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We consider the sequences $ B = \{ b_m \colon b_m = 6m+1, m \in \mathbf{N} \} $ and $A = \{ a_m \colon a_m = 6m-1, m \in \mathbf{N} \}$ and their difference $T$: $$ \begin{array}{rrrrrrrrrrrrrrrrrrr} B = &{7,}&{13,}&{19,}&{25,} &{31,}&{37,}&{43,}&{49,} &{55,} &{61,}&{67,}&{73,}&{79,}&{85,} &{91,} &{97,}&{103,}&{109, \, \ldots} \\ A = &{5,}&{11,}&{17,}&{23,}&{29,}&{35,} &{41,}&{47,}&{53,}&{59,}&{65} &{71,} &{77,} &{83,}&{89,}&{95,} &{101,}&{107, \, \ldots} \\ \hline T = &{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2,}&{2, \, \ldots} \end{array} $$

We will perform the process of sifting the composite numbers of the sequence $B$ and map the result to the sequence $T$: $$ \begin{array}{rrrrrrrrrrrrrrrrrrr} \mathcal B = &{7,}&{13,}&{19,}&{0,} &{31,}&{37,} &{43,}&{0,} &{0,} &{61,}&{67,}&{73,}&{79,} &{0,} &{0,} &{97,} &{103,}&{109, \, \ldots} \\ A = &{5,}&{11,}&{17,}&{23,}&{29,}&{35,} &{41,}&{47,}&{53,}&{59,}&{65} &{71,}&{77,} &{83,}&{89,}&{95,} &{101,}&{107, \, \ldots} \\ \hline \mathcal T = &{2,}&{2,} &{2,} &{0,} &{2,} &{2,} &{2,} &{0,} &{0,} &{2,} &{2,} &{2,} &{2,} &{0,} &{0,} &{2,} &{2,} &{2, \, \ldots} \end{array} $$

Then, we will perform the process of sifting the composite numbers of the sequence $A$ and also map the result to the sequence $T$: $$ \begin{array}{rrrrrrrrrrrrrrrrrrr} \mathcal B = &{7,}&{13,}&{19,}&{0,} &{31,}&{37,}&{43,}&{0,} &{0,} &{61,}&{67,}&{73,}&{79,}&{0,} &{0,} &{97,}&{103,}&{109, \, \ldots} \\ \mathcal A = &{5,}&{11,}&{17,}&{23,}&{29,}&{0,} &{41,}&{47,}&{53,}&{59,}&{0,} &{71,}&{0,} &{83,}&{89,}&{0,} &{101,}&{107, \, \ldots} \\ \hline \mathcal T = &{2,}&{2,} &{2,} &{0,} &{2,} &{0,} &{2,} &{0,} &{0,} &{2,} &{0,} &{2,} &{0,} &{0,} &{0,} &{0,} &{2,} &{2, \ldots} \end{array} $$ Thus, the sequence $T$ is sifted for the first time together with the sequence $B$ and the second time together with the sequence $A$, that is, it is sifted twice.

Cutting out one progression with the difference $p$ from each of the sequences $B$ and $A$, we cut out two progressions with the difference $p$ from the sequence $T$.

As a result, we obtained two sequences $\mathcal B$ and $\mathcal A$ consisting of primes and a sequence $\mathcal T$, each nonzero element of which defines one pair of twins. The sequence $ \mathcal T $ is a double sieve for twin-primes.

For the following discussion, we take $n=6m$. Let $\mathcal A_{m}$, $\mathcal B_{m}$, and $\mathcal T_{m}$ denotes the segment of length $m$ of sequences $\mathcal A$, $\mathcal B$, and $\mathcal T$ respectively. Let $\pi (a,n)$ denote the number of primes not exceeding $n$ that are of the form $6i-1$, and let $\pi (b,n)$ denote the number of primes not exceeding $n$ that are of the form $6i+1$.

