Sexy prime numbers are pairs of prime numbers that differ from 6. The real sexy prime numbers are pairs of successive prime numbers that differ from 6 [1] .

Remark:

The interestiong thing about sexy primes is that not only do they form pairs, but also triplets and quardruplets.

As for what concerns us for the remainder of this article, we will focus on pairs of sexy primes.

According to the Euclidean division theorem for positive integers, we have

∀ ( a , b ) ∈ N × N * , ∃ q , r ∈ N / a = b q + r and r < b

N = { a ∈ N } = { b q + r     with   b ∈ N * ;   q , r ∈ N     and   r < b }

We then write:

N = { a n + b   with   a ∈ N *   and   n , b ∈ N   suchthat   b < a   and   b   isbetween   0   and   a − 1 }

We can write:

N = ∪ b = 0 a − 1 { a n + b   with   a ∈ N * , n ∈ N }

Example:

• If a = 1

N = ∪ 0 0 { n , n ∈ N } = { n , n ∈ N }

• If a = 2

N = { 2 n , n ∈ N } ∪ { 2 n + 1, n ∈ N }

• If a = 3

N = { 3 n , n ∈ N } ∪ { 3 n + 1, n ∈ N } ∪ { 3 n + 2, n ∈ N }

• If a = 4

N = { 4 n , n ∈ N } ∪ { 4 n + 1, n ∈ N } ∪ { 4 n + 2, n ∈ N } ∪ { 4 n + 3, n ∈ N }

• If a = 5

N = { 5 n , n ∈ N } ∪ { 5 n + 1, n ∈ N } ∪ { 5 n + 2, n ∈ N }     ∪ { 5 n + 3, n ∈ N } ∪ { 5 n + 4, n ∈ N }

• If a = 6

N = { 6 n , n ∈ N } ∪ { 6 n + 1, n ∈ N } ∪ { 6 n + 2, n ∈ N } ∪ { 6 n + 3, n ∈ N }     ∪ { 6 n + 4, n ∈ N } ∪ { 6 n + 5, n ∈ N }

Consider the set

N = { 6 n , n ∈ N } ∪ { 6 n + 1, n ∈ N } ∪ { 6 n + 2, n ∈ N } ∪ { 6 n + 3, n ∈ N }     ∪ { 6 n + 4, n ∈ N } ∪ { 6 n + 5, n ∈ N }

Note1:

The elements of { 6 n , n ∈ N } are even.

The elements of { 6 n + 2, n ∈ N } are even.

The elements of { 6 n + 3, n ∈ N } are multiples of 3.

The elements of { 6 n + 4, n ∈ N } are even.

When we eliminate these four previous sets in N we are left with the following two sets:

{ 6 n + 1, n ∈ N } and { 6 n + 5, n ∈ N } .

Consequently the set { 6 n + 1, n ∈ N } ∪ { 6 n + 5, n ∈ N } contains all the prime numbers except 2 and 3.

Note2:

{ 6 n + 1, n ∈ N } = { 1 } ∪ { 6 n + 1, n ∈ N * } .

Demonstrate that { 6 n + 1, n ∈ N * } = { 6 n + 7, n ∈ N } Let p = n − 1 with n ∈ N * so p ∈ N and n = p + 1

6 n + 1 = 6 ( p + 1 ) + 1 = 6 p + 6 + 1 = 6 p + 7 , p ∈ N

Then { 6 n + 1, n ∈ N } = { 1 } ∪ { 6 n + 1, n ∈ N * } .

So: { 6 n + 1, n ∈ N } = { 1 } ∪ { 6 n + 1, n ∈ N * } = { 1 } ∪ { 6 n + 7, n ∈ N }

{ 6 n + 1 ; 6 n + 5, n ∈ N } = { 1 } ∪ { 6 n + 5 ; 6 n + 7, n ∈ N }

We note that the set { 6 n + 5 ; 6 n + 7, n ∈ N } contains all prime numbers except 2 and 3. Name E_sp: The Set of supposedly prime numbers, E s p = { 2 ; 3 ; 6 n + 5 ; 6 n + 7   with   n ∈ N } contains all prime numbers [2] .

