Some Properties and Algorithms for Twin Primes

Abstract

In this article, we study some new properties of twin primes and algorithms for their generation. We find the necessary conditions to generate a pair of twins. These conditions seem to indicate that the conjecture is true, namely, there are infinitely many twin primes. Furthermore, we developed some algorithms that are very useful from a computer science point of view, which can be applied in cryptography and data encryption.

1. Introduction

Arithmetic is the queen of Mathematics and prime numbers are the building blocks of Arithmetic. It is well known that every natural composed number greater than 1 can be represented uniquely as a product of prime numbers, up to the order of the factors. The physical world is founded through a finite number of bricks, namely quarks, leptons and bosons, while the Arithmetic world is built with an infinite number of bricks. Indeed, Euclid, in his “Elements”, proved in a wonderfully simple way that prime numbers are infinite. How they are distributed within natural numbers is not very easy to understand [1,2,3,4]. The distribution is so complex that it seems random even if, from a theoretical point of view, we can say that it is not [5,6,7]. Indeed, Wilson’s theorem tells us that $n>1$ is a prime number if and only if $\left(n-1\right)!\equiv -1\left(modn\right).$ The distribution of prime numbers is not only of mathematical interest but also, for example, in the cryptographic and computer science field [8,9,10]. The aim of this paper is, instead, to use twin primes to apply them in the cryptography framework. The mathematical literature on prime numbers is so vast that it is not easy to report a complete bibliography [11,12,13,14]. Another unsolved property concerns the number of twin primes. The twin prime conjecture assumes that they are infinite, but there is still no proof of it officially accepted by mathematicians [15,16,17,18,19]. We know that the sum of the reciprocals does not diverge but tends towards the so-called Brun’s constant [20]

$$B_2=\left({\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{5}}\right)+\left({\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{7}}\right)+\left({\displaystyle \frac{1}{11}}+{\displaystyle \frac{1}{13}}\right)+\dots ..\approx 1.902160578$$

(1)

This fact, unfortunately, does not allow us to deduce anything about the number of twins. In fact, as is well known, a finite sum can be obtained either with an infinite number of terms or with a finite number.

2. Twin Primes

This section is devoted to finding a method to generate twin primes. We start to observe that, except for $2$, all prime numbers are odd numbers. For this reason, a prime number is necessarily of the type

$$2k+1,\forall k\in \mathbb{N}$$

(2)

or

$$2k-1,\forall k\in \mathbb{N}-\left\{1\right\}.$$

(3)

From the previous relations (2) and (3), we have the following

Table 1. Sequences of 2k − 1, 2k + 1 with $k\in \mathsf{{\rm N}}$.

$\mathit{k}$	$2\mathit{k}-1$	$2\mathit{k}+1$
1	1	3
2	3	5
3	5	7
4	7	9
5	9	11
6	11	13
7	13	15
8	15	17
9	17	19
10	19	21
11	21	23
12	23	25
13	25	27
14	27	29
15	29	31
16	31	33
17	33	35
18	35	37
19	37	39
…	…	…

.

It is trivial to observe, in the previous table, that there are values of $k$ to which a pair of twins is associated. This occurs for values of $k$ that generate a prime number in both columns. For example, for each value of $k$ equal to 2, 3, 6, 9, 15, we achieve twin primes. We do not know by what rule $k$ varies to return a prime number in the two columns, but we can understand when it generates a composite number. Recalling that the composite numbers are the product of two odd numbers different from 1, we have that the composite numbers of (2) are generated by the following product

$$\left(2x+1\right)\left(2y+1\right)=2\left(2xy+x+y\right)+1,\forall x,y\in \mathbb{N}.$$

(4)

It is also possible to consider $\left(2x-1\right)\left(2y-1\right)$ with $x,y\in \mathbb{N}-\left\{1\right\}$ achieving the same result. Therefore, we obtain a prime number in the second column, if and only if $k\ne 2xy+x+y.$ Instead, from (4), we deduce that a composite odd number of the second column of the table is generated by the following values

$$k=2xy+x+y,\forall x,y\in \mathbb{N}.$$

(5)