We first estimate $\pi (a,n)$ and $\pi (b,n)$ as a function of $m$ and $\prod_{5 \leqslant p \leqslant n} {\left( {1 - 1/p} \right)}$. Combining the inequalities (3.26) and (3.1) from 'Approximate formulas for some functions of prime numbers' of Rosser and Schoenfeld, we obtain
$$ \pi (x) > \frac {x} {\log x} \left( 1 + \frac{1} {2 \log x} \right) > \frac{ x} {\log x} \left( 1 + \frac{1} {2 (\log x)^2} \right) > x e^{ \gamma } \prod\limits_{p \leqslant x} {\left( {1 - \frac{1} {p}} \right)}.$$ This inequality holds for all sufficiently large $x$. From this inequality and the Dirichlet's theorem on primes in arithmetic progressions it follows that inequalities $\pi (a,n) > m e^{ \gamma } \prod_{5 \leqslant p \leqslant n} (1 - 1/p)$ and $\pi (b,n) > m e^{ \gamma } \prod_{5 \leqslant p \leqslant n} (1 - 1/p)$ holds for all sufficiently large $m$.

Formally, as a result of repeated sifting, we must obtain an estimate

$$ {\pi}_2 (n) > m {e^{ 2\gamma }} \prod\limits_{5 \leqslant p \leqslant n} {\left( {1 - \frac{1} {p}} \right)^2}. $$ But this is not true; as shown above, the result of the repeated sieving is a double sieve. To proceed to double sieve we use relation $$ C (n) = { \prod_{5 \leqslant p \leqslant n}} ( 1 - 2/p ) \Big{/} { \prod_{5 \leqslant p \leqslant n}} ( 1 - 1/p )^2 = { \prod \limits_{5 \leqslant p \leqslant n}} \frac{p(p-2)}{(p-1)^2}. $$

Finally, we obtain

$$ \pi_2 (n) > m { {e^{2 \gamma }} \prod \limits_{5 \leqslant p \leqslant n}} \left( {1 - \frac{2}{p}} \right) = { C (n) m {e^{2 \gamma }} \prod \limits_{5 \leqslant p \leqslant n}} \left( {1 - \frac{1}{p}} \right)^2. $$

$C (n) \to \frac {4} {3} C_2$ as $n \to \infty$, where $C_2 = 0.66016 \ldots $ is the twin prime constant.
It is easy to show that $\pi_2 (n) \to \infty$ as $n \to \infty$.

I know that everyone knows that it is impossible to solve the problem of twins. I would like to know where in this reasoning the error is. If someone thinks that there is no mistake, then I would also like to hear it.

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  • $\begingroup$ Proving the weak form isn't all that difficult. It takes a little elbow grease to do it, and a team of experts that have no time to dilly dally playing board games to get it published. Your work is a strong start and I'm looking forward to more. $\endgroup$ – JustKevin Aug 4 '17 at 2:31
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It's quite easy to show that there are infinitely many primes. And it follows very easily that there are either infinitely many primes of the form 6k-1, or infinitely many primes of the form 6k+1.

The problem is: Let's say there are infinitely many primes of the form 6k-1. How can you show that for the infinitely many of those primes greater than for example $10^{100}$, there must be one where p+2 is also prime? Yes, it's very likely. It's extremely likely. But can you prove it?

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  • $\begingroup$ That's not an answer, though. $\endgroup$ – JustKevin Aug 4 '17 at 2:23
  • $\begingroup$ I also recently learned that there are infinitely many prime numbers. And I also heard that there are infinitely many primes both of the form 6k-1 and of the form 6k + 1. Not either... or. And moreover, it seems that $\pi (a,n) / \pi (a,n) \to 1$ as $n \to \infty$. How can I show that for an infinite number of these primes, greater than, for example, $10^100$, there is one where $p + 2$ is also prime? Such numbers is called twin primes. Each nonzero element of the sequence $\mathcal T$ defines one pair of twins. $\endgroup$ – Andrei Allakhverdov Aug 4 '17 at 21:40
  • $\begingroup$ We are talking about the fact that there are infinitely many such elements. Infinity is slightly larger than any number that you can write by a finite number of digits. $\endgroup$ – Andrei Allakhverdov Aug 4 '17 at 21:40

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