# Distribution of twin primes in repeating sequences of prime factors

John K Sellers

###### Abstract

We present a deterministic relationship between relative primes and twin
primes in successively larger sequences of the natural numbers. This enables
setting a finite lower limit on the occurrence of actual twin primes in an
unbounded selection of domains.

##  1 Natural Numbers as Repeating Sequences of Prime Factors

The formulation presented here differs from other approaches to prime pairs as
outlined in [8] and which are covered in detail in [1, 2, 5, 6, 10]. Here we
determine the locations of relative primes, prime to $P\leq P_{k}$, in
successively larger sequences of the natural numbers. Those are the only
locations where actual prime numbers $P>P_{k}$ can appear. Therefore we refer
to numbers in those locations as prospective prime numbers.

This approach is more direct and less abstract where the key is tracking
prospective prime numbers not prime numbers themselves. In defining a finite
set covering all combinations of prime factors $P_{1}$, $P_{2}$, $P_{3}$,
$\cdots,P_{k}$ relative to each other, we can count the number of prospective
prime numbers and prospective twin primes in the set, specify their locations,
and determine the uniformity of their distribution.

The tie to actual twin primes occurs when a prospective twin prime, prime to
$P\leq P_{k}$, occurs between $P_{k}$ and $P_{k+1}^{2}$. Then they are actual
twin primes. Given this, we can place a non-vacant lower bound on the number
of such twin primes for an unbounded succession of specific choices of
$P_{k}$.

###  1.1 Numbers in Relation to Prime Factors $2$ and $3$

If one considers a sequence of natural numbers greater than $5$ in terms of
its relationships to prime factors $2$ and $3$ only, one sees a repeating
pattern of six numbers relative to those factors as shown in Figure 1. The
bottom row shows a useful representation, which we refer to as ”sequence
notation,” that explicitly shows the count, $n$, of the repetitions of the
$6$-number sequence and a count, ”$i$”, within each repeated sequence. The
count within the repeating sequence shows an arbitrary but convenient choice
for its start that has the two numbers prime to $2$ and $3$ as the first and
third numbers in the repeating sequence.

N | N+1 | N+2 | N+3 | N+4 | N+5
---|---|---|---|---|---
$\widetilde{P}$ |  $2|$, $3|$ | $\widetilde{P}$ | $2|$ | $3|$ | $2|$
$N_{n,1}$ | $N_{n,2}$ | $N_{n,3}$ | $N_{n,4}$ | $N_{n,5}$ | $N_{n,6}$
Figure 1:  Underlying Sequence of Natural Numbers Relative to Prime Factors
$2$ and $3$, The figure as an example shows an arbitrary starting point of
$N=6n+5$. $\widetilde{P}$ represents a prospective prime number because it
marks the only location where a prime number $>3$ may occur.

Each sequence repeats with the identical pattern of prime factors $2$ and $3$
because: $N_{n,i}=N_{n^{\prime},i}+6(n-n^{\prime}$) shows that both $N_{n,i}$
and $N_{n^{\prime},i}$ share the same combination of the prime factors $2$ and
$3$. Therefore the two locations occupied by numbers prime to $2$ and $3$ in
each $6$-number sequence represent the only locations where prime numbers can
occur and each occurrence has the potential to be a twin prime. Whether they
are prime numbers or twin primes of course depends on where they occur in the
overall sequence of natural numbers in relation to all other prime factors.
Since we retain the notation citing repetitions of the $6$-number sequence and
the position within a given repetition of that sequence even as we consider
larger sequences with other prime factors, we refer to the $6$-number sequence
as the elemental sequence, where the first ($zero^{th}$), such elemental
sequence starts at the number $5$, i.e.,

$N_{0,1}=5,N_{0,2}=6,N_{0,3}=7,N_{0,4}=8,N_{0,5}=9,N_{0,6}=10$

An immediate observation from this formulation is that any prime number,
$P>3$, may be represented as either $5+6n$, Progression 1, or $7+6n$,
Progression 2, where $(n\geq 0)$ or equivalently as: $6n^{\prime}\pm
1,(n^{\prime}\geq 1)$.

###  1.2 Conversion in and out of sequence notation

As just introduced we use the representation of natural numbers, $N_{n,i}$,
starting at $N_{0,1}=5$, where $i$ represents the number’s position in the
elemental sequence and $n$ represents which elemental sequence. For example,
$5,6,7,8,9,10$ represents the ${zero}^{th}$ elemental sequence. Then other
natural numbers are represented by

$N=N_{n,i}=5+6n+(i-1)\quad\textrm{with}\quad i=1,2,3,4,5,6$ (1)

Given any natural number, $N$, one can find its representation in terms of its
elemental sequence and part of that sequence by choosing $t=0,1,2,3,4,5$ so
that $6|(N-5-t)$, giving:

$n=\frac{(N-5-t)}{6}\quad\textrm{and}\quad i=t+1$ (2)

Testing for values of $t$ is easy, but one can also solve for $n$ and $i$ in
terms of $N$ as follows:

Note that $N-5=\left\lfloor\frac{N-5}{6}\right\rfloor 6+(N-5)\bmod{6}$. Then
comparing this with Equation (1) gives:

$n=\left\lfloor\frac{N-5}{6}\right\rfloor\quad\textrm{and}\quad
i=(N-5)\bmod{6}+1$ (3)

in turn giving:

$N=N_{\left\lfloor\frac{N-5}{6}\right\rfloor,(N-5)\bmod{6}+1}$ (4)

Numbers in the first or third position of the elemental sequence may be
represented as the following arithmetic progressions:

$N_{n,1}=6n+5\quad\textrm{and}\quad N_{n,3}=6n+7$

###  1.3 Products of Numbers Represented in the Sequence Notation

Multiplication rules for Numbers in the first and third positions of the
elemental sequence:

$\displaystyle N_{n_{i},1}N_{n_{j},1}$
$\displaystyle=N_{[6n_{i}n_{j}+5(n_{i}+n_{j})+3],3}$ $\displaystyle
N_{n_{i},1}N_{n_{j},3}$ $\displaystyle=N_{[6n_{i}n_{j}+7n_{i}+5n_{j}+5],1}$
(5) $\displaystyle N_{n_{i},3}N_{n_{j},3}$
$\displaystyle=N_{[6n_{i}n_{j}+7(n_{i}+n_{j})+7],3}$

Define $P_{n,1}=N_{n,1}$ and $P_{n,3}=N_{n,3}$ when $N_{n,1}$ and $N_{n,3}$
respectively are prime numbers. Then In terms of products of prime numbers the
multiplication equations become:

$\displaystyle P_{n_{i},1}P_{n_{j},1}$
$\displaystyle=N_{[6n_{i}n_{j}+5(n_{i}+n_{j})+3],3}$ $\displaystyle
P_{n_{i},1}P_{n_{j},3}$ $\displaystyle=N_{[6n_{i}n_{j}+7n_{i}+5n_{j}+5],1}$
(6) $\displaystyle P_{n_{i},3}P_{n_{j},3}$
$\displaystyle=N_{[6n_{i}n_{j}+7(n_{i}+n_{j})+7],3}$

Summarizing in words:

1) Any product of prime numbers $>3$ that involves an odd number of prime
numbers from Progression 1 (all of the form $P_{n,1}$) produces a number in
Progression 1: $N_{n^{\prime},1}$.

2) Any other product of prime numbers $>3$ produces a number in Progression 2:
$N_{n^{\prime},3}$.

##  2 Prospective Prime Numbers and Prospective twin primes in Successively
Larger Sequences of The Natural Numbers

###  2.1 Definitions and Properties

In this section we add some formalism to a discussion on the sequences of
prime factors underlying the natural numbers. Our intention is to add some
precision and clarity to this view as we consider successively larger
sequences of natural numbers and their implications for the occurrence of
prime numbers and twin primes.

It is key to this perspective of the natural numbers to note that the number
$N$ and $N+n\prod_{i=1}^{k}P_{i}$, where $n$ is an integer, $0\leq n\leq
P_{k+1}-1$, have the same occurrence of distinct prime factors among $P\leq
P_{k}$. Therefore the sequences

$N,N+1,N+2,\cdots$

and

$N+n\prod_{i=1}^{k}P_{i}\;,\;N+n\prod_{i=1}^{k}P_{i}+1\;,\;N+n\prod_{i=1}^{k}P_{i}+2,\cdots$

have the same sequence of prime factors $P\leq P_{k}$ and that sequence
repeats at a frequency of every $\prod_{i=1}^{k+1}P_{i}$ numbers.

Define $S_{k}$ as the following ordered sequence of natural numbers:

$S_{k}=\left\\{N_{i}:5\leq N_{i}\leq 4+\prod_{i=1}^{k}P_{i}\quad\&\quad
N_{i+1}=N_{i}+1\right\\}$ (7)

Note that sets $S_{k}$ each contain $\prod_{i=1}^{k}P_{i}$ successive natural
numbers, all starting with the number $5$, but ending on successively larger
numbers as we increment $k$.

Define $\psi_{k}$ as the underlying ordered sequence of prime factors $P\leq
P_{k}$ found in $S_{k}$. For example, given that $S_{2}=\\{5,6,7,8,9,10\\}$,
then $\psi_{2}=\\{\widetilde{P},2\cdot 3,\widetilde{P},2,3,2\\}$, where
$\widetilde{P}$ indicates a number prime relative to all $P\leq P_{2}$ (i.e.,
$P\leq 3$). The values in $\psi_{2}$ are the distinct prime factors $P\leq
P_{2}$ found in the numbers of $S_{2}$ that are in the same sequencial
position as the corresponding values in $\psi_{2}$.

When we extend $S_{k}$ to $S_{k+1}$, the underlying sequence of prime factors
$\psi_{k}$ of $S_{k}$ repeats $P_{k+1}$ times in $S_{k+1}$, giving:

$\widetilde{\psi}_{k+1}=\\{\psi_{k}^{(0)},\psi_{k}^{(1)},\psi_{k}^{(2)},\cdots,\psi_{k}^{(P_{k+1}-1)}\\}$
(8)

The tilde on $\psi_{k+1}$ indicates the tentative representation of the
sequence $\psi_{k+1}$ because the locations of prime factor $P_{k+1}$ are not
represented in $\psi_{k}$. Each sub-sequence $\psi_{k}^{(i)}$, the $i^{th}$
repetition of $\psi_{k}$, is identical in each of it’s repetitions. The
occurrence $\widetilde{P}$ in in the repeated sequences of $\psi_{k}$ may or
may not be prime numbers, but they are the only locations where new prime
numbers, i.e., $P>P_{k}$, can occur, because every other position in the
sequence and its repetitions has one or more prime factors $P\leq P_{k}$.

For example:

$S_{3}=\\{/5,6,7,8,9,10/11,12,13,14,15,16/17,18,19,20,21,22/23,24,25,26,27,28/\\\
29,30,31.32.33.34/\\}$ (9)

The slash is used merely to indicate underlying groups of numbers
corresponding to repetitions of $\psi_{2}^{(i)}$. The underlying prime factor
sequence for $S_{3}$ is then:

$\psi_{3}=\\{5,2\cdot 3,\widetilde{P},2,3,2\cdot 5/\widetilde{P},2\cdot
3,\widetilde{P},2,3\cdot 5,2/\widetilde{P},2\cdot 3,\widetilde{P},2\cdot
5,3,2/\widetilde{P},2\cdot 3,5,2,3,3/\\\ \widetilde{P},2\cdot 3\cdot
5,\widetilde{P},2,3,2/\\}$ (10)

and $\psi_{3}$111 One could think of $\psi_{3}$ as a 30-number elemental
sequence relative to prime factors $2,3,5$ analogous to the 6-number sequence
defined in Figure 1. The numbers would be represented as $N=N_{5n,i}$, where
$1\leq i\leq 30$, with prospective prime numbers, prime to $P\leq 5$ occuring
at: $i=3,7,9,13,15,19,25,\;\textrm{and}\;27$. The conversion would then be:
$N_{5n,i}=5+6\cdot 5n+(i-1)$.  has the underlying sequence $\psi_{2}$:

$\widetilde{\psi}_{3}=\\{/\widetilde{P},2\cdot
3,\widetilde{P},2,3,2/\widetilde{P},2\cdot
3,\widetilde{P},2,3,2/\widetilde{P},2\cdot
3,\widetilde{P},2,3,2/\widetilde{P},2\cdot 3,\widetilde{P},2,3,3/\\\
\widetilde{P},2\cdot 3,\widetilde{P},2,3,2/\\}$ (11)

One can see from this example that the transition
$\psi_{2}\rightarrow\psi_{3}$ involves repeating $\psi_{2}$ $P_{3}=5$ times in
succession then include occurrences of prime factor $5$.

When we extend a sequence using repetitions of the underlying sequence of
prime factors, not all numbers prime to those factors will be actual prime
numbers. Therefore we refer to locations in the sequence where prime factors
must appear as prospective prime numbers and when it is necessary to make that
distinction we represent prospective prime numbers as $\widetilde{P}$.
Prospective prime numbers in $S_{k}$ are all numbers prime to $P\leq P_{k}$.
We do not consider a prime number prime to itself. Therefore prospective prime
numbers represent the only locations in $S_{k}$ where prime numbers $P>P_{k}$
can occur.

Define $S_{k}^{(i)}$ as subsets of $S_{k}$ representing each successive group
of $\prod_{j=1}^{k-1}P_{j}$ numbers, which correspond to instances of
$\psi_{k-1}^{(i)}$ found in $\psi_{k}$. Then, if we represent $S_{k}^{(i)}+N$
as adding the number $N$ to each number within $S_{k}^{(i)}$, then given $i>j$
we have:

$S_{k}^{(i)}=S_{k}^{(j)}+(i-j)\prod_{l=1}^{k-1}P_{l}$ (12)

Note that this relationship between $S_{k}^{(i)}$ and $S_{k}^{(j)}$ means that
numbers corresponding to the same locations within $S_{k}^{(i)}$ and
$S_{k}^{(j)}$ have the same prime factors from among $P\leq P_{k-1}$. That is,
if $x_{n}^{(i)}$ is the $n^{th}$ number in $S_{k}^{(i)}$ and $x_{n}^{(j)}$ is
the $n^{th}$ number in $S_{k}^{(j)}$ then for all $P\leq P_{k-1}$:
$P|x_{n}^{(i)}$ if and only if $P|x_{n}^{(j)}$. The fact that $0\leq(i-j)\leq
P_{k}-1$ is consistent with this rule, because $(i-j)$ can only have prime
factors $P\leq P_{k-1}$.

We can extend the definition of $S_{k}^{(i)}$ to successively smaller subsets
of $S_{k}$ using:

$\displaystyle S_{k}$
$\displaystyle=\left\\{S_{k}^{(0)},S_{k}^{(1)},\cdots,S_{k}^{(P_{k}-1)}\right\\}$
$\displaystyle=\left\\{S_{k-1}^{(0)},S_{k-1}^{(1)},\cdots,S_{k-1}^{(P_{k}P_{k-1}-1)}\right\\}$
$\displaystyle\qquad\vdots$
$\displaystyle=\left\\{S_{2}^{(0)},S_{2}^{(1)},\cdots,S_{2}^{(\prod_{j=3}^{k}P_{j}-1)}\right\\}$

Note that for $l>k$, $S_{k}\subset S_{l}$.

This concept of nested subsets can be applied to $\psi_{k}$ as well, however
there is some loss of information since the lower level sequences represent
fewer prime factors. Never-the-less it is important to realize the repetitive
nature of prime factors associated with any set or subset of prime numbers.
The nesting for $\psi_{k}$ looks like the following:

$\displaystyle\psi_{k}$
$\displaystyle=\left\\{\psi_{k-1}^{(0)},\psi_{k-1}^{(1)},\cdots,\psi_{k-1}^{(P_{k}-1)}\right\\}$
$\displaystyle=\left\\{\psi_{k-2}^{(0)},\psi_{k-2}^{(1)},\cdots,\psi_{k-2}^{(P_{k}P_{k-1})-1}\right\\}$
$\displaystyle\qquad\vdots$
$\displaystyle=\left\\{\psi_{2}^{(0)},\psi_{2}^{(1)},\cdots,\psi_{2}^{(\prod_{j=3}^{k}P_{j}-1)}\right\\}$

Given this nesting of subsets, any such prime factor sequence represented by
$\psi_{k}\subset\psi_{l}$, where $k<l$, continues to repeat throughout the
natural numbers, e.g., as $l\rightarrow\infty$. While this is important from
the fact that any combinations of prime factors produces an infinitely
repeating sequence of those factors within the natural numbers, however the
location of prospective prime numbers differ between the different levels of
$\psi_{k}$.