Remark:

Prime numbers other than 2 and 3 are generated by the formulas U n = 6 n + 5 and V n = 6 n + 7 with n ∈ N .

A prime number other than 2 and 3 can be written in the form 6 n + 5 or 6 n + 7 with n ∈ N .

U n = 6 n + 5 with n ∈ N then U n − 5 6 = n with n ∈ N

V n = 6 n + 7 with n ∈ N then V n − 7 6 = n with n ∈ N

Consequence:

A supposed prime number N is of class U if and only if N − 5 6 = n with

n ∈ N

A supposedly prime number N is of class V if and only if N − 7 6 = n with n ∈ N

U n = 6 n + 5 with n ∈ N and V n = 6 n + 7 with n ∈ N

U n + 1 = 6 ( n + 1 ) + 5 = 6 n + 11 with n ∈ N

V n + 1 = 6 ( n + 1 ) + 7 = 6 n + 13 with n ∈ N

We have: U n < V n < U n + 1 < V n + 1 , for fixed n.

U n = 6 n + 5 with n ∈ N

U n + 1 − U n = 6 ( n + 1 ) + 5 − ( 6 n + 5 ) = 6

U n < V n < U n + 1 < V n + 1

Consequence:

( U n ; U n + 1 ) is prime sexy if and only if U n is prime and U n + 1 is prime ( U n ; U n + 1 ) is real prime sexy if and only if U n is prime, U n + 1 is prime and V n not prime.

Statement:

Let U n = 6 n + 5 with n ∈ N and V n = 6 n + 7 with n ∈ N sexy primes of class U are the couples ( U n ; U n + 1 ) such that U n is prime, U n + 1 is prime.

The Real sexy primes of class U are pairs ( U n ; U n + 1 ) such that U n is prime, U n + 1 is prime and V n is not prime.

Example:

a) Determine the class of the couple (47; 53)

b) Show that it is sexy

c) Show that it is a real sexy

Solution

a) 47 − 5 6 = 7 with 7 ∈ N so 47 is class U

53 − 5 6 = 8 with 8 ∈ N so 53 is class U

Hence the couple (47; 53) is class U

b) 47 is prime and 53 is prime (1)

53 − 47 = 6 (2)

(1) and (2) so (47; 53) is prime sexy

c) n = 47 − 5 6 = 7 then V 7 = 6 × 7 + 7 = 49

But 49 is a non-prime number so (47; 53) is a real sexy (47 and 53 follow one another) and class U.

Counter-example:

a) Determine the class of the couple (5; 11)

b) Show that it is sexy

c) Show that it is not a real sexy

Solution:

a) 5 − 5 6 = 0 with 0 ∈ N so 5 is class U

11 − 5 6 = 1 with 1 ∈ N so 11 is class U

Hence the couple (5; 11) is of class U.

b) 5 is prime and 11 is prime (1)

11 − 5 = 6 (2)

(1) and (2) so (5; 11) is prime sexy

c) n = 5 − 5 6 = 0 then V 0 = 6 × 0 + 7 = 7

But 7 is a prime number so (5; 11) is not a real sexy prime because 5 and 11 do not follow each other (there is 7 in the middle).

V n = 6 n + 7 with n ∈ N

V n + 1 − V n = 6 ( n + 1 ) + 7 − ( 6 n + 7 ) = 6

U n < V n < U n + 1 < V n + 1

Consequence:

( V n ; V n + 1 ) is prime sexy if and only if V n is prime and V n + 1 is prime.

( V n ; V n + 1 ) is real prime sexy if and only if V n is prime and V n + 1 is prime and U n + 1 not prime.

