Prime Potential Density Function

Jackson Capper | 19 March 2021

Consider that primes are the dissonant instances within harmonic sets of those dissonant instances. That is, the n^th 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 n^th 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 n^th 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: