$ approximates to $|\mathcal{A}_d|$, and we can write

$$ |\mathcal{A}_d| = \frac{w(d)}{d} X + R_d, $$

where $R_d$ is the remainder term. Substituting this result into (1) gives

$$ S(\mathcal{A}, \mathcal{P}, z) = \sum_{d|P(z)} \mu(d) \left( \frac{w(d)}{d} X + R_d \right) = \sum_{d|P(z)} \mu(d) \frac{w(d)}{d} X + \sum_{d|P(z)} \mu(d) R_d. \qquad (2) $$

On the other hand, since $w$ is a multiplicative function,

$$ \sum_{d|P(z)} \mu(d) \frac{w(d)}{d} = \prod_{\substack{p \in \mathcal{P} \\ p < z}} \left( 1 - \frac{w(p)}{p} \right), $$

and substituting this result into (2) we have

$$ S(\mathcal{A}, \mathcal{P}, z) = X \prod_{\substack{p \in \mathcal{P} \\ p < z}} \left(1 - \frac{w(p)}{p}\right) + \sum_{d|P(z)} \mu(d) R_d. $$

Thus, taking $X = x$ we obtain

$$ S(\mathcal{A}, \mathcal{P}, z) = x \prod_{\substack{p \in \mathcal{P} \\ p < z}} \left(1 - \frac{1}{p}\right) + \sum_{d|P(z)} \mu(d) R_d, \qquad (3) $$

since in the case of the sieve of Eratosthenes $w(p) = 1$.

However, for our purposes we need to find a lower estimate on the size of the subset of twin primes between $\sqrt{x}$ and $x$, for every $x$ sufficiently large. We can certainly modify the sieve so that the sifting function gives us the number of twin primes between $\sqrt{x}$ and $x$. One possible option is to use a different set $\mathcal{A}$ such that after the sifting process the elements surviving are twin primes (see [FRI, Chapter 1, Examples 1.3 and 1.4] and [HAL, Chapter 1, Introductory Remarks]); then we should apply the usual sieve techniques for obtaining a lower estimate for the sifting function of this modified sieve. Nevertheless, obtaining a lower bound for the sifting function by the usual procedures of the sieve theory does not seem to be possible due to the parity problem, as we have seen in the last paragraph of [BAR, Introduction, subsection 1.1]. Thus far all attempts to solve the Twin prime problem by this way have not succeeded.

Starting in the next subsection we will formulate the sieve of Eratosthenes by means of a sequence of k-tuples of remainders, and then we will use techniques associated with this class of sequences to isolate the subset of the twin primes and obtain a lower bound for the size of this subset.

From now on we shall use throughout in the present paper concepts, definitions, conventions and results from the author's work [BAR].
## 1.2 Formulation of the sieve of Eratosthenes using a sequence of k-tuples of remainders

Let $\mathcal{P}$ be the sequence of the primes. Henceforth, for convenience, we take $x$ to be a real number greater than $p_4^2 = 49$ unless specified otherwise. Let $z = \sqrt{x}$ and $p_k$ be the greatest prime less than $z$; furthermore let $m_k = p_1p_2p_3\cdots p_k$ (see [BAR, Definition 2.4]). So, every real number $x > 49$ satisfies $p_k^2 < x \le p_{k+1}^2 < m_k$.

<figure><img src="image_2.png" /><figcaption>Figure 1</figcaption></figure>

We formulate the sieve of Eratosthenes by means of a sequence of $k$-tuples (see [BAR, Subsection 1.3]) as follows. 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 the $k$-tuples of the sequence, all the zeroes are defined as selected remainders. Furthermore, as we have seen in [BAR, Subsection 1.3], the sequence of $k$-tuples of remainders (arranged from top to bottom) can be seen as a matrix formed by $k$ columns and infinitely many rows, where each column is a periodic sequence of remainders modulo $p_h$ ($1 \le h \le k$). Thus, within the periods of every sequence of remainders modulo $p_h$ the element 0 is always a selected remainder. Let $\mathcal{A}$ be the set consisting of the indices $n$ of the sequence of $k$-tuples that lie in the interval $[1, x]$. For each $p = p_h \in \mathcal{P}$, $p < z$, the set $\mathcal{A}_p \subset \mathcal{A}$ 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 $\mathcal{A}$ are sifted out, and the indices of the permitted $k$-tuples lying in $\mathcal{A}$ remain unsifted. The sifting function is given by the number of permitted $k$-tuples whose indices lie in the interval $[1, x]$. In Section 2, we more formally define the formulation of this sieve based on a sequence of $k$-tuples.

**Remark 1.1.** Note that given a $k$-tuple whose index is $n \le x$, if $n \equiv 0 \pmod{p}$ for at least one $p < \sqrt{x}$, then it is a prohibited $k$-tuple, and if $n \not\equiv 0 \pmod{p}$ for every $p < \sqrt{x}$, then it is a permitted $k$-tuple.
We can see that given a real number $x > 49$ ($p_k^2 < x \le p_{k+1}^2$), 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 of Eratosthenes, described in the preceding subsection; clearly, the indices of the permitted $k$-tuples lying in the interval $[\sqrt{x}, x)$ are the primes in this interval. 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.

## 1.3 Brief discussion about the main ideas

Recall the notation $I[n, m]$ for the ordered set of $k$-tuples (an interval of $k$-tuples) whose indices lie in the interval $[n, m]$ (see [BAR, Definition 1.1]). Furthermore, recall that for a given interval $I[n, m]$, the average number of permitted $k$-tuples in subintervals of size $p_k$ is called the density of permitted $k$-tuples (see [BAR, Definition 3.3]).

Let us consider the sieve of Eratosthenes in the form of a sequence of $k$-tuples, in horizontal position (see [BAR, Definition 3.1]), for $x$ sufficiently large. Recall that we can consider the first period of the sequence of $k$-tuples (in horizontal position) as a matrix of $k$ rows and $m_k$ columns; furthermore, for every $h$ ($1 \le h < k$) the rows from 1 to $h$ are part of a sequence of $h$-tuples.

Let $I_x \subset \mathcal{A}$ be the set of indices of the permitted $k$-tuples that lie in the interval $[1, x]$; furthermore, let us consider a given subset $I \subseteq I_x$.

Now, as a first case assume that the elements of $I$ are the indices of the permitted $k$-tuples that lie in $[1, x]$, that is $I = I_x$, and suppose that we want to bound from below the size of $I$. Regarding this case, an important fact is that for an interval whose size is the period $m_k$ of the sequence of $k$-tuples we can compute the exact number of permitted $k$-tuples within this interval. Note that the indices of the permitted $k$-tuples in the interval $I[1, m_k]$ are clearly the integers in $\{n : 1 \le n \le m_k\}$ that are coprimes to $m_k$. Thus, the number of permitted $k$-tuples in the interval $I[1, m_k]$ of the sequence of $k$-tuples is given by

$$ \phi(m_k) = m_k \prod_{\substack{p \le p_k \\ p \in \mathcal{P}}} \left(1 - \frac{1}{p}\right). \qquad (4) $$

Furthermore, for every $h$ ($1 \le h < k$) we can also compute precisely the number of permitted $h$-tuples in each interval $I[1, m_k]_h$, since the period $m_h$ divides $m_k$.

We can see that the product

$$ \prod_{\substack{p \le p_k \\ p \in \mathcal{P}}} \left(1 - \frac{1}{p}\right) \qquad (5) $$

appears in both formulas (3) and (4). In the formula (4) the factors (arranged in the order of primes) represent the precise fraction of permitted $h$-tuples in the interval $I[1, m_k]_h$ of $S_h$ that remain unsifted as permitted $(h+1)$-tuples in the interval $I[1, m_k]_{h+1}$ of $S_{h+1}$, after every stage of the sifting process from $h=1$ to $h=k$. On the other hand, given a level $h$ ($1 \le h < k$), the proportion of permitted $h$-tuples whose indices lie in $[1, x]$ and remain unsifted as permitted $(h+1)$-tuples, clearly, is not equal to the proportion of permitted $h$-tuples in $I[1, m_k]_h$ that remain unsifted as permitted $(h+1)$-tuples. For this reason, in the formula (3) the ‘true’ factors (corresponding to the real fraction of permitted $h$-tuples whose indices lie in $[1, x]$ and remain unsifted after every stage of the sifting process) are replaced by the ‘average’ factors that appear in (5), and then an error term appears in (3). (We have repeated here the explanation given for the similar case in [BAR, Subsection 1.8], below the formula [BAR, (7)].)

The usual approach in sieve theory consist in estimating $S(\{n : 1 \le n \le x\}, \mathcal{P}, z)$ by using (3) or some other formula derived from the inclusion-exclusion principle as starting point. However, the limitation imposed by the parity problem appears when we apply the usual methods developed so far in sieve theory. In this paper we proceed in a way quite different from that is usual in sieve theory; we explain this as follows.

It is an obvious fact that after every stage of the sifting process from $h=1$ to $h=k$, the number of permitted $h$-tuples in $I[1, p_k^2]_h$ decreases. This fact makes very difficult (maybe impossible) to directly obtain a lower bound for the number of permitted $k$-tuples in $I[1, p_k^2]$ after the whole sifting process has ended (the motivation for the choice of the interval $[1, p_k^2]$ will be clear later). For this reason it is better to obtain first a lower bound for the density of permitted $k$-tuples in $I[1, p_k^2]$ starting from (4), and then to use this lower bound for computing the lower bound for the number of permitted $k$-tuples in $I[1, p_k^2]$. Thus, suppose that we rewrite the right-hand side of (4) in the expanded form

$$ m_k \prod_{\substack{p \le p_k \\ p \in \mathcal{P}}} \left(1 - \frac{1}{p}\right) = m_k \left(1 - \frac{1}{p_1}\right) \left(1 - \frac{1}{p_2}\right) \left(1 - \frac{1}{p_3}\right) \cdots \left(1 - \frac{1}{p_k}\right). $$
Then, if we shift denominators to the right we obtain

$$m_k \prod_{\substack{p \le p_k \\ p \in \mathcal{P}}} \left(1 - \frac{1}{p}\right) = \frac{m_k}{p_k} \left(\frac{p_2 - 1}{p_1}\right) \left(\frac{p_3 - 1}{p_2}\right) \left(\frac{p_4 - 1}{p_3}\right) \cdots \left(\frac{p_k - 1}{p_{k-1}}\right).$$

Clearly, the quotient $m_k/p_k$ is the number of subintervals of size $p_k$ within the period of the sequence of $k$-tuples, and the product

$$\left(\frac{p_2 - 1}{p_1}\right) \left(\frac{p_3 - 1}{p_2}\right) \left(\frac{p_4 - 1}{p_3}\right) \cdots \left(\frac{p_k - 1}{p_{k-1}}\right)$$

is the density of permitted $k$-tuples within the period of the sequence, denoted by $\delta_k$; furthermore, every partial product in the preceding formula is $\delta_h$ ($2 \le h \le k$).

Now, since there are $p_k$ subintervals of size $p_k$ in the interval $I[1, p_k^2]$ of the sequence of $k$-tuples, multiplying by $p_k$ the density of permitted $k$-tuples in $I[1, p_k^2]$ we obtain the number of permitted $k$-tuples within this interval. Thus, if we find a lower bound for the density of permitted $k$-tuples in $I[1, p_k^2]$, multiplying by $p_k$ this lower bound we can obtain a lower estimate for the size of $I$ (the sifting function of the sieve of Eratosthenes), since $I[1, p_k^2] \subset I[1, x]$. (Note that the density of permitted $k$-tuples in the interval $I[1, p_k^2]$ of the sequence of $k$-tuples should be more or less close to $\delta_k$, for sufficiently large $k$.)

However, we are not interested in all the permitted $k$-tuples whose indices lie in the interval $[1, x]$; indeed, we are interested only in the pairs of permitted $k$-tuples whose indices $n, n + 2$ differ by two and lie in $[1, x]$. (In general, a pair of permitted $k$-tuples whose indices differ by two is called a $k$-doubleton. See Definition 2.3.) Clearly, if there is a pair of permitted $k$-tuples in the sequence, whose indices are $n, n + 2$ ($n > 1$) and lie in $[\sqrt{x}, x)$, we have a pair of twin primes.

Suppose that the elements of $I$ are now the indices of permitted $k$-tuples forming $k$-doubletons that lie in the interval $[1, x]$. In particular, in this case there is an important issue in 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 $[\sqrt{x}, x)$ are all the primes in this interval; so, these indices can be twin primes or other primes in $[\sqrt{x}, x)$. However, given the sequence of $k$-tuples, the exact number of permitted $k$-tuples which form $k$-doubletons within the period of the sequence (that is within $I[1, m_k]$) can be computed separately from the others permitted $k$-tuples. We shall prove that

$$2 \prod_{h=2}^{k} (p_h - 2) \qquad (6)$$

counts the permitted $k$-tuples which form $k$-doubletons within the interval $[1, m_k]$, where $k \ge 2$ (Proposition 2.5). In addition, for every $h$ ($2 \le h < k$), since the number of permitted $h$-tuples forming $h$-doubletons in a period of the sequence of $h$-tuples can be computed using (6) and $m_h|m_k$, we can also compute exactly the number of permitted $h$-tuples forming $h$-doubletons in the interval $I[1, m_k]_h$. On the other hand, from the number of permitted $k$-tuples forming $k$-doubletons within $I[1, m_k]$ given by (6), we can compute the density of permitted $k$-tuples forming $k$-doubletons within the period $I[1, m_k]$ of the sequence of $k$-tuples. We proceed in a similar way to the first case but starting now from (6); thus, multiplying and dividing (6) by $m_k$, we obtain

$$2 \prod_{h=2}^{k} (p_h - 2) = m_k \left(\frac{p_2 - 2}{p_2}\right) \left(\frac{p_3 - 2}{p_3}\right) \left(\frac{p_4 - 2}{p_4}\right) \cdots \left(\frac{p_k - 2}{p_k}\right),$$

and shifting denominators to the right,

$$2 \prod_{h=2}^{k} (p_h - 2) = \frac{m_k}{p_k} \left(\frac{p_3 - 2}{p_2}\right) \left(\frac{p_4 - 2}{p_3}\right) \cdots \left(\frac{p_k - 2}{p_{k-1}}\right). \qquad (7)$$

For $k = 2$, the period of the sequence of 2-tuples is $m_2 = p_1p_2 = 2 \times 3 = 6$, and it is easy to check that in every period there are two permitted 2-tuples forming 2-doubletons, at positions $n = 1$ and $n = 5$ in the first period (see Remark 2.2). The density of permitted 2-tuples forming 2-doubletons is 1, since there is one permitted 2-tuple forming a 2-doubleton in every subinterval of size $p_2$ (see Table 1). For $k > 2$, since $m_k/p_k$ is the number of subintervals of size $p_k$ within the period of the sequence of $k$-tuples, the product
$$ \left( \frac{p_3 - 2}{p_2} \right) \left( \frac{p_4 - 2}{p_3} \right) \cdots \left( \frac{p_k - 2}{p_{k-1}} \right) $$

at the right-hand side of (7) is the density of permitted $k$-tuples forming $k$-doubletons within the period of the sequence of $k$-tuples.

We denote by $c_k^d$ and $\delta_k^d$ (where the symbol ‘d’ in the superscript makes reference to the word ‘doubleton’) the number and density of permitted $k$-tuples forming $k$-doubletons within the period $I[1, m_k]$ of the sequence of $k$-tuples, respectively (see Definition 2.6 and Definition 3.3). For some values of $k$, Table 1 gives $c_k^d$, the ratio $c_k^d/m_k$ and $\delta_k^d$.

<table><caption>Table 1: Values of $c_k^d$, $c_k^d/m_k$ and $\delta_k^d$.</caption><thead><tr><th>k</th><th>p<sub>k</sub></th><th>m<sub>k</sub></th><th>c<sub>k</sub><sup>d</sup></th><th>c<sub>k</sub><sup>d</sup>/m<sub>k</sub></th><th>&delta;<sub>k</sub><sup>d</sup></th></tr></thead><tbody><tr><td>2</td><td>3</td><td>6</td><td>2</td><td>0.333</td><td>1.000</td></tr><tr><td>3</td><td>5</td><td>30</td><td>6</td><td>0.200</td><td>1.000</td></tr><tr><td>4</td><td>7</td><td>210</td><td>30</td><td>0.142</td><td>1.000</td></tr><tr><td>5</td><td>11</td><td>2310</td><td>270</td><td>0.116</td><td>1.286</td></tr><tr><td>6</td><td>13</td><td>30030</td><td>2970</td><td>0.098</td><td>1.286</td></tr><tr><td>7</td><td>17</td><td>510510</td><td>44550</td><td>0.088</td><td>1.484</td></tr><tr><td>8</td><td>19</td><td>9699690</td><td>757350</td><td>0.078</td><td>1.484</td></tr><tr><td>9</td><td>23</td><td>223092870</td><td>15904350</td><td>0.072</td><td>1.640</td></tr><tr><td>10</td><td>29</td><td>-</td><td>-</td><td>0.066</td><td>1.924</td></tr><tr><td>11</td><td>31</td><td>-</td><td>-</td><td>0.062</td><td>1.924</td></tr><tr><td>12</td><td>37</td><td>-</td><td>-</td><td>0.058</td><td>2.174</td></tr><tr><td>13</td><td>41</td><td>-</td><td>-</td><td>0.056</td><td>2.290</td></tr><tr><td>14</td><td>43</td><td>-</td><td>-</td><td>0.054</td><td>2.290</td></tr><tr><td>15</td><td>47</td><td>-</td><td>-</td><td>0.050</td><td>2.398</td></tr><tr><td>16</td><td>53</td><td>-</td><td>-</td><td>0.048</td><td>2.602</td></tr><tr><td>17</td><td>59</td><td>-</td><td>-</td><td>0.048</td><td>2.798</td></tr><tr><td>18</td><td>61</td><td>-</td><td>-</td><td>0.046</td><td>2.798</td></tr><tr><td>19</td><td>67</td><td>-</td><td>-</td><td>0.044</td><td>2.980</td></tr><tr><td>20</td><td>71</td><td>-</td><td>-</td><td>0.044</td><td>3.070</td></tr><tr><td>21</td><td>73</td><td>-</td><td>-</td><td>0.042</td><td>3.070</td></tr><tr><td>22</td><td>79</td><td>-</td><td>-</td><td>0.040</td><td>3.238</td></tr><tr><td>23</td><td>83</td><td>-</td><td>-</td><td>0.040</td><td>3.320</td></tr><tr><td>24</td><td>89</td><td>-</td><td>-</td><td>0.038</td><td>3.480</td></tr><tr><td>.</td><td>.</td><td>.</td><td>.</td><td>.</td><td>.</td></tr></tbody></table>

We shall prove that the density of permitted $k$-tuples forming $k$-doubletons in $I[1, m_k]$, denoted by $\delta_k^d$, increases and tends to infinity with $k$ (see Lemma 3.2 and Theorem 3.4). Furthermore we shall prove that the density of permitted $k$-tuples forming $k$-doubletons in $I[1, p_k^2]$ also increases and tends to infinity as $k \to \infty$. Thus, multiplying by $p_k$ the density of permitted $k$-tuples forming $k$-doubletons in $I[1, p_k^2]$ it follows immediately that the number of permitted $k$-tuples forming $k$-doubletons in $I[1, p_k^2]$ tends also to infinity with $k$. Since $I[1, p_k^2] \subset [1, x]$ it follows that the size of $I$ (in this case the number of permitted $k$-tuples forming $k$-doubletons in the interval $I[1, x]$ of the sequence of $k$-tuples) tends also to infinity as $k \to \infty$; this is clearly what is required to prove the main theorem.

## 2 The sieve of Eratosthenes formulated as a sequence of $k$-tuples

In this section we formally define the formulation of the sieve of Eratosthenes based in a sequence of $k$ tuples of remainders.

**Definition 2.1.** Let $S_k$ be a partial sum of the series $\sum s_k$ (see [BAR, Section 2]). Let $s_h$ ($1 \le h \le k$) be one of the sequences of remainders that form the partial sum $S_k$. In every period of the sequence $s_h$, the remainder 0 will be the unique selected remainder. See Figure 1.

Now we can formally define the formulation of Sieve E (Eratosthenes) based on a sequence of $k$-tuples, as follows.

**Definition 2.2.** Let $x > 49$ be a real number and let $z = \sqrt{x}$. Let $\mathcal{P}$ be the sequence of the primes and let $k$ be the index of the greatest prime less than $z$. Let $\mathcal{A}$ be the set consisting of the indices $n$ of the partial sum $S_k$ that lie in the interval $[1, x]$. For each $p = p_h \in \mathcal{P}$ ($1 \le h \le k$), the subset $\mathcal{A}_p$ of $\mathcal{A}$ consists of the indices $n$ of $S_k$ such
that the remainder of dividing $n$ by the modulus $p$ is the selected remainder $0$. The indices of the prohibited $k$-tuples lying in $\mathcal{A}$ are sifted out, and the indices of the permitted $k$-tuples lying in $\mathcal{A}$ remain unsifted. See Remark 1.1. The sifting function

$$S(\mathcal{A}, \mathcal{P}, z) = \left| \mathcal{A} \setminus \bigcup_{\substack{p \in \mathcal{P} \\ p < z}} \mathcal{A}_p \right|,$$

is given by the number of permitted $k$-tuples whose indices belong to the set $\mathcal{A}$.

**Definition 2.3.** Given a partial sum $S_k$ ($k > 1$), 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 $S_1$ the indices of the permitted 1-tuples are the odd numbers; so, the concept of 1-doubleton does not make sense.

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

*Proof.* Let $s_h$ ($1 \le h \le k$) be the sequences of remainders that form the partial sum $S_k$. Since $1 \not\equiv 0 \pmod{p_h}$ for every $p_h \in \{p_1, p_2, p_3, \dots, p_k\}$, 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, we can 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.

**Remark 2.1.** Note that given a period of the partial sum $S_k$ ($k \ge 2$), 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 $S_k$, the first $k$-tuple and the penultimate $k$-tuple of this period are counted as part of a $k$-doubleton.

**Remark 2.2.** Let $S_2$ be the partial sum of level $k = 2$. The period of $S_2$ is $m_2 = p_1p_2 = 6$; and there are two permitted 2-tuples at positions $n = 1$ and $n = 5$ in the first period, by Proposition 2.1.

**Proposition 2.2.** Let $S_k$ be a given partial sum, where $k > 1$. Let $s_{k+1}$ be the sequence of remainders modulo $p_{k+1}$. Let $n$ and $n+2$ be the indices of the First $k$-tuple and the Last $k$-tuple of a given $k$-doubleton of $S_k$. Then, when we juxtapose the elements of the sequence $s_{k+1}$ to each $k$-tuple of $S_k$, if the First $k$-tuple is removed, the Last $k$-tuple is not removed; and vice versa.

*Proof.* In the sequence $s_{k+1}$, the only selected remainder is 0. Assume that both the First $k$-tuple and the Last $k$-tuple are removed; this means that $n \equiv 0 \pmod{p_{k+1}}$, and $n+2 \equiv 0 \pmod{p_{k+1}}$, by [BAR, Proposition 2.2]. It follows that $2 \equiv 0 \pmod{p_{k+1}}$. But this is not possible since $p_{k+1} > 2$; it follows that if the First $k$-tuple is removed, the Last $k$-tuple is not removed; and vice versa.

**Definition 2.5.** When we juxtapose the elements of the sequence $s_{k+1}$ to each $k$-tuple of $S_k$ ($k > 1$), if either the First $k$-tuple or the Last $k$-tuple of a $k$-doubleton is removed, we say that at the level transition $k \to k+1$ one $k$-doubleton is removed.

To get a period of the partial sum $S_{k+1}$, we first take $p_{k+1}$ periods of the partial sum $S_k$ (see the paragraph above [BAR, Proposition 2.4]). Let us consider the permitted $k$-tuples that form $k$-doubletons and are within the first $p_{k+1}$ periods of the partial sum $S_k$. The following proposition shows that the distribution of these permitted $k$-tuples forming $k$-doubletons over the residue classes modulo $p_{k+1}$ is uniform; the proof is almost the same as that of [BAR, Proposition 2.4].

**Proposition 2.3.** The permitted $k$-tuples that form $k$-doubletons and are 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.* Step 1. Let $[y] = [0], [1], [2], \dots, [p_{k+1} - 1]$ be the residue classes modulo $p_{k+1}$. Let $n$ be the index of a given permitted $k$-tuple within the first period of the partial sum $S_k$. Thus, within the first $p_{k+1}$ periods of the partial sum $S_k$ there are $p_{k+1}$ permitted $k$-tuples with indices $n' = m_kx + n$, where $x = 0, 1, 2, 3, \dots, p_{k+1} - 1$ represents each period. Because $(m_k, p_{k+1}) = 1$, for each residue class $[y]$ the congruence $m_kx + n \equiv y \pmod{p_{k+1}}$ has a unique solution $x$.
Step 2. Let $c$ be the number of permitted $k$-tuples which form $k$-doubletons, within a period of $S_k$. So, there are $p_{k+1}c$ of these permitted $k$-tuples within the first $p_{k+1}$ periods of the partial sum $S_k$. Now, from Step 1 it follows that there are $c$ permitted $k$-tuples forming $k$-doubletons within each residue class modulo $p_{k+1}$. Thus, the $p_{k+1}c$ permitted $k$-tuples which form $k$-doubletons within the first $p_{k+1}$ periods of the partial sum $S_k$ are uniformly distributed over the residue classes modulo $p_{k+1}$.

**Corollary 2.4.** If there are $m'$ consecutive periods of the partial sum $S_k$, where $m'$ is a multiple of $p_{k+1}$, the permitted $k$-tuples that form $k$-doubletons within these $m'$ periods are also uniformly distributed over the residue classes modulo $p_{k+1}$.

