The Pattern of Prime Numbers
Abstract: The prime numbers P≥5 obey a pattern that can be described by two forms or geometric progressions or that facilitates obtaining them sequentially, being possible also to calculate the quantity of primes that are in the geometric progressions as it is described in this document.

1. Introduction

Since ancient times when humans discovered the counting system and natural numbers, prime numbers immediately attracted their attention, were numbers whose only divisors are 1 and the same number. The problem to find them was not to be able to describe by means of an equation. There are countless publications about the properties of prime numbers that can be found in all languages and theorems have been created in different ways, seeking always to find a pattern of ordering   . The inability to find an order has been eloquently documented, such as in Havil’s book:

“The succession of primes is unpredictable. We don’t know if they will obey any rule or order that we have not been able to discover still. For centuries, the most illustrious minds tried to put an end to this situation, but without success. Leonhard Euler commented on one occasion: mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers and we have reason to believe that it is a mystery into which the human mind will never penetrate. In a lecture given by D. Zagier in 1975, he said: “There are two facts about the distribution of prime numbers of which I hope to convince you so overwhelmingly that they will be permanently engraved in your hearts. The first is that, [they are] the most arbitrary and ornery objects studied by mathematicians: they grow like weeds among the natural numbers, seeming to obey no other law than that of chance, and nobody can predict where the next one will sprout. The second fact is even more astonishing, for it states just the opposite: that the prime numbers exhibit stunning regularity, that there are laws governing their behavior, and that they obey these laws with almost military precision.” (Havil, 2003  )

To put prime numbers into context, let’s begin by saying anecdotally, as late as 20,000 years ago humans marked the bone of Ishango with 19, 17, 13, 11  and 2300 years ago Euclid proved that there are infinitely many prime numbers (e.g. Williamson, 1782)  . Later, Euler made another formal proof of it (e.g. Hardy and Woodgold, 2009)  .

Until now, there is no known efficient formula for primes, nor a recognizable pattern or sequence the primes follow. All recent publications dealing with this issue established that primes are distributed at random and looked more to a white noise distribution  . Here will be shown that prime numbers are not random, they obey mathematic rules and can be expressed by equations.

2. Form of Prime Numbers

Porras Ferreira and Andrade (2014)  had found that all prime numbers ${p}_{n}\ge 7$ have the following form:

${p}_{n}=\left[31,7,11,13,17,19,23,29\right]+30n,\text{where}n\in {ℤ}^{+}$ (1)

Or all prime numbers ${p}_{m}\ge 5$ have the following form:

${p}_{m}=\left[5,7\right]+6m,\text{where}m\in {ℤ}_{0}^{+}$ (2)

It can be seen that Equation (2) is a derivation of Equation (1), as demonstrated in  , although Equation (2) includes, the prime 5 which is not included in Equation (1).

Equation (2) can be transformed to a simpler form where $k=1+m$ :

${p}_{k}=6k±1,\text{where}k\in {ℤ}^{+}$ (3)

It means, the form of all prime numbers only have these three equations: Equation (1) which does not include primes 2, 3 and 5, and Equation (2) and Equation (3) which does not include primes 2 and 3, therefore Equation (2) and Equation (3) are equivalent:

$\left[{p}_{m}=\left[5,7\right]+6m\to m\in {ℤ}_{0}^{+}\right]\equiv \left[{p}_{k}=6k±1\to k\in {ℤ}^{+}\right]$ (4)

Taking the Equation (3) a table is constructed (Figure 1), where all the primes ${p}_{k}\ge 5$ are formed only in two columns, the column $6k-1$ and the column $6k+1$ fore $k\ge 1$ . The cells highlighted in yellow correspond to composite numbers and those that are not highlighted to prime numbers.

Figure 1. Formation of prime and composite numbers.

3. Analysis of How Composite and Prime Numbers Are Formed in $6k±1$ Columns

In the following analysis, Figure 1 is used as reference.