Now, we consider $x$ as a row and $y$ as a column of a matrix. If $x=1$, from (5) we have

$$k=3y+1,\forall y\in \mathbb{N}.$$

Therefore, in the first row we have 4, 7, 10, 13, … and so on. If $x=2$, the relation (5) is

$$k=5y+2,\forall y\in \mathbb{N}.$$

Therefore, in the second row we have 7, 12, 17, 22, … and so on. It is easy to verify that the third row is $k=7y+3$, the fourth is $k=9y+4$ and so on. In conclusion, we can write the following infinite matrix

$$A=\left(\begin{array}{cccccccccccc}4& 7& 10& 13& 16& 19& 22& 25& 28& 31& 34& \dots \\ 7& 12& 17& 22& 27& 32& 37& 42& 47& 52& 57& \dots \\ 10& 17& 24& 31& 38& 45& 52& 59& 66& 73& 80& \dots \\ 13& 22& 31& 40& 49& 58& 67& 76& 85& 94& 103& \dots \\ 16& 27& 38& 49& 60& 71& 82& 93& 104& 115& 126& \dots \\ 19& 32& 45& 58& 71& 84& 97& 110& 123& 136& 149& \dots \\ \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \end{array}\right)$$

(6)

All numbers that are not in $A$ correspond to values of $k$ which, when substituted in $2k+1$, give us a prime number. They are infinite in number since there are infinitely many primes. Now, with the same reasoning, we deduce the law that generates composite odd numbers in the first column of the table. Indeed, the composite numbers will be

$$\left(2x-1\right)\left(2y+1\right)=2\left(2xy+x-y\right)-1,\forall x\in \mathbb{N}-\left\{1\right\},\forall y\in \mathbb{N}.$$

(7)

Therefore, we obtain a prime number, if and only if $k\ne 2xy+x-y.$ We obtain a composite odd number if

$$k=2xy+x-y, \forall x,y\in \mathbb{N}.$$

(8)

Like before, we built a matrix with $x$ as row and $y$ as column. If $x=2$, the relation (8) becomes

$$k=3y+2, \forall y\in \mathbb{N},$$

achieving the sequence 5, 8, 11, 14, … and so on. If $x=3$, we have

$$k=5y+3, \forall y\in \mathbb{N},$$

obtaining the sequence 8, 13, 18, 23, … and so on. It is easy to verify that the third row is $k=7y+4$, the fourth is $k=9y+5$. In conclusion, we can write the following infinite matrix

$$B=\left(\begin{array}{cccccccccccc}5& 8& 11& 14& 17& 20& 23& 26& 29& 32& 35& \dots \\ 8& 13& 18& 23& 28& 33& 38& 43& 48& 53& 58& \dots \\ 11& 18& 25& 32& 39& 46& 53& 60& 67& 74& 81& \dots \\ 14& 23& 32& 41& 50& 59& 68& 77& 86& 95& 104& \dots \\ 17& 28& 39& 50& 61& 72& 83& 94& 105& 116& 127& \dots \\ 20& 33& 46& 59& 72& 85& 98& 111& 124& 137& 150& \dots \\ \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \end{array}\right)$$

(9)

All numbers that are not in $B$ correspond to values of $k$ which, when substituted in $2k-1$, give us a prime number. They are infinite in number since there are infinitely many primes. In summary, the values of $k$ which are in (6) but not in (9) (as for example 4) generate a prime in $2k-1$ and a composite number in $2k+1$. The values which are in (9) but not in (6) (as for example 5) generate, instead, a composite number in $2k-1$ and a prime number in $2k+1$. The values of $k$ that are in both sequences (as for example 17) generate composite numbers in both $2k-1$ and $2k+1$. Finally, there are values of $k$ which are neither in (6) nor in (9). These values of $k$ generate prime numbers in both columns of the table and, therefore, generate twins. To understand how many twin primes there are, we must find how many numbers are missing in the union of (6) and (9). Since we are interested in twin primes and not prime numbers, we merge the matrices (6) and (9) achieving the following matrix, which will allow us to identify twin prime numbers

$$C=\left(\begin{array}{cccccccccccc}4& 5& 7& 8& 10& 11& 13& 14& 16& 17& 19& \dots \\ 7& 8& 12& 13& 17& 18& 22& 23& 27& 28& 32& \dots \\ 10& 11& 17& 18& 24& 25& 31& 32& 38& 39& 45& \dots \\ 13& 14& 22& 23& 31& 32& 40& 41& 49& 50& 58& \dots \\ 16& 17& 27& 28& 38& 39& 49& 50& 60& 61& 71& \dots \\ 19& 20& 32& 33& 45& 46& 58& 59& 71& 72& 84& \dots \\ \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \end{array}\right)$$