The following proposition gives us an important formula for the number of permitted $k$-tuples forming $k$-doubletons within a period of $S_k$. First, a definition.

**Definition 2.6.** For a given partial sum $S_k$ ($k > 1$), we denote by $c_k^d$ the number of permitted $k$-tuples which form $k$-doubletons within a period of $S_k$.

**Proposition 2.5.** Let $S_k$ be a given partial sum ($k > 1$). We have: $c_k^d = 2(p_3 - 2)(p_4 - 2) \cdots (p_k - 2)$.

*Proof.* Let $s_h$ ($1 \le h \le k$) be the sequences that form $S_k$. By definition, we have $S_k = S_{k-1} + s_k$.

Step 1. By definition, the number of permitted ($k-1$)-tuples that form ($k-1$)-doubletons within a period of the partial sum $S_{k-1}$ is $c_{k-1}^d$. To construct the period of the partial sum $S_k$, we first take $p_k$ periods of $S_{k-1}$; so, there are $p_k c_{k-1}^d$ permitted ($k-1$)-tuples that form ($k-1$)-doubletons within the first $p_k$ periods of $S_{k-1}$.

Step 2. Now, when we juxtapose the elements of the sequence $s_k$ to each ($k-1$)-tuple of $S_{k-1}$, by [BAR, Proposition 2.2], the permitted ($k-1$)-tuples whose indices are included in the residue class 0 modulo $p_k$ are removed, since 0 is a selected remainder within the period of $s_k$, 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 $p_k$ periods of $S_{k-1}$ are distributed uniformly over the residue classes modulo $p_k$, a fraction $1/p_k$ 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 ($k-1 \to k$) is $2(p_k c_{k-1}^d)1/p_k = 2c_{k-1}^d$. Since by Step 1 the original number of permitted ($k-1$)-tuples that form ($k-1$)-doubletons within the first $p_k$ periods of $S_{k-1}$ is $p_k c_{k-1}^d$, it follows that $c_k^d = p_k c_{k-1}^d - 2c_{k-1}^d = c_{k-1}^d(p_k - 2)$ 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 $S_k$.

Step 3. Let $S_2$ be the partial sum of level $k=2$. In a period of $S_2$ 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 $S_2$ is $c_2^d = 2$. From this and Step 2 it follows that the number of permitted $k$-tuples forming $k$-doubletons within a period of the partial sum $S_k$ is given recursively by the formula

$$c_2^d = 2, \\ c_k^d = c_{k-1}^d (p_k - 2).$$

It follows that $c_k^d = 2(p_3 - 2)(p_4 - 2) \cdots (p_k - 2)$.

# 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. Hereafter we consider that every partial sum $S_k$ is in horizontal position unless we specifically state otherwise.

**Definition 3.1.** Let $S_k$ ($k > 1$) be a given partial sum of the series $\sum s_k$; let $I[m, n]$ be a given interval of $k$-tuples. We denote by $c_k^{d, I[m, n]}$ the number of permitted $k$-tuples forming $k$-doubletons within $I[m, n]$.
**Remark 3.1.** Note that given a k-doubleton, the First k-tuple could be inside the interval $I[m, n]$ while the Last k-tuple is outside; so, only the First k-tuple is counted in $c_k^{d,I[m,n]}$. On the other hand, the Last k-tuple could be inside the interval $I[m, n]$ while the First k-tuple is outside; in this case only the Last k-tuple is counted in $c_k^{d,I[m,n]}$.

The following two definitions are the same as [BAR, Definition 3.3] and [BAR, Definition 3.4] respectively, applied here to permitted k-tuples forming k-doubletons.

**Definition 3.2.** Let $S_k (k > 1)$ be a partial sum of the series $\sum s_k$; 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 which form k-doubletons in the interval $I[m, n]$ by

$$ \delta_k^{d, I[m,n]} = \frac{c_k^{d, I[m,n]}}{|I[m,n]|/p_k}. $$

**Definition 3.3.** Let $S_k (k > 1)$ be a given partial sum of the series $\sum s_k$; let $m_k$ be the period of $S_k$. Recall that we have used the notation $c_k^d = c_k^{d,I[1,m_k]}$ for the number of permitted k-tuples forming k-doubletons within the interval $I[1, m_k]$ (the first period of $S_k$). We normally use the notation $\delta_k^d = \delta_k^{d,I[1,m_k]}$ for the density of permitted k-tuples which form k-doubletons 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 $S_k$, we define

$$ \delta_k^d = \frac{c_k^d}{m_k/p_k}. $$

The following lemma gives a formula for computing $\delta_k^d$; the proof is similar to the proof of [BAR, Lemma 3.1].

**Lemma 3.1.** Let $S_k (k > 1)$ be a given partial sum of the series $\sum s_k$. We have

$$ \delta_k^d = 2 \left( \frac{p_1 - 1}{p_1} \right) \left( \frac{p_2 - 2}{p_2} \right) \left( \frac{p_3 - 2}{p_3} \right) \cdots \left( \frac{p_{k-1} - 2}{p_{k-1}} \right) (p_k - 2). $$

*Proof.* The size of the period of the partial sum $S_k$ is equal to $m_k = p_1p_2p_3 \cdots p_{k-1}p_k$ (see [BAR, Proposition 2.1]). Therefore, the number of subintervals of size $p_k$ is equal to $(p_1p_2p_3 \cdots p_{k-1}p_k)/p_k = p_1p_2p_3 \cdots p_{k-1}$. On the other hand, the number of permitted k-tuples forming k-doubletons within a period of $S_k$ is equal to $c_k^d = 2(p_3 - 2)(p_4 - 2) \cdots (p_{k-1} - 2)(p_k - 2)$ (see Proposition 2.5). Consequently, by definition, we obtain

$$ \begin{aligned} \delta_k^d &= \frac{2(p_3 - 2)(p_4 - 2)\cdots(p_{k-1} - 2)(p_k - 2)}{p_1 p_2 p_3 \cdots p_{k-1}} = \frac{2(p_1 - 1)(p_2 - 2)(p_3 - 2)\cdots(p_{k-1} - 2)(p_k - 2)}{p_1 p_2 p_3 \cdots p_{k-1}} = \\ &= 2\left(\frac{p_1 - 1}{p_1}\right)\left(\frac{p_2 - 2}{p_2}\right)\left(\frac{p_3 - 2}{p_3}\right)\cdots\left(\frac{p_{k-1} - 2}{p_{k-1}}\right)(p_k - 2). \end{aligned} $$

The next lemma proves that $\delta_k^d$ is increasing; this lemma is almost a copy of [BAR, Lemma 3.2], after replacing $\delta_h$ by $\delta_h^d$ and ‘permitted h-tuples’ by ‘permitted h-tuples which form h-doubletons’ throughout.

**Lemma 3.2.** Let $S_k$ and $S_{k+1}$ be consecutive partial sums of the series $\sum s_k$, where $k > 1$. If $\delta_k^d$ denotes the density of permitted k-tuples which form k-doubletons within a period of $S_k$, and $\delta_{k+1}^d$ denotes the density of permitted $(k+1)$-tuples which form $(k+1)$-doubletons within a period of $S_{k+1}$, then

$$ \delta_{k+1}^d = \delta_k^d \left( \frac{p_{k+1} - 2}{p_k} \right). $$

*Proof.* Taking the quotient $\delta_{k+1}^d/\delta_k^d$ and simplifying, the proof follows immediately.

**Corollary 3.3.** By Lemma 3.2,

$$ 1. p_{k+1} - p_k = 2 \implies \delta_{k+1}^d = \delta_k^d. $$

$$ 2. p_{k+1} - p_k > 2 \implies \delta_{k+1}^d > \delta_k^d. $$
In the following theorem we prove that $\delta_k^d \to \infty$ as $k \to \infty$; this theorem is almost a copy of [BAR, Theorem 3.4]; one just replaces $\delta_h$ by $\delta_h^d$, and ‘permitted h-tuples’ by ‘permitted h-tuples which form h-doubletons’; and the proof is also a copy of the proof of [BAR, Theorem 3.4].

First, we make a definition.

**Definition 3.4.** Let $p_k > 2$ and $p_{k+1}$ be consecutive primes. We denote by $\theta_k$ the difference $p_{k+1} - p_k - 2$.

**Theorem 3.4.** Let $S_k$ ($k > 1$) be a given partial sum. Let $\delta_k^d$ be the density of permitted k-tuples which form k-doubletons within a period of $S_k$. As $k \to \infty$, we have $\delta_k^d \to \infty$.

