Consider that primes are the dissonant instances within harmonic sets of those dissonant instances. That is, the nth prime begins a harmonic set across \({Z}\). Suppose:
- \({Z}=\{1, 2, 3, 4, 5, 6, 7, 8, 9\dots\}\) the positive integer set,
- \({P}\) is the prime number set,
- \(p(n)\) is the nth prime, and
- \(H(p) = {H} \subset {Z}\) where \(p\) is a prime number, and \(H(p)\) is the set of all products of \(p\) in \({Z}\).
When a harmonic set of \(p(n)\) encounters a number that is dissonant (not within that harmonic set), this number becomes the next element of \({P}\).
For example, starting at \(2\):
\( h(p(1)) = {h}(2)=\{ 2, 4, 6, 8, 10 \dots\} \)
and,
\( {h}(p(2)) = {h}(3) = \{3, 6, 9, 12, 15 \dots\} \)
So,
\(\vert {H}(n) \vert = \infty \frac{ 1 }{ p(n) }\), e.g. \(\vert {H}(1) \vert = \infty \frac{ 1 }{ 2 }\)
and,
\(\vert {Z} – {H}(n) \vert =\infty \frac{p(n) – 1}{p(n)}\), e.g. \(\vert {Z} – {H}(2) \vert = \infty \frac{ 2 }{ 3 }\)
We see that as we progress through the primes, harmonic sets progressively eliminate all subsequent factors of that prime as potential primes. Therefore, the potential of encountering a dissonant real (and consequently, a prime number) becomes less frequent as \(n \to \infty\). We will refer to this quality as Prime Potential Density (\(d\)). Thus, as \(n \to \infty, d \to 0\).
- \(d(p)\) is the Prime Density Function
If we are looking for \(d(n)\) at the nth prime, then all primes prior must be factored into the prime potential density to ensure all harmonic sets are eliminated. So,
\(d(n) = \vert {Z} \vert \frac{p(n)-1}{p(n)} \frac{p(n-1)-1}{p(n-1)} \frac{p(n-2)-1}{p(n-2)} \dots\), e.g. \(d(3) = \infty \frac{ 5 – 1 }{ 5 } \frac{ 3 -1 }{ 3 } \frac{ 2 – 1 }{ 2 } \dots\)
Formally, the Prime Potential Density Function:
\(d(n) = \infty \prod\limits_{i=0}^{n} \frac{ p(n – i) – 1 }{ p(n – i) } \)