We refer to $\psi_{2}=\\{\widetilde{P},2\cdot 3,\widetilde{P},2,3,2\\}$ as the
elemental sequence of prime factors and correspondingly refer to
$S_{2}^{(i)}$, where $0\leq i\leq P_{k}-1$, as elemental sequences of any set
$S_{k}$. We can locate individual numbers precisely by their location within
an elemental sequence representing them as: $N_{n,i}\in S_{2}^{(n)}$, where:

$S_{2}^{(n)}=\left\\{N_{n,1},N_{n,2},N_{n,3},N_{n,4},N_{n,5},N_{n,6}\right\\}$

Using this sequence notation, note that prime numbers can only occur at
$N_{n,1}$ or $N_{n,3}$, because all other $N_{n,i}$ have a prime factor of
$2$, $3$ or both.

For our purposes we start the progressions at $N_{0,1}=5$ and $N_{0,3}=7$.
Then given that $N_{n+1,i}=N_{n,i}+6$, the numbers $N_{n,1}$ and $N_{n,3}$
represent the following progressions of numbers:

$N_{n,1}=6n+5\quad\textrm{and}\quad N_{n,3}=6n+7$

All prime numbers $P>3$ therefore fall in one or the other of these two
progressions although clearly, not all numbers in each progression is a prime
number.

It is useful to define $P_{m,1}=N_{m,1}$ and $P_{m,3}=N_{m,3}$ when $N_{m,1}$
and $N_{m,3}$ respectively, are prime numbers. Likewise, it is also useful to
define $\widetilde{P}_{m,1}=N_{m,1}$ and $\widetilde{P}_{m,3}=N_{m,3}$ when
$N_{m,1}$ and $N_{m,3}$ respectively, are prospective prime numbers.

###  2.2 Propagation of Prospective Prime Numbers

When we extend $S_{k}$ to $S_{k+1}$ we also extend $\psi_{k}$ to $\psi_{k+1}$,
first by $P_{k+1}$ repetitions of $\psi_{k}$,
$\widetilde{\psi}_{k+1}=\left\\{\psi_{k}^{(0)},\psi_{k}^{(1)}\psi_{k}^{(2)},\cdots,\psi_{k}^{(P_{k+1}-1)},\right\\}$,
where each $\psi_{k}^{(i)}$ represents the identical repeating sequence of
prime factors $P\leq P_{k}$ throughout $S_{k+1}$. Then we identify the
locations of prime factor $P_{k+1}$, whereby
$\widetilde{\psi}_{k+1}\longrightarrow\psi_{k+1}$.

$S_{k}$ has $\prod_{i=1}^{k}P_{i}$ numbers: $N\in S_{k}\longrightarrow 5\leq
N\leq 4+\prod_{i=1}^{k}P_{i}$. When we extend $S_{k}$ to $S_{k+1}$ to include
all combinations of prime factors $P\leq P_{k+1}$ relative to each other we
need to incude $P_{k+1}$ repetitions of $\prod_{i=1}^{k}P_{i}$ numbers.

Given

$n\prod_{i=1}^{k}P_{i}=\left\lfloor\frac{n\prod_{i=1}^{k}P_{i}}{P_{k+1}}\right\rfloor
P_{k+1}+\left(n\prod_{i=1}^{k}P_{i}\right)\bmod{P_{k+1}}$

$\left(n\prod_{i=1}^{k}P_{i}\right)\bmod{P_{k+1}}=0$ if and only if $n$ is a
multiple of $P_{k+1}$ and $P_{k+1}$ itself is the least such multiple.

Given any $N\in S_{k}$ and and any integer $m\geq 0$, then for all $P\leq
P_{k}$:

$P|N\leftrightarrow P|\left(N+m\prod_{i=1}^{k}P_{i}\right)$

Therefore the sequence of occurrence of all prime factors $P\leq P_{k}$ in
$S_{k}$ (i.e. $\psi_{k}$) is repeated identically in each successive sequence
of $\prod_{i=1}^{k}P_{i}$ numbers.

If we limit $m$ to $0\leq m\leq P_{k+1}-1$ then
$\left(N+m\prod_{i=1}^{k}P_{i}\right)\in S_{k+1}$.

Also, if we represent $S_{k+1}$ as:

$S_{k+1}=\left\\{S_{k+1}^{(0)},S_{k+1}^{(1)},\cdots,S_{k+1}^{(P_{k+1}-1)}\right\\}$

where each $S_{k+1}^{(i)}$ represents one of a succession of
$\prod_{i=1}^{k}P_{i}$ numbers, then:

$S_{k+1}^{(0)}=S_{k}$

and for any $N\in S_{k}$ and $N^{\prime}=N+m\prod_{i=1}^{k}P_{i}$, then
$N^{\prime}\in S_{k+1}^{(m)}$, where for all $P\leq P_{k}$:

$P|N\longleftrightarrow P|N^{\prime}$

Each repetition of the subsequence $\psi_{k}^{(i)}$ in $\psi_{k+1}$ represents
the fact that numbers prime to all $P\leq P_{k}$ occur at the same locations
within each instance of $S_{k+1}^{(i)}$.

Numbers in $S_{k}$ that are prime to all $P\leq P_{k}$ are what we have
defined as prospective prime numbers in $S_{k}$. They can only occur at
$N_{m,1}$ or $N_{m,3}$ in their respective elemental sequences. Those
prospective prime number locations in $\psi_{k}$ are repeated $P_{k+1}$ times
as an underlying sequence to $\psi_{k+1}$. All of those repetitions of
prospective prime numbers from $S_{k}$ going to $S_{k+1}$ remain prime to
$P\leq P_{k}$. However, the prime factor $P_{k+1}$ will occur in exactly one
instance of the $P_{k+1}$ repetitions of each prospective prime number from
$S_{k}$. Exactly one instance, because in the $P_{k+1}$ repetitions of
$\psi_{k}$ the prime factor $P_{k+1}$ occurs exactly once in each position of
$\psi_{k}$ creating a sequene of numbers of length $\prod_{i=1}^{k+1}P_{i}$ in
$S_{k+1}$.

The fact that $\frac{\prod_{i=1}^{k+1}P_{i}}{P_{k+1}}=\prod_{i=1}^{k}P_{i}$
requires that $P_{k+1}$ occurs exactly once in each position of the
$\prod_{i=1}^{k}P_{i}$-number sequence spread throughout its $P_{k+1}$
repetitions in $S_{k+1}$. If $P_{k+1}$ occured as a prime factor within
$S_{k+1}$ at two instances of the same repeated prospective prime number,
$\widetilde{P}\in S_{k}$, i.e.,

$P_{k+1}|\left(\widetilde{P}+m\prod_{i=1}^{k}P_{i}\right)\quad\textrm{and\quad}P_{k+1}|\left(\widetilde{P}+m^{\prime}\prod_{i=1}^{k}P_{i}\right)$

where $0\leq m,m^{\prime}\leq P_{k+1}-1$, then necessarily
$P_{k+1}|(m^{\prime}-m)$ which contradicts the condition that
$m,m^{\prime}<P_{k+1}$.

Also, given $N\in S_{k}$ where for some $P\leq P_{k}$ if $P|N$ then
$P|\left(N+m\prod_{i=1}^{k}P_{i}\right)$, therefore, $N$ cannot be a
prospective prime number when extended into $S_{k+1}$. Therefore, all
prospective prime numbers in $S_{k+1}$, i.e. numbers in $S_{k+1}$ prime to all
$P\leq P_{k+1}$, result from extensions of prospective prime numbers
$\widetilde{P}\in S_{k}$, via $\widetilde{P}+m\prod_{i=1}^{k}P_{i}$, where
$0\leq m\leq P_{k+1}-1$.

We can now state the following theorem:

######  Theorem 2.1.

Given any prime number, $P_{k}>2$, let $n_{k}^{\widetilde{p}}$ be the number
of prospective prime numbers $\widetilde{P}$, prime relative to all $P\leq
P_{k}$, where $P_{k}<\widetilde{P}<4+\prod_{i=1}^{k}P_{i}$.222Note that
$4+\prod_{i=1}^{k}P_{k}$ cannot be a prime number, because it is an even
number. Also, we assume that a prime number is not prime relative to itself.
Then defining $n_{1}^{\widetilde{p}}=1$:
$n_{k}^{\widetilde{p}}=(P_{k}-1)n_{k-1}^{\widetilde{p}}=\prod_{i=2}^{k}(P_{i}-1)$.

###### Proof.

: Consider the sequence of natural numbers $5\rightarrow
4+\prod_{i=1}^{k}P_{i}$. This sequence consists of $\prod_{i=1}^{k}P_{i}$
sequential natural numbers that includes all combinations of prime factors
$P\leq P_{k}$ situated relative to each other. This sequence of those prime
factors repeats in each successive group of $\prod_{i=1}^{k}P_{i}$ natural
numbers without end throughout the natural numbers. Assume this sequence has,
$n_{k}^{\widetilde{p}}$ numbers prime to prime numbers $P\leq P_{k}$. Then
consider the extension of this sequence to the sequence of numbers
$5\longrightarrow 4+\prod_{i=1}^{k+1}P_{i}$. This involves $P_{k+1}$
repetitions of the underlying prime factor sequence associated with the
numbers $5\rightarrow\prod_{i=1}^{k}P_{i}$. Therefore there are
$P_{k}n_{k}^{\widetilde{p}}$ numbers prime to $P<P_{k}$ in the extended
sequence. Since the prime factor $P_{k+1}$ occurs exactly once in each
position of the underlying $\prod_{i=1}^{k}P_{i}$ prime factor sequence spread
throughout its $P_{k+1}$ repetitions, each of the numbers prime to $P\leq
P_{k}$ has the prime factor $P_{k+1}$ exactly once during the $P_{k+1}$
iterations of the underlying sequence. Therefore there are
$n_{k+1}^{\widetilde{p}}=P_{k+1}n_{k}^{\widetilde{p}}-n_{k}^{\widetilde{p}}=(P_{k}-1)n_{k}^{\widetilde{p}}$
numbers prime to all $P\leq P_{k+1}$ within the sequence $5\rightarrow
4+\prod_{i=1}^{k+1}P_{i}$. If then we define $n_{1}^{\widetilde{p}}=1$, to
initialize the succession of prime factor sequences, we have:
$n_{k}^{\widetilde{p}}=\prod_{i=2}^{k}(P_{i}-1)$. Starting with $k=2$ we have
the sequence $5,6,7,8,9,10$ where only $5$ and $7$ are prime to $2$ and $3$
giving $n_{2}^{\widetilde{p}}=2$. Then extending this to include prime factor
$5$ we get the sequence $5,6,\cdots,34$, where only $7,11,13,17,19,23,29,$ and
$31$ are prime to $2,3,$ and $5$, giving $n_{3}^{\widetilde{p}}=8$, which
verifies the initial cases of the formula
$n_{k}^{\widetilde{p}}=\prod_{i=2}^{k}(P_{i}-1)$ and completes the proof. ∎

We see from Theorem 2.1 that $n_{k}^{\widetilde{p}}$ is the number of
prospective prime numbers in $S_{k}=\left\\{5\longrightarrow
4+\prod_{i=1}^{k}P_{i}\right\\}$. That is the number of numbers in positions
$N_{n,1}$ and $N_{n,3}$ in $S_{k}$ that are prime to all $P\leq P_{k}$.

Note that $n_{k}^{\widetilde{p}}=\prod_{i=2}^{k}(P_{i}-1)$ alternatively
results from using the number $N=\prod_{i=1}^{k}P_{i}$ in Euler’s totient
function:[3]

$\varphi(N)=N\prod_{P|N}\left(1-\frac{1}{P}\right)\longrightarrow\prod_{i=1}^{k}P_{i}\prod_{i=1}^{k}\left(1-\frac{1}{P_{i}}\right)=\prod_{i=1}^{k}P_{i}\left(1-\frac{1}{P_{i}}\right)=\prod_{i=2}^{k}(P_{i}-1)$

which has he same interpretation.

The average density of prospective prime numbers among all numbers in the set
of natural numbers $S_{k}$ is :

$\rho^{\widetilde{P}}_{k}=\frac{\prod_{i=1}^{k}(P_{i}-1)}{\prod_{j=1}^{k}P_{j}}=\prod_{i=1}^{k}\left(1-\frac{1}{P_{k}}\right)$
(13)

This can be viewed as a partial or finite reciprocal of Euler’s product for
the Riemann Zeta function with $s=1$:

$\zeta_{k}(s=1)=\prod_{i=1}^{k}\left(1-\frac{1}{P_{k}}\right)^{-1}$ (14)

This reciprocal version of Equation (13) represents the average number of
natural numbers per prospective prime number in $S_{k}$ or the average spacing
of prospective prime numbers among all numbers in $S_{k}$.

Letting $k\rightarrow\infty$ we get the full Riemann Zeta function for $s=1$
which diverges:

$\zeta(1)=\prod_{i=1}^{\infty}\left(1-\frac{1}{P_{k}}\right)^{-1}\longrightarrow\infty$
(15)

which means, like with actual prime numbers, the average density of
prospective prime numbers becomes vanishingly small in $S_{k}$ as
$k\rightarrow\infty$.

Next consider the fraction of prospective prime numbers in $S_{k}$ that are
actual prime numbers:

$\boldsymbol{\pi}_{k}^{\widetilde{P}}=\frac{\boldsymbol{\pi}\left(4+\prod_{i=1}^{k}P_{i}\right)-k}{n_{k}^{\widetilde{P}}}$
(16)

$k$ appears in the numerator because $P\leq P_{k}$ are not considered as
prospective prime numbers in $S_{k}$, because they are already established
prime numbers within this concept and are not considered prime to themselves.

Table 1 gives numerical examples of Equation (16) for $k=2\rightarrow 10$.

Table 1: Numerical evaluation of $\boldsymbol{\pi}_{k}^{\widetilde{P}}$ for small $k$. $k$ | $\boldsymbol{\pi}\left(4+\prod_{i=1}^{k}P_{i}\right)-k$ | $n_{k}^{\widetilde{P}}$ | $\boldsymbol{\pi}_{k}^{\widetilde{P}}$
---|---|---|---
2 | $\boldsymbol{\pi}(10)-2=2$ | $\prod_{i=1}^{2}(P_{i}-1)=2$ | 1
3 | $\boldsymbol{\pi}(34)-3=8$ | $\prod_{i=1}^{3}(P_{i}-1)=8$ | 1
4 | $\boldsymbol{\pi}(214)-4=43$ | $\prod_{i=1}^{4}(P_{i}-1)=48$ | .896
5 | $\boldsymbol{\pi}(2314)-5=339$ | $\prod_{i=1}^{5}(P_{i}-1)=480$ | .706
6 | $\boldsymbol{\pi}(30,034)-6=3,242$ | $\prod_{i=1}^{6}(P_{i}-1)=5,760$ | .563
7 | $\boldsymbol{\pi}(510,514)-7=42,204$ | $\prod_{i=1}^{7}(P_{i}-1)=92,160$ | .458
8 | $\boldsymbol{\pi}(9,699,694)-8=646,021$ | $\prod_{i=1}^{8}(P_{i}-1)=1,658,880$ | .389
9 | $\boldsymbol{\pi}(223,092,874)-9=12,283,522$ | $\prod_{i=1}^{9}(P_{9}{i}-1)=36,495,360$ | .337
10 | $\boldsymbol{\pi}(6,469,693,234)-10=300,369,786$ | $\prod_{i=1}^{10}(P_{i}-1)=1,021,870,080$ | .294

In order to evaluate the trend in $\boldsymbol{\pi}_{k}^{\widetilde{P}}$ for
large $k$ we use the approximation
$\boldsymbol{\pi}(N)\approx\frac{N}{\ln{N}}$ in Equation (16) giving:

$\displaystyle\boldsymbol{\pi}_{k}^{\widetilde{P}}\approx$
$\displaystyle\left[\frac{4+\prod_{i=1}^{k}P_{i}}{\ln{\left(4+\prod_{i=1}^{k}P_{i}\right)}}-k\right]\cdot\frac{1}{\prod_{i=1}^{k}(P_{i}-1)}$
$\displaystyle\approx$
$\displaystyle\frac{1}{\ln{\left(4+\prod_{i=1}^{k}P_{i}\right)}}\cdot\left[\frac{4+\prod_{i=1}^{k}P_{i}}{\prod_{i=1}^{k}(P_{i}-1)}\right]-\frac{k}{\prod_{i=1}^{k}(P_{i}-1)}$
(17)

Ignoring the 4 relative to the product term gives:

$\displaystyle\boldsymbol{\pi}_{k}^{\widetilde{P}}\approx$
$\displaystyle\frac{1}{\sum_{i=1}^{k}\ln{P_{i}}}\cdot\prod_{i=1}^{k}\left(1-\frac{1}{P_{i}}\right)^{-1}-\frac{k}{\prod_{i=1}^{k}(P_{i}-1)}$
(18)

Given that prime numbers have no upper bound,
$\boldsymbol{\pi}_{k}^{\widetilde{P}}$ must remain positive as
$k\rightarrow\infty$ and the negative term clearly tends toward zero. It turns
out the negative term decreases much faster than the first term and therefore
can be ignored. Ignoring the negative term is equivalent to redefining
$\boldsymbol{\pi}_{k}^{\widetilde{P}}$ as the ratio of all prime numbers in
$S_{k}$, i.e. including $P\leq P_{k}$, to all prospective prime numbers in
$S_{k}$. However, the number of prospective prime numbers grows multiplicative
while the number of prime numbers less than or equal to $P_{k}$, i.e. $k$,
grows linearly and therefore $P\leq P_{k}$ can be ignored.