(10)

Indeed, any number that does not fit in $C$ (as for example 6) corresponds to a value of $k$ which gives us a pair of twins. Let us call $r_n$ the complement rows. In other words

$$\begin{array}{l}{r}_1=1,2,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,\dots .\\ r_2=1,2,3,4,5,6,9,10,11,14,15,16,19,20,21,24,25,26,29,30,31,34,35,36,39,40,41,44,\dots ..\\ r_3=1,2,3,4,5,6,7,8,9,12,13,14,15,16,19,20,21,22,23,26,27,28,29,30,\dots ..\\ r_4=1,2,3,4,5,6,7,8,9,10,11,12,15,16,17,18,19,20,21,24,\dots ..\\ r_5=1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,18,19,20,21,22,23,24,25,26,29,\dots ..\\ r_6=1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,21,22,23,24,25,26,27,28,29,30,31,\dots ..\\ \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\end{array}$$

If $k$ is a natural number with

$$k\in {\cap }_{n=1}^{\infty }{r}_n$$

(11)

then $2k-1$ and $2k+1$ are twin primes. We can write the twin prime conjecture in the following equivalent form

Twin Prime Conjecture.

Twin primes are infinite in number if and only if $\left|{\cap }_{n=1}^{\infty }{r}_n\right|=\left|\mathbb{N}\right|,$ where the vertical bar on each side is the cardinality.

First of all, we observe that $r_1$, except 2, only has multiples of 3. Therefore, thanks to relation (11), for $k$ to belong to the intersection of all lines, it must have a value that is a multiple of 3. Otherwise, from the definition of intersection between sets, it would not verify (11). In other words, the values of $k$ that generate twins are necessarily multiples of 3. If one wants to reduce the computational cost, it is possible also to observe that, by construction, the row $r_2$ cannot have multiples of three which have the last digit equal to 2,3,7,8. Therefore, we have that a necessary condition for k to generate a pair of twins is that it is a multiple of three with the last digit equal to 0, 1, 4, 5, 6, 9. In [21], we verified the validity of the second condition for the first 10⁴ pairs of twins. It is possible to note that the $nth$ multiple of three is certainly present from $r_{n-1}$ onwards and, therefore, the rows to be checked are from $r_2$ to $r_{n-2}$. Indeed, the generic row $r_n$ has the first numbers from 1 to $3n$, it does not have the numbers $3n+1$ and $3n+2$ and then has $2n-1$ consecutive numbers starting from $3n+3$; successively it does not have two numbers and so on. Obviously, since it is not possible to control for infinite numbers, it is necessary to find a new strategy. Let us observe that

$$r_1=\left\{1\right\}\cup \left\{2\right\}\cup \left\{3n\right\}$$

$$r_2=\left\{1\right\}\cup \left\{2\right\}\cup \left\{3\right\}\cup \left\{5n-1\right\}\cup \left\{5n\right\}\cup \left\{5n+1\right\}$$

$$r_3=\left\{1\right\}\cup \left\{2\right\}\cup \left\{3\right\}\cup \left\{4\right\}\cup \left\{7n-2\right\}\cup \left\{7n-1\right\}\cup \left\{7n\right\}\cup \left\{7n+1\right\}\cup \left\{7n+2\right\}$$

$$r_4=\left\{1\right\}\cup \left\{2\right\}\cup \left\{3\right\}\cup \left\{4\right\}\cup \left\{5\right\}\cup \left\{9n-3\right\}\cup \left\{9n-2\right\}\cup \left\{9n-1\right\}\cup \left\{9n\right\}\cup \left\{9n+1\right\}\cup \left\{9n+2\right\}\cup \left\{9n+3\right\}$$

$$\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots$$

$$r_n=\left[{\cup }_{i=1}^{n+1}\left\{i\right\}\right]\cup \left[{\cup }_{i=1}^{n-1}\left\{\left(2i+1\right)n\pm {\gamma }_{i-1}\right\},{\gamma }_{i-1}=i-1,i=n-1,\forall n\in \mathbb{N}\right].$$

(12)

Definition. 

