Classification proposal based on multiples of 6
ES | EN

About possible primes (Pp)

-

6n - 1

6n

+

6n + 1

Pp are always at a distance of 1 from a multiple of 6.

That is why we take multiples of 6 as a reference to identify Pp

We assign the notation to Pp in 6x - 1 "Ppa"

We assign the notation to Pp in 6x + 1 "Ppb"

...

6n

6

12

+

6n + 1

7

13

6n + 2

8

14

6n + 3

9

15

6n + 4

10

16

-

6n2 - 1

11

17

6n2

12

18

...

We can establish the types of numbers that may exist between two multiples of 6

Apart from possible primes, we always find two even numbers and one divisible by 3

That is, every integer is: Multiple of 6, Ppa, multiple of 2, multiple of 3 or Ppb

6n1
-
1
*
6n2
+
1
=
6n3
-
1
5 * 13 = 65
6n1
-
1
*
6n2
-
1
=
6n3
+
1
5 * 11 = 55
6n1
+
1
*
6n2
+
1
=
6n3
+
1
7 * 13 = 91

Application of sign law

This notation among possible primes gives us information about the characteristics of their possible products. Since they are multiplications, they follow the sign law

23 24 25 26 27 28 29 30 31 32 33 34 35 36 37
Ppa1 - Npb1 - - - Ppa2 Br Ppb2 - - - Npa1 - Ppb3
+
-
-
35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
Npa2 - Ppb3 - - - Ppa4 Br Ppb4 - - - Ppa5 - Npb2
-
+
+

Relative bases

Np are also equidistant to multiples of 6.
They do so as follows:
Br = 6 · Pp;

Ppa = 5, Npb1 = (6 · Ppa) · i - Ppa
Ppb = 7, Npb2 = (6 · Ppb) · i + Ppb
Ppb = 7, Npa2 = (6 · Ppb) · i - Ppb
Ppa = 5, Npa1 = (6 · Ppa) · i + Ppa

Ppb - Ppa = 2 => Npa1 = Npa2
Ppb - Ppa ≠ 2 => Npa1 ≠ Npa2

Npb1 = (6 · (6n1 - 1)) · i - (6n1 - 1)
i = Npb1 + (6n1 - 1) / (36n1 - 6)

Npa2 = (6 · (6n1 + 1)) · i - (6n1 + 1)
i = Npa2 + (6n1 + 1) / (36n1 + 6)

Npa1 = (6 · (6n1 - 1)) · i + (6n1 - 1)
i = Npa1 - (6n1 - 1) / (36n1 - 6)

Npb2 = (6 · (6n1 + 1)) · i + (6n1 + 1)
i = Npb2 - (6n1 + 1) / (36n1 + 6)