Mystery about the set of all prime numbers solved

PRIME NUMBER SERIES 16

With

$\mathbb{P}$ \ {2} = {p | p = $2n\prod\limits_{k=0} ^{\lfloor{\frac{|\sqrt{2n+3}-3|} {2}\rfloor}} \lceil1-\frac{k+\lfloor\frac{n-k}{2k+3}\rfloor(2k+3)}{n}\rceil+3$ ʌ k ∈ $\mathbb{N}$ ʌ n ∈ $\mathbb{N}$ \ {0}}

all prime numbers greater than 2 are given.

Proof: see PRIME NUMBER SERIES 9 THE DUAL PRIM NUMBER FUNCTION

There is another solution for the set of all prime numbers:

$\mathbb{P}$ \ {2} = {p | p = $2n\prod\limits_{k=0} ^{\lfloor{\frac{\sqrt{2n+3}-3} {2}\rfloor}} \lceil|\sin(\frac{n-k}{2k+3}\pi)|\rceil+3$ ʌ k,n ∈ $\mathbb{N}$ }

Proof: see PRIME NUMBER SERIES 12 THE DUAL SINE PRIME FUNCTION

Munich, March 22, 2023

Gottfried Färberböck

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THE PRIME NUMBER SERIES

Mystery about the set of all prime numbers solved