3.1. The Composite Numbers Are Formed as Follows

Theorem 1:

If $p=6k±1$ in $k={k}_{1}$ , them all the composite numbers ${N}_{k}=6k±1$ with $k={k}_{1}+pm$ where $m\ge 1$ will contain $p$ .

Demonstration:

Let $p=6k±1$ in $k={k}_{1}$ , therefore $p=6{k}_{1}±1$ is a number that can be prime or not and ${N}_{k}=6k±1$ in $k={k}_{1}+pm$ being $m\ge 1$ , therefore:

${N}_{k}=6k±1=6\left({k}_{1}+pm\right)±1=6{k}_{1}±1+6pm=p+6pm$

Factoring $p$ then:

${N}_{k}=p\left(6m+1\right)$

which must be a composite number with two factors $p$ and $6m+1$ .

The theorem is proved.

Corollary 1:

The composite numbers ${N}_{k}=6k±1$ contain two equal or different factors that come from the columns $6k-1$ and/or $6k+1$ therefore if ${N}_{k}=6k±1$ $=p\left(6m+1\right)$ is decomposed into their two factors $p$ and $\left(6m+1\right)$ ; there will be no prime numbers according to Theorem 1, in the rows $k$ of Equation (5) where ${k}_{1}$ is the row where $p$ and $6m+1$ appears for the first time where $\left[m,n\right]\ge 1$ .

$k=\left[\begin{array}{c}{k}_{1}+pn\\ {k}_{1}+\left(6m+1\right)n\end{array}\right]$ (5)

Corollary 2:

There are not identical composite numbers, one from column $6k-1$ and another from column $6k+1$ .

Corollary 3:

All the composite numbers ${N}_{k}$ from column $6k-1$ always have a factor $6m-1$ coming from that column and a factor $6n+1$ coming from the other column, which means ${N}_{k}=6k+1=\left(6m-1\right)\left(6n+1\right)$ where $\left[m,n\right] .

Corollary 4:

All the composite numbers Nk of column 6k + 1 may have two factor from the column 6k − 1, that means ${N}_{k}=6k+1=\left(6m-1\right)\left(6n-1\right)$ , where $m\le n$ and $\left[m,n\right] or two factors from the same column, that means ${N}_{k}=\left(6m+1\right)\left(6n+1\right)$ , where $m and $\left[m,n\right] .

Corollary 5:

Eliminating all k, product of Equation (5), the rest k will only contain primes of the given form of Equation (3) and as it is shown in Figure 1. Note that the initial pattern is given by primes 5 and 7, leaving the cells in rows      from column 6k − 1 with primes and rows    from column 6k + 1 with primes. Subsequently these primes give rise to the composite numbers ${N}_{k}$ according to Equation (5), leaving other different rows where there will be prime numbers, this pattern continues until infinity.

3.1.1. Examples of Composite and Prime Numbers from $6k-1$ Column

1) The prime $p=5$ appears for the first time in the row ${k}_{1}=1$ , applying Equation (5) there are no primes but composite numbers that have the 5 as one of its factors in the rows $k=1+5n$ for $n\ge 1$ . These compound numbers have the form ${N}_{k}=6k-1=6\left(1+5n\right)-1=5\left(6n+1\right)$ . The second factor of the composite number ${N}_{k}$ having as the first factor, the prime 5 always has the form $6n+1$ (Corollary 3).

2) The prime $p=7$ from column $6k+1$ appears for the first time in the row ${k}_{1}=6$ , applying Equation (5) there are no primes but composite numbers that have the 7 as one of its factors in the rows $k=6+7n$ for $n\ge 1$ . These compound numbers have the form ${N}_{k}=6k-1=6\left(6+7n\right)-1=7\left(6n-1\right)$ . The second factor of the composite number ${N}_{k}$ having as the first factor, the prime 7 always has the form $6n-1$ (Corollary 3).

3) The prime $p=11$ appears for the first time in the row ${k}_{1}=2$ , applying Equation (5) there are no primes but composite numbers that have the 11 as one of its factors in the rows $k=2+11n$ for $n\ge 1$ . These compound numbers have the form ${N}_{k}=6k-1=6\left(2+11n\right)-1=11\left(6n+1\right)$ . The second factor of the composite number ${N}_{k}$ having as the first factor, the prime 5 always has the form $6n+1$ (Corollary 3).