*Proof.* By Lemma 3.1,

$$ \delta_k^d = 2 \left( \frac{p_1-1}{p_1} \right) \left( \frac{p_2-2}{p_2} \right) \left( \frac{p_3-2}{p_3} \right) \left( \frac{p_4-2}{p_4} \right) \left( \frac{p_5-2}{p_5} \right) \cdots \left( \frac{p_{k-1}-2}{p_{k-1}} \right) (p_k-2). $$

If we shift denominators to the right, we obtain

$$ \delta_k^d = 2(p_1 - 1) \left( \frac{p_2 - 2}{p_1} \right) \left( \frac{p_3 - 2}{p_2} \right) \left( \frac{p_4 - 2}{p_3} \right) \left( \frac{p_5 - 2}{p_4} \right) \cdots \left( \frac{p_{k-1} - 2}{p_{k-2}} \right) \left( \frac{p_k - 2}{p_{k-1}} \right). $$

By definition, $\theta_k = p_{k+1} - p_k - 2 \implies p_{k+1} - 2 = p_k + \theta_k$. Consequently, we can write

$$ \begin{aligned} \delta_k^d &= \left( \frac{p_2 + \theta_2}{p_2} \right) \left( \frac{p_3 + \theta_3}{p_3} \right) \left( \frac{p_4 + \theta_4}{p_4} \right) \cdots \left( \frac{p_{k-2} + \theta_{k-2}}{p_{k-2}} \right) \left( \frac{p_{k-1} + \theta_{k-1}}{p_{k-1}} \right) = \\ &= \left( 1 + \frac{\theta_2}{p_2} \right) \left( 1 + \frac{\theta_3}{p_3} \right) \left( 1 + \frac{\theta_4}{p_4} \right) \cdots \left( 1 + \frac{\theta_{k-2}}{p_{k-2}} \right) \left( 1 + \frac{\theta_{k-1}}{p_{k-1}} \right) = \\ &= \frac{2}{3} \left[ \left( 1 + \frac{1}{p_1} \right) \left( 1 + \frac{\theta_2}{p_2} \right) \left( 1 + \frac{\theta_3}{p_3} \right) \left( 1 + \frac{\theta_4}{p_4} \right) \cdots \left( 1 + \frac{\theta_{k-1}}{p_{k-1}} \right) \left( 1 + \frac{\theta_k}{p_k} \right) \right] \frac{p_k}{p_k + \theta_k}. \end{aligned} $$

Then

$$ \lim_{k \to \infty} \delta_k^d = \frac{2}{3} \left[ \left( 1 + \frac{1}{p_1} \right) \prod_{k=2}^{\infty} \left( 1 + \frac{\theta_k}{p_k} \right) \right] \lim_{k \to \infty} \frac{p_k}{p_k + \theta_k}. \quad (8) $$

The infinite product between square brackets diverges if the series

$$ \frac{1}{p_1} + \sum_{k=2}^{\infty} \frac{\theta_k}{p_k} \qquad (9) $$

diverges. In the series (9), if $p_k$ is the first of a pair of twin primes, by definition we have $\theta_k = 0$; otherwise we have $\theta_k \ge 2$. Let $\sum_{j=1}^{\infty} 1/q_j$ 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 [BRU], the series $\sum_{j=1}^{\infty} 1/q_j$ also converges. Therefore, the series $\sum_{k=1}^{\infty} 1/p_k - \sum_{j=1}^{\infty} 1/q_j$ diverges, because $\sum_{k=1}^{\infty} 1/p_k$ diverges. By comparison with the series $\sum_{k=1}^{\infty} 1/p_k - \sum_{j=1}^{\infty} 1/q_j$, it follows that the series (9) diverges, because $\theta_k/p_k > 1/p_k$ for the terms where $\theta_k > 0$. Thus, the infinite product in (8) tends to $\infty$ as well. On the other hand, by the Bertrand-Chebyshev theorem, $p_k < p_{k+1} < 2p_k \implies \theta_k < p_k \implies p_k/(p_k + \theta_k) > 1/2$. Consequently, $\delta_k^d \to \infty$ as $k \to \infty$. ■

Let $S_k$ ($k > 1$) be a partial sum of the series $\sum s_k$; recall that $p_k$ denotes its characteristic prime modulus and $m_k$ denotes its period. In addition, recall that $c_k^d$ denotes the number of permitted k-tuples which form k-doubletons and $\delta_k^d$ denotes the density of permitted k-tuples forming k-doubletons within the period of $S_k$.

Let $I[1, n]$ ($n \ge m_k$) be a given interval of k-tuples of the partial sum $S_k$. Let $c_k^{d,I[1,n]}$ be the number of permitted k-tuples forming k-doubletons and let $\delta_k^{d,I[1,n]}$ be the density of permitted k-tuples which form k-doubletons within $I[1, n]$. The following proposition is almost the same as [BAR, Proposition 5.3] replacing $\delta_k^{I[1,n]}$ by $\delta_k^{d,I[1,n]}$ and $\delta_k$ by $\delta_k^d$.

**Proposition 3.5.** The value of $\delta_k^{d,I[1,n]}$ converges to $\delta_k^d$ as $n \to \infty$.
Proof. By definition,

$$ \delta_k^{d,I[1,n]} = \frac{c_k^{d,I[1,n]}}{\frac{n}{p_k}}. $$

Since $n \ge m_k$, we have a first part of the interval $I[1, n]$ whose size is a multiple of $m_k$. Let us denote by $c_k^d$ the number of permitted $k$-tuples which form $k$-doubletons within this part of $I[1, n]$. So, $\lfloor n/m_k \rfloor m_k/p_k$ is the number of subintervals of size $p_k$ 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 $S_k$, we obtain

$$ c_{\eta}^{d} = \frac{\lfloor \frac{n}{m_k} \rfloor m_k}{p_k} \delta_{k}^{d}. $$

On the other hand, let us denote by $c_{\epsilon}^d$ the number of permitted $k$-tuples forming $k$-doubletons within the remainder part of $I[1, n]$. Since $c_k^{d,I[1,n]}$ is the number of permitted $k$-tuples which form $k$-doubletons within $I[1, n]$, we have $c_k^{d,I[1,n]} = c_{\eta}^d + c_{\epsilon}^d$. Then

$$ \delta_k^{d,I[1,n]} = \frac{c_k^{d,I[1,n]}}{\frac{n}{p_k}} = \frac{c_{\eta}^d + c_{\epsilon}^d}{\frac{n}{p_k}} = \frac{\frac{\lfloor \frac{n}{m_k} \rfloor}{p_k} \delta_k^d + c_{\epsilon}^d}{\frac{n}{p_k}} = \frac{\lfloor \frac{n}{m_k} \rfloor m_k}{n} \delta_k^d + \frac{p_k c_{\epsilon}^d}{n}. $$

Now, as $n \to \infty$, the values $m_k, p_k, c_k^d$ and $\delta_k^d$ are constants; and $c_{\epsilon}^d$ is bounded by $c_k^d$. Therefore, in the expression of $\delta_k^{d,I[1,n]}$, the factor that multiplies $\delta_k^d$ tends to one, and the second term tends to zero. This implies that $\delta_k^{d,I[1,n]}$ converges to $\delta_k^d$ as $n \to \infty$. ■

# 4 The behaviour of the density of permitted $h$-tuples forming $h$-doubletons between $h=2$ and $h=k$

Suppose given a partial sum $S_k$ ($k > 1$) of the series $\sum s_k$. Let $s_h$ ($1 \le h \le k$) be the periodic sequences of remainders that form the partial sum $S_k$. Recall the notation $m_h$ for the period of the partial sum $S_h$; recall the notation $c_h^d$ for the number of permitted $h$-tuples forming $h$-doubletons and $\delta_h^d$ for the density of permitted $h$-tuples forming $h$-doubletons within the period of every partial sum $S_h$ ($2 \le h \le k$).

The following lemma gives us the number of permitted $h$-tuples forming $h$-doubletons within the interval $I[1, m_k]_h$ (whose size is the period $m_k$ of $S_k$) of every partial sum $S_h$, where $2 \le h < k$; the proof is almost a copy of the proof of [BAR, Lemma 6.1].

**Lemma 4.1.** Let $S_k$ ($k \ge 4$) be the partial sum of the series $\sum s_k$ associated to the Sieve E. Let us consider the interval $I[1, m_k]_h$ in every partial sum $S_h$ ($2 \le h < k$). The number of permitted $h$-tuples forming $h$-doubletons within every interval $I[1, m_k]_h$ ($2 \le h < k$) is equal to $c_h^d p_{h+1} p_{h+2} \cdots p_k$.