Statement:

Let U n = 6 n + 5 with n ∈ N and V n = 6 n + 7 with n ∈ N sexy primes of class V are the couples ( V n ; V n + 1 ) such that V n is prime, V n + 1 is prime.

The Real sexy primes of class V are pairs ( V n ; V n + 1 ) such that V n is prime, V n + 1 is prime and U n + 1 is not prime.

Noticed:

U n < V n < U n + 1 < V n + 1 , for fixed n.

Example:

1. Determine the class of the couple (61; 67)

2. Show that it is sexy

3. Show that it is a real sexy

Solution:

a) 61 − 7 6 = 9 with 9 ∈ N so 61 is class V

67 − 7 6 = 10 with 10 ∈ N so 67 is class V

Hence the couple (61; 67) is class V

b) 61 is prime and 67 is prime (1)

67 - 61 = 6 (2)

(1) and (2) so (61; 67) is prime sexy

c) n = 61 − 7 6 = 9 then U 9 + 1 = U 10 = 6 × 10 + 5 = 65

But 65 is a non-prime number so (61; 67) is a real sexy prime of class V.

Counter-example:

a) Determine the class of the couple (7; 13)

b) Show that it is sexy

c) Show that it is not a real sexy

Solution:

a) 7 − 7 6 = 0 with 0 ∈ N so 7 is class V

13 − 7 6 = 1 with 1 ∈ N so 13 is class V

Hence the couple (7; 13) is class V

b) 7 is prime and 13 is prime (1)

13 − 7 = 6 (2)

(1) and (2) so (7; 13) is prime sexy of class V

c) n = 7 − 7 6 = 0 then U 0 + 1 = U 1 = 6 × 1 + 5 = 11

But 11 is a prime number so (7; 13) is not real sexy prime because 7 and 13 do not follow each other (there is 11 in the middle).

Cousin prime numbers are pairs of prime numbers that differ from 4 [3] .

U n = 6 n + 5 with n ∈ N and V n = 6 n + 7 with n ∈ N

U n + 1 = 6 ( n + 1 ) + 5 = 6 n + 11 with n ∈ N

U n + 1 − V n = 6 ( n + 1 ) + 11 − ( 6 n + 7 ) = 4

U n < V n < U n + 1 < V n + 1

Consequence:

( V n ; U n + 1 ) is cousin prime if and only if V n is prime and U n + 1 is prime.

Statement:

Let U n = 6 n + 5 with n ∈ N and V n = 6 n + 7 with n ∈ N cousin primes are pairs ( V n ; U n + 1 ) such that V n is prime, U n + 1 is prime.

Noticed :

U n < V n < U n + 1 < V n + 1 , for n fixed therefore V n and U n + 1 are two supposedly prime which therefore follow each other:

If ( V n ; U n + 1 ) is cousin prime then V n and U n + 1 are two prime numbers that follow each other.

Example:

The following three pairs are cousin prime pairs (7; 11), (13; 17), (19; 23).

Counter-example:

When two prime numbers do not follow each other they cannot be cousins. To give counter examples, simply choose two non-successive prime numbers.

Consider E s p the set of supposed prime numbers.

E s p = { 2 ; 3 ; 6 n + 5 ; 6 n + 7   with   n ∈ N }

Supposedly prime numbers other than 2 and 3 are generated by the following two formulas:

U n = 6 n + 5 with n ∈ N and V n = 6 n + 7 with n ∈ N

Noticed:

• Let p = n − 1 with n ∈ N * therefore p ∈ N and n = p + 1

6 n + 1 = 6 ( p + 1 ) + 1 = 6 p + 6 + 1 = 6 p + 7 ,   p ∈ N

so:

{ 6 n + 1, n ∈ N * } = { 6 n + 7, n ∈ N }

• Let p = n − 1 with n ∈ N * therefore p ∈ N and n = p + 1

6 n − 1 = 6 ( p + 1 ) − 1 = 6 p + 6 − 1 = 6 p + 5 ,   p ∈ N

therefore { 6 n − 1, n ∈ N * } = { 6 n + 5, n ∈ N } .

So { 6 n + 5 ; 6 n + 7   with   n ∈ N * } = { 6 n − 1 ; 6 n + 1   with   n ∈ N * }

From where

E s p = { 2 ; 3 ; 6 n + 5 ; 6 n + 7   with   n ∈ N } or E s p = { 2 ; 3 ; 6 n − 1 ; 6 n + 1   with   n ∈ N * } Noticed:

U n = 6 n − 1 with n ∈ N * and V n = 6 n + 1 with n ∈ N *

We have: U n + 1 6 = n and V n − 1 6 = n with n ∈ N * .

Consequence :

A number N is a supposed prime number other than 2 and 3 if and

Only if N + 1 6 ∈ N * or N − 1 6 ∈ N *

We found through calculations that:

1. The formula 2 n + 3 m with n , m ∈ N * only gives supposedly prime numbers.

2. The formula | 2 n − 3 m | with n , m ∈ N * , with the exception of 1, only gives supposedly prime numbers.

3. Formulas | 2 n ± 3 m | with n , m ∈ N * do not all give supposed prime numbers.

NB:

This observation was made through in-depth calculations up to the supposedly very high first ones but nothing proves it to us through demonstrations.

Let us accept that this hypothesis is true, we will therefore have

{ 2 n + 3 m ; | 2 n − 3 m |   with   n , m ∈ N * } \ { 1 } ⊂ { 2 ; 3 ; 6 n − 1 ; 6 n + 1   with   n ∈ N * }

{ 2 n + 3 m ; | 2 n − 3 m |   with   n , m ∈ N * } \ { 1 } is a subset of the set of presumed prime numbers denoted by E s p .

Result :

1. 2 n + 3 m with n , m ∈ N * are supposed primes other than 2 and 3 so 2 n + 3 m with n , m ∈ N * is written in the form 6 k − 1 with k ∈ N * or 6 k + 1 with k ∈ N * .

2. | 2 n − 3 m | with n , m ∈ N * are supposed prime numbers other than 2 and 3 more | 2 n − 3 m | gives 1 for n = 1, m = 1 or n = 3, m = 2 and 6n + 1 gives 1 for n = 0 therefore | 2 n − 3 m | with n , m ∈ N * can be written as 6 k − 1 with k ∈ N * or 6 k + 1 with k ∈ N .

States :

• For all non-zero integers n and m, there exists a non-zero integer k verifying the equation:

2 n + 3 m = 6 k − 1 or 2 n + 3 m = 6 k + 1

• For all non-zero integers n and m, there exists a non-zero integer k and an integer p verifying the equation:

| 2 n − 3 m | = 6 k − 1 or | 2 n − 3 m | = 6 p + 1

These two pairs of equations are made available to researchers in the field for their demonstrations.

Noticed:

• The formula 2 n + 3 m with n , m ∈ N * gives an infinity of prime numbers and an infinity of non-prime numbers.

• The formula | 2 n − 3 m | with n , m ∈ N * , with the exception of 1, gives an infinity of prime numbers and an infinity of non-prime numbers.

• Not all prime numbers are written in the form | 2 n ± 3 m | with n , m ∈ N * .

NB: (for verification)

1. To verify that the formula 2 n + 3 m with n , m ∈ N * , gives only supposed primes, just choose two integers n 1 and n 2 then:

• Calculate 2 n 1 + 3 n 2 which gives M 1

• Check that: M 1 + 1 6 ∈ N * or M 2 − 1 6 ∈ N *

Example:

if n = 2; m = 2

| 2 2 − 3 2 | = | 4 − 9 | = 5

5 − 5 6 = 0 ∈ N so 5 is a supposedly prime

if n = 2; m = 3

| 2 2 − 3 3 | = | 4 − 27 | = 23

23 − 5 6 = 9 ∈ N so 23 is a supposedly prime

if n = 2; m = 4

| 2 2 − 3 4 | = | 4 − 81 | = 77

77 − 5 6 = 12 ∈ N so 77 is a supposedly prime.