The reciprocal summation term, which is equal to
$\ln{\left(\prod_{i=1}^{k}P_{i}\right)}$, tends to zero while the
multiplication term tends to infinity as $k\rightarrow\infty$. Ignoring the
negative term, look at the ratio:

$\frac{\boldsymbol{\pi}_{k+1}^{\widetilde{P}}}{\boldsymbol{\pi}_{k}^{\widetilde{P}}}\approx\left(\frac{P_{k+1}}{P_{k+1}-1}\right)\cdot\left[\frac{1}{1+\frac{\ln{P_{k+1}}}{\sum_{i=1}^{k}\ln{P_{i}}}}\right]$

The fraction in the denominator in the square brackets is clearly less than 1,
so let $\frac{\ln{P_{k+1}}}{\sum_{i=1}^{k}\ln{P_{i}}}=\frac{1}{\alpha_{k}}$
where $\alpha_{k}>1$, giving:

$\frac{\boldsymbol{\pi}_{k+1}^{\widetilde{P}}}{\boldsymbol{\pi}_{k}^{\widetilde{P}}}\approx\left(\frac{P_{k+1}}{P_{k+1}-1}\right)\cdot\left(\frac{\alpha_{k}}{\alpha_{k}+1}\right)$
(19)

Where:

$\alpha_{k}\ln{P_{k+1}}=\sum_{i=1}^{k}\ln{P_{i}}=\ln{\left(\prod_{i=1}^{k}P_{i}\right)}\longrightarrow
P_{k+1}^{\alpha_{k}}=\prod_{i=1}^{k}P_{i}$

In Equation (19) the first fractional factor is asymtotic to 1 from above
while the second fractional factor is asymtotic to 1 from below and their
product is asymtotic to 1 from below (see Table 2). Therefore we always have:
$\boldsymbol{\pi}_{k+1}^{\widetilde{P}}<\boldsymbol{\pi}_{k}^{\widetilde{P}}$,
while
$\frac{\boldsymbol{\pi}_{k+1}^{\widetilde{P}}}{\boldsymbol{\pi}_{k}^{\widetilde{P}}}\longrightarrow
1$ as $k\rightarrow\infty$.

As an example consider: $P_{9591}=99,989$, where
$\boldsymbol{\pi}_{9591}^{\widetilde{P}}=.000102892$ and $P_{9592}=99,991$,
where $\boldsymbol{\pi}_{9592}^{\widetilde{P}}=.000102882$, giving
$\frac{\boldsymbol{\pi}_{9592}^{\widetilde{P}}}{\boldsymbol{\pi}_{9591}^{\widetilde{P}}}=.999902811$.

If we then look at $P_{11176}=118633$ where
$\boldsymbol{\pi}_{11176}^{\widetilde{P}}=.0000881401$ and $P_{11177}=118661$
where $\boldsymbol{\pi}_{11177}^{\widetilde{P}}=.0000881322$ giving
$\frac{\boldsymbol{\pi}_{11177}^{\widetilde{P}}}{\boldsymbol{\pi}_{11176}^{\widetilde{P}}}=.9999103700$.

$\boldsymbol{\pi}_{\widetilde{P}}^{k}$ appears to asymtotically approach zero
as $k\rightarrow\infty$, where:

$\sum_{i=1}^{k}\ln{P_{i}}>\prod_{i=1}^{k}\left(1-\frac{1}{P_{i}}\right)^{-1}=\prod_{i=1}^{k}\left(\frac{P_{i}}{P_{i}-1}\right)$

Table 2: Evaluation of Equation (19) $k$ | $P_{k}$ | $P_{k+1}$ | $\frac{P_{k+1}}{P_{k+1}-1}$ | $\prod_{i=1}^{k}P_{i}$ | $\alpha_{k}$ | $\frac{\alpha_{k}}{\alpha_{k}+1}$ | $\frac{\boldsymbol{\pi}_{k+1}^{\widetilde{P}}}{\boldsymbol{\pi}_{k}^{\widetilde{P}}}$
---|---|---|---|---|---|---|---
2 | 3 | 5 | 1.25 | 6 | 1.11 | .526 | .658
3 | 5 | 7 | 1.16 | 30 | 1.75 | .636 | .742
4 | 7 | 11 | 1.1 | 210 | 2.23 | .690 | .759
5 | 11 | 13 | 1.083 | 2,310 | 3.02 | .751 | .814
6 | 13 | 17 | 1.062 | 30,030 | 3.64 | .784 | .833
7 | 17 | 19 | 1.05 | 510,510 | 4.46 | .816 | .862
8 | 19 | 23 | 1.045 | 9,699,690 | 5.13 | .837 | .875
9 | 23 | 29 | 1.035 | 22,3092,870 | 5.71 | .851 | .881

###  2.3 Propagation of Prospective twin primes

Prospective twin primes, where $\widetilde{P}_{i+1}=\widetilde{P}_{i}+2$
follow the same propagation concept as prospective prime numbers except they
are paired occurrences within the same elemental sequence, appearing as
$\widetilde{P_{i}}=\widetilde{P}_{m,1}$ and
$\widetilde{P}_{i+1}=\widetilde{P}_{m,3}$.

When extending $S_{k}$ to $S_{k+1}$ each prospective twin prime
$(\widetilde{P}_{m,1},\widetilde{P}_{m,3})\in S_{k}$, prime relative to $P\leq
P_{k}$ corresponds to a location of paired prospective prime number locations
in the underlying sequence $\psi_{k}$. That sequence is repeated $P_{k+1}$
times in $S_{k+1}$ and each instance retains that paired set of locations
corresponding to numbers in $S_{k+1}$ that remain prime to all $P\leq P_{k}$.

These repetitions in $S_{k+1}$ occur as:

$\left(\widetilde{P}_{i}+m_{k}\prod_{j=1}^{k}P_{j}\;,\widetilde{P}_{i+1}+m_{k}\prod_{j=1}^{k}P_{j}\right)=\left(\widetilde{P}_{[m+m_{k}\prod_{j=3}^{k}P_{j}],1}\;,\widetilde{P}_{[m+m_{k}\prod_{j=3}^{k}P_{j}],3}\right)$
(20)

As with the propagation of prospective prime numbers, the prime factor
$P_{k+1}$ occurs in each position of $\psi_{k}$ exactly once throughout its
$P_{k+1}$ repetitions in $S_{k+1}$ and it cannot occur twice in the same
elemental sequence. Therefore, one instance of prime factor $P_{k+1}$ occurs
in two of the repeated locations of each prospective twin prime in $S_{k}$
extended to $S_{k+1}$, corresponding to the two components of the prospective
twin primes.

This leads to the following theorem, using $\widetilde{t}$ to represent a
prospective twin prime:

######  Theorem 2.2.

Given any prime number, $P_{k}>2$, let $n_{k}^{\widetilde{t}}$ be the number
of prospective twin primes $(\widetilde{P}_{i},\widetilde{P}_{i+1})$, where
$\widetilde{P}_{i+1}=\widetilde{P}_{i}+2$ and $\widetilde{P}_{i}$ and
$\widetilde{P}_{i+1}$ are prime relative to all $P\leq P_{k}$, and where
$P_{k}<\widetilde{P}_{i}$ and $\widetilde{P}_{i+1}<4+\prod_{i=1}^{k}P_{k}$.
Then defining $n_{1}^{\widetilde{t}}=1$,:
$n_{k}^{\widetilde{t}}=(P_{k}-2)n_{k-1}^{\widetilde{t}}=\prod_{i=2}^{k}(P_{i}-2)$.

###### Proof.

Consider the sequence of natural numbers $5\rightarrow
4+\prod_{i=1}^{k-1}P_{i}$. This sequence consists of $\prod_{i=1}^{k-1}P_{i}$
sequential natural numbers that includes all combinations of prime factors
$P<P_{k-1}$ situated relative to each other. That sequence of prime factors
repeats without end throughout the natural numbers for each successive
sequence of $\prod_{i=1}^{k-1}P_{i}$ numbers. Assume this sequence has,
$n_{k-1}$ prospective twin primes prime to all prime numbers $P\leq P_{k-1}$.
Then consider the extension of this sequence to the number
$4+\prod_{i=1}^{k}P_{i}$. This involves $P_{k}$ repetitions of the underlying
prime factor sequence associated with the numbers
$5\rightarrow\prod_{i=1}^{k-1}P_{i}$. Therefore there are $P_{k}n_{k-1}$
prospective twin primes prime to $P<P_{k-1}$ in the extended sequence. Since
the prime factor $P_{k}$ occurs exactly once in each of the
$\prod_{i=1}^{k-1}P_{i}$ positions of the underlying prime factor sequence,
each of the twin primes prime to $P\leq P_{k-1}$ has the prime factor $P_{k}$
exactly twice during the $P_{k}$ iterations of the underlying sequence, once
for each number in each twin prime, and the prime factor $P_{k}$ never occurs
more than once in the same 6-number elemental sequence. Therefore there are
$n_{k}=P_{k}n_{k-1}-2n_{k-1}=(P_{k}-2)n_{k-1}$ twin primes prime to all $P\leq
P_{k}$ within the sequence $5\rightarrow 4+\prod_{i=1}^{k}P_{i}$. If then we
define $n_{1}=1$, to initialize the succession of prime factor sequences, we
have: $n_{k}=\prod_{i=2}^{k}(P_{i}-2)=\prod_{i=3}^{k}(P_{i}-2)$.

Starting with $k=2$ we have the sequence $5,6,7,8,9,10$ where $5$ and $7$ are
the only twin prime prime to $2$ and $3$ giving $n_{2}^{\widetilde{t}}=1$.
Then extending this to include prime factor $5$ we get the sequence
$5,6,\cdots,34$, where only $(11,13),(17,19),$ and $(29,31)$ are twin primes
prime to $2,3,$ and $5$, giving $n_{3}^{\widetilde{t}}=3$, which verifies the
initial cases of the formula $n_{k}^{\widetilde{t}}=\prod_{i=2}^{k}(P_{i}-2)$
and completes the proof. ∎

The theorem shows the number of prospective twin primes in

$S_{k}=\left\\{N:5\leq N\leq 4+\prod_{i=1}^{k}P_{i}\right\\}$ is given by:

$n_{k}^{\widetilde{t}}=\prod_{i=2}^{k}(P_{i}-2)$ (21)

The density of prospective twin primes relative to all numbers in $S_{k}$ is:

$\rho_{k}^{\widetilde{t}}=\frac{\prod_{i=2}^{k}(P_{i}-2)}{\prod_{j=1}^{k}P_{j}}=\frac{1}{2}\prod_{i=2}^{k}\frac{P_{i}-2}{P_{i}}=\frac{1}{2}\prod_{i=2}^{k}\left(1-\frac{2}{P_{i}}\right)$
(22)

The density of prospective twin primes relative to prospective prime numbers
in $S_{k}$ is:

$\sigma_{k}^{\widetilde{t}}=\frac{\prod_{i=2}^{k}\left(P_{i}-2\right)}{\prod_{i=2}^{k}\left(P_{i}-1\right)}=\prod_{i=2}^{k}\frac{P_{i}-2}{P_{i}-1}$
(23)

Table 3: Numbers ($n_{k}^{\widetilde{t}}$) of prospective twin primes and their associated densities with respects to all numbers ($\rho_{k}^{\widetilde{t}}$) in a set $S_{k}$ and with respects to prospective prime numbers ($\sigma_{k}^{\widetilde{t}}$) in that set. $k$ | $P_{k}$ | $4+\prod_{i=1}^{k}P_{i}$ | $n_{k}^{\widetilde{t}}$ | $\rho_{k}^{\widetilde{t}}$ | $\sigma_{k}^{\widetilde{t}}$
---|---|---|---|---|---
3 | 5 | 34 | 3 | .1 | .375
4 | 7 | 214 | 15 | .0714 | .313
5 | 11 | 2,314 | 135 | .05834 | .281
6 | 13 | 30,034 | 1,485 | .0495 | .258
7 | 17 | 510,514 | 22,275 | .0436 | .242
8 | 19 | 9,699,694 | 378,675 | .0390 | .228
9 | 23 | 223,092,874 | 7,952,175 | .0356 | .218
10 | 29 | 6,469,693,234 | 214,708,725 | .0332 | .210
11 | 31 | 200,560,490,134 | 6,226,553,035 | .0310 | .203
12 | 37 | 7,420,738,134,814 | 217,929,355,875 | .0294 | .197

Given that all prospective prime numbers $\widetilde{P}\in S_{k}$ occur in the
interval, $P_{k}<\widetilde{P}<4+\prod_{i=1}^{k}P_{i}$, one might think using
that interval would give a more accurate density for prospective prime numbers
and prospective twin primes, however:

$\widehat{n}_{k}^{\widetilde{p}}=\frac{\prod_{i=1}^{k}(P_{i}-1)}{\prod_{i=1}^{k}P_{i}-P_{k}}=\left(\frac{1}{1-\frac{1}{\prod_{i=1}^{k-1}P_{i}}}\right)\frac{\prod_{i=1}^{k}(P_{i}-1)}{\prod_{i=1}^{k}P_{i}}\approx
n_{k}^{\widetilde{p}}$

and

$\widehat{n}_{k}^{\widetilde{t}}=\frac{\prod_{i=1}^{k}(P_{i}-2)}{\prod_{i=1}^{k}P_{i}-P_{k}}=\left(\frac{1}{1-\frac{1}{\prod_{i=1}^{k-1}P_{i}}}\right)\frac{\prod_{i=1}^{k}(P_{i}-2)}{\prod_{i=1}^{k}P_{i}}\approx
n_{k}^{\widetilde{t}}$

Given Theorem 2.2 we also have the following theorem.

######  Theorem 2.3.

Given any prime number $P_{k}$, there are an infinite number of prospective
twin primes, prime to $P\leq P_{k}$

###### Proof.

Given Theorem 2.2, there are $n_{k}^{\widetilde{t}}=\prod_{i=3}^{k}(P_{i}-2)$
prospective twin primes relative to prime factors $P\leq P_{k}$ in the
sequence $5\rightarrow 4+\prod_{i=1}^{k}P_{i}$. Since this sequence of natural
numbers includes all possible combinations of prime factors $P\leq P_{k}$
situated relative to each other, the underlying sequence of prime factors
$P\leq P_{k}$ repeats without end throughout the natural numbers.

Then if $(\widetilde{P}_{i},\widetilde{P}_{i+1})\in S_{k}$ is such a
prospective twin prime relative to all $P\leq P_{k}$, then so is:

$\left(\widetilde{P}_{i}+n\prod_{i=1}^{k}P_{i}\;,\widetilde{P}_{i+1}+n\prod_{i=1}^{k}P_{i}\right)$

where $n$ is any positive integer.

Therefore given that $n_{k}^{\widetilde{t}}$ is not zero for any $P_{k}$,
there are an infinite number of prospective twin primes relative to $P\leq
P_{k}$. ∎

###  2.4 Systematic Generation of Prospective Prime Numbers and Prospective
twin primes

The initial elemental 6-number sequence contains prime numbers $N_{0,1}=5$ and
$N_{0,3}=7$. Then numbers that remain prime to $2$ and $3$ occur at:

$N_{0,1}+m\prod_{i=1}^{2}P_{i}\quad\textrm{and}\quad
N_{0,3}+m\prod_{i=1}^{2}P_{i}$ (24)

where $m$ has any integer value.