*Proof.* Choose a level $h$ ($2 \le h < k$). By definition, we have $m_k = p_1 p_2 p_3 \cdots p_h p_{h+1} p_{h+2} \cdots p_k = m_h p_{h+1} p_{h+2} \cdots p_k$. That is, the size of the interval $I[1, m_k]_h$ of the partial sum $S_h$ is equal to $p_{h+1} p_{h+2} \cdots p_k$ times the period $m_h$ of the partial sum $S_h$. Consequently, the number of permitted $h$-tuples forming $h$-doubletons within the interval $I[1, m_k]_h$ is equal to $c_h^d p_{h+1} p_{h+2} \cdots p_k$. ■

Now, let us denote by $c_h^{d'}$ the number of permitted $h$-tuples forming $h$-doubletons within the interval $I[1, m_k]_h$ of every partial sum $S_h$ ($2 \le h \le k$), which is computed using Proposition 2.5 and Lemma 4.1. The behaviour of $c_h^{d'}$ as $h$ goes from level 2 to level $k$ is given by the following lemma (the proof is almost the same as that of the [BAR, Lemma 6.2]).

**Lemma 4.2.** Let $S_k$ ($k \ge 4$) be the partial sum of the series $\sum s_k$ associated to the Sieve E. Let us consider the interval $I[1, m_k]_h$ in every partial sum $S_h$, from level $h=2$ to level $h=k$. Let $h$ and $h+1$ be consecutive levels, where $2 \le h < k$. We have

$$ c_{h+1}^{d'} = c_h^{d'} \left( \frac{p_{h+1}-2}{p_{h+1}} \right). $$
**Proof.** Given a partial sum $S_h$ ($2 \le h < k$), suppose that we juxtapose the elements of the sequence $s_{h+1}$ to each $h$-tuple of $S_h$. By [BAR, Proposition 2.2], the permitted $h$-tuples forming $h$-doubletons within the interval $I[1, m_k]_h$ whose indices are included in the residue class 0 modulo $p_{h+1}$ 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 $I[1, m_k]_{h+1}$ of the partial sum $S_{h+1}$ but is not part of any $(h+1)$-doubleton in $I[1, m_k]_{h+1}$. The other permitted $h$-tuples forming $h$-doubletons whose indices are not included in the residue class 0 modulo $p_{h+1}$ are transferred to level $h+1$ as permitted $(h+1)$-tuples forming $(h+1)$-doubletons within the interval $I[1, m_k]_{h+1}$ of the partial sum $S_{h+1}$. Since for every level $h < k$ the size of the interval $I[1, m_k]_h$ is a multiple of $p_{h+1}$, the permitted $h$-tuples forming $h$-doubletons within the interval $I[1, m_k]_h$ of $S_h$ are distributed uniformly over the residue classes modulo $p_{h+1}$, by Proposition 2.3 and Corollary 2.4. Therefore, a $2/p_{h+1}$ fraction of the permitted $h$-tuples forming $h$-doubletons within the interval $I[1, m_k]_h$ of $S_h$ have been removed, and a $(p_{h+1}-2)/p_{h+1}$ fraction have been transferred to level $h+1$ as permitted $(h+1)$-tuples forming $(h+1)$-doubletons within the interval $I[1, m_k]_{h+1}$ of $S_{h+1}$. The lemma is proved.

Let us now examine the behaviour of $\delta_h^d$ as $h$ goes from level 2 to level $k$. Since the selected remainder of the sequence $s_{h+1}$ remove permitted $h$-tuples forming $h$-doubletons within the interval $I[1, m_k]_h$ of the partial sum $S_h$, at each level transition $h \to h+1$, the number of permitted $h$-tuples forming $h$-doubletons decreases as the level increases from $h=2$ to $h=k$ (the factor by which we must multiply $c_h^{d'}$ to obtain $c_{h+1}^d$ 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 $I[1, m_k]_h$ of the partial sum $S_h$ grows at each transition $p_h \to p_{h+1}$ of order greater than 2; and if $p_h \to p_{h+1}$ is a level transition of order 2, the density of permitted $h$-tuples forming $h$-doubletons within $I[1, m_k]_h$ does not change.

**Remark 4.1.** We repeat here the arguments given in [BAR, Remark 6.4], replacing $\delta_h$ by $\delta_h^d$ and 'permitted $h$-tuples' by 'permitted $h$-tuples forming $h$-doubletons': if $p_h \to p_{h+1}$ is a level transition of order greater than 2, the value of $\delta_h^d$ increases because to compute $\delta_h^d$ we count the permitted $h$-tuples forming $h$-doubletons within subintervals of size $p_h$, which grow by more than 2, overcompensating for the removed permitted $h$-tuples forming $h$-doubletons. On the other hand, if $p_h \to p_{h+1}$ is a level transition of order 2, the value of $\delta_h^d$ does not change because the increase in $p_h$ 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

$$ \delta_{h+1}^d = \delta_h^d \left( \frac{p_{h+1}-2}{p_{h+1}} \right) \left( \frac{p_{h+1}}{p_h} \right), \quad (10) $$

where the factor $(p_{h+1}-2)/p_{h+1}$ is related to the permitted $h$-tuples forming $h$-doubletons removed from $I[1, m_k]_h$ (see Lemma 4.2), and the factor $p_{h+1}/p_h$ is related to the increase of the size of the subintervals.

**Definition 4.1.** For a given partial sum $S_h$ ($2 \le h \le k$), we use the notation $c_h^{d,L_k}$ to denote the number of permitted $h$-tuples that form $h$-doubletons within the Left interval $I[1, p_k^2]_h$, and we use the notation $c_h^{d,R_k}$ to denote the number of permitted $h$-tuples that form $h$-doubletons within the Right interval $I[p_k^2+1, m_k]_h$ (see [BAR, Definition 6.3]). We use the notation $\delta_h^{d,L_k}$ to denote the density of permitted $h$-tuples that form $h$-doubletons within the Left interval $I[1, p_k^2]_h$, and we use the notation $\delta_h^{d,R_k}$ to denote the density of permitted $h$-tuples that form $h$-doubletons within the Right interval $I[p_k^2+1, m_k]_h$.

Recall that we use the notation $\delta_h^d$ for the density of permitted $k$-tuples that form $k$-doubletons within the interval $I[1, m_k]$. Now, choose a level $h$ such that $2 \le h < k$. As we have seen in the proof of Lemma 4.1, we have $m_k = m_h p_{h+1} p_{h+2} \cdots p_k$; in other words, the size of the interval $I[1, m_k]_h$ is equal to $p_{h+1} p_{h+2} \cdots p_k$ times the period $m_h$ of the partial sum $S_h$. Consequently, the density of permitted $h$-tuples that form $h$-doubletons within the interval $I[1, m_k]_h$ is equal to $\delta_h^d$. Hence, for every level $h$ ($2 \le h \le k$), we have

$$ \begin{aligned} \delta_h^{d,L_k} &> \delta_h^d &\iff & \delta_h^{d,R_k} < \delta_h^d, \\ \delta_h^{d,L_k} &< \delta_h^d &\iff & \delta_h^{d,R_k} > \delta_h^d. \end{aligned} \quad (11) $$

# 5 Proof of the Main theorem

For every partial sum $S_h$ from level $h=2$ to level $h=k$ ($k \ge 4$), let us consider the interval $I[1, m_k]_h$, the Left interval $I[1, p_k^2]_h$ and the Right interval $I[p_k^2+1, m_k]_h$. Recall the notation $\delta_h^d$ for the density of permitted $h$-tuples forming $h$-doubletons within the period of $S_h$, and recall the notation $\delta_h^{d,L_k}$ and $\delta_h^{d,R_k}$ for the density of permitted $h$-tuples forming $h$-doubletons within the intervals $I[1, p_k^2]_h$ and $I[p_k^2+1, m_k]_h$, respectively, where $2 \le h \le k$.

In this section, we first prove that the density of permitted $k$-tuples forming $k$-doubletons within the Left interval $I[1, p_k^2]$ of the partial sum $S_k$ tends to infinity with $k$; from this result we shall prove the main theorem.
By the same arguments as at the begin of [BAR, Section 7] we can see that also in the present case the quantity of permitted k-tuples forming k-doubletons within a given interval $I[1, n]_k$ of $S_k$ depends on the relationship between $k$ and the size of $I[1, n]_k$; in the case of the Left interval $I[1, p_k^2]_k$ this relationship is given by the sequence of the primes.

Now, $\delta_h^d$ increases between $h = 2$ and $h = k$, by Lemma 3.2 and Corollary 3.3, for sufficiently large $k$. On the other hand, it seems reasonable to conjecture that, for every sufficiently large $k$, the behaviour of $\delta_h^{d,L_k}$ as $h$ goes from level 2 to level $k$ is similar to the behaviour of $\delta_h^d$. Thus, $\delta_k^{d,L_k}$ should also increase to infinity with $k$ (see Theorem 3.4); before proving this statement, we need the following remarks.

