The dual sine prime function

The sine prime function ps(n) provides for n ∈ ℕ greater than 0 always a prime greater than 2.

The sine prime function ps(n) is designated dual because

the result is always prime 3 for predictable n ∈ ℕ greater than 0

and

ps(n) provides all prime numbers greater than 3 for the residual n ∈ ℕ greater than 0.

The dual sine prime function reads:

ps(n)=2n*ts(n)+3

Proof:

With the aid of the deterministic test function ts(n) the ts(n) is determined for a choosed n. Because ts(n) can only have the values 0 for not prime and 1 for prime the results for  ps(n) are:

ts(n)=0 → ps(n)=3 → The choosed n gives in 2n+3 a composite uneven number

ts(n)=1 → ps(n) gives a prime > 3.

Because

2n+3 for n ∈ ℕ

provides all uneven numbers > 1 which are comprised only of all composite uneven numbers and all not composite uneven numbers (= all prime numbers > 2), all n which are in ts(n)=1 must be in ps(n) all prime numbers > 2.

The predictable n ∈ ℕ greater than 0 which have always in function ps(n) prime 3 as result, of course must satify formula

n=n_km = 2k²+6k+3+m(2k+3) ʌ k, m ∈ ℕ

What had to be proved.

Munich, 22 January 2023

Gottfried Färberböck