Considering the succession of sets $S_{2}\rightarrow S_{3}\rightarrow
S_{3}\rightarrow\cdots\rightarrow S_{k}$, we can write Equation (24) as:

For $N_{n,1}\;\&\;N_{n,3}\in S_{k}$:

$\displaystyle\left(\begin{array}[]{c}N_{n_{1},1}\\\
N_{n_{3},3}\end{array}\right)_{S_{k}}$
$\displaystyle=\left(\begin{array}[]{c}N_{0,1}\\\
N_{0,3}\end{array}\right)+\left(\begin{array}[]{c}m_{3}^{(1)}\\\
m_{3}^{(3)}\end{array}\right)\prod_{i=1}^{2}P_{i}+\left(\begin{array}[]{c}m_{4}^{(1)}\\\
m_{4}^{(3)}\end{array}\right)\prod_{i=1}^{3}P_{i}+\cdots$ (33)
$\displaystyle+\sum_{j=3}^{k}\left(\begin{array}[]{c}m_{j}^{(1)}\\\
m_{j}^{(3)}\end{array}\right)\prod_{i=1}^{j-1}P_{i}$ (36)
$\displaystyle\left(\begin{array}[]{c}N_{n_{1},1}\\\
N_{n_{3},3}\end{array}\right)_{S_{k}}$
$\displaystyle=\left(\begin{array}[]{c}N_{0,1}\\\
N_{0,3}\end{array}\right)+\sum_{j=3}^{k}\left(\begin{array}[]{c}m_{j}^{(1)}\\\
m_{j}^{(3)}\end{array}\right)\prod_{i=1}^{j-1}P_{i}$ (43)

Where the upper numbers go together and the lower numbers go together
representing two separate equations.

Also: $0\leq m_{j}\leq P_{j}-1$ at each stage and:

$n_{1}=\sum_{j=3}^{k}m_{j}^{(1)}\prod_{i=3}^{j-1}P_{i}\quad\textrm{and}\quad
n_{3}=\sum_{j=3}^{k}m_{j}^{(3)}\prod_{i=3}^{j-1}P_{i}$

Numbers prime to $P\leq P_{k}$ in $S_{k}$ are produced using this equation by
including terms up to and including $m_{k}$, while noting that one value of
$m_{j}$ in each successive stage of the summation must be omitted separately
for each of $N_{n_{1},1}$ and $N_{n_{3},3}$ because those cases produce
numbers divisible by $P_{j}$. The reason only one value is omitted for each
set of $m_{j}$ is because when extending $S_{j-1}$ to $S_{j}$ the underlying
sequence $\psi_{j-1}$ is repeated $P_{j}$ times in $\psi_{j}$ and the prime
factor $P_{j}$ occurs exactly once in each position of $\psi_{j-1}$ spread
throughout it’s $P_{j}$ repetitions in $\psi_{j}$. Therefore exactly one
instance of $N_{n_{1},1}$ and $N_{n_{3},3}$ for each $m_{j}$ will have prime
factor $P_{j}$ and those correspond to different values of $m_{j}$ in the case
of $N_{n_{1},1}$ and $N_{n_{3},3}$, because the prime factor $P_{j}>6$ cannot
occur twice in the same elemental sequence.

Therefore the supplemental condition for Equations (33) is that for each
successive stage of summation:

$P_{l}\nmid\left[\left(\begin{array}[]{c}N_{0,1}\\\
N_{0,3}\end{array}\right)+\sum_{j=3}^{l}\left(\begin{array}[]{c}m_{j}^{(1)}\\\
m_{j}^{(3)}\end{array}\right)\prod_{i=1}^{j-1}P_{i}\right]$ (44)

This condition can also be stated as:

$\left[\left(\begin{array}[]{c}N_{0,1}\\\
N_{0,3}\end{array}\right)+\sum_{j=3}^{l}\left(\begin{array}[]{c}m_{j}^{(1)}\\\
m_{j}^{(3)}\end{array}\right)\prod_{i=1}^{j-1}P_{i}\right]\bmod{P_{l}}\neq 0$

Starting with prospective prime numbers in $S_{k}$ and extending them to
$S_{k+1}$, using all values of $m_{k+1}$ produces all numbers in $S_{k+1}$
that remain prime to $P\leq P_{k}$. Elimiating the one value of $m_{k+1}$ for
each $\widetilde{P}\in S_{k}$ then gives the prospective prime numbers in
$S_{k+1}$ that are prime to $P\leq P_{k+1}$.

Numbers in $S_{k}$ that have a prime factor from among $P\leq P_{k}$ cannot be
extended into $S_{k+1}$ as a prospective prime number because, given $N\in
S_{k}$, if $P|N$ then $P|\left(N+m_{k+1}\prod_{i=1}^{k}P_{i}\right)$ for all
values of $m_{k+1}$. That is why the existence of prospective prime numbers
$\widetilde{P}>P_{k}$ in $S_{k+1}$ depend on extension from prospective prime
numbers in $S_{k}$ and actual prime numbers $P>P_{k}$ can only occur at
locations of prospective prime numbers and in this respect they also depend on
extention from the prospective prime numbers in $S_{k}$.

Calculation of prospective prime numbers is best done iteratively, formulating
Equation (33) for $S_{k}\longrightarrow S_{k+1}$ as:

$\left(\begin{array}[]{c}\widetilde{P}_{n_{1},1}\\\
\widetilde{P}_{n_{3},3}\end{array}\right)_{S_{k+1}}=\left(\begin{array}[]{c}\widetilde{P}_{n^{\prime}_{1},1}\\\
\widetilde{P}_{n^{\prime}_{3},3}\end{array}\right)_{S_{k}}+\left(\begin{array}[]{c}m_{k+1}^{(1)}\\\
m_{k+1}^{(3)}\end{array}\right)\prod_{i=1}^{k}P_{i}$ (45)

where $0\leq m_{k+1}^{(1)},m_{k+1}^{(3)}\leq P_{k+1}-1$ and the condition from
equation (44) gives: $m_{k+1}^{(1)}\neq\widehat{m}_{k+1}^{(1)}$ and
$m_{k+1}^{(3)}\neq\widehat{m}_{k+1}^{(3)}$, where $\widehat{m}_{k+1}^{(1)}$
and $\widehat{m}_{k+1}^{(3)}$ are disallowed values (since they give results
divisible by $P_{k+1}$) given by:

$\displaystyle\widehat{m}_{k+1}^{(1)}=$
$\displaystyle\frac{\alpha_{1}P_{k+1}-\widetilde{P}_{n^{\prime}_{1},1}\bmod{P_{k+1}}}{\left(\prod_{i=1}^{k}P_{i}\right)\bmod{P_{k+1}}}$
and (46) $\displaystyle\widehat{m}_{k+1}^{(3)}=$
$\displaystyle\frac{\alpha_{3}P_{k+1}-\widetilde{P}_{n^{\prime}_{3},3}\bmod{P_{k+1}}}{\left(\prod_{i=1}^{k}P_{i}\right)\bmod{P_{k+1}}}$

where $\alpha_{1}$ and $\alpha_{3}$ are the smallest integers making
$\widehat{m}_{k+1}^{(1)}$ and $\widehat{m}_{k+1}^{(3)}$ integers; and
$\alpha_{1}=0$ if $\widetilde{P}_{n^{\prime}_{1},1}\bmod{P_{k+1}}=0$ and
$\alpha_{3}=0$ if $\widetilde{P}_{n^{\prime}_{3},1}\bmod{P_{k+1}}=0$.

Note that using the maximum value of $m_{j}=P_{j}-1$ at each stage in
Equations (33) gives:

$\displaystyle\left(\begin{array}[]{c}N_{n_{1},1}\\\
N_{n_{3},3}\end{array}\right)_{S_{k}}$
$\displaystyle=\left(\begin{array}[]{c}N_{0,1}\\\
N_{0,3}\end{array}\right)+\sum_{j=3}^{k}(P_{j}-1)\prod_{i=1}^{j-1}P_{i}$ (51)
$\displaystyle=\left(\begin{array}[]{c}N_{0,1}\\\
N_{0,3}\end{array}\right)+\sum_{j=3}^{k}\prod_{i=1}^{j}P_{i}-\sum_{j=3}^{k}\prod_{i=1}^{j-1}P_{i}$
(54) $\displaystyle=\left(\begin{array}[]{c}5\\\
7\end{array}\right)+\prod_{i=1}^{k}P_{i}-6$ (57)
$\displaystyle=\prod_{i=1}^{k}P_{i}+\left(\begin{array}[]{c}-1\\\
+1\end{array}\right)$ (60)

$\prod_{i=1}^{k}P_{i}-1$ and $\prod_{i=1}^{k}P_{i}+1$ are prime to $P\leq
P_{k}$. they are therefore largest prospective prime numbers and prospective
twin prime in $S_{k}$.333However, they are not alway an actual prime numbers,
e.g., $\prod_{i=1}^{4}P_{i}-1=209=11\cdot 19$; and
$\prod_{i=1}^{6}P_{i}+1=30031=59\cdot 509$. Therefore using the maximum value
of $m_{j}$ at every stage is always a valid choice since it always satisfies
the supplemental condition (44). However, the maximum value of $m_{j}$ may not
be valid if a value less than the maximum value was used at a prior stage.

In extending a prospective prime number $\widetilde{P}\in S_{k}$ to $S_{k+1}$,
the choice of $m_{j}=0$ is valid unless $\widetilde{P}$ itself contains the
prime factor $P_{k+1}$.

We can now state the following theorem:

######  Theorem 2.4.

Given $S_{k}=\left\\{N:5\leq N\leq 4+\prod_{i=1}^{k}P_{i}\right\\}$, then let
$\widetilde{P}_{n,1}$ and $\widetilde{P}_{n^{\prime},3}$ represent numbers in
$S_{k}$ from the arithmetic progressions $6n+5$ and $6n^{\prime}+7$
respectively ($n\geq 0$) that are also prime to all $P\leq P_{k}$. Then there
are equal numbers of $\widetilde{P}_{n,1}$ and $\widetilde{P}_{n^{\prime},3}$
in $S_{k}$.

###### Proof.

This follows from Equation (33), because in each summation term there are
equal numbers of valid $m_{j}$ for both $N_{n,1}$ and $N_{n^{\prime},3}$. This
is true even when considering the supplemental condition Equation (44) which
eliminates one value of $m_{j}$ for each of $N_{n,1}$ and $N_{n^{\prime},3}$
at each summation stage. Equation (33) together with its supplemental
condition yields numbers in $S_{k}$ that are prime to all $P\leq P_{k}$. ∎

The generation of prospective twin primes follows the same scheme except at
each stage we must have:

$m_{j}^{(1)}=m_{j}^{(3)}=m_{j}$

where both $m_{j}\neq\widehat{m}_{j}^{(1)}$ and
$m_{j}\neq\widehat{m}_{j}^{(3)}$ as defined by Equation (2.4).

Therefore for prospective twin primes, Equation (33) becomes:

$\left(\begin{array}[]{c}N_{n,1}\\\
N_{n,3}\end{array}\right)_{S_{k}}=\left(\begin{array}[]{c}N_{0,1}\\\
N_{0,3}\end{array}\right)+\sum_{j=3}^{k}(m_{j})\prod_{i=1}^{j-1}P_{i}$ (61)

which we can abbreviate as:

$\widetilde{t}_{k}=\left(\begin{array}[]{c}N_{0,1}\\\
N_{0,3}\end{array}\right)+\sum_{j=3}^{k}(m_{j})\prod_{i=1}^{j-1}P_{i}$ (62)

And iterative Equation (45) becomes:

$\widetilde{t}_{k+1}=\widetilde{t}_{k}+(m_{k+1})\prod_{i=1}^{k}P_{i}$ (63)

The parenthesis around coefficient $m$ represents that $m$ has a range of
values.

As prospective prime numbers in set $S_{k}$ are used to generate prospective
prime numbers in $S_{k+1}$, so are prospective twin primes in $S_{k}$ used to
generate prospective twin primes in $S_{k+1}$. An individual prospective prime
number, not part of a prospective twin prime, in $S_{k}$ cannot be used to
generate prospective twin prime, because its other component necessarily has a
prime factor $P\leq P_{k}$ and cannot be used to generate larger prospective
twin primes.

The next two subsections give simple numerical examples of calculating
prospective prime numbers and twin primes.

Numerical Example for Numbers Prime to $P\leq P_{3}=5$

Starting with the prospective prime numbers (actual primes in this case) in
the first elemental sequence: $N_{0,1}=5$ and $N_{0,3}=7$, which are prime to
$2$ and $3$, then in order to calculate other prospective prime numbers, prime
to $2$, $3$, and $5$, we have:

$N_{n,1}=N_{0,1}+\sum_{j=3}^{3}m_{j}\prod_{i=1}^{j-1}P_{i}=5+m_{3}\cdot 6$
(64)

and

$N_{n,3}=N_{0,3}+\sum_{j=3}^{3}m_{j}\prod_{i=1}^{j-1}P_{i}=7+m_{3}\cdot 6$
(65)

Then using Equations (2.4) find the disallowed values of $m_{j}$:

$\widehat{m}_{j}^{(1)}=\frac{\alpha_{1}\cdot 5-5\bmod{5}}{6\bmod{5}}=0$
$\widehat{m}_{j}^{(3)}=\frac{\alpha_{3}\cdot 5-7\bmod{5}}{6\bmod{5}}=3$

We then get:

$\left(\begin{array}[]{c}N_{0,1}\\\ N_{1,1}\\\ N_{2,1}\\\ N_{3,1}\\\
N_{4,1}\end{array}\right)=5+\left(\begin{array}[]{c}-\\\ 1\\\ 2\\\ 3\\\
4\end{array}\right)\cdot 6=\left(\begin{array}[]{c}-\\\ 11\\\ 17\\\ 23\\\
29\end{array}\right)$ (66)

and

$\left(\begin{array}[]{c}N_{0,3}\\\ N_{1,3}\\\ N_{2,3}\\\ N_{3,3}\\\
N_{4,3}\end{array}\right)=7+\left(\begin{array}[]{c}0\\\ 1\\\ 2\\\ -\\\
4\end{array}\right)\cdot 6=\left(\begin{array}[]{c}7\\\ 13\\\ 19\\\ -\\\
31\end{array}\right)$ (67)

Note also that the number pairs, with same value of $n$, are twin primes
relative to $P\leq 5$ and in this instance are actual twin primes:

$\left(N_{1,1},N_{1,3}\right)$ $\left(N_{2,1},N_{2,3}\right)$
$\left(N_{4,1},N_{4,3}\right)$

and the following number pairs continue as prospective twin primes, prime to
$P\leq 5$, for all values of $m_{j}\geq 0$:

$\left(N_{1,1}+\sum_{j=4}^{k}m_{j}\prod_{i=1}^{j-1}P_{i}\quad,\quad
N_{1,3}+\sum_{j=4}^{k}m_{j}\prod_{i=1}^{j-1}P_{i}\right);$
$\left(N_{2,1}+\sum_{j=4}^{k}m_{j}\prod_{i=1}^{j-1}P_{i}\quad,\quad
N_{2,3}+\sum_{j=4}^{k}m_{j}\prod_{i=1}^{j-1}P_{i}\right);$
$\left(N_{4,1}+\sum_{j=4}^{k}m_{j}\prod_{i=1}^{j-1}P_{i}\quad,\quad
N_{4,3}+\sum_{j=4}^{k}m_{j}\prod_{i=1}^{j-1}P_{i}\right);$

Numerical Example for Numbers prime to $P\leq P_{4}=7$

$(N_{n_{1},1})=\left(\begin{array}[]{c}N_{1,1}\\\ N_{2,1}\\\ N_{3,1}\\\
N_{4,1}\end{array}\right)+(m_{4})\prod_{i=1}^{3}P_{i}=\left(\begin{array}[]{c}11\\\
17\\\ 23\\\ 29\end{array}\right)+(m_{4})\cdot 30$ (68)

and

$(N_{n_{3},3})=\left(\begin{array}[]{c}N_{0,3}\\\ N_{1,3}\\\ N_{2,3}\\\
N_{4,3}\end{array}\right)+(m_{4})\prod_{i=1}^{3}P_{i}=\left(\begin{array}[]{c}7\\\
13\\\ 19\\\ 31\end{array}\right)+(m_{4})\cdot 30$ (69)

Here we use the parenthesis around $m_{4}$ because it is an array of numbers
which vary for each $N_{n^{\prime}_{1},1}$ and $N_{n^{\prime}_{3},3}$.