Given the sequence $r_i$, we call the first hole of $r_i$ and denote it by ${\dot{r}}_i$ the first missing element so that $r_i$ does not coincide with ℕ.

Theorem of the localizations of the first hole.

Given the sequence $r_n$, we have that ${\dot{r}}_n=3n+1\forall n\in \mathbb{N}$.

Proof. 

We want to show that ${\dot{r}}_{n+1}=3+{\dot{r}}_n.$ Let

$$P\left(n\right):{\dot{r}}_n=3n+1.$$

The base case

$$P\left(1\right)={\dot{r}}_1=4$$

is true, as you can verify. For the inductive step, we assume that $P\left(n\right)$ is true for some integer $n\ge 1$ and use this to prove that $P\left(n+1\right)$ is true. We have that

$$P\left(n+1\right)=3\left(n+1\right)+1=3n+1+3={\dot{r}}_n+3.$$

Then, we may conclude that $P\left(n\right)$ is true for all integers $n\ge 1.$ □

It is easy to deduce the following result.

Corollary. 

For the sequences $r_i$ is valid that $\underset{n\to +\infty }{\mathit{lim}}{r}_n=\mathbb{N}.$

Proof. 

From the theorem of the localizations of the first hole, we have that ${\dot{r}}_{n+1}=3+{\dot{r}}_n\forall n\in \mathbb{N}.$ Therefore,

$$\underset{n\to +\infty }{\lim }{\dot{r}}_n=\underset{n\to +\infty }{\lim }3n+1=+\infty$$

and this means that $r_n\to \mathbb{N}.$ □

Definition. 

We define the jump extension ${\dot{r}}_n^e$ as the number of missing elements in $r_n$ between two successive subsequences starting from ${\dot{r}}_n.$ That is

$${\dot{r}}_n^e=\left[\left(2k_n+1\right)m+\left(n-1\right)\right]-\left[\left(2k_n+1\right)m-\left(n-1\right)\right],\forall n\in \mathbb{N}.$$

Definition. 

We call the set of jumps for level $n$ the set $r_n^j$ defined as

$$r_n^j=\mathbb{N}/r_n.$$

For example, $r_1^j=\left\{4,5,7,8,10,11,\dots \right\}.$

Definition. 

We define jump size ${\dot{r}}_n^s$ the distance between a subsequence $r_n$ and its successive. In other words, the distance between one hole and the successive corresponding one

$${\dot{r}}_n^s=\left\{\left[\left(2k_n+1\right)m+\left(n-1\right)\right]+4\right\}-\left\{\left[\left(2k_n+1\right)m-\left(n-1\right)\right]+1\right\}.$$

(13)

We observe that the jump size of $r_1$ is 3. In fact, ${\dot{r}}_1^s=7-4=10-7=\dots =3.$ Instead, for $r_2$ is 5. Indeed, ${\dot{r}}_2^s=12-7=5.$ From (12) it is possible to verify that the following is valid

Theorem of the invariance of the jump extension.

The jump extension ${\dot{r}}_n^e$ is invariant for each $n$ and it is equal to 3, then there are two missing successive elements in $r_n$ for each jump size.

Proof. 

As a direct consequence of (12), we have that $r_n$ is made up of two types of terms, and that is, singletons $\left\{i\right\}$ and subsets centered in $\left(2k_n+1\right)m$ with subsets from $\left[\left(2k_n+1\right)m-\left(n-1\right)\right]$ to $\left[\left(2k_n+1\right)m+\left(n-1\right)\right].$ The case $n=1$ is trivial. Indeed, $r_1=\left\{\left\{1\right\}\cup \left\{2\right\}\cup \left\{2k_1+1\right\}m,\forall m\in \mathbb{N},k_1=1\right\}=\left\{1,2,3,6,9,\dots \right\}.$

Then, by using (13) we have

$${\dot{r}}_1^e=6-3=9-6=\dots =3.$$

For this reason, there are two missing elements between two successive subsequences. With the same reasoning, we can write

$${\dot{r}}_n^e=\left[\left(2k_n+1\right)\left(m+1\right)-\left(n-1\right)\right]-\left[\left(2k_n+1\right)m+\left(n-1\right)\right]=2k_n+1-2\left(n-1\right).$$

The element $k_n$ at the nth level coincides with the value of $n$ and we have

$${\dot{r}}_n^e=2n+1-2n+2=3.$$

Therefore, there are two missing elements between two successive subsequences. □

The following theorem characterizes the jump size.