4) The prime $p=13$ from column $6k+1$ appears for the first time in the row ${k}_{1}=11$ , applying Equation (5) there are no primes but composite numbers that have the 13 as one of its factors in the rows $k=11+13n$ for $n\ge 1$ . These compound numbers have the form ${N}_{k}=6k-1=6\left(11+13n\right)-1=13\left(6n-1\right)$ . The second factor of the composite number ${N}_{k}$ having as the first factor the prime 13, always has the form $6n-1$ (Corollary 3).

5) From the analysis of the previous 4 points and applying Theorem 1 we can conclude the following with respect to Equation (5) being the geometric progression $p=6m-1$ :

With ${k}_{1}=m$ , $p=6m-1$ , we have:

${k}_{m,n}=m+\left(6m-1\right)n\to \left[m,n\right]\ge 1$ (6)

They contain composite numbers of the form $6{k}_{m.n}-1$

And

$k\ne {k}_{m,n}=m+\left(6m-1\right)n\to \left[m,n\right]\ge 1$ (7)

They contain prime numbers of the form $6k-1.$

In Equation (6) there exist symmetry of cells where ${k}_{1+5m,n}={k}_{1,1+7n}$ , ${k}_{11,n}={k}_{1,1+14n}$ , ${k}_{m,4}={k}_{1,4+5\left(n-1\right)}$ and so on, this is important to take into account in order to calculate the number of primes in this column as will be seen later. Table 1 gives an example of the above.

Table 1. Examples where there are symmetry of cells (in yellow) ${k}_{1+5m,n}={k}_{1,1+7n}$ (in blue) ${k}_{11,n}={k}_{1,1+14n}$ and (in green) ${k}_{m,4}={k}_{1,4+5\left(n-1\right)}$ , in Equation (6).

Table 2 shows examples where there are composite and prime numbers of the form $6k-1$ in the rows $1\le k\le 51$ applying the Equation (6) and Equation (7).

In Table 2, two cells ${k}_{m,n}$ which are the same, but with different $\left[m,n\right]$ , are highlighted in yellow and green. This occurs when the compound number in ${k}_{m,n}$ contains primes with powers greater than 1, Example

${k}_{1,8}={k}_{6,1}=41$ , ${N}_{k=41}=6×41-1=245={7}^{2}×5$

and

${k}_{1,9}={k}_{2,4}=46$ , ${N}_{k=46}=6×46-1=275={5}^{2}×11$ .

The number of times the above occurs in ${N}_{k}=6k-1$ , where ${p}_{m}=6m-1$ ; ${p}_{n}=6n+1$ ; ${p}_{m1}=6{m}_{1}-1$ ; ${p}_{m2}=6{m}_{2}-1$ ; ${p}_{n1}=6{n}_{1}+1$ ; ${p}_{n2}=6{n}_{2}+1$ with $\left[m,n,{m}_{1},{m}_{2},{n}_{1},{n}_{2}\right] (m1 ≠ m2 ${n}_{1}$ and ${n}_{2}$ can be the same or different), can be calculated by

$\left({\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{n1}{p}_{n2}{p}_{m}}|\ge 1+{\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{m1}{p}_{m2}{p}_{m}}|\ge 1\right)$ .

This is important for calculating the number of primes smaller or equal to Nk, as will be seen later.

3.1.2. Examples of Composite and Prime Numbers from $6k+1$ Column

1) The prime $p=7$ appears for the first time in the row ${k}_{1}=1$ , applying Equation (5) there are no primes but composite numbers that have the 7 as one of its factors in the rows $k=1+7n$ for $n\ge 1$ . These compound numbers have the form ${N}_{k}=6k+1=6\left(1+7n\right)-1=7\left(6n+1\right)$ . The second factor of the composite number ${N}_{k}$ having the first factor, the prime 5 always has the form $6n+1$ (Corollary 4).

