The sine primality test

PRIME NUMBER SERIES 10

This primality test is not a probability test but an indirect test to determine clearly if a choosed uneven number greater than 1 is prime or not.

The test is based on definition 6 of prime number series 1:

Because of the existence of an infinite quantity of primes and only one even prime namely 2 follows that an infinite quantity of uneven primes must exist. Because of this the definition of all primes greater then 2 is:

A prime greater than 2 is an uneven number U, that is not divisible integrally by any uneven number u in the domain $3 \leq u \leq \sqrt{U}$ .

This test is valid for all uneven numbers $U > 1$ .

With

$U=2n+3$ ᴧ n ∈ ℕ

all uneven numbers > 1 are given, which can be selected for this test.

Solved for n gives

$n=\frac{U-3}{2}$

The determined n is put into the following test function ts(n):

$ts(n)=\prod\limits_{k=0} ^{k_{max}} \lceil|\sin(\frac{n-k}{2k+3}\pi)|\rceil$

∧ n, k ∈ ℕ

with

$k_{max}=\lfloor{\frac{\sqrt{2n+3}-3} {2}\rfloor}$

Legend:

$\prod\limits_{k=0} ^{k_{max}}$ product in which the run variable k runs from 0 to $k_{max}$

|x| absolute value of any value x

⌊x⌋ lower Gaussian bracket, x rounded to the nearest integer

⌈x⌉ upper Gaussian bracket, x rounded up to the nearest integer

ts(n) is either 0 or 1 for a choosed n from the set of the natural numbers.

If ts(n)=0, then 2n+3 is not prime for the choosed n.

If ts(n)=1, then 2n+3 is prime for the choosed n.

Proof:

In prime number series 6 it has been proofed, that the formula

n_km= 2k²+6k+3+m(2k+3) ʌ k, m ∈ ℕ

provides those n, which give all composite uneven numbers greater than 2 in the formula 2n+3.

Let’s suppose a choosed n equals n_km meaning

n = n_km

then n in

sin((n-k)/(2k+3)π)

can be replaced by

2k²+6k+3+m(2k+3)

This gives

sin((2k²+6k+3+m(2k+3)-k)/(2k+3)π)

and further

sin(((2k²+5k+3)/(2k+3)+m)π)

Because

(2k²+5k+3)/(2k+3)=k+1

follows

sin((k+1+m)π)

And this term is always 0, because k, 1 and m are natural numbers, which summed up give a natural number and the sine of π multiplied by whole numbers equals always 0.

But if

n≠n_km

follows

sin((n-k)/(2k+3)π)≠0

because all uneven composite numbers > 1 are given by n_km in 2n+3 and all other n in 2n+3 must give prime numbers.

But how great is k_max?

According to definition 6, prime number series 1 is the following valid:

A prime greater than 2 is an uneven number U, that is not divisible integrally by any uneven number u in the domain $3 \leq u \leq \sqrt{U}$ .

For the Sine Test, must be U > 1, because all primes are greater than 1. Otherwise the test would go down to a probabilistic test.

Putting

U=2n+3 ᴧ n ∈ ℕ

and

u=2k+3 ᴧ k ∈ ℕ

into

u ≤ √ U

gives

2k+3 ≤ √ (2n+3)

The solutions for k and the selected n spread from k=0 up to k=k_max for n, k ∈ ℕ.

k_maxshall be determined as a natural number. But solving for k we come into the domain of real numbers. Therefore we write for better differentiation:

2k_maxr+3 = √ (2n+3)

Solved for k_maxr gives:

k_maxr=(√(2n+3)-3)/2

The real solution can be lead back into the domain of natural numbers by rounding up or down.

According to definition 6, prime number series 1 it has to be rounded down.

Thus

k_max=⌊√(2n+3)-3/2⌋

The term

⌈|sin((n-k)/(2k+3)π)|⌉

can take on only the values 0 or 1, if

n=n_km→ sin((k+1+m)π) =0

n≠n_km→ ⌈|sin((n-k)/(2k+3)π)|⌉=1

Is this term 0 the choosed n gives no prime in the formula 2n+3 but a composite uneven number.

Is this term 1 the choosed n can give a prime in the formula 2n+3 but not forcible.

But forming the product for all k’s from 0 up to k_max and is’nt this product 0 but 1, then it is ensured cearly with this deterministic primality test, that 2n+3 is prime. But is in this product at least one factor 0, then the entire product is 0 and it is ensured cearly with this deterministic primality test, that 2n+3 is not prime.

What had to be proved.

Examples

n=1 → ts(1)=1 → 2*1+3=5 → prime

n=10 → ts(10)=1 → 2*10+3=23 → prime

n=97 → ts(97)=1 → 2*97+3=197 → prime

n=1002 → ts(1001)=0 → 2*1001+3=2005 → not prime

n=30997 → ts(30997)=0 → 2*30997+3=61997=13*19*251 → not prime

n=30103 → ts(30103)=1 → 2*30103+3=60209 → prime

Munich, 11 November 2022

Gottfried Färberböck

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THE PRIME NUMBER SERIES

The sine primality test