Next we find the disallowed values of $m_{4}$ using Equations (2.4):

$\displaystyle\widehat{m}_{4}^{(1)}(11)$ $\displaystyle=\frac{\alpha_{1}\cdot
7-11\bmod{7}}{30\bmod{7}}=\frac{\alpha_{1}\cdot 7-4}{2}=5;\;\alpha_{1}=2$
$\displaystyle\widehat{m}_{4}^{(1)}(17)$ $\displaystyle=\frac{\alpha_{1}\cdot
7-17\bmod{7}}{30\bmod{7}}=\frac{\alpha_{1}\cdot 7-3}{2}=2;\;\alpha_{1}=1$
$\displaystyle\widehat{m}_{4}^{(1)}(23)$ $\displaystyle=\frac{\alpha_{1}\cdot
7-23\bmod{7}}{30\bmod{7}}=\frac{\alpha_{1}\cdot 7-2}{2}=6;\;\alpha_{1}=2$
$\displaystyle\widehat{m}_{4}^{(1)}(29)$ $\displaystyle=\frac{\alpha_{1}\cdot
7-29\bmod{7}}{30\bmod{7}}=\frac{\alpha_{1}\cdot 7-1}{2}=3;\;\alpha_{1}=1$ and
$\displaystyle\widehat{m}_{4}^{(3)}(7)$ $\displaystyle=\frac{\alpha_{3}\cdot
7-7\bmod{7}}{30\bmod{7}}=\frac{\alpha_{3}\cdot 7-0}{2}=0;\;\alpha_{1}=0$
$\displaystyle\widehat{m}_{4}^{(3)}(13)$ $\displaystyle=\frac{\alpha_{3}\cdot
7-13\bmod{7}}{30\bmod{7}}=\frac{\alpha_{3}\cdot 7-6}{2}=4;\;\alpha_{1}=2$
$\displaystyle\widehat{m}_{4}^{(3)}(19)$ $\displaystyle=\frac{\alpha_{3}\cdot
7-19\bmod{7}}{30\bmod{7}}=\frac{\alpha_{3}\cdot 7-5}{2}=1;\;\alpha_{1}=1$
$\displaystyle\widehat{m}_{4}^{(3)}(31)$ $\displaystyle=\frac{\alpha_{3}\cdot
7-31\bmod{7}}{30\bmod{7}}=\frac{\alpha_{3}\cdot 7-3}{2}=2;\;\alpha_{1}=1$

The allowed values of $m$ are then:

$\left(m_{4}^{(1)}\right)=\left(\begin{array}[]{ccccccc}0&1&2&3&4&-&6\\\
0&1&-&3&4&5&6\\\ 0&1&2&3&4&5&-\\\ 0&1&2&-&4&5&6\\\ \end{array}\right)$ (70)

and in the equation for $N_{n,3}$:

$\left(m_{4}^{(3)}\right)=\left(\begin{array}[]{ccccccc}-&1&2&3&4&5&6\\\
0&1&2&3&-&5&6\\\ 0&-&2&3&4&5&6\\\ 0&1&-&3&4&5&6\\\ \end{array}\right)$ (71)

Using these results in Equations 68 and 69 gives:

$(N_{n_{1},1})=\left(\begin{array}[]{c}11\\\ 17\\\ 23\\\
29\end{array}\right)+\left(\begin{array}[]{ccccccc}0&1&2&3&4&-&6\\\
0&1&-&3&4&5&6\\\ 0&1&2&3&4&5&-\\\ 0&1&2&-&4&5&6\\\ \end{array}\right)\cdot 30$
(72)

giving:

$\left(N_{n_{1},1}\right)=\left(\begin{array}[]{ccccccc}11&41&71&101&131&-&191\\\
17&47&-&107&137&167&197\\\ 23&53&83&113&143^{*}&173&-\\\
29&59&89&-&149&179&209^{*}\\\ \end{array}\right)$ (73)

and

$\left(N_{n_{3},3}\right)=\left(\begin{array}[]{c}7\\\ 13\\\ 19\\\
31\end{array}\right)+\left(\begin{array}[]{ccccccc}-&1&2&3&4&5&6\\\
0&1&2&3&-&5&6\\\ 0&-&2&3&4&5&6\\\ 0&1&-&3&4&5&6\\\ \end{array}\right)\cdot 30$
(74)

giving:

$\left(N_{n_{3},3}\right)=\left(\begin{array}[]{ccccccc}-&37&67&97&127&157&187^{*}\\\
13&43&73&103&-&163&193\\\ 19&-&79&109&139&169^{*}&199\\\
31&61&-&121^{*}&151&181&211\\\ \end{array}\right)$ (75)

Where numbers with an asterix are prime to $p\leq 7$, but are not prime
numbers, having multiple prime factors $>7$. Note that the two array solutions
for $N_{n_{1},1}$ and $N_{n_{3},3}$ contain every prospective prime number $P$
(i.e., prime to all $P\leq 7$), where $7<P<4+\prod_{i=1}^{4}P_{i}=214\;$.

These results can be rearranged as follows correspond to the next stage
$S_{4}\longrightarrow S_{5}$:

$\left(N_{n,1}\right)=\left(\begin{array}[]{c}N_{1,1}\\\ N_{2,1}\\\ N_{3,1}\\\
N_{4,1}\\\ N_{6,1}\\\ N_{7,1}\\\ N_{8,1}\\\ N_{9,1}\\\ N_{11,1}\\\ N_{13,1}\\\
N_{14,1}\\\ N_{16,1}\\\ N_{17,1}\\\ N_{18,1}\\\ N_{21,1}\\\ N_{22,1}\\\
N_{23,1}^{*}\\\ N_{24,1}\\\ N_{27,1}\\\ N_{28,1}\\\ N_{29,1}\\\ N_{31,1}\\\
N_{32,1}\\\ N_{34,1}\end{array}\right)=\left(\begin{array}[]{c}11\\\ 17\\\
23\\\ 29\\\ 41\\\ 47\\\ 53\\\ 59\\\ 71\\\ 83\\\ 89\\\ 101\\\ 107\\\ 113\\\
131\\\ 137\\\ 143^{*}\\\ 149\\\ 167\\\ 173\\\ 179\\\ 191\\\ 197\\\
209^{*}\end{array}\right)\quad\textrm{and}\quad\left(N_{n,3}\right)=\left(\begin{array}[]{c}N_{1,3}\\\
N_{2,3}\\\ N_{4,3}\\\ N_{5,3}\\\ N_{6,3}\\\ N_{9,3}\\\ N_{10,3}\\\ N_{11,3}\\\
N_{12,3}\\\ N_{15,3}\\\ N_{16,3}\\\ N_{17,3}\\\ N_{19,3}\\\ N_{20,3}\\\
N_{22,3}\\\ N_{24,3}\\\ N_{25,3}^{*}\\\ N_{26,3}\\\ N_{27,3}\\\ N_{29,3}\\\
N_{30,3}\\\ N_{31,3}\\\ N_{32,3}\\\
N_{34,3}\end{array}\right)=\left(\begin{array}[]{c}13\\\ 19\\\ 31\\\ 37\\\
43\\\ 61\\\ 67\\\ 73\\\ 79\\\ 97\\\ 103\\\ 109\\\ 121^{*}\\\ 127\\\ 139\\\
151\\\ 157\\\ 163\\\ 169^{*}\\\ 181\\\ 187^{*}\\\ 193\\\ 199\\\
211\par\par\end{array}\right)$ (76)

These give the following 15 prospective twin primes:

$(N_{1,1},N_{1,3});(N_{2,1},N_{2,3});(N_{4,1},N_{4,3});(N_{6,1},N_{6,3});(N_{9,1},N_{9,3});(N_{11,1},N_{11,3});$
$(N_{16,1},N_{16,3});(N_{17,1},N_{17,3});(N_{22,1},N_{22,3});(N_{24,1},N_{24,3});(N_{27,1},N_{27,3});(N_{29,1},N_{29,3});$
$(N_{31,1},N_{31,3});(N_{321,1},N_{32,3});(N_{34,1},N_{34,3})$

In proceeding to the next stage $S_{4}\longrightarrow S_{5}$, one must keep
the prospective prime numbers with asterix, which are not actual prime
numbers, since they also generate prospective prime numbers.

Prospective prime numbers in $S_{5}$ like $121=11^{2}$ and $143=11\cdot 13$
which contain prime factor $11$ generate prospective prime numbers in $S_{5}$
prime to $P\leq 11$, as long as one excludes $m_{5}=0$, ($\widehat{m}_{5}=0$).

This leads to a general rule: starting with prospective prime numbers in
$S_{k}$ to generate prospective prime numbers in $S_{k+1}$. Any prospective
prime number in $S_{k}$ that has prime factor $P_{k+1}$ generates a
prospective prime number in $S_{k+1}$ using any value of $m_{k+1}$ except
$m_{k+1}=0$. Therefore given any $\widetilde{P}\in S_{k}$, where
$P_{k+1}|\widetilde{P}$ one can generate a subset of the prospective prime
numbers in $S_{k+1}$ using:

$\widetilde{P}+\left(\begin{array}[]{c}1\\\ 2\\\ 3\\\ 4\\\ \cdots\\\
P_{k+1}-1\end{array}\right)\cdot\prod_{i=1}^{k}P_{i}$

We have shown above that in generating prospective prime numbers one can sort
out the prospective twin primes. However, one can generate all prospective
twin primes directly using Equation (62) or Equation (63).

$\widetilde{t}_{3}=\left(\begin{array}[]{c}5\\\
7\end{array}\right)+(m_{j})\prod_{i=1}^{2}P_{i}=\left(\begin{array}[]{c}5\\\
7\end{array}\right)+\left(\begin{array}[]{ccccc}-&1&2&3&4\\\
0&1&2&-&4\end{array}\right)\cdot 6$

were we used the disallowed values of $m_{j}$ calculated above for the
prospective prime number example, giving:

$\widetilde{t}_{3}=\left(\begin{array}[]{ccccc}-&11&17&23&29\\\
7&13&19&-&31\end{array}\right)=\left(\begin{array}[]{ccc}11&17&29\\\
13&19&31\end{array}\right)$

Then these prospective twin primes (actual twin primes in this example) can be
used to generate all the prospective twin primes in $S_{4}$, etc. See Figure
2.

Figure 2: Branching families of prospective twin primes in the progression of
sets: $S_{2}\rightarrow S_{3}\rightarrow S_{4}\rightarrow S_{5}$, where in
each set $S_{k}$, prospective twin primes are prime to $P\leq P_{k}$.

##  3 Distribution of Prospective Prime Numbers and Prospective Twin Primes

###  3.1 Distribution of prospective prime numbers

We can represent a generic prospective prime number in $S_{k}$ as:

$\widetilde{P}_{\left\\{k\right\\}}=\widetilde{P}_{\left\\{k-1\right\\}}+(m_{k})\prod_{i=1}^{k-1}P_{i}$
(77)

where there are $n_{k-1}^{\widetilde{P}}=\prod_{i=1}^{k-1}(P_{i}-1)$ separate
$\widetilde{P}_{\left\\{k-1\right\\}}$ in $S_{k-1}$ and where $(m_{k})$
represents a spectrum of $P_{k}-1$ numbers out of the $P_{k}$ numbers ranging
from $0$ to $P_{k}-1$ thus making Equation (77) an abrieviation for $P_{k}-1$
separate equations for each $\widetilde{P}_{\left\\{k-1\right\\}}\in S_{k-1}$.

Looking at $S_{k}$ as a sequence of subsets
$S_{k}=\left\\{S_{k}^{(0)},S_{k}^{(1)},S_{k}^{(2)},\cdots,S_{k}^{(P_{k}-1)}\right\\}$,
where $S_{k}^{(j)}=\left\\{5+j\cdot\prod_{i=1}^{k-1}P_{i}\longrightarrow
4+(j+1)\cdot\prod_{i=1}^{k-1}P_{i}\right\\}$ one can see that for a specific
value $(m_{k})=m$:

${\widetilde{P}}_{\\{k-1\\}}+m\prod_{i=1}^{k-1}P_{i}\in S_{k}^{(m)}$

Therefore equation (77) represents that each
$\widetilde{P}_{\left\\{k-1\right\\}}\in S_{k-1}$ generates $P_{k}-1$
prospective prime numbers $\widetilde{P}_{\\{k\\}}\in S_{k}$, distributed one
each to all but one subset $S_{k}^{(m_{k})}$.

Let the one disallowed value of $(m_{k})$ corresponding to
$\widetilde{P}_{\\{k\\}}\bmod{P_{k}}=0$ be $\widehat{m}_{k}$, where:

$\left[\widetilde{P}_{\\{k-1\\}}+\widehat{m}_{k}\prod_{i=1}^{k-1}P_{i}\right]\bmod{P_{k}}=0$
(78)

and we can express this as:

$\widetilde{P}_{\\{k-1\\}}\bmod{P_{k}}+\widehat{m}_{k}\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}=\alpha
P_{k}$

where $\alpha$ is the smallest integer such that $\widehat{m}_{k}$ is also an
integer, giving:

$\widehat{m}_{k}=\frac{\alpha
P_{k}-\widetilde{P}_{\\{k-1\\}}\bmod{P_{k}}}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}$
(79)

We can let $\widetilde{P}_{\\{k-1\\}}\bmod{P_{k}}=\beta$ and let
$\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}=\gamma$, where:
$0\leq\alpha,\beta\leq P_{k}-1$ and $1\leq\gamma\leq P_{k}-1$. Giving:

$\widehat{m}_{k}=\frac{\alpha P_{k}-\beta}{\gamma}$

which requires that: $\alpha P_{k}\equiv\beta\bmod{\gamma}$. This has a unique
solution for $\alpha$ dependent on $\beta$ given that $\gamma$ is fixed for
$S_{k}$.[3, 7] Note that $\widehat{m}_{k}=0$ only if $\alpha=0=\beta$, which
occurs only when $P_{k}|\widetilde{P}_{\\{k-1\\}}$.

$\widehat{m}_{k}$ is the disallowed value of $(m_{k})$ in equation (77) which
is unique for distinct $\beta=\widetilde{P}_{\\{k-1\\}}\bmod{P_{k}}$. Any two
distinct instances of $\widetilde{P}_{\\{k-1\\}}$ that differ by a multiple of
$P_{k}$ will therefore have the same value of $\widehat{m}_{k}$. Meaning they
have the same disallowed subset and that they each generate one prospective
prime number to all the other immediate subsets of $S_{k}$.

Equation (77) gives:

$\widetilde{P}_{\left\\{k\right\\}}\bmod{P_{k}}=\left[\widetilde{P}_{\left\\{k-1\right\\}}\bmod{P_{k}}+(m_{k})\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}\right]\bmod{P_{k}}$

The only variable on the right hand side given a choice of
$\widetilde{P}_{\\{k-1\\}}$ is $(m_{k})$, which takes on $P_{k}-1$ distinct
values in the range $0\leq m_{k}\leq P_{k-1}$. Therefore the set of
$\widetilde{P}_{\\{k\\}}$ generated by one instance of
$\widetilde{P}_{\\{k-1\\}}$ represents $P_{k}-1$ distinct residue classes mod
$P_{k}$. The only residue class not represented is that corresponding to
$(m_{k})=\widehat{m}_{k}$ given in Equation (79).

In generating the set of all $\widetilde{P}_{\\{k\\}}$ we start with
$n_{k-1}^{\widetilde{P}}=\prod_{i=1}^{k-1}(P_{i}-1)$ prospective prime numbers
in $S_{k-1}$. These represent $n_{k-2}^{\widetilde{P}}$ sets of
$\widetilde{P}_{\\{k-1\\}}$ each with $P_{k-1}-1$ members generated by a
single $\widetilde{P}_{\\{k-2\\}}$. Each of these sets, which we represent as
$\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$,
represents $P_{k-1}-1$ distinct residue classes mod $P_{k-1}$ and therefore
the same number of residue classes mod $P_{k}$.444
$N=\left\lfloor\frac{N}{P}\right\rfloor\cdot
P+N\bmod{P}=\left\lfloor\frac{N}{P^{\prime}}\right\rfloor\cdot
P^{\prime}+N\bmod{P^{\prime}}\longrightarrow
N\bmod{P^{\prime}}=\left\lfloor\frac{N}{P}\right\rfloor\cdot
P-\left\lfloor\frac{N}{P^{\prime}}\right\rfloor\cdot P^{\prime}+N\bmod{P}$.
Therefore, given $P^{\prime}>P$ there is a one-to-one correspondence of the
residue classes mod $P$ to a subset of the residue classes mod $P^{\prime}$.