Table 2. Examples where there are composite and prime numbers of the form $6k-1$ in the rows $1\le k\le 51$ applying the Equation (6) and Equation (7).

2) The prime $p=5$ from column 6k − 1 appears for the first time in the row ${k}_{1}=4$ , applying Equation (5) there are no primes but composite numbers that have the 5 as one of its factors in the rows $k=4+5\left(n-1\right)$ for $n\ge 1$ . These compound numbers have the form ${N}_{k}=6k+1=6\left(4+5\left(n-1\right)\right)+1=5\left(6n-1\right)$ . The second factor of the composite number ${N}_{k}$ having as the first factor, the prime 5 always has the form $6n-1$ (Corollary 4).

3) The prime $p=13$ appears for the first time in the row ${k}_{1}=2$ , applying Equation (5) there are no primes but composite numbers that have the 13 as one of its factors in the rows $k=2+13n$ for $n\ge 1$ . These compound numbers have the form ${N}_{k}=6k+1=6\left(2+13n\right)+1=13\left(6n+1\right)$ . The second factor of the composite number ${N}_{k}$ having as the first factor, the prime 13 always has the form $6n+1$ (Corollary 4).

4) The prime p = 11 from column 6k − 1 appears for the first time in the row ${k}_{1}=9$ , applying Equation (5) there are no primes but composite numbers that have the 11 as one of its factors in the rows $k=9+11\left(n-1\right)$ for $n\ge 1$ . These compound numbers have the form ${N}_{k}=6k+1=6\left(9+11\left(n-1\right)+1=11\left(6n-1\right)$ . The second factor of the composite number ${N}_{k}$ having as the first factor, the prime 11 always has the form $6n-1$ (Corollary 4).

5) From the analysis of the previous 4 points and applying Theorem 1 we can conclude the following with respect to Equation (5) being the geometric progression $p=6m-1$ or $p=6m+1$ :

With ${k}_{1}=m+pn$ , p = 6m + 1, them ${K}_{m,n}=m+\left(6m+1\right)n$ , where $\left[m,n\right]\ge 1$ , and ${N}_{k}=6{k}_{m,n}+1$ contains composite numbers of the form $\left(6m+1\right)\left(6n+1\right)$ .

With ${k}_{1}=4+5\left(m-1\right)+p\left(n-1\right)$ , $p=6m-1$ , then

${k}_{m,n}=4+5\left(m-1\right)+\left(6m-1\right)\left(n-1\right)$ ,

where $\left[m,n\right]\ge 1$ , and ${N}_{k}=6{k}_{m,n}+1$ contains composite numbers of the form $\left(6m-1\right)\left(6n-1\right)$ .

Therefore:

${k}_{m,n}=\left[\begin{array}{c}m+\left(6m+1\right)n\to \left[m,n\right]\ge 1\\ 4+5\left(m-1\right)+\left(6m-1\right)\left(n-1\right)\to \left[m,n\right]\ge 1\end{array}\right]$ (8)

They contain composite numbers of the form $6{k}_{m.n}+1$

$k\ne {k}_{m,n}=\left[\begin{array}{c}m+\left(6m+1\right)n\to \left[m,n\right]\ge 1\\ 4+5\left(m-1\right)+\left(6m-1\right)\left(n-1\right)\to \left[m,n\right]\ge 1\end{array}\right]$ (9)

They contain prime numbers of the form $6k+1.$

In ${k}_{m,n}=m+\left(6m+1\right)n$ , there exists symmetry of equal cells in ${k}_{1,n}={k}_{m,1}$ , ${k}_{2,\left(n+1\right)}={k}_{\left(m+1\right),2}$ and so on, i.e. ${k}_{a,\left(n+a-1\right)}={k}_{\left(m+a-1\right),a}$ , where $\left[a,m,n\right]\ge 1$ .