Theorem of the jump size.

The jump size

$${\dot{r}}_n^s=2n+1$$

(14)

and, therefore, is strictly increasing.

Proof. 

The truth of this proposition follows immediately from the structure (12) of $r_n$. Indeed, we see that it is made up of the union of $n+1$ singletons and $2n-1$ subsets of the type $\left\{\left(2k_n+1\right)m\pm \gamma \right\}$ with $\gamma =0,..,n-1.$ If $\left(2k_n+1\right)m=\left(2n+1\right)m$ is the central term, as γ varies, there will be $n-1$ subsets $\left(2k_n+1\right)m+\gamma$ and $n-1$ subsets $\left(2k_n+1\right)m-\gamma$. Thus, they, with the addition of the central term, are in total $2n-1$. The difference between the two following holes is given from $2n-1$ by adding the two holes, that is $\left(2n-1\right)+2=2n+1$ as we would prove. In addition, for the symmetries of the structure, the same result is obtained by making the difference between the two following centroids $\left(i.e\left(2k_n+1\right)m\right)$ for each subset. For example, we have $c_2\left(r_1\right)-c_1\left(r_1\right)=6-3$, where $c_i\left(r_j\right)$ is the i-th centroid of $r_j$. Moreover, $c_2\left(r_2\right)-c_1\left(r_2\right)=6,$ $c_2\left(r_3\right)-c_1\left(r_3\right)=7$, etc. Finally, from (14), we deduce that ${\dot{r}}_n^s$ is an increasing function. □

Now, we deduce the following property.

Remark. 

By observing that ${\dot{r}}_n^s$ is strictly increasing, the holes will be increasingly rarer. In other words, in $r_1$ we have that the distance between holes is 3, in $r_2$ is 5 and so on. This means that as you progress with the lines there are fewer and fewer holes. Therefore, $r_n\to \mathbb{N}$ if $n\to \infty .$

Corollary. 

The rows $r_n,\forall n\in \mathbb{N},$ are made up of a number of subsets equal to $3n$.

Proof. 

It is possible to note, from (12), that each $r_n$ is made up of a number of subsets equal to $3n$. Indeed, we have $n+1$ singleton and $2n-1$ of the type $\left(2k_n+1\right)m\pm \gamma$. Therefore, we have $n+1+2n-1=3n.$ □

3. Smart Algorithm for Finding Twin Primes

In this section, we develop an algorithm using the definition of twin primes, i.e., $p=q+2$ with $p$ and $q$ prime numbers, in a way that is easier from a computational point of view. We have seen that, once $k$ is fixed, two numbers $p$ and $q$ are twins if they are of the type $p=2k-1$ and $q=2k+1$ with $k\in {\cap }_{i=1}^{\infty }{r}_i$. As we have seen, from the first and second condition, we deduce that a necessary condition for $k$ to generate a pair of twins is that it is a multiple of 3 with the last digit equal to 0, 1, 4, 5, 6, 9. Condition (11), therefore, can be simplified in order to obtain an effective algorithm. Indeed, rather than considering the infinite intersection, it will be sufficient to stop at a certain $r_j$ to check whether $\tilde{k}$ generates a pair of twins. This is possible because, if $\tilde{k}$ is contained in the first $r_j$, it will certainly also belong to all the subsequent ones. Indeed, it is possible to note that the nth multiple of 3 is certainly present from $r_{n-1}$ onwards and, therefore, the rows to be checked are from $r_2$ to $r_{n-2}$. The determination of $j$ is obtained, as an immediate consequence of (12), in the following way

$$j_{max}={\displaystyle \frac{\tilde{k}}{3}}-1.$$

(15)

Furthermore, the maximum column $m$ of each row $r_i$ can be determined by

$$m_{max}=⌈{\displaystyle \frac{\tilde{k}}{3}}⌉$$

(16)

by using the condition

$$m_{max}=⌈{\displaystyle \frac{\tilde{k}}{5}}⌉$$

(17)

where the symbol [a] means the biggest integer smaller than a.