Equation (79) shows there is a one-to-one correspondence between the residue
class of $\widetilde{P}_{\\{k-1\\}}$ mod $P_{k}$ and $\widehat{m}_{k}$. Each
member of set
$\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$ used
to generate $\widetilde{P}_{\\{k\\}}$ has a corresponding disallowed value
$\widehat{m}_{k}$ unique among the members of that set. Therefore, among the
$P_{k}$ subsets $S_{k}^{(m_{k})}$ of $S_{k}$, $P_{k-1}-1$ have $P_{k-1}-2$
prospective prime numbers, $\widetilde{P}_{\\{k\\}}$, generated by the set
$\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$ and
$P_{k}-(P_{k-1}-1)$ have $P_{k-1}-1$ prospective prime numbers generated by
that same set. So, every immediate subset of $S_{k}$ has a minimum of
$P_{k-1}-2$ prospective prime numbers $\widetilde{P}_{\\{k\\}}$.

Then given that there are $n_{k-2}^{\widetilde{P}}$ subsets
$\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$, each
subset $S_{k}^{(m_{k})}$ of $S_{k}$ has a number of prospective prime numbers
$\widetilde{P}_{\\{k\\}}$ at least equal to:

$\displaystyle n_{S_{k}^{(m_{k})}}^{\widetilde{P}}$ $\displaystyle\geq
n_{k-2}^{\widetilde{P}}\cdot(P_{k-1}-2)$
$\displaystyle=(P_{k-1}-2)\prod_{i=1}^{k-2}(P_{i}-1)$
$\displaystyle=[(P_{k-1}-1)-1]\prod_{i=1}^{k-2}(P_{i}-1)$
$\displaystyle=\prod_{i=1}^{k-1}(P_{i}-1)-\prod_{i=1}^{k-2}(P_{i}-1)$
$\displaystyle=n_{k-1}^{\widetilde{P}}-n_{k-2}^{\widetilde{P}}$ (80)

Given that result we can state the following theorem:

######  Theorem 3.1.

Given the set $S_{k}=\left\\{N:5\leq N\leq 4+\prod_{i=1}^{k}P_{i}\right\\}$
and subsets $S_{k}^{(m)}=\left\\{5+m\prod_{i=1}^{k-1}P_{i}\leq N\leq
4+(m+1)\prod_{i=1}^{k-1}P_{i}\right\\}$, $0\leq m\leq(P_{k}-1)$, then every
subset $S_{k}^{(m)}$ has $n_{S_{k}^{(m)}}^{\widetilde{P}}\geq
n_{k-1}^{\widetilde{P}}-n_{k-2}^{\widetilde{P}}$ prospective prime numbers
$\widetilde{P}\in S_{k}^{(m)}$, where $\widetilde{P}=N\in S_{k}$ and
$P|N\longrightarrow P>P_{k}$.

###### Proof.

Let $\widetilde{P}_{\\{k\\}}$ represent a generic prospective prime number in
$S_{k}$. Then consider the subset
$\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$ of
prospective prime numbers in $S_{k-1}$ generated by a single
$\widetilde{P}_{\\{k-2\\}}$ via the relation:

$\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}=\widetilde{P}_{\\{k-2\\}}+(m_{k-1})\prod_{i=1}^{k-2}P_{i}$

where the different $\widetilde{P}_{k-1}$ are each distinguished by their
values of $m_{k-1}$, where $m_{k-1}$ has $P_{k-1}-1$ valid values among,
$0\leq m_{k-1}\leq P_{k-1}-1$. Therefore, each
$\widetilde{P}_{\\{k-1\\}}\in\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$
belongs to a unique subset $S_{k-1}^{(m_{k-1})}\subset S_{k-1}$ meaning
$\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$
consists of one prospective prime number in each of $P_{k-1}-1$ of the
$P_{k-1}$ subsets of $S_{k-1}^{(m)}\subset S_{k-1}$.

Note that each
$\widetilde{P}_{\\{k-1\\}}\in\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$
belongs to a unique residue class mod $P_{k-1}$ as determined by its
associated value of $m_{k-l}$ and therefore also belongs to a unique residue
class mod $P_{k}$, given $P_{k}>P_{k-1}$.

Now consider:

$\left\\{\widetilde{P}_{\\{k\\}}\right\\}=\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}+(m_{k})\prod_{i=1}^{k-1}P_{i}$

where we are representing multiple equations, for the $(P_{k-1}-1)$ values of
$\widetilde{P}_{\\{k-1\\}}\in\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$
and the $P_{k}-1$ valid values of $m_{k}$, where $m_{k}$ takes the values
$0\leq m_{k}\leq P_{k}-1$ with one disallowed value given by:

$\widehat{m}_{k}=\frac{\alpha\cdot
P_{k}-\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}\bmod{P_{k}}}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}$

where again we represent multiple equations. $\alpha$ is the least integer in
each case that yields an integer for $\widehat{m}_{k}$.

One can see from the equation for $\widehat{m}_{k}$, where each
$\widetilde{P}_{\\{k-1\\}}\in\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$
represents a unique residue class mod $P_{k}$ and therefore each
$\widetilde{P}_{\\{k\\}}$ derived from that $\widetilde{P}_{\\{k-1\\}}$ has a
unique disallowed subset $S_{k}^{(\widehat{m}_{k})}$. Therefore, for each
$\widetilde{P}_{\\{k-2\\}}$ there are $(P_{k-1}-1)$ individual
$\widetilde{P}_{\\{k-1\\}}\in\left\\{\widetilde{P}_{\\{k-1\\}}\right\\}_{\widetilde{P}_{\\{k-2\\}}}$
each which generates associated $\widetilde{P}_{k}$ distributed to all subsets
of $S_{k}^{(m)}\subset S_{k}$ except for its unique disallowed subset. This
leads to a minimum of $(P_{k-1}-2)$ individual $\widetilde{P}_{\\{k\\}}$ in
each subset of $S_{k}^{(m)}\subset S_{k}$ for each $\widetilde{P}_{\\{k-2\\}}$
in $S_{k-2}$.

Given that there are $n_{k-2}^{\widetilde{P}}=\prod_{i=1}^{k-2}(P_{i}-1)$
individual $\widetilde{P}_{\\{k-2\\}}\in S_{k-2}$ we get the number total
number of $\widetilde{P}_{\\{k\\}}$ in each subset $S_{k}^{(m)}$ given by:

$n_{S_{k}^{(m)}}^{\widetilde{P}}\geq
n_{k-2}^{\widetilde{P}}\cdot(P_{k-1}-2)=n_{k-1}^{\widetilde{P}}-n_{k-2}^{\widetilde{P}}$
(81)

∎

###  3.2 Distribution of prospective twin primes

We know from Theorem 2.2 that $S_{k-1}$ contains
$n_{k-1}^{\widetilde{t}}=\prod_{i=1}^{k-1}(P_{i}-2)$ prospective twin primes
prime to all $P\leq P_{k-1}$ and $S_{k}$ contains
$n_{k}^{\widetilde{t}}=\prod_{i=1}^{k}(P_{k}-2)$ prospective twin primes prime
to all $P\leq P_{k}$. Let
$\widetilde{t}_{k-1}=\left(\begin{array}[]{c}\widetilde{P}_{n_{k-1},1}\\\
\widetilde{P}_{n_{k-1},3}\end{array}\right)$ be a generic prospective twin
prime in $S_{k-1}$ and let
$\widetilde{t}_{k}=\left(\begin{array}[]{c}\widetilde{P}_{n_{k},1}\\\
\widetilde{P}_{n_{k},3}\end{array}\right)$ be a corresponding prospective twin
prime in $S_{k}$ given by:

$\widetilde{t}_{k}=\widetilde{t}_{k-1}+(m_{k})\prod_{i=1}^{k-1}P_{i}$

where this actually represents separate equations relating
$\widetilde{P}_{n_{k},1}$ to $\widetilde{P}_{n_{k-1},1}$ and
$\widetilde{P}_{n_{k},3}$ to $\widetilde{P}_{n_{k-1},3}$ both using the same
value of $m_{k}$ among the $P_{k}-2$ allowed values of $(m_{k})$, where:

$0\leq m_{k}\leq P_{k}-1$

and where additionally:

$m_{k}\neq\widehat{m}_{k}^{(1)}=\frac{\alpha_{1}P_{k}-\widetilde{P}_{n_{k-1},1}\bmod{P_{k}}}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}\quad\textrm{and}\quad
m_{k}\neq\widehat{m}_{k}^{(3)}=\frac{\alpha_{3}P_{k}-\widetilde{P}_{n_{k-1},3}\bmod{P_{k}}}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}$

where $\widehat{m}_{k}^{(1)}$ and $\widehat{m}_{k}^{(3)}$ specify disallowed
sets and $\alpha_{1}$ and $\alpha_{3}$ represent the lowest integer values
yielding integer solutions for $\widehat{m}_{k}^{(1)}$ and
$\widehat{m}_{k}^{(3)}$.

Each prospective twin prime, prime to all $P\leq P_{k-1}$, in $S_{k-1}$ can be
used to generate a prospective twin prime, prime to all $P\leq P_{k}$, in each
subset $S_{k}^{(m_{k})}$ except for the subsets
$S_{k}^{\left(\widehat{m}_{k}^{(1)}\right)}$ and
$S_{k}^{\left(\widehat{m}_{k}^{(3)}\right)}$, where the latter are specific to
each $\widetilde{t}_{k-1}\in S_{k-1}$.

Given that
$\widetilde{P}_{n_{k-1},3}\bmod{P_{k}}=\left[\widetilde{P}_{n_{k-1},1}\bmod{P_{k}}+2\right]\bmod{P_{k}}$,
we always have $\widehat{m}_{k}^{(1)}\neq\widehat{m}_{k}^{(3)}$ starting with
any specific $\widetilde{t}_{k-1}$. Therefore, each prospective twin prime of
$S_{k-1}$ projects a prospective twin prime into $P_{k}-2$ of the $P_{k}$
subsets $S_{k}^{(m_{k})}$ of $S_{k}$.

Analogous to the approach in the last section we consider generating all the
$\widetilde{t}_{k}\in S_{k}$ from the $\widetilde{t}_{k-1}\in S_{k-1}$ where
the latter consist of $n_{k-2}^{\widetilde{t}}=\prod_{i=2}^{k-2}(P_{i}-2)$
sets of $\left\\{\widetilde{t}_{k-1}\right\\}_{\widetilde{t}_{k-2}}$. Where
each of the sets $\left\\{\widetilde{t}_{k-1}\right\\}_{\widetilde{t}_{k-2}}$
consist of $(P_{k-1}-2)$ prospective twin primes in $S_{k-1}$ that are
generated from the same $\widetilde{t}_{k-2}$.

The problem in this case is that while the sets of $\widehat{m}_{k-1}^{(1)}$
and $\widehat{m}_{k-1}^{(3)}$ are unique within each set, being derived from
one specific $\widetilde{t}_{k-2}$, there can be common values between the two
sets for two distinct
$\widetilde{t}_{k-1}\in\left\\{\widetilde{t}_{k-1}\right\\}_{\widetilde{t}_{k-2}}$.
This can occur when:

$\widetilde{t}^{\prime}_{k-1}=\widetilde{t}_{k-1}\pm(nP_{k}\pm 2)$

This condition implies that
$P_{n_{k-1},1}\bmod{P_{k}}=P_{n^{\prime}_{k-1},3}\bmod{P_{k}}$ or
$P_{n^{\prime}_{k-1},1}\bmod{P_{k}}=P_{n_{k-1},3}\bmod{P_{k}}$ and
consequently: $\widehat{m}_{k}^{(1)}=\widehat{m}_{k}^{{}^{\prime}(3)}$ or
$\widehat{m}_{k}^{(3)}=\widehat{m}_{k}^{{}^{\prime}(1)}$.

Also, given that the two $\widetilde{t}_{k-1}$ are derived from the same
$\widetilde{t}_{k-2}$, we have:

$\widetilde{t}_{k-1}=\widetilde{t}_{k-2}+m_{k-1}\prod_{i=1}^{k-2}P_{i}$

and

$\widetilde{t}^{\prime}_{k-1}=\widetilde{t}_{k-2}+m^{\prime}_{k-1}\prod_{i=1}^{k-2}P_{i}$

giving:

$\Delta\widetilde{t}_{k-1}=\Delta m_{k-1}\prod_{i=1}^{k-2}P_{i}=\pm(nP_{k}\pm
2)$

Therefore in generating the set of
$\left\\{\widetilde{t}_{k-1}\right\\}_{\widetilde{t}_{k-2}}$ instead of having
four excluded subsets between a pair of $\widetilde{t}_{\\{k-1\\}}$ there will
be three excluded subsets whenever that pair of prospective twin primes are
generated with:

$\Delta m_{k-1}=\frac{n^{\pm}P_{k}\pm 2}{\prod_{i=1}^{k-2}P_{i}}$

That is $\widetilde{t}_{k-1}$ and $\widetilde{t}^{\prime}_{k-1}$ may share an
excluded subset, $S_{k-1}^{(m)}$. They cannot share the same two excluded
subsets.

Letting: $\Delta m_{k-1}^{+}=\frac{n^{+}P_{k}+2}{\prod_{i=1}^{k-2}P_{i}}$ and
$\Delta m_{k-1}^{-}=\frac{n^{-}P_{k}-2}{\prod_{i=1}^{k-2}P_{i}}$, Gives:

$\Delta m_{k-1}^{+}+\Delta
m_{k-1}^{-}=\frac{n^{+}P_{k}+2}{\prod_{i=1}^{k-2}P_{i}}+\frac{n^{-}P_{k}-2}{\prod_{i=1}^{k-2}P_{i}}=\frac{(n^{+}+n^{-})P_{k}}{\prod_{i=1}^{k-2}P_{i}}$

Since $\Delta m_{k-1}^{+}+\Delta m_{k-1}^{-}$ must be an integer, this
requires that $n^{+}+n^{-}=c\cdot\prod_{i=1}^{k-2}P_{i}$, were $c$ is an
integer, giving:

$n^{+}+n^{-}=c\prod_{i=1}^{k-2}P_{i}$

giving:

$\Delta m_{k-1}^{+}+\Delta m_{k-1}^{-}=cP_{k}$

Additionally, there are $P_{k-1}$ subsets in $S_{k-1}$ and to be valid $\Delta
m_{k-1}^{\pm}<P_{k-1}$, therefore $c=1$.

As in the last section dealing with the distribution of prospective prime
numbers in $S_{k}$ we would like to determine the number of prospective twin
primes in the subsets of $S_{k}$. The number of cases where redundancies
occur, e.g. $\widehat{m}_{k}^{(1)}=\widehat{m}_{k}^{{}^{\prime}(3)}$, which
may differ, means we will look for a minimum number.

As shown in the last section for prospective prime numbers, the sets of
$\widehat{m}_{k}^{(1)}$ and $\widehat{m}_{k}^{(3)}$ each have internally
unique values when generating $\widetilde{t}_{k}$ from the set of
$\left\\{\widetilde{t}_{k-1}\right\\}_{\widetilde{t}_{k-2}}$. Therefore for
the purposes of counting $\widetilde{t}_{k}\in S_{k}^{(m)}$ we can use Table 4
as a guide for calculating the distribution of prospective twin primes to the
subsets of $S_{k}^{(m)}\in S_{k}$.