Also in ${k}_{m,n}=4+5\left(m-1\right)+\left(6m-1\right)\left(n-1\right)$ , there exist symmetries of equal cells in ${k}_{1,n}={k}_{m,1}$ , ${k}_{2,\left(n+1\right)}={k}_{\left(m+1\right),2}$ and so on, i.e. ${k}_{a,\left(n+a-1\right)}={k}_{\left(m+a-1\right),a}$ , where $\left[a,m,n\right]\ge 1$ .

Table 3 shows examples of the above, where the horizontal and vertical cells having the same value have been colored with the same color.

Table 3. Examples of symmetry in Equation (8), where horizontal and vertical cells having the same value have been colored with the same color.

The number of cells repeated in each column $m$ is $m-1$ and the sum until

$m$ of all columns of a matrix $m×n$ , being $m=n$ would be $\frac{{m}^{2}-m}{2}$ both in

${k}_{m,n}=m+\left(6m+1\right)n$ , as in ${k}_{m,n}=4+5\left(m-1\right)+\left(6m-1\right)\left(n-1\right)$ . This is important to find the number of primes less than or equal to k in column $6k+1$ , as will be seen later.

The cell symmetry in Equation (8) can be eliminated by taking only the values of $n\ge m$ , so Equation (8) and Equation (9) would be:

${k}_{m,n}=\left[\begin{array}{c}m+\left(6m+1\right)n\to \left[n\ge m\right]\ge 1\\ 4+5\left(m-1\right)+\left(6m-1\right)\left(n-1\right)\to \left[n\ge m\right]\ge 1\end{array}\right]$ (10)

They contain composite numbers of the form $6{k}_{m.n}+1$

$k\ne {k}_{m,n}=\left[\begin{array}{c}m+\left(6m+1\right)n\to \left[n\ge m\right]\ge 1\\ 4+5\left(m-1\right)+\left(6m-1\right)\left(n-1\right)\to \left[n\ge m\right]\ge 1\end{array}\right]$ (11)

They contain prime numbers of the form $6k+1.$

Table 4 shows examples where there are composite and prime numbers in rows $1\le k\le 51$ applying Equation (10) and Equation (11).

In some cases ${k}_{m,n}=m+\left(6m+1\right)n$ can be equal to

${k}_{m,n}=4+5\left(m-1\right)+\left(6m-1\right)n$ , example ${k}_{1,4}={k}_{1,6}=29$ , value highlighted in the yellow cells in Table 4. These cases correspond to cells containing a composite number ${N}_{k}=6k+1={p}_{m1}{p}_{m2}{p}_{n}$ , where ${p}_{\text{m}1}=6{m}_{1}-1$ and ${p}_{m2}=6{m}_{2}-1$ for $\left[{m}_{1},{m}_{2}\right]\ge 1$ . Example cells 29, 64, 141 are repeated in ${k}_{m,n}=m+\left(6m+1\right)n$ and en ${k}_{m,n}=4+5\left(m-1\right)+\left(6m-1\right)\left(n-1\right)$ . This is important in order to calculate the number of primes in $6k+1$ column, as will be seen later.

4. How to Recognize the Prime Numbers of $6k±1$ Columns

Regardless of how to find primes with Equation (7) and Equation (11), there is the traditional test for whether or not a number is prime, using Theorem 2.

1) For $6k-1$ column

Theorem 2:

If ${N}_{k}>5$ is a composite integer of the form ${N}_{k}=6k-1$ , then ${N}_{k}$ has a prime factor $p$ , with $p\le \sqrt{{N}_{k}}$ .

Demonstration:

Let Nk > 5 be a composite integer of the form ${N}_{k}=6k-1$ . Then ${N}_{k}=a\cdot b$ , with $5\le a<{N}_{k}$ and $5\le b<{N}_{k}$ . We affirm that one of them $a$ or $b$ is less than or equal to $\sqrt{{N}_{k}}$ . As $a>5$ , from the previous lemma, there exists a prime $p$ such that $p|a$ .

Table 4. Examples where there are composite and prime numbers in rows $1\le k\le 51$ applying Equation (10) and Equation (11).

As $a|{N}_{k}$ , it follows that $p|{N}_{k}$ and as $p|a$ , we also have $p\le a\le \sqrt{{N}_{k}}$ . The theorem has being proved.