Thanks to these conditions, we can write relation (11) as

$$k\in {\bigcap }_{n=1}^{j_{max}}{r}_i$$

(18)

where $\forall r_i,m\le m_{max}.$ This twin prime number search algorithm can be further optimized by choosing $m_{max}=f\left(i\right).$ In particular, we have that

$$m_{max}\left(i\right)={\displaystyle \frac{\tilde{k}}{2i+1}}\forall k_i\in \mathbb{N},\forall r_i.$$

(19)

In this way, instead of considering $j_{max}·m_{max}$ elements we will obtain a smaller set of variable length $m=m_{min}\left(i\right).$ Let us analyze the Algorithm 1:

Algorithms 1 Algorithm for finding and verifying the twin prime generator $\tilde{k}$

(1) Accept input $\tilde{k}$ (2) Compute $j_{max}$ following (15) (3) Once $i$ is fixed, calculate $m_{max}\left(i\right)$ through the relation (19) (4) If $k_{m_{max}\left(i\right)}=\tilde{k},$ - increase $i$ - repeat (4) If $k_{m_{max}\left(i\right)}<\tilde{k},$ - calculate $k_{m_{max}\left(i\right)+1}$ until it is greater than or equal to $\tilde{k}$ - if $k_{m_{max}\left(i\right)+r}=\tilde{k}$ then increase $i$ - if $k_{m_{max}\left(i\right)+r}>\tilde{k}$ then $\tilde{k}$ does not generate twins If $k_{m_{max}\left(i\right)}>\tilde{k}$, - calculate $k_{m_{max}\left(i\right)-1}$ until it is minor than or equal to $\tilde{k}$ - if $k_{m_{max}\left(i\right)-r}=\tilde{k}$ then increase $i$ - $k_{m_{max}\left(i\right)-r}<\tilde{k}$ then $\tilde{k}$ does not generate twins (5) If $i=j_{max}$, calculate, save and display $\left(\tilde{k},2\tilde{k}-1,2\tilde{k}+1,\#\left(k\right)\right)$ where #($k$) is the total number of analyzed $k$.

The advantage of this smart algorithm compared to traditional algorithms is that the latter would generate a matrix $r_{im}$ with at least $j_{max}·m_{max}$ elements. Instead, the algorithm just presented, calculates only some elements of $r_{im}$. In particular, for each row, with $1\le i\le j_{max}$, it calculates only the element $k$ in column $m_{max}$ and possibly its neighbors. Below we show a conceptual scheme of what was said

$$r_{ij}=\left(\begin{array}{ccccccccccc}& & & & & & & & & .& \\ & & & & & & & & ..& & \\ & & & & & & & .& & & \\ & & & & & & ..& & & & \\ & & & & & .& & & & & \\ & & & & ..& & & & & & \\ & & & \dots & & & & & & & \end{array}\right)$$

There are many classic sieve algorithms and they play an important role in finding prime numbers. These classical algorithms have applications in many contexts including cryptography. For example, we can report some important sieves such as sieve of Eratosthenes, linear sieve, sieve of Sundaram, sieve of Atkin, sieve of Sorenson, segmented sieve of Eratosthenes with Wheel factorization [22,23,24,25].

The experiments conducted concerned $1\le \tilde{k}\le {10}^{10}$. Below, we report the comparison

Table 2. Comparison of classical and smart algorithms across different values of $\tilde{k}$—the table presents a side-by-side comparison of the computational load required by the classical algorithm versus the smart algorithm for various values of $\tilde{k}$ ranging from 1 to 10¹⁰.

$\tilde{\mathit{k}}$	Classical Algorithm	Smart Algorithm
1	$1\times 1=1$	$1$
10	136	29
${10}^2$	136,567	3653
${10}^3$	112,464,651	333,385
…
${10}^{10}$	1.1 × ${10}^{26}$	$3.3\times {10}^{19}$

.

As shown in Table 2, the classical algorithm requires the generation of all matrix cells, which leads to a significantly higher computational burden. In contrast, our smart algorithm optimizes the process by intelligently navigating through the matrix without needing to construct every cell. This approach drastically reduces the number of operations required, especially as $\tilde{k}$ increases. The efficiency gains are particularly evident at higher values of $\tilde{k}$, where the smart algorithm demonstrates a reduction in computational load by several orders of magnitude compared to the classical approach.