**Remark 5.1.** Let $S_k$ be a given partial sum of the series $\sum s_k$. Given a partial sum $S_h$ ($2 \le h \le k$), when the interval $I[1, m_k]_h$ is subdivided into the Left interval $I[1, p_k^2]_h$ and the Right interval $I[p_k^2 + 1, m_k]_h$, the proportion of permitted $h$-tuples forming $h$-doubletons within the Left interval $I[1, p_k^2]_h$ will be greater than the proportion of permitted $h$-tuples forming $h$-doubletons within the Right interval $I[p_k^2 + 1, m_k]_h$, or vice versa, since $p_k^2$ is not a multiple of $m_h$.

**Remark 5.2.** From the preceding remark, it follows that the density of permitted $h$-tuples forming $h$-doubletons within the Left interval $I[1, p_k^2]_h$ will be greater than the density of permitted $h$-tuples forming $h$-doubletons within the Right interval $I[p_k^2 + 1, m_k]_h$; or vice versa (see (11)). Therefore, considering the Left intervals $I[1, p_k^2]_h$ from $h = 2$ to $h = k$ (left block of the partition of the first period of $S_k$), there must be levels for which $\delta_h^{d,L_k} < \delta_h^d$ mixed with levels for which $\delta_h^{d,L_k} > \delta_h^d$, for $k$ sufficiently large.

**Definition 5.1.** Let $S_k$ be a given partial sum of the series $\sum s_k$. Given the level transition $p_h \to p_{h+1}$ ($2 \le h < k$), we denote by $F_{h,h+1}^{d,L_k}$ the fraction of the permitted $h$-tuples forming $h$-doubletons within the Left interval $I[1, p_k^2]_h$ of $S_h$ that are transferred to the Left interval $I[1, p_k^2]_{h+1}$ of $S_{h+1}$ as permitted ($h+1$)-tuples forming ($h+1$)-doubletons.

**Remark 5.3.** Suppose that in every level transition $p_h \to p_{h+1}$ ($2 \le h < k$) the value of $F_{h,h+1}^{d,L_k}$ is very close to the fraction $(p_{h+1} - 2)/p_{h+1}$, given by Lemma 4.2 for the intervals $I[1, m_k]_h$. In this case, if $k$ is sufficiently large $\delta_h^{d,L_k}$ must increase between $h = 2$ and $h = k$, by the same arguments given in Remark 4.1 for $\delta_h^d$.

Now, suppose that there exists a level $h = h' (2 < h' < k)$, where the difference $k - h'$ is sufficiently large, such that $\delta_h^{d,L_k}$ does not increase between $h = h'$ and $h = k$. Clearly, this can only happen if there are level transitions $p_h \to p_{h+1}$ between $h = h'$ and $h = k$ for which $F_{h,h+1}^{d,L_k}$ is too far below the fraction $(p_{h+1} - 2)/p_{h+1}$ given by Lemma 4.2.

**Remark 5.4.** Let $S_k$ be a partial sum of the series $\sum s_k$, and consider a given level $h = h' (2 < h' < k)$. Given consecutive primes $p_h$ and $p_{h+1}$, we denote $g_h = p_{h+1} - p_h$. Assume that in each period (of size $p_h$) of every sequence $s_h$ from $h = h' + 1$ to $h = k$ of $S_k$ there are $g_h/2$ selected remainders (the same in every period of the sequence). As we already know, the selected remainders in every sequence $s_h$ ($h' < h \le k$) remove permitted ($h-1$)-tuples from the partial sum $S_{h-1}$ (see [BAR, Definition 6.2]). However, we can check that under the preceding assumption the fraction of permitted $h$-tuples forming $h$-doubletons within the interval $I[1, m_k]_h$ of $S_h$ that are transferred to the interval $I[1, m_k]_{h+1}$ of $S_{h+1}$ as permitted ($h+1$)-tuples forming ($h+1$)-doubletons is equal to $(p_{h+1} - g_h)/p_{h+1}$, where $h' \le h < k$ (see the proof of Lemma 4.2). So, for sufficiently large $k$, the whole number of permitted $h$-tuples forming $h$-doubletons removed between $h = h'$ and $h = k$ is rather greater than in the case where there is only one selected remainder in every period of the sequences $s_h$ ($h' < h \le k$). Furthermore, in every level transition $p_h \to p_{h+1}$ ($h' \le h < k$), the increase of $\delta_h$ under this assumption would be given by

$$ \delta_{h+1} = \delta_h \left( \frac{p_{h+1} - g_h}{p_{h+1}} \right) \left( \frac{p_{h+1}}{p_h} \right) = \delta_h $$

(compare this formula with (10)). This means that under the preceding assumption, the density of permitted $h$-tuples forming $h$-doubletons within the interval $I[1, m_k]_h$ of $S_h$ does not increase between $h = h'$ and $h = k$.

The proof of the following lemma is almost the same as that of the [BAR, Lemma 7.1], with a few minor changes.

**Lemma 5.1.** Let $S_k$ ($k > 4$) be the partial sum associated to the Sieve E. Let $I[1, p_k^2]_h$ be the Left interval for every partial sum $S_h$ from $h = 2$ to $h = k$. Furthermore, let $\delta_h^{d,L_k}$ be the density of permitted $h$-tuples forming $h$-doubletons within every Left interval $I[1, p_k^2]_h$ ($2 \le h \le k$).

For every real number $M > 0$ there is a positive integer $K$ such that $\delta_k^{d,L_k} > M$ for every $k > K$.

*Proof.* Step 1. Let us consider the density of permitted $h$-tuples forming $h$-doubletons within every interval $I[1, m_k]_h$ from $h = 2$ to $h = k$, denoted by $\delta_h^d$. Recall that, as the level increases from $h = 2$ to $h = k$, the value of $\delta_h^d$ increases at each level transition $p_h \to p_{h+1}$ of order greater than 2, and $\delta_h^d$ does not change if $p_h \to p_{h+1}$ is a level transition of order 2, by Lemma 3.2 and Corollary 3.3.

The fact that $I[1, p_k^2]_h \subset I[1, m_k]_h$ together with Remark 5.2 suggests that for every $h$ ($2 \le h \le k$) the value of $\delta_h^{d,L_k}$ is relatively close to $\delta_h^d$, if $k$ is sufficiently large. Thus, it seems reasonable to assume that $\delta_k^{d,L_k} > M$ for every sufficiently large $k$, since $\delta_k^d \to \infty$ with $k$ by Theorem 3.4.
Step 2. Let us now examine the behaviour of the density of permitted h-tuples forming h-doubletons within every Left interval $I[1, p_k^2]_h$ (denoted by $\delta_h^{d,L_k}$) as $h$ goes from level 2 to level $k$. For sufficiently large $k$, the positions of the permitted 2-tuples show an almost regular pattern along the Left interval $I[1, p_k^2]_2$ of the partial sum $S_2$ (see Remark 2.2). Nevertheless, when we add the sequences $s_h$ from level $h=3$ to level $h=k$, the selected remainders in every sequence $s_h$ remove permitted $(h-1)$-tuples from the partial sum $S_{h-1}$ (see [BAR, Definition 6.2]). Thus, the number and proportion of permitted h-tuples in general (and permitted h-tuples forming h-doubletons in particular) within the Left interval $I[1, p_k^2]_h$ decrease slowly as the level increases from $h=2$ to $h=k$.

Now, assume that for every level transition $p_h \to p_{h+1}$ ($2 \le h < k$), the value of $F_{h,h+1}^{d,L_k}$ is sufficiently close to the average fraction $(p_{h+1}-2)/p_{h+1}$, given by Lemma 4.2. Then, for $k$ sufficiently large $\delta_h^{d,L_k}$ must increase as $h$ goes from level 2 to level $k$, as we have seen in the first paragraph of Remark 5.3; in other words, the behaviour of $\delta_h^{d,L_k}$ between $h=2$ and $h=k$ will be very similar to the behaviour of $\delta_h^d$.

Step 3. Let us consider the partial sum $S_k$ ($k > 5$) and choose a fixed level $h = h' (4 < h' < k)$. As $k \to \infty$, the following two things occur at the same time:

(a) For the fixed level $h'$, the size of the Left interval $I[1, p_k^2]_h$ increases, so $\delta_{h'}^{d,L_k}$ converges to $\delta_h^d$, by Proposition 3.5. Thus, given any small number $\varepsilon > 0$, for a level $k$ sufficiently large it must be

$$\delta_{h'}^d - \varepsilon < \delta_{h'}^{d,L_k} < \delta_h^d + \varepsilon.$$

(b) The number and proportion of the level transitions $h \to h+1$ of order greater than 2 increase between $h=h'$ and $h=k$.