As it is known that for $k=\left[1,2,3,4,5\right]$ , $p=6k-1$ are primes, then for $k\ge 6$ and $p=6k-1$ is a prime greater than 5 according to Theorem 2, as long as:

$\left(6m-1\right)\nmid p\text{Or}\left(6m+1\right)\nmid p\text{where}m\le |\sqrt{k/6}|\text{and}k\ge 6$ (12)

Example: primes (11, 17, 23, 29, and 41). The problem with Equation (12) is that its application is more difficult for very large primes; one would have to know all primes smaller than $\sqrt{p}$ , and the application of Equation (8) and Equation (9), is simpler; all $6k-1$ primes can be sequentially obtained.

2) For the $6k+1$ column

Applying the same theorem shown for the $6k-1$ column, we have that $p=6k+1$ is a prime greater than 7 as long as:

$\left(6m-1\right)\nmid p\text{Or}\left(6m+1\right)\nmid p\text{where}m\le 1+|\sqrt{k/6}|\text{and}k\ge 4$ (13)

Example: primes (13, 19, 31, and 37). Similarly, the application of Equation (13), is more difficult for very large primes, while the application of Equation (10) and Equation (11), is simpler to obtain all $6k+1$ primes sequentially.

5. Numbers of Primes in $6k±1$ Columns

1) The total of primes $\Pi$ including primes 2 and 3 will be $\Pi =2+{\Pi }_{1}+{\Pi }_{2}$ , where ${\Pi }_{1}$ is the total of primes in $6k-1$ column and ${\Pi }_{2}$ the total of primes in $6k+1$ column for the same k in both columns. The remarkable thing about this is that ${\Pi }_{1}\ge {\Pi }_{2}$ , by the way the composite numbers that have 5 and 7 appear as one of their prime factors in each column, therefore $2{\Pi }_{2}\le \Pi -2\le 2{\Pi }_{1}$ .

2) The number of primes ${\Pi }_{1}$ in $6k-1$ column, smaller than or equal to ${N}_{k}=6k-1;$ where ${p}_{m}=6m-1$ ; ${p}_{n}=6n+1$ ; ${p}_{m1}=6{m}_{1}-1$ ; ${p}_{m2}=6{m}_{2}-1$ ; ${p}_{{n}_{1}}=6{n}_{1}+1;$ ${p}_{2}=6{n}_{2}+1$ with $\left[m,n,{m}_{1},{m}_{2},{n}_{1},{n}_{2}\right] (m1 ≠ m2, ${n}_{1}$ and ${n}_{2}$ can be the same or different), can be calculated by Equation (14):

${\Pi }_{1}=k-{\sum }^{\text{​}}|\frac{k-m}{{p}_{m}}|\ge 1+\left({\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{n1}{p}_{n2}{p}_{m}}|\ge 1+{\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{m1}{p}_{m2}{p}_{m}}|\ge 1\right)$ (14)

The term ${\sum }^{\text{​}}|\frac{k-m}{{p}_{m}}|\ge 1$ corresponds to the number of composite numbers in ${N}_{k}$ , including cells that have equal numbers in different cells as shown in Table 1 and Table 2 and the term $\left({\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{n1}{p}_{n2}{p}_{m}}|\ge 1+{\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{m1}{p}_{m2}{p}_{m}}|\ge 1\right)$ corresponds to the number of cells that are repeated with the same value in ${\sum }^{\text{​}}|\frac{k-m}{{p}_{m}}|\ge 1$ .

Example of ${\Pi }_{1}$ for $k=16,{N}_{k}=95$ , with ${p}_{1}=5$ and ${p}_{2}=11$ , therefore

${\sum }^{\text{​}}|\frac{k-m}{{p}_{m}}|\ge 1=3+1=4$ and

$\begin{array}{l}\left({\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{n1}{p}_{n2}{p}_{m}}|\ge 1+{\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{m1}{p}_{m2}{p}_{m}}|\ge 1\right)=\left[\left({\sum }^{\text{​}}|\frac{95}{7×7×5}|\ge 1\right)=0\right]\\ \text{}+\left[\left({\sum }^{\text{​}}|\frac{95}{5×11×{5}^{}}|\ge 1\right)=0\right]=0,\end{array}$

then

${\Pi }_{1}=16-4=12$ .