Figure 1 shows, as $\tilde{k}$ varies, the curve of $k$ calculated with both the traditional and the smart algorithms. The successive figure shows, as $\tilde{k}$ varies, the useful fraction graph of $k$ calculated with respect to $j_{max}·m_{max}$ and that is

$${\displaystyle \frac{\#\left(k\right)}{j_{max}·m_{max}}}.$$

This comparison is made between the traditional algorithm and the smart algorithm, demonstrating the computational efficiency of the smart algorithm in reducing the number of operations required as k increases. The logarithmic scale on both axes emphasizes the substantial difference in performance between the two methods, with the smart algorithm consistently requiring fewer computations.

Furthermore, it shows the saved fraction which is

$$1-{\displaystyle \frac{\#\left(k\right)}{j_{max}·m_{max}}}$$

The graph in Figure 2 depicts as k varies, highlighting the proportion of computational effort saved by the smart algorithm compared to the traditional approach. As k increases, the smart algorithm demonstrates a significant reduction in computational complexity, effectively illustrating the efficiency gains achieved by avoiding unnecessary calculations. The logarithmic scale accentuates how the savings become more pronounced for larger values of k.

The algorithm exhibits a time complexity of O(k) relative to the parameter k. This linear complexity is highly efficient, especially considering that it avoids the explicit construction of a matrix, which would likely result in a higher computational cost, potentially O(k²) or greater depending on the matrix dimensions involved. The approach taken is therefore well-optimized for the problem at hand, balancing computational efficiency with the necessity of accurately simulating the matrix traversal process. This makes the algorithm particularly suitable for scenarios where memory usage and processing time are critical considerations.

Theorem. 

From the twin matrix, we can extract infinite sequences.

Proof. 

We start by writing

$$r_1=\left\{1\right\}\cup \left\{2\right\}\cup \left\{3n\right\}=\left\{1\right\}\cup \left\{2\right\}\cup a_1$$

with $a_1=\left(2k_1+1\right)n\forall n\in \mathbb{N}$ and $k_1=1.$

$$r_2=\left\{1\right\}\cup \left\{2\right\}\cup \left\{3\right\}\cup \left\{5n-1\right\}\cup \left\{5n\right\}\cup \left\{5n+1\right\}=\left\{1\right\}\cup \left\{2\right\}\cup \left\{3\right\}\cup \left\{a_2-1\right\}\cup \left\{a_2\right\}\cup \left\{a_2+1\right\}$$

with $a_2=\left(2k_2+1\right)n\forall n\in \mathbb{N}$ and $k_2=2.$

$$r_3=\left\{1\right\}\cup \left\{2\right\}\cup \left\{3\right\}\cup \left\{4\right\}\cup \left\{a_3-2\right\}\cup \left\{a_3-1\right\}\cup \left\{a_3\right\}\cup \left\{a_3+1\right\}\cup \left\{a_3+2\right\}$$

with $a_3=\left(2k_3+1\right)n\forall n\in \mathbb{N}$ and $k_3=3$ and so on.

Therefore, we have

$$a_m\subset r_m\forall m\in \mathbb{N}\mathrm{with}{a}_m=\left(2k_m+1\right)n\forall n\in \mathbb{N}\mathrm{and}{k}_m=m.$$

(20)

Now, from the sequence $a_1=\left\{3n_1\right\}=\left(2k_1+1\right)n_1\forall n_1\in \mathbb{N}$ and $k_1=1$, we can consider the subsequence

$${\tilde{a}}_2={\left\{3n_1\right\}}_{n_1=a_2}=\left\{\left(2k_1+1\right)\left(2k_2+1\right)n_2\right\}\forall n_2\in \mathbb{N},k_1=1,k_2=2.$$

From ${\tilde{a}}_2$ we can consider the subsequence

$${\tilde{a}}_3={\left\{3·5n_2\right\}}_{n_2={\tilde{a}}_2}=\left\{\left(2k_1+1\right)\left(2k_2+1\right)\left(2k_3+1\right)n_3\right\}\forall n_3\in \mathbb{N},k_1=1,k_2=2,k_3=3.$$

Proceeding in this way, we obtain

$${\tilde{a}}_w={\prod }_{i=1}^w\left(2k_i+1\right)n\forall n\in \mathbb{N},k_i=i.$$

(21)