Table 4: The table shows an allocation of prospective twin primes to subsets where the sets of $\widehat{m}_{k}^{(1)}$ and $\widehat{m}_{k}^{(3)}$ are each separately unique. It also arbitrarily shows a case where $\Delta m=2$. Such a table for an arbitrary $S_{k}$ would have $P_{k}$ subsets across the top and $P_{k-1}-2$ individual $\widetilde{t}_{k-1}$ down the left-hand column. | $S_{k}^{(0)}$ | $S_{k}^{(1)}$ | $S_{k}^{(2)}$ | $S_{k}^{(3)}$ | $S_{k}^{(4)}$ | $S_{k}^{(5)}$ | $S_{k}^{(6)}$ | $S_{k}^{(7)}$ | $\cdots$ | $S_{k}^{(P_{k}-2)}$ | $S_{k}^{(P_{k}-1)}$
---|---|---|---|---|---|---|---|---|---|---|---
$\widetilde{t}_{k-1}$ | $\widehat{m}_{k}^{(1)}$ | $\widetilde{t}_{k}$ | $\widehat{m}_{k}^{(3)}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\cdots$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$
$\widetilde{t}_{k-1}$ | $\widetilde{t}_{k}$ | $\widehat{m}_{k}^{(1)}$ | $\widetilde{t}_{k}$ | $\widehat{m}_{k}^{(3)}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\cdots$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$
$\widetilde{t}_{k-1}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widehat{m}_{k}^{(1)}$ | $\widetilde{t}_{k}$ | $\widehat{m}_{k}^{(3)}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\cdots$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$
$\widetilde{t}_{k-1}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widehat{m}_{k}^{(1)}$ | $\widetilde{t}_{k}$ | $\widehat{m}_{k}^{(3)}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\cdots$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\cdots$ | $\vdots$ | $\vdots$
$\widetilde{t}_{k-1}$ | $\widehat{m}_{k}^{(3)}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\widetilde{t}_{k}$ | $\cdots$ | $\widehat{m}_{k}^{(1)}$ | $\widetilde{t}_{k}$

The excluded subsets for a given $\widetilde{t}_{k-1}$ are represented where
the $\widehat{m}_{k}^{(1)}$ and $\widehat{m}_{k}^{(3)}$ are shown. With some
constraints, the actual distribution to specific subsets doesn’t matter for
calculating the minimum number of prospective twin primes per subset. Rows may
be interchanged and the $\widehat{m}_{j}$ may be shifted along a row, subject
to the following considerations: no two $\widehat{m}_{k}^{(1)}$ or two
$\widehat{m}_{k}^{(3)}$ can occur in the same column;
$\widehat{m}_{j}^{(3)}=\left[\widehat{m}_{j}^{(1)}+\Delta
m\right]\bmod{P_{k}}$; and
$\widehat{m}_{j}^{(1)}=\widehat{m^{\prime}}_{j}^{(3)}$ can occur.

The separation of $\widehat{m}_{k}^{(1)}$ and $\widehat{m}_{k}^{(3)}$ affects
the distribution to subsets and that can be calculated:

$\displaystyle\widetilde{t}_{k}=$
$\displaystyle\widetilde{t}_{k-1}+(m_{k})\prod_{i=1}^{k-1}P_{i}$ expanded
gives: $\displaystyle\left(\begin{array}[]{c}P_{n_{k,1}}\\\
P_{n_{k},3}\end{array}\right)=$
$\displaystyle\left(\begin{array}[]{c}\widetilde{P}_{n_{k-1},1}\\\
\widetilde{P}_{n_{k-1},3}\end{array}\right)+\left(\begin{array}[]{c}m_{k}^{(1)}\\\
m_{k}^{(3)}\end{array}\right)\prod_{i=1}^{k-1}P_{i}$ (88)

Then relating the two disallowed values of $m_{k}$:

$\widehat{m}_{k}^{(1)}=\frac{\alpha_{1}P_{k}-\widetilde{P}_{n_{k-1},1}\bmod{P_{k}}}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}$

and

$\displaystyle\widehat{m}_{k}^{(3)}=$
$\displaystyle\frac{\alpha_{3}P_{k}-\widetilde{P}_{n_{k-1},3}\bmod{P_{k}}}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}$
$\displaystyle=$
$\displaystyle\frac{\alpha_{3}P_{k}-(\widetilde{P}_{n_{k-1},1}+2)\bmod{P_{k}}}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}$
$\displaystyle=$
$\displaystyle\frac{\alpha^{\prime}_{3}P_{k}-\widetilde{P}_{n_{k-1},1}\bmod{P_{k}}-2}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}$
$\displaystyle=$
$\displaystyle\frac{\alpha_{1}P_{k}-\widetilde{P}_{n_{k-1},1}\bmod{P_{k}}}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}+\frac{\Delta\alpha\cdot
P_{k}-2}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}$
$\displaystyle\widehat{m}_{k}^{(3)}=$
$\displaystyle\widehat{m}_{k}^{(1)}+\frac{\Delta\alpha\cdot
P_{k}-2}{\left(\prod_{i=1}^{k-1}P_{i}\right)\bmod{P_{k}}}$

where $\Delta\alpha=\alpha^{\prime}_{3}-\alpha_{1}$ are integers chosen to
give an integer result, and $\alpha^{\prime}_{3}$ is modified $\alpha_{3}$ to
account for removing the $2$ from the mod function.

One can see that the separation of the two disallowed sets is the same for
each prospective twin prime in $S_{k}$, therefore justifying the fixed
separation shown in Figure 4.

For the set $S_{k}$ we have $P_{k}$ subsets and $P_{k-1}-2$ prospective twin
primes generated in $S_{k-1}$ for each $\widetilde{t}_{k-2}$. Therefore the
left hand column of the corresponding table for $S_{k}$ would have $P_{k-1}-2$
rows of $\widetilde{t}_{k-1}$. Columns (subsets) with no occurrence of
$\widehat{m}_{k}^{(1)}$ or $\widehat{m}_{k}^{(3)}$ have $P_{k-1}-2$
prospective twin primes $\widetilde{t}_{k}$. Columns with one occurrence of
one or the other of $\widehat{m}_{k}^{(1)}$ or $\widehat{m}_{k}^{(3)}$ have
$P_{k-1}-3$ prospective twin primes $\widetilde{t}_{k}$ and columns with
occurrences of both $\widehat{m}_{k}^{(1)}$ and $\widehat{m}_{k}^{(3)}$ have
$P_{k-1}-4$ prospective twin primes $\widetilde{t}_{k}$.

######  Theorem 3.2.

Given the set of sequential numbers $S_{k}=\left\\{N:5\leq N\leq
4+\prod_{i=1}^{k}P_{i}\right\\}$ and its $P_{k}$ subsets $S_{k}^{(m)}\subset
S_{k}$, where
$S_{k}^{(m)}=\left\\{N:5+m\prod_{i=1}^{k-1}P_{i}\leq N\leq
4+(m+1)\prod_{i=1}^{k-1}P_{i}\right\\}$, $0\leq m\leq P_{k}-1$, and letting
$n_{S_{k}^{(m)}}^{\widetilde{t}}$ represent the number of prospective twin
primes, $\widetilde{t}_{k}$, in each subset, then:
$n_{S_{k}^{(m)}}^{\widetilde{t}}\geq
n_{k-1}^{\widetilde{t}}-2n_{k-2}^{\widetilde{t}}$ and where
$P_{k}\leq\widetilde{t}_{k}\leq 4+\prod_{i=1}^{k}P_{i}$ and
$P|\widetilde{t}_{k}\longrightarrow P>P_{k}$.

###### Proof.

Given the preceding discussion we know that for each $\widetilde{t}_{k-2}\in
S_{k-2}$ we have a minimum of $P_{k-1}-4$ prospective twin primes in each of
the subsets $S_{k}^{(m)}$ which are prime to all $P\leq P_{k}$. Then given
that there are $n_{k-2}^{\widetilde{t}}=\prod_{i=1}^{k-2}(P_{i}-2)$
prospective twin primes in $S_{k-2}$ and representing the number of prospetive
twin primes in $S_{k}^{(m)}$ as $n_{S_{k}^{(m)}}^{\widetilde{t}}$, we have:

$\displaystyle n_{S_{k}^{(m)}}^{\widetilde{t}}$ $\displaystyle\geq
n_{k-2}^{\widetilde{t}}\cdot(P_{k-1}-4)$
$\displaystyle=(P_{k-1}-4)\prod_{i=1}^{k-2}(P_{i}-2)$
$\displaystyle=[(P_{k-1}-2)-2]\prod_{i=1}^{k-2}(P_{i}-2)$
$\displaystyle=\prod_{i=1}^{k-1}(P_{i}-2)-2\prod_{i=1}^{k-2}(P_{i}-2)$
$\displaystyle n_{S_{k}^{(m)}}^{\widetilde{t}}$ $\displaystyle\geq
n_{k-1}^{\widetilde{t}}-2n_{k-2}^{\widetilde{t}}$ (89)

∎

Given that $S_{k}$ has $P_{k}$ subsets $S_{k}^{(m)}$, the stated minimum
number of prospective twin primes in each subset accounts for
$P_{k}\cdot(P_{k-1}-4)\prod_{i=3}^{k-2}(P_{i}-2)$ of the
$n_{k}^{\widetilde{t}}$ total prospective twin primes in $S_{k}$. Therefore we
have:

$\frac{P_{k}\cdot
n_{S_{k}^{(m)}}^{\widetilde{t}}}{n_{k}^{\widetilde{t}}}\geq\frac{P_{k}(P_{k-1}-4)\prod_{i=3}^{k-2}(P_{i}-2)}{\prod_{i=3}^{k}(P_{i}-2)}=\frac{P_{k}(P_{k-1}-4)}{(P_{k}-2)(P_{k-1}-2)}$

Note that $\frac{P_{k}}{P_{k}-2}>1$ and $\frac{P_{k-1}-4}{P_{k-1}-2}<1$ and
their product is less than $1$, but is asymptotic to $1$ from below as $P_{k}$
increases.555 Let $P_{k}=P_{k-1}+\Delta$, then
$\frac{P_{k}(P_{k-1}-4)}{(P_{k}-2)(P_{k-1}-2)}=\frac{1}{1+\frac{2(\Delta+2)}{P_{k}[P_{k}-(\Delta+4)]}}$.
Then from the prime number theorem we can approximate the average spacing of
prime numbers as $\Delta=\ln{N}$, which shows that the fraction approaches $1$
for large $P_{k}$.  This is evident in Table 5.

Table 5: Trend of $\frac{P_{k}(P_{k-1}-4)}{(P_{k}-2)(P_{k-1}-2)}$ as $P_{k}$ increases. $P_{k}$ | $P_{k-1}$ | $\frac{P_{k}}{P_{k}-2}$ | $\frac{P_{k-1}-4}{P_{k-1}-2}$ | $\frac{P_{k}(P_{k-1}-4)}{(P_{k}-2)(P_{k-1}-2)}$
---|---|---|---|---
$11$ | $7$ | $\frac{11}{9}$ | $\frac{3}{5}$ | $.73$
$23$ | $19$ | $\frac{23}{21}$ | $\frac{15}{17}$ | $.966$
$53$ | $47$ | $\frac{53}{51}$ | $\frac{43}{45}$ | $.993$
$103$ | $101$ | $\frac{103}{101}$ | $\frac{97}{99}$ | $.991$
$1577$ | $1573$ | $\frac{1577}{1575}$ | $\frac{1569}{1571}$ | $.999992$

Therefore we can say that prospective twin primes are fairly evenly
distributed between the subsets of $S_{k}$ and the minimum distribution of
prospective twin primes to each subset $S_{k}^{(m)}$ approaches
$\frac{n_{k}^{\widetilde{t}}}{P_{k}}$ as $P_{k}$ increases, where
$n_{k}^{\widetilde{t}}$ is the total number of prospective twin primes in
$S_{k}$ and $P_{k}$ is the number of subsets, $S_{k}^{(m)}$ in $S_{k}$.

Consider the set of prospective twin primes in $S_{k}$:
$\widetilde{T}_{k}\subset S_{k}$, where:666 We consider that $P|\widetilde{t}$
if $P$ divides either of the two components of
$\widetilde{t}=\left(\begin{array}[]{c}P_{n,1}\\\ P_{n,3}\end{array}\right)$

$\widetilde{T}_{k}=\left\\{\widetilde{t}:P_{k}<\widetilde{t}\leq
4+\prod_{i=1}^{k}P_{i}\quad\&\quad P|\widetilde{t}\rightarrow
P>P_{k}\right\\}$

If $\widetilde{t}\in\widetilde{T}_{k}$ then $\widetilde{t}=\widetilde{t}_{k}$.
We know from the discussion on the distribution of prospective twin primes
that the subset $S_{k}^{(0)}$ contains $n_{S_{k}^{(0)}}^{\widetilde{t}}$
prospective twin primes $\widetilde{t}_{k}$. Also since $S_{k}^{(0)}=S_{k-1}$
we know $\widetilde{T}_{k-1}$ contains $n_{S_{k}^{(0)}}^{\widetilde{t}}$
prospective twin primes $>P_{k}$ that are prime to all $P\leq P_{k}$; i.e.,
there are $n_{S_{k}^{(0)}}^{\widetilde{t}}$ $\quad\widetilde{t}_{k-1}$ where
$\widetilde{t}_{k-1}=\widetilde{t}_{k}$.

The set of prospective twin primes in $S_{k-1}$ is equal to:

$\widetilde{T}_{k-1}=\left[\widetilde{T}_{k}\cap\widetilde{T}_{k-1}\right]\cup\left\\{\widetilde{t}_{k-1}:P_{k-1}<\widetilde{t}_{k-1}<4+\prod_{i=1}^{k-1}P_{i}\quad\&\quad
P_{k}|\widetilde{t}_{k-1}\right\\}$

The first set involving the intersection of $\widetilde{T}_{k}$ and
$\widetilde{T}_{k-1}$ includes those $\widetilde{t}_{k-1}$ where
$\widetilde{t}_{k-1}=\widetilde{t}_{k}$, i.e. where
$\widetilde{t}_{k-1}>P_{k}$ and $P_{k}\nmid\widetilde{t}_{k-1}$. The second
set includes those $\widetilde{t}_{k-1}$ that contain the prime factor $P_{k}$
including $(P_{k},P_{k+1})$ if it is a twin prime. $\widetilde{t}_{k-1}$ that
has $P_{k}$ as a prime factor does generate prospective twin primes in $S_{k}$
via $\widetilde{t}_{k-1}+(m_{k})\prod_{i=1}^{k-1}P_{i}$, but not for $m_{k}=0$
and therefore not in $S_{k}^{(0)}$. We can therefore place a limit on the
number of $\widetilde{t}_{k-1}$ that contain prime factor $P_{k}$ by:

$\displaystyle n_{k-1}^{P_{k}|\widetilde{t}}=$ $\displaystyle
n_{k-1}^{\widetilde{t}}-n_{S_{k}^{(m)}}^{\widetilde{t}}$ $\displaystyle\leq$
$\displaystyle
n_{k-1}^{\widetilde{t}}-[n_{k-1}^{\widetilde{t}}-2n_{k-2}^{\widetilde{t}}]$
$\displaystyle\leq$ $\displaystyle 2n_{k-2}^{\widetilde{t}}$

If we look at the fraction of prospective twin primes in $S_{k-1}$ with prime
factor $P_{k}$ we get:

$\frac{n_{k-1}^{P_{k}|\widetilde{t}}}{n_{k-1}^{\widetilde{t}}}\leq\frac{2n_{k-2}^{\widetilde{t}}}{n_{k-1}^{\widetilde{t}}}=2\frac{\prod_{i=3}^{k-2}(P_{i}-2)}{\prod_{i=3}^{k-1}(P_{i}-2)}=\frac{2}{P_{k-1}-2}$

While $n_{k-1}^{P_{k}|\widetilde{t}}$ grows if we pick a larger $P_{k}$, the
fraction it represents of all prospective twin primes in $S_{k-1}$ becomes
negligible.

This gives an interpretation to the formula in Theorem 3.2:

$n_{S_{k}^{(m)}}^{\widetilde{t}}\geq
n_{k-1}^{\widetilde{t}}-2n_{k-2}^{\widetilde{t}}$

The first term on the right is the number of prospective twin primes
$\widetilde{t}_{k-1}\in S_{k-1}$ which if there were no disallowed subsets,
contribute equally to all subsets $S_{k}^{(m)}$. The negative term then
represents subtracting the number of resulting
$\widetilde{t}_{k}=\widetilde{t}_{k-1}+m\prod_{i=1}^{k-1}P_{i}$ in
$S_{k}^{(m)}$ where $P_{k}|\widetilde{t}_{k}$. Therefore the negative term
accounts for the disallowed subsets which is uniform across all subsets with
respects to determining the minimum number of prospective twin primes in each
$S_{k}^{(m)}$. Note also, since each subset of $S_{k-1}$ has a uniform minimum
of $n_{S_{k-1}^{(m)}}^{\widetilde{t}}$ prospective twin primes, in this
respect each subset of $S_{k-1}$ contributes equally to the minimum number of
prospective twin primes in each subset of $S_{k}$. This latter point is key to
proving Theorem 3.3.

Given Theorems 2.2 and 3.2 we can now prove the following theorem.