Example of ${\Pi }_{1}$ for $k=51,$ ${N}_{k}=305$ , with ${p}_{1}=5$ , ${p}_{2}=11$ , ${p}_{3}=17$ , ${p}_{4}=23$ , ${p}_{5}=29$ , ${p}_{6}=35$ y

${p}_{7}=41{\sum }^{\text{​}}|\frac{k-m}{{p}_{m}}|\ge 1=10+4+2+2+1+1+1=21$

and

$\left({\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{n1}{p}_{n2}{p}_{m}}|\ge 1+{\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{m1}{p}_{m2}{p}_{m}}|\ge 1\right)=\left(|\frac{305}{7×7×5}|=1\right)+\left(|\frac{305}{5×11×5}|=1\right)=2$

therefore ${\Pi }_{1}=51-21+2=32$ (see Table 2).

3) The number of primes ${\Pi }_{2}$ in $6k+1$ column, less than or equal to ${N}_{k}=6k+1$ , where ${p}_{m}=6m-1$ and ${p}_{n}=6n+1$ with $\left[m,n\right] , is a little more complicated than Equation (14) because of the large number of cells that are repeated as shown in Table 3 and Table 4 and can be calculated by Equation (15):

$\begin{array}{l}{\Pi }_{2}=k-|\frac{k+1}{5}|-{\sum }^{\text{​}}{a}_{m}-{\sum }^{\text{​}}{a}_{n}+{\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{m1}{p}_{m2}{p}_{n}}|\ge 1\\ {a}_{m}=|\frac{k-\left(4+5\left(m-1\right)\right)}{{p}_{m\ge 2}}|-\left(m-2\right)\to |\frac{k-\left(4+5\left(m-1\right)\right)}{{p}_{m\ge 2}}|\ge \left(m-2\right)\\ {a}_{n}=|\frac{k-n}{{p}_{n}}|-\left(n-1\right)\to |\frac{k-n}{{p}_{n}}|\ge \left(n-1\right)\end{array}$ (15)

The term $|\frac{k+1}{5}|$ contains the total number of composite having 5 as one of

its factors, the term ${\sum }^{\text{​}}{a}_{m}$ contains the total number of composite numbers having ${p}_{m\ge 2}$ as one of its Factors after eliminating the repeated cells in ${k}_{m,n}=4+5\left(m-1\right)+\left(6m-1\right)\left(n-1\right)$ , the term ${\sum }^{\text{​}}{a}_{n}$ contains the total number of composite numbers having ${p}_{n}$ as one of its factors after eliminating the re-

peated cells in ${k}_{m,n}=m+\left(6m+1\right)n$ and the term ${\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{m1}{p}_{m2}{p}_{n}}|\ge 1$ corresponds

to the number of repeated cells where ${k}_{m,n}=m+\left(6m+1\right)n$ is equal to ${k}_{m,n}=4+5\left(m-1\right)+\left(6m-1\right)\left(n-1\right)$ where its factors are pm1pm2pn with ${p}_{m1}=6{m}_{1}-1$ , ${p}_{m2}=6{m}_{2}-1$ and ${p}_{n}=6n+1\to \left[{m}_{1},{m}_{2},n\right]\ge 1$ .

Example of ${\Pi }_{2}$ where $k=16$ and ${N}_{k}=97$ :

With ${p}_{m}=\left[5,11\right]$ and ${p}_{n}=\left[7,13\right]$ then

$|\frac{k+1}{5}|=|\frac{16+1}{5}|=3$ ,

${\sum }^{\text{​}}{a}_{m}=|\frac{k-\left(4+5\left(m-1\right)\right)}{{p}_{m\ge 2}}|-\left(m-2\right)=\left[\left(|\frac{16-9}{11}|\ge 0\right)=0\right]-0=0$ ,

$\begin{array}{c}{\sum }^{\text{​}}{a}_{n}={\sum }^{\text{​}}\left(|\frac{k-n}{{p}_{n}}|\ge \left(n-1\right)\right)-\left(n-1\right)\\ =\left[\left\{\left(|\frac{16-1}{7}|\ge 0\right)=2\right\}-0=2\right]+\left[\left\{\left(|\frac{16-2}{13}|\ge 1\right)=1\right\}-1=0\right]=2,\end{array}$

$\left({\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{m1}{p}_{m2}{p}_{n}}|\ge 1\right)=\left(|\frac{97}{5×5×7}|\ge 1\right)=0.$ Therefore ${\Pi }_{2}=16-3-0-2+0=11$ .

Example of ${\Pi }_{2}$ where $k=51$ and ${N}_{k}=307$ :

With ${p}_{m}=\left[5,11,17,23\right]$ and ${p}_{n}=\left[7,13,19\right]$ then

$|\frac{k+1}{5}|=|\frac{51+1}{5}|=10$ ,

$\begin{array}{l}{\sum }^{\text{​}}{a}_{m}=|\frac{k-\left(4+5\left(m-1\right)\right)}{{p}_{m\ge 2}}|-\left(m-2\right)\\ \text{}=\left[\left\{\left(|\frac{51-9}{11}|\ge 0\right)=3\right\}-0=3\right]+\left[\left\{\left(|\frac{51-14}{17}|\ge 3\right)=2\right\}-1=1\right]\\ \text{}+\left[\left\{\left(|\frac{51-19}{23}|\ge 4\right)=2\right\}-2=0\right]=4,\end{array}$

$\begin{array}{l}{\sum }^{\text{​}}{a}_{n}={\sum }^{\text{​}}\left(|\frac{k-n}{{p}_{n}}|\ge \left(n-1\right)\right)-\left(n-1\right)\\ \text{}=\left[\left\{\left(|\frac{51-1}{7}|\ge 0\right)=7\right\}-0=7\right]+\left[\left\{\left(|\frac{51-2}{13}|\ge 1\right)=3\right\}-1=2\right]\\ \text{}+\left[\left\{\left(|\frac{51-3}{19}|\ge 2\right)=2\right\}-2=0\right]=9,\end{array}$

$\left({\sum }^{\text{​}}|\frac{{N}_{k}}{{p}_{m1}{p}_{m2}{p}_{n}}|\ge 1\right)=\left(|\frac{307}{5×5×7}|\ge 1\right)=1$ . Therefore ${\Pi }_{2}=51-10-4-9+1=29$ (see Table 4).

4) With $k=16$ then $\Pi =2+{\Pi }_{1}+{\Pi }_{2}=2+12+11=25$ and with $k=51$ then $\Pi =2+{\Pi }_{1}+{\Pi }_{2}=2+32+29=63$ . In both cases, $2{\Pi }_{2}\le \Pi -2\le 2{\Pi }_{1}.$

6. Conclusions

All prime numbers $p\ge 5$ have a pattern or form described by Equation (4).

Verification of whether a number is prime can be done in $6k-1$ column by means of Equation (6) and Equation (7) or Equation (12) and in $6k+1$ column by means of Equation (10) and Equation (11) or Equation (13).

The primes do not appear in random form, their sequence is determined by the Equation (6) and Equation (7) in $6k-1$ column and Equation (10) and Equation (11) for $6k+1$ column.

The number of primes ${\Pi }_{1}$ less than or equal to ${N}_{k}=6k-1$ can be calculated exactly by Equation (14) without needing to know all primes having that form equally for ${\Pi }_{2}$ with Equation (15).

Acknowledgements

To all my professors from Escuela Naval de Cadetes Colombia and the Naval Postgraduate School of the United States (Naval Postgraduate School, Monterey California USA).

Cite this paper: Ferreira, J. (2017) The Pattern of Prime Numbers. Applied Mathematics, 8, 180-192. doi: 10.4236/am.2017.82015.
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