The number sequences 6n+5 (for all n0) and 6n+1 (for all n0) cover all odd natural numbers that are not divisible by 3. For example, 6n+5 leads to 5, 11, 17, ... and 6n+1 leads to 7, 13, 19, ... . Because all primes larger than 3 are odd natural numbers and not divisible by 3, they must be of type 6n+5 or 6n+1. All potential prime numbers larger than 3 can therefore be shown as zeros of the following two sine functions: f1(x) = sin (( for all numbers of type 6n+5 and f2(x) = sin (( for all numbers of type 6n+1. The allowed range for x reaches for both sine functions from 0 to infinite.

Now the question arises, which natural numbers of type 6n+5 and 6n+1 are not prime?

__Under what conditions are numbers of type 6n+5 not prime and what regularity do these follow?__

6n+5 is not a prime number if: (6n+5) : a = b ↔ a · b = 6n+5 ↔ = n a,b,n N and a,b

A natural number is divisible by 6 if it is divisible by 2 and by 3. For the divisibility by 2, a and b must be odd natural numbers because an odd natural number multiplied by another odd natural number leads to an odd natural number. And an odd natural number larger than or equal to 7 minus 5 leads to an even natural number. For the divisibility by 3, it is necessary that a · b has the structure (3x+1)(3y-1) for all x,y N because

= = = 3xy-x+y – 2

Therefore, for a nonprime number of type 6n+5, the following must hold for the index n:

for all n and x, y even natural numbers

For x = 2 and y = 2, 4, 6, 8, 10, ... after inserting n in 6n+5 this leads to the numbers 35, 77, 119, 161, 203, ... as you can see this is every 42nd number starting at 35.

For x = 4 and y = 2, 4, 6, 8, 10, ... after inserting n in 6n+5 this leads to the numbers 65, 143, 221, 299, 377, ... as you can see this is every 78th number starting at 65. Note, that 65 equals 35+30.

For x = 6 and y = 2, 4, 6, 8, 10, ... after inserting n in 6n+5 this leads to the numbers 95, 209, 323, 437, 551, ... as you can see this is every 114th number starting at 95. Note, that 95 equals 35+2 times 30.

This pattern continues indefinitely and can be represented by (35 + 30k) + (42 + 36k)m with k = 0, 1, 2, 3, ... and m = 0, 1, 2, 3, ...

All nonprimes of type 6n+5 can thus be represented as zeros of the following parametric sine function:

f3(x) = sin(()

The allowed range for x in the parametric sine function f3(x) ranges from 35+30t to infinite, and for t, t = 0, 1, 2, 3, 4, ...

If x and y are interchanged, the same numbers are produced, but in a different order. Therefore, the interchange of the variables x and y can be ignored in this context. Nevertheless, I would like to point out that swapping x and y also leads to regular columns of numbers, which can be represented as zeros of the parametric sine function: f4(x) = sin(()(x – 35 – 42t)). The allowed range for x reaches in f4(x) from 35+42t to infinite and for t applies t = 0, 1, 2, 3, 4, ...

__Under what conditions are numbers of type 6n+1 not prime and what regularity do these follow?____ __6n+1 is not a prime number if: (6n+1) : a = b ↔ a · b = 6n+1 ↔ = n a,b,n N and a,b1

Following the proof idea from above, a and b must be odd natural numbers to ensure divisibility by 2. For the divisibility by 3, it is necessary that a times b has either the form (3x+1)(3y+1) or the form (3x-1)(3y-1) because = = = 3xy+x+y and

= = = 3xy-x-y for all x,y N

Therefore, for a nonprime number of type 6n+1, the following must hold: either or for all n and x, y even natural numbers .

By using the values for x and y the same way as above, the first of those two terms produces the following columns of numbers:

49, 91, 133, 175, 217, ... then 91, 169, 247, 325, 403, ... then 133, 247, 361, 475, 589, ...

This pattern continues indefinitely and can be represented by (49 + 42k) + (42 + 36k)m with k = 0, 1, 2, 3, ... and m = 0, 1, 2, 3, ... .

This first part of the nonprime numbers of type 6n+1 can therefore be represented as zeros of the following parametric sine function:

f5(x) = sin((

The allowed range for x in the parametric sine function f5(x) ranges from 49+42t to infinite, and for t, t = 0, 1, 2, 3, 4, ...

The second term produces the following columns of numbers:

25, 55, 85, 115, 145, ... then 55, 121, 187, 253, 319, ... then 85, 187, 289, 391, 493, ...

Also this pattern continues indefinitely. It can be represented by (25 + 30k) + (30 + 36k)m with k = 0, 1, 2, 3, ... and m = 0, 1, 2, 3, ... .

This second part of nonprime numbers of type 6n+1 can therefore be represented as zeros of the following parametric sine function:

f6(x) = sin((

The allowed range for x in the parametric sinus function f6(x) ranges from 25+30t to infinite, and for t, t = 0, 1, 2, 3, 4, ...

To make all prime numbers appear in the correct order, only the functions f1(x), f2(x), f3(x), f5(x) and f6(x) have to be entered into a common coordinate system. The following figures are made with MatheGrafix 11. The zeros of the characteristic green functions show all potential prime numbers of type 6n+5 or type 6n+1. Every green zero (f1(x) and f2(x)), which is not crossed by the graph of one of the “forbidding” parametric sine functions (f3(x), f5(x), f6(x)), represents a prime number. The software used does not allow a parameter-dependent definition set. Therefore, these two figures have the same rigid definition range and because of these different parametric settings. Increasing the parameters stepwise is necessary when using this software because the “forbidding” parametric sinus functions logically cross out wrong numbers when the functions are under their definition range.

__A consequential possibility to ____prove____ whether a natural number ____larger____ than 3 is prime or not__

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- Proof of whether a natural number A under investigation is of type 6n+5 or of type 6n+1.

If (A-5) : 6 equals a natural number, then A is of type 6n+5. If (A-1) : 6 equals a natural number, then A is of type 6n+1. If both calculations do not equal a natural number, then the number under investigation is not of type 6n+5 or of type 6n+1. Such a natural number larger than 3 cannot be prime.

- a) If the investigated number A is of type 6n+5, calculate if f
3(x) = sin(()

has a zero at this number A. Then, A is not a prime number. To do this, it is enough to prove if A – 35 – 30t = 0 can be solved with a natural number value of t (including t=0) or if can be solved with a natural number result when using a natural number value of t (including t=0). If the first term shows a natural number for the value of t, f3(x) shows zero at 0; if the second term shows a natural number result for a natural value of t, f3(x) shows zero at an integer multiple of .

b) If the investigated number A is of type 6n+1, calculate if f5(x) = sin(( or f6(x) = sin(( has a zero at this number A. If one of these two functions shows a zero at this number A, then A is not prime.

This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.