######  Theorem 3.3.

Given integer $l\geq 4$, define
$P_{k}=P_{\boldsymbol{\pi}\left(\sqrt{\prod_{i=1}^{l}P_{i}}\right)}$, then let
$n_{P_{k}\rightarrow P_{k+1}^{2}}^{t}$ be the number of twin primes between
$P_{k}$ and $P_{k+1}^{2}$, then:

$n_{P_{k}\rightarrow
P_{k+1}^{2}}^{t}\geq\left(\prod_{j=l}^{k-1}\frac{P_{j}-4}{P_{j}-2}\right)\prod_{i=3}^{l}(P_{i}-2)=\left(\prod_{j=l}^{k-1}\frac{P_{j}-4}{P_{j}-2}\right)n_{l}^{\widetilde{t}}\\\
$

###### Proof.

Given $l$ and
$P_{k}=P_{\boldsymbol{\pi}\left(\sqrt{\prod_{i=1}^{l}P_{i}}\right)}$ consider
the set of sequential natural numbers $S_{k}=\left\\{5\longrightarrow
4+\prod_{i=1}^{k}P_{i}\right\\}$. We will show that $S_{k}$ always contains
prospective twin primes, $\widetilde{t}_{k}\in S_{k}$ prime to all $P\leq
P_{k}$ where $P_{k}<\widetilde{t}_{k}<P_{k+1}^{2}$ and consequently
$\widetilde{t}_{k}=t_{k}$ is an actual twin prime and the number of such twin
primes meets the stated minimum.

Note that while $\prod_{i=1}^{l}P_{i}+1$ is the largest prospective prime
number in $S_{l}=\left\\{5\longrightarrow 4+\prod_{i=1}^{l}P_{i}\right\\}$ in
that it is prime to all $P\leq P_{l}$, as a consequence of the multiplication
rules $\prod_{i=1}^{l}P_{i}\pm 1$ cannot be the square of a prime
number.777$\prod_{i=1}^{l}P_{i}-1=P_{n,1}$ and
$\prod_{i=1}^{l}P_{i}+1=P_{n,3}$, where
$n=\left[\prod_{i=3}^{l}P_{i}-1\right]$ is even. The multiplication rules
(1.3) show that the square of any prime number, where $n_{i}+n_{j}=2n$, has an
odd number for its elemental sequence number.

With the definition of $P_{k}$ we have:

$P_{k}^{2}<\prod_{i=1}^{l}P_{i}\longrightarrow P_{k}^{2}\in S_{l}$

and given

$P_{k+1}^{2}=\left(P_{\boldsymbol{\pi}\left(\sqrt{\prod_{i=1}^{l}P_{i}}\right)+1}\right)^{2}$,
we have:

$\prod_{i=1}^{l}P_{i}<P_{k+1}^{2}<\prod_{i=1}^{l+1}P_{i}\longrightarrow
P_{k+1}^{2}\in S_{l+1}\quad\&\quad P_{k+1}^{2}\notin S_{l}$

Note that $P_{k+1}$ is the smallest prime number whose square is greater than
$4+\prod_{i=1}^{l}P_{i}$ and $P_{k}$ is the largest prime number whose square
is less than $\prod_{i=1}^{l}P_{i}$. Therefore all prospective prime numbers
and prospective twin primes in $S_{l}$ are less than $P_{k+1}^{2}$. It remains
to show that some $\widetilde{t}_{l}$ are greater than $P_{k}$ and are prime
to all $P\leq P_{k}$ which means some $\widetilde{t}_{l}=\widetilde{t}_{k}$ or
equivalently $\widetilde{T}_{l}\cap\widetilde{T}_{k}\neq\emptyset$, where
$\widetilde{T}_{k}$ is the set of all $\widetilde{t}_{k}\in S_{k}$ and
$\widetilde{T}_{l}$ is the set of all $\widetilde{t}_{l}\in S_{l}$. In doing
this we will show the inequality for $n_{P_{k}\rightarrow P_{k+1}^{2}}^{t}$
holds.

To prove $\widetilde{T}_{l}\cap\widetilde{T}_{k}\neq\emptyset$ we must trace:
$\widetilde{t}_{l}\in S_{l}\rightarrow\widetilde{t}_{l}=\widetilde{t}_{l+1}\in
S_{l+1}\rightarrow\cdots\rightarrow\widetilde{t}_{l}=\widetilde{t}_{k}\in
S_{k}$. This can only be done through the zeroth subset of each set:
$\widetilde{t}_{l}\in S_{l}\rightarrow\widetilde{t}_{l}=\widetilde{t}_{l+1}\in
S_{l+1}^{(0)}\rightarrow\cdots\rightarrow\widetilde{t}_{l}=\widetilde{t}_{k}\in
S_{k}^{(0)}$, requiring $m_{j}=0$ at each stage of:
$\widetilde{t}_{k}=\widetilde{t}_{l}+\sum_{j=l+1}^{k}(m_{j})\prod_{i=j}^{k-1}P_{i}$,
otherwise the resulting $\widetilde{t}_{k}$ may be larger than $P_{k+1}^{2}$.
This is straightforward because:

$S_{l}=S_{l+1}^{(0)}\subset S_{l+2}^{(0)}\subset S_{l+3}^{(0)}\cdots\subset
S_{k}^{(0)}$

Based on our definitions, $S_{l+1}^{(0)}$ contains
$n_{S_{l+1}^{(0)}}^{\widetilde{t}}$ prospective twin primes, prime to $P\leq
P_{l+1}$, where $\widetilde{t}_{l+1}=\widetilde{t}_{l}$. We also know that
$S_{l+2}^{(0)}=S_{l+1}$ has $n_{S_{l+2}^{(0)}}^{\widetilde{t}}$ prospective
twin primes, prime to $P\leq P_{l+2}$. However all subsets of $S_{l+1}$ have
contributed prospective twin primes to $S_{l+2}^{(0)}$.

Given that all subsets of $S_{l+1}$ contribute prospective twin primes
uniformly to all subsets to $S_{l+2}$ with respects to the minimum number in
each subset, the fraction of prospective twin primes in $S_{l+2}^{(0)}$
generated from $\widetilde{t}_{l}=\widetilde{t}_{l+1}\in S_{l+1}^{(0)}$ is
given by:

$\frac{n_{S_{l+1}^{(0)}}^{\widetilde{t}}}{n_{l+1}^{\widetilde{t}}}n_{S_{l+2}^{(0)}}^{\widetilde{t}}\quad\approx\quad\textrm{number
of}\quad\widetilde{t}_{l+2}=\widetilde{t}_{l}$

where we have divided the number of prospective twin primes in $S_{l+1}^{(0)}$
by the total number of prospective twin primes in $S_{l+1}$ as the fraction of
prospective twin primes in $S_{l+2}^{(0)}$ generated by prospective twin
primes in $S_{l+1}^{(0)}$.

Then we have $n_{S_{l+3}^{(0)}}^{\widetilde{t}}$ prospective twin primes,
$\widetilde{t}_{l+3}\in S_{l+3}^{(0)}$ derived from all
$\widetilde{t}_{l+2}\in S_{l+2}$. The fraction of those derived from the set
of $\widetilde{t}_{l+2}=\widetilde{t}_{l}\in S_{l+2}^{(0)}$ is:

$\frac{n_{S_{l+1}^{(0)}}^{\widetilde{t}}}{n_{l+1}^{\widetilde{t}}}\frac{n_{S_{l+2}^{(0)}}^{\widetilde{t}}}{n_{l+2}^{\widetilde{t}}}n_{S_{l+3}^{(0)}}^{\widetilde{t}}\quad\approx\quad\textrm{number
of}\quad\widetilde{t}_{l+3}=\widetilde{t}_{l}$

Carrying this process forward up to the number of
$\widetilde{t}_{k}=\widetilde{t}_{l}$, where then $P_{k}<\widetilde{t}_{l}\leq
P_{k+1}^{2}$, gives:

$n_{p_{k}\rightarrow P_{k+1}^{2}}^{t}\approx
n_{S_{k}^{(0)}}^{\widetilde{t}}\prod_{j=l+1}^{k-1}\frac{n_{S_{j}^{(0)}}^{\widetilde{t}}}{n_{j}^{\widetilde{t}}}$
(90)

Expanding this using Theorem 3.2 and Theorem 2.2 we get:

$\displaystyle n_{p_{k}\rightarrow P_{k+1}^{2}}^{t}\geq$
$\displaystyle(P_{k-1}-4)\prod_{i=3}^{k-2}(P_{i}-2)\cdot\prod_{j=l+1}^{k-1}\frac{(P_{j-1}-4)\prod_{i=3}^{j-2}(P_{i}-2)}{\prod_{i=3}^{j}(P_{i}-2)}$
$\displaystyle=$
$\displaystyle(P_{k-1}-4)\prod_{i=3}^{k-2}(P_{i}-2)\cdot\left(\prod_{j=l+1}^{k-1}\frac{P_{j-1}-4}{P_{j-1}-2}\right)\left(\prod_{j=l+1}^{k-1}\frac{1}{P_{j}-2}\right)$
$\displaystyle=$
$\displaystyle\left(\frac{P_{k-1}-4}{P_{k-1}-2}\right)\left(\prod_{j=l}^{k-2}\frac{P_{j}-4}{P_{j}-2}\right)\prod_{i=3}^{l}(P_{i}-2)$
$\displaystyle=$
$\displaystyle\left(\prod_{j=l}^{k-1}\frac{P_{j}-4}{P_{j}-2}\right)\prod_{i=3}^{l}(P_{i}-2)=\left(\prod_{j=l}^{k-1}\frac{P_{j}-4}{P_{j}-2}\right)n_{l}^{\widetilde{t}}$
(91)

The first factor is less than one because its individual factors approach $1$
from below, but there is a disproportionate increase in $k$ when $l$ is
increased so there are more terms in the product as $l$ and $k$ increase. The
last factor is just the number of prospective twin primes in $S_{l}$, which
grows by the factor $(P_{l+1}-2)$ as $l$ increases to $l+1$. Table 6 evaluates
inequality (3.2) for a few low values of $l$ showing growing accuracy with $l$
where the second factor appears to grow fast enough to keep a finite and
growing product.888Numbers in the ”Actual” column from:
https://www.hugin.com.au/prime/twin.php. We will now show formally that
$n_{p_{k^{\prime}}\rightarrow
P_{k^{\prime}+1}^{2}}^{t}\geq\;n_{p_{k}\rightarrow P_{k+1}^{2}}^{t}$.

Table 6: Calculated minimum number of twin primes $P_{k}\leq t\leq P_{k+1}^{2}$ compared to actual. $l$ | $k$ | $\begin{subarray}{c}n_{p_{k}\rightarrow P_{k+1}^{2}}^{t}\\\ >\end{subarray}$ | $\begin{subarray}{c}Actual\>t\\\ P_{k}\leq t_{k}<P_{k+1}^{2}\end{subarray}$ | $\frac{n_{p_{k}\rightarrow P_{k+1}^{2}}^{t}}{Actual\>t}>$
---|---|---|---|---
$4$ | $6$ | $7$ | $16$ | $.44$
$5$ | $15$ | $43$ | $74$ | $.58$
$6$ | $40$ | $350$ | $480$ | $.73$
$7$ | $127$ | $3,988$ | $4,653$ | $.86$
$8$ | $443$ | $52,432$ | $57,529$ | $.91$

Now consider Equation (3.2) letting $l\rightarrow l+1$ and $k\rightarrow
k^{\prime}$, where
$k=\boldsymbol{\pi}\left(\sqrt{\prod_{i=1}^{l}P_{i}}\right)$ and
$k^{\prime}=\boldsymbol{\pi}\left(\sqrt{\prod_{i=1}^{l+1}P_{i}}\right)$.

$\displaystyle n_{p_{k^{\prime}}\rightarrow P_{k^{\prime}+1}^{2}}^{t}\geq$
$\displaystyle\left(\prod_{j=l+1}^{k^{\prime}-1}\frac{P_{j}-4}{P_{j}-2}\right)n_{l+1}^{\widetilde{t}}$
$\displaystyle\geq$
$\displaystyle\left(\prod_{j=l}^{k-1}\frac{P_{j}-4}{P_{j}-2}\right)n_{l}^{\widetilde{t}}\cdot\left(\frac{P_{l}-2}{P_{l}-4}\right)\left(\prod_{j=k}^{k^{\prime}-1}\frac{P_{j}-4}{P_{j}-2}\right)(P_{l+1}-2)$
which gives: $\displaystyle n_{p_{k^{\prime}}\rightarrow
P_{k^{\prime}+1}^{2}}^{t}\geq$ $\displaystyle\;n_{p_{k}\rightarrow
P_{k+1}^{2}}^{t}\cdot\frac{(P_{l}-2)(P_{l+1}-2)}{(P_{l}-4)}\left(\prod_{j=k}^{k^{\prime}-1}\frac{P_{j}-4}{P_{j}-2}\right)$
(92)

Noting that $P_{k}+2\leq P_{k+1}$ we can use:

$\displaystyle\prod_{j=k}^{k^{\prime}-1}\frac{P_{j}-4}{P_{j}-2}=\prod_{j=k}^{k^{\prime}-1}\left(1-\frac{2}{P_{j}-2}\right)\geq$
$\displaystyle\prod_{j=0}^{k^{\prime}-k}\left(1-\frac{2}{P_{k}+2j-2}\right)$
$\displaystyle\geq$ $\displaystyle
1-2\sum_{j=0}^{k^{\prime}-k}\frac{1}{P_{k}+2(j-1)}$ $\displaystyle\geq$
$\displaystyle 1-\frac{2}{P_{k}-2}-(k^{\prime}-k)\frac{2}{P_{k}}$ (93)

Then using the definition of $k^{\prime}$:

$k^{\prime}=\boldsymbol{\pi}\left(\sqrt{\prod_{1}^{l+1}P_{i}}\right)\approx\frac{\sqrt{\prod_{1}^{l+1}P_{i}}}{\ln{\sqrt{\prod_{1}^{l+1}P_{i}}}}=\frac{\sqrt{P_{l+1}}\cdot\sqrt{\prod_{1}^{l}P_{i}}}{\ln{\sqrt{P_{l+1}}}+\ln{\sqrt{\prod_{1}^{l}P_{i}}}}\\\
=\frac{\sqrt{\prod_{1}^{l}P_{i}}}{\ln{\sqrt{\prod_{1}^{l}P_{i}}}}\left(\frac{\sqrt{P_{l+1}}}{1+\frac{\ln{\sqrt{P_{l+1}}}}{\ln{\sqrt{\prod_{1}^{l}P_{i}}}}}\right)$
(94)

giving:

$k^{\prime}\approx k\sqrt{P_{l+1}}$

Using this in result (3.2) gives:

$\prod_{j=k}^{k^{\prime}-1}\left(1-\frac{2}{P_{j}-2}\right)\geq
1-\frac{2}{P_{k}-2}-k\left(\sqrt{P_{l+1}}-1\right)\frac{2}{P_{k}}$

Then ignoring the $-1$ in the last term and using
$k\approx\frac{P_{k}}{\ln{P_{k}}}$ gives:

$\prod_{j=k}^{k^{\prime}-1}\left(1-\frac{2}{P_{j}-2}\right)\geq
1-\frac{2}{P_{k}-2}-\frac{2\sqrt{P_{l+1}}}{\ln{P_{k}}}$

Using this in the inequality (3.2) gives:

$n_{p_{k^{\prime}}\rightarrow
P_{k^{\prime}+1}^{2}}^{t}\geq\;n_{p_{k}\rightarrow
P_{k+1}^{2}}^{t}\cdot\frac{(P_{l}-2)(P_{l+1}-2)}{(P_{l}-4)}\left[1-\frac{2}{P_{k}-2}-\frac{2\sqrt{P_{l+1}}}{\ln{P_{k}}}\right]$

The factor in front of the square brackets is $>(P_{l+1}-2)$. Since $P_{k}$ is
the largest prime whose square is less than $\prod_{i=1}^{l}P_{i}$, if we take
$P_{k}\approx\sqrt{\prod_{i=1}^{l}P_{i}}$ then the last term in the square
brackets is seen to be less than $1$ for $l>5$ and gets ever smaller for
larger $l$ and the second term in the square brackets is much less than $1$
and gets ever smaller as $l$ and $P_{k}$ increase. Therefore, given Table 6
showing growing values of $n_{p_{k}\rightarrow P_{k+1}^{2}}^{t}$ one can see
that for $l>4$:

$n_{p_{k^{\prime}}\rightarrow P_{k^{\prime}+1}^{2}}^{t}>n_{p_{k}\rightarrow
P_{k+1}^{2}}^{t}$

This completes the proof.

∎

Given Theorem 3.3 we can prove the following theorem:

######  Theorem 3.4.

Given any natural number, $N$ there is always a twin prime greater than $N$.

###### Proof.

Pick integer $l$ so that
$P_{k}=P_{\boldsymbol{\pi}\left(\sqrt{\prod_{i=1}^{l}P_{i}}\right)}>N$.Then we
know from Theorem (3.3) that there is always a twin prime $>P_{k}$. ∎

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