Step 4. Now, given any level $h$, where $2 \le h < h'$, as the size of the Left interval $I[1, p_k^2]_h$ increases (as $k \to \infty$), $\delta_h^{d,L_k}$ also converges to $\delta_h^d$, by Proposition 3.5. Note that, as the size of the Left interval $I[1, p_k^2]_h$ increases, the number of times that $m_h$ (the period of $S_h$) fits in $I[1, p_k^2]_h$ increases more and more, and this fact is underlying the result $\delta_h^{d,L_k} \to \delta_h^d$. Therefore, for a level $k$ sufficiently large, if $\delta_{h'}^d - \varepsilon < \delta_{h'}^{d,L_k} < \delta_{h'}^d + \varepsilon$ (see Step 3 (a)), then $\delta_h^d - \varepsilon < \delta_h^{d,L_k} < \delta_h^d + \varepsilon$ for every $h$ ($2 \le h < h'$), since $m_{h+1} = m_h p_{h+1}$. Thus $\delta_h^{d,L_k}$ will be very close to $\delta_h^d$ for every level from $h=2$ to $h=h'$ if $\varepsilon$ is sufficiently small and $k$ sufficiently large. This means that the behaviour of $\delta_h^{d,L_k}$ is very similar to the behaviour of $\delta_h^d$, so $\delta_h^{d,L_k}$ increases between $h=2$ and $h=h'$.

Step 5. Consider the interval $I[1, m_k]_h$ in every partial sum $S_h$ ($h' \le h \le k$). Under the condition assumed in Remark 5.4 (with regard to the selected remainders in the sequences $s_h$), for every level transition $p_h \to p_{h+1}$ ($h' \le h < k$), the fraction of permitted h-tuples forming h-doubletons within the interval $I[1, m_k]_h$ of $S_h$ that are transferred to the interval $I[1, m_k]_{h+1}$ of $S_{h+1}$ as permitted $(h+1)$-tuples forming $(h+1)$-doubletons is $(p_{h+1}-g_h)/p_{h+1}$. Hence, for every level transition $p_h \to p_{h+1}$ ($h' \le h < k$) of order greater than 2 (that is $g_h > 2$), the number of permitted h-tuples forming h-doubletons removed from $I[1, m_k]_h$ is rather greater than in the case where there is only one selected remainder in every period of the sequences $s_h$ ($h' < h \le k$), so, the density of permitted h-tuples forming h-doubletons within $I[1, m_k]_h$ does not increase as $h$ goes from level $h'$ to level $k$.

In our case the condition assumed in Remark 5.4 is not present (see Definition 2.1), so, for every level transition $p_h \to p_{h+1}$ ($h' \le h < k$), the fraction of permitted h-tuples forming h-doubletons within the interval $I[1, m_k]_h$ of $S_h$ that are transferred to the interval $I[1, m_k]_{h+1}$ of $S_{h+1}$ as permitted $(h+1)$-tuples forming $(h+1)$-doubletons is $(p_{h+1}-2)/p_{h+1}$, by Lemma 4.2. Thus, for sufficiently large $k$, the density of permitted h-tuples forming h-doubletons within $I[1, m_k]_h$ increases between $h=h'$ and $h=k$, as we have seen in Section 4. Now, with regard to the Left intervals $I[1, p_k^2]_h$ ($h' \le h \le k$), for every level transition $p_h \to p_{h+1}$ ($h' \le h < k$) clearly the fraction $F_{h,h+1}^{d,L_k}$ must be much closer to $(p_{h+1}-2)/p_{h+1}$ than to $(p_{h+1}-g_h)/p_{h+1}$. Thus, in view of Step 3 (b) it follows that $\delta_h^{d,L_k}$ must increase between $h=h'$ and $h=k$, for sufficiently large $k$. (See the second paragraph in Step 2.)

Step 6. By Step 4 and Step 5 we can see that $\delta_h^{d,L_k}$ increases between $h=2$ and $h=k$ for sufficiently large $k$, as we have assumed in the second paragraph of Step 2.

Now, given $M$, suppose we choose a level $h=h'$ such that $\delta_{h'}^d > M$ (see Theorem 3.4). By the arguments given in Step 3 (a), as $k$ increases $\delta_{h'}^{d,L_k}$ tends to $\delta_h^d$, thus, since $\delta_{h'}^d > M$ there must exist $K'$ such that $\delta_{h'}^{d,L_k} > M$ for every $k > K'$.

On the other hand, by Step 5 there must exist $K > K'$ such that $\delta_k^{d,L_k} > \delta_{h'}^{d,L_k}$ for every $k > K$, so, $\delta_k^{d,L_k} > M$ by the preceding paragraph. The lemma is proved.

<figure><img src="image_3.png" /></figure>
Finally, we prove the main theorem.

**Theorem 5.2. Main theorem**

Let $x > 49$ be a real number, and let $p_k$ be the greatest prime less than $z = \sqrt{x}$. Let $S_k$ ($k \ge 4$) be the partial sum associated to the Sieve E. Recall that we have used the notation $\pi_2(x)$ for the number of twin primes in $[1, x]$ (see Subsection 1.1).

We have $\pi_2(x) \to \infty$ as $x \to \infty$.

*Proof.* Step 1. Recall the notation $\mathcal{P}$ for the sequence of the primes. Let $\mathcal{A} = \{n : 1 \le n \le x\}$. By definition, the integers $n \in \mathcal{A}$ that have survived the sifting process are the indices of permitted $k$-tuples that lie in $[1, x]$. As we know, the indices $n$ of permitted $k$-tuples that lie in $[z, x)$ are all the primes in this interval (see Remark 1.1 and the paragraph below Remark 1.1). It follows that the indices of the permitted $k$-tuples forming $k$-doubletons that lie in $[z, x)$ are twin primes.

Step 2. On the other hand, recall that $c_k^{d,L_k}$ denote the number of permitted $k$-tuples forming $k$-doubletons within the Left interval $I[1, p_k^2]$ of the partial sum $S_k$. Therefore, the number of permitted $k$-tuples forming $k$-doubletons whose indices lie in $[z, p_k^2]$ is $c_k^{d,L_k} - 1$, since $n = 1$ is the index of a permitted $k$-tuple forming a $k$-doubleton, by definition. Hence, by Step 1 the number of twin primes in $[z, x)$ is greater than or equal to $c_k^{d,L_k} - 1$, since $[z, p_k^2] \subset [z, x)$. Thus, we can write

$$ \pi_2(x) > c_k^{d,L_k} - 1, \qquad (12) $$

since there are twin primes less than $z$.

Step 3. Now, since the size of the Left interval $I[1, p_k^2]$ of the partial sum $S_k$ is equal to $p_k^2$, there are $p_k$ subintervals of size $p_k$ in this interval; it follows that $c_k^{d,L_k} = \delta_k^{d,L_k} p_k$. Replacing $c_k^{d,L_k}$ by $\delta_k^{d,L_k} p_k$ in the right-hand side of (12), we obtain

$$ \pi_2(x) > \delta_k^{d,L_k} p_k - 1. \qquad (13) $$

Since $x \to \infty$ implies $\sqrt{x} \to \infty$ and this implies $p_k \to \infty$, by Lemma 5.1 it follows that the right-hand side of (13) tends to $\infty$ as $x \to \infty$. Therefore also $\pi_2(x) \to \infty$ as $x \to \infty$. The main 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.

## References

[HAL] H. Halberstam and H.-E. Richert, *Sieve Methods*, Academic Press, London, 1974.

[FRI] J. Friedlander and H.Iwaniec, *Opera de Cribro*, American Mathematical Society, 2010.

[COJ] A. C. Cojocaru and M. Ram Murty, *An Introduction to Sieve Methods an Their Applications*, Cambridge University Press, London, 2005.

[BRU] V. Brun, La série $1/5 + 1/7 + 1/11 + 1/13 + 1/17 + 1/19 + 1/29 + 1/31 + 1/41 + 1/43 + 1/59 + 1/61 + \dots$, où les dénominateurs sont nombres premieres jumeaux est convergente où finie, *Bull. Sci. Math.*, **43**, (1919), 124-128.

[ZHA] Y. Zhang, Bounded gaps between primes, *Ann. of Math.*, **179**, (2014), 1121-1174.

[POL] Polymath8, Bounded gaps between primes, http://michaelnielsen.org/polymath1/index.php

[TAO] 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/.

[BAR] Ricardo Barca, Every sufficiently large even number is the sum of two primes (preprint), https://hal.archives-ouvertes.fr/hal-02075531.

Ricardo G. Barca

Universidad Tecnológica Nacional

Buenos Aires (Argentina)

E-mail address: rbarca@frba.utn.edu.ar