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The aim is to find a recurrence of prime numbers. So we'll take the trigonometric functions as they're standard periodic functions, like $\sin(x)$: we'll take as an argument of $\sin(x)$ the angle found before:

$$y=\sin(πx/n) \mid n ∈ ℕ, n>1.$$

Fixed $n$ the function gives as result 0 exactly when $x=n$ for obvious reasons and the other zeros are given for every k*n, where $k=1,2,3,...$

Then take $n_1 ∈ ℕ, n_1 > 1$, for all $n ≠ n_1$ and $ n_1 ∉ $ set of all zeros of the harmonics, $n_1$ is prime because there is only one harmonic that has the first zero in $x=n_1$. This harmonic is called prime harmonic.

The set of all zeros of prime harmonics are ℕ/{1} then we can build a function that preserves these zeros. The best method is the multiplication. If the first prime harmonic has n=2, the next number without the zero is 3. Then take the second harmonic that has n=3, the next number without zero is 5 and so on. With this method we build gradually a function $p(x)$:

$$p(x) = \prod_{n=2}^\infty \sin (πx/n)$$ where $x, n ∈ ℕ$ for all $n \in \sin (πx/n).$ This function $(p(X))$ is called prime function.

Klangen
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Fernando Di Lillo
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1 Answers1

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There are formulas to generate primes in that fashion. In this other MSE question you can find the same topic. As per my answer before, there is a formula for primes based on sines, and probably the theoretical base of yours is another way of writing it. It is based on Wilson's theorem:

$\displaystyle f(j)=\frac{\sin^2\left(\pi\frac{(j-1)!^2}{j}\right)}{\sin^2\left(\frac{\pi}{j}\right)} \cdot j$

So if you run a loop it should provide the list of primes and $0$'s as follows:

$[1,2,3,4,5,6,7,8,9,10,11...] \to [0,2,3,0,5,0,7,0,0,0,11...]$

More information in this previous question.

iadvd
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