As they are defined, we have

$${\tilde{a}}_w\subset {\bigcap }_{i=1}^w{r}_i,\forall w\in \mathbb{N}$$

(22)

and, being a subsequence,

$$\left|{\tilde{a}}_w\right|=\left|\mathbb{N}\right|,\forall w\in \mathbb{N}.$$

(23)

Ultimately,

$$\underset{w\to \infty }{\lim }{\tilde{a}}_w={\tilde{a}}_{\infty }\subset {\bigcap }_{i=1}^{\infty }{r}_i$$

(24)

with

$$\left|{\tilde{a}}_{\infty }\right|=\left|\mathbb{N}\right|$$

(25)

□

The previous theorem seems to indicate that the twin conjecture is true which is that there are infinite pairs of twins.

4. Another Algorithm for Twin Primes

In Appendix A Section [1], we start by writing the following Algorithm 2 for twin primes:

Algorithms 2 Algorithm for twin primes

(1) Start with a list of initial known primes, P = {2,3,5}, if the list is empty (2) Define the initial generator value G and the generator index I. (3) Compute the next generator value $G_{next}$ using the current generator G and the next prime in the list P $$G_{next}=G\times G_{I+1}$$ (4) Construct a set R of primes such that $p\in P$ and ${p^2<G}_{next}$ (5) Generate a new list of potential primes: - Generate a set F of potential new primes using the following families: - From existing primes: - For each $p\in P$ and $p\ne P_i$, calculate potential primes: For $j=1,2\dots$: $F=\left(G\times j-p\right)if(G\times j-p<G_{next})$ - From previously removed numbers: - For each $r\in R$, calculate potential primes: For $j=1,2\dots$: $F=\left(G\times j-r\right)if(G\times j-r<G_{next}$) - With adjustments of −1 and +1: $$F=\left(G\times j\mp 1\right)$$ (6) Optimize the list of potential primes: - Remove numbers from $F$ that are multiples of any prime in $R$ $$F=F\left\{f\in F\right|\exists p\in R:f \mathrm{m}\mathrm{o}\mathrm{d} p=0$$ (7) Merge the optimized list of potential primes into the main list of known primes (8) Clean the list of known primes to remove any invalid entries (9) Repeat the process with updated parameters to continue generating primes.

We generate a total of 896.062 pairs of twin prime numbers as in

Table 3. Generated twin prime pairs—the table shows a selection of the 896,062 twin prime pairs identified in the experiment, including the smallest (3, 5) and the largest (22,309,2671, 22,309,2673). The dataset highlights the algorithm’s efficiency in identifying twin primes across a wide numeric range. Consequently, the first column is the value k, while the second and the third columns are the first and the second elements of a couple of twin prime numbers.

$\mathit{I}\mathit{n}\mathit{d}\mathit{e}\mathit{x}$	P1	P2
0	3	$5$
1	5	7
$2$	11	13
3	17	19
4	29	31
5	41	43
…	…	…
896,056	223,091,501	223,091,503
896,057	223,091,549	223,091,551
896,058	223,091,621	223,091,623
896,059	223,092,047	223,092,049

.

Below in Figure 3, we report the graph that relates the prime numbers generated to the time spent generating them, where the prime generated scale is multiplied to 1 × 10⁷.

The function presents a logarithmic behavior, which could be an excellent field of application for a Polynomial Regression in search of a possible Fit. The following work is performed in Python in a Jupyter-lab environment (the details are in Appendix A Section [2]).

Following a series of trials, as in Figure 4 we found that the level 15 configuration emerged as the most optimal and, in Appendix A Section [3], we proceed to generate the actual model by exploiting the degree just identified.

To do this, we use, in the following application, three evaluation systems, that is MSE, RMSE, and R² to obtain the following output

R²: 0.9994752721646499

MSE: 1,831,498,334.0579565

RMSE: 42,796.00838931075

Base model MSE (average): 4,087,907,127,997.419

The polynomial regression model is better than the baseline model.

5. Conclusions

In this paper, we described some new properties of twin primes. Since a prime number is necessarily of the type $2k-1$ and $2k+1$, we described some necessary conditions for k to generate twin primes. We built a matrix and by studying its rows we have proved some of their properties. In addition, we developed a smart algorithm to find twin primes and to confirm the twin primes’ conjecture in a very efficient way.