We can continue the verifications by assigning larger numbers to n and m, and we will see that we will always have supposedly prime numbers.

2. To verify that the formula | 2 n − 3 m | with n , m ∈ N * , gives only supposed primes, just choose two integers n 1 and n 2 then:

• Calculate | 2 n 1 − 3 n 2 | which gives M 2

• Check that: M 2 + 1 6 ∈ N * or M 2 − 1 6 ∈ N *

Example:

if n = 1, m = 1

2 1 + 3 1 = 2 + 3 = 5 and 5 is a supposedly prime number

if n = 2, m = 2

2 2 + 3 2 = 4 + 9 = 13

13 − 7 6 = 1 ∈ N so 13 is a supposedly prime number

if n = 2, m = 3

2 2 + 3 3 = 4 + 27 = 31

31 − 7 6 = 3 ∈ N so 31 is a supposedly prime number

if n = 2,m = 4

2 2 + 3 4 = 4 + 81 = 85

85 − 7 6 = 13 ∈ N so 85 is a supposedly prime number.

We can continue the verifications by assigning larger numbers to n and m, and we will see that we will always have supposedly prime numbers.

In mathematics and more precisely in arithmetic, a Mersenne number is a number of the form 2 n − 1 (often denoted M n ), where n is a natural number not zero; a Mersenne prime (or prime Mersenne number) is therefore a prime number of this form. These numbers owe their name to the erudite French religious and mathematician Marin Mersenne of the 17th century; but, nearly 2,000 years earlier, Euclid was already using them to study perfect numbers. Before Mersenne, and even for some time after him, the search for Mersenne primes is intrinsically linked to that of perfect numbers [4] .

Consider the formula | 2 n − 3 m | with n , m ∈ N * .

Let | 2 n − 3 m | with n , m ∈ N .

So this formula generates an infinity of supposed prime nombers and an infinity of multiple numbers of 2 or/and 3.

If we set m = 0, we will have | 2 n − 3 0 | = | 2 n − 1 | .

2 n − 1 ≥ 0 for n ∈ N therefore | 2 n − 1 | = 2 n − 1 with n ∈ N , this is the Mersenne number.

A Fermat number is a number that can be written in the form 2 2 n + 1 , with n a natural number. The Fermat number of rank n, 2 2 n + 1 , is denoted F n .

The sequence ( F n ), which begins with 3, 5, 17, 257, 65537, 4294967297, 18446744073709551617 is listed as sequence A000215 in the EIS.

These numbers owe their name to Pierre de Fermat, who made the conjecture that all of these numbers were prime. This conjecture turned out to be false, F5 being composite, as are all the following ones up to F 33 . It is not known whether the numbers from F33 onwards are prime or composite. Thus, the only known Fermat primes are five in number, namely the first five F 0 , F 1 , F 2 , F 3 and F 4 , which are respectively 3, 5, 17, 257 and 65,537.

Fermat numbers have interesting properties, generally arising from modular arithmetic. In particular, the Gauss-Wantzel theorem establishes a link between these numbers and the construction with a ruler and compass of regular polygons: a regular polygon with n sides can be constructed with a ruler and compass if and only if n is a power of 2, or the product of a power of 2 and distinct Fermat primes [5] .

Consider the formula 2 n + 3 m with n , m ∈ N * .

Let 2 n + 3 m with n , m ∈ N .

So this formula generates an infinity of supposed prime nombers and an infinity of multiple numbers of 2 or/and 3.

If we set m = 0 and we choose n = 2k with k ∈ N .

We will have: 2 2 k + 3 0 = 2 2 k + 1 with k ∈ N , this is Fermat’s number.

Noticed:

Concerning the origins of other formulas of the past, it will be necessary to make a limited development of | 2 n ± 3 m | with n , m ∈ N * .