Complete k-Filtering Characterization - noir

Part I. – K index Filtering Model

1. General Framework

Let a, b, c, d be fixed integers. Consider an arithmetic progression:
N = ak + b

Suppose the index k can be expressed in the form:
k = axy + cx + dy
for some integers x, y.

We seek to find conditions under which N factors based on this structure of k.
Let’s substitute the expression for k into N:
N = a(axy + cx + dy) + b
N = a²xy + acx + ady + b

Now, consider the potential factorization:
P = (ax + c)(ay + d) = a²xy + adx + acy + cd

For N to be equal to this product P, we must have:
a²xy + acx + ady + b = a²xy + adx + acy + cd
acx + ady + b = adx + acy + cd
b – cd = adx – acx + acy – ady
b – cd = a(d-c)x – a(d-c)y
b – cd = a(d-c)(x-y)

Multiplying by -1, we get the identity:
cd – b = a(c-d)(x-y)

➤ Implication:
If the parameters a, b, c, d and the variables x, y satisfy the identity cd – b = a(c-d)(x-y), then the number N = ak + b (where k = axy + cx + dy) admits the factorization:
N = (ax+c)(ay+d)

If both factors |ax+c| > 1 and |ay+d| > 1, then N is composite.

Note on the Identity: The identity cd – b = a(c-d)(x-y) implies a constraint. If a(c-d) \neq 0, then x-y = (cd-b) / (a(c-d)), meaning x-y must be a specific constant. This restricts the independence of x and y. The framework is most powerful when this constraint is trivially satisfied or leads to a useful simplification, as seen below.

2. Simplified Case: c = d

A significant simplification occurs when c = d.
The identity cd – b = a(c-d)(x-y) becomes:
c² – b = a(c-c)(x-y)
c² – b = a(0)(x-y)
c² – b = 0
This implies b = c².

In this scenario:

The condition on x,y vanishes, allowing x and y to be independent integers.
The index expression for k becomes:
k = axy + cx + cy = axy + c(x+y)
The arithmetic progression value N = ak+b becomes:
N = ak + c²
The factorization becomes:
N = (ax+c)(ay+c)

This simplified form guarantees compositeness for N = ak+c² whenever k = axy+c(x+y), provided |ax+c| > 1 and |ay+c| > 1. This is a robust way to generate composites.

3. Applications to Specific Cases

Let k_{orig} be the original index in a progression N = A k_{orig} + B. We will use k to denote the parameterized form axy+c(x+y) which, if k_{orig}=k, implies compositeness.

Case 1: Filtering for Universal form N = k+1 (also expressible as N = n² + 1)

We can write this as N = K + 1, where K = n². So, the arithmetic progression parameters are A=1, B=1.
We use the simplified case: c=d. We require B=c². Since B=1, we have c²=1. Let c=1.
The corresponding a in the k = axy+c(x+y) formula is A=1.
The k-parameterization that generates composites is:
k = (1)xy + 1(x+y) = xy+x+y
If n² (our K) can be expressed as xy+x+y, then N = n²+1 is composite:
N = (1)(xy+x+y) + 1² = (x+1)(y+1)
Compositeness is ensured if |x+1|>1 and |y+1|>1. For N>0, we typically take x,y \ge 1.

Case 2: Filtering for an adapted form, e.g., N = 2k+1

Consider numbers N = 2K+1. This fits ak_{orig}+b with a=2 and b=1.
Using the simplified case: c=d. We need b=c². Since b=1, we choose c=1.
The a in k_{param} = axy+c(x+y) is a=2.
The k-parameterization is:
k_{param} = 2xy + 1(x+y) = 2xy+x+y.
If an index K can be expressed as k_{param} = 2xy+x+y, then N = 2K+1 is composite:
N = 2k_{param} + 1² = 2(2xy+x+y) + 1 = 4xy+2x+2y+1 = (2x+1)(2y+1).
For compositeness with factors greater than 1, if N > 0, we typically require x, y \ge 1 (so 2x+1 \ge 3, 2y+1 \ge 3).
This shows that any odd number 2K+1 where K = ( (2x+1)(2y+1)-1 )/2 for x,y \ge 1 is composite. This directly relates to the K-Filtering Model’s Case 2 (n=2k’+1, where k’=K).

Case 3: Filtering for numbers N = 6k±1 (using symmetry of {|6k-1|}={|6k+1|} in symmetrical ranges around 0)

This case involves numbers that can be prime (if >3). We aim to identify composite numbers within this family using a single k-parameterization.

Let a=6 (from the 6k_{orig} part).
Using the simplified case b=c², let c=1. So, b=1.
This naturally addresses the 6k_{orig}+1 form.
The k-parameterization is:
k_{param} = 6xy + 1(x+y) = 6xy+x+y
The corresponding number from the simplified framework is N_{framework} = ak_{param} + c² = 6k_{param} + 1.
This yields the factorization N_{framework} = (6x+1)(6y+1).

To ensure non-trivial factors (6x+1) and (6y+1), we must have 6x+1 \neq \pm 1 and 6y+1 \neq \pm 1. This means x \neq 0, -1/3 and y \neq 0, -1/3. Since x, y are integers, we require:

x, y \in \mathbb{Z} \setminus \{0\}.

Now, let’s connect this k_{param} to filtering composites in 6k_{orig} \pm 1:

Filtering 6k_{orig}+1 numbers:
If k_{orig} = k_{param} = 6xy+x+y, then
N = 6k_{orig}+1 = 6(6xy+x+y)+1 = (6x+1)(6y+1).
This N is composite.
- If x,y > 0, then k_{param} > 0. N = (6x+1)(6y+1) is a positive composite \equiv 1 \pmod 6.
  (e.g., x=1,y=1 \implies k_{param}=8, N=6(8)+1=49=(7)(7)).
- If x,y < 0 (let x’=-x, y’=-y with x’,y’>0), then k_{param} = 6x’y’ – x’ – y’.
  N = (6x+1)(6y+1) = (-6x’+1)(-6y’+1) = (6x’-1)(6y’-1).
  This N is also a positive composite \equiv 1 \pmod 6. k_{param} can be positive (e.g., x=-1,y=-1 \implies k_{param}=4, N=6(4)+1=25=(-5)(-5)) or negative or zero (but we excluded x,y=0).
Filtering 6k_{orig}-1 numbers (using negative k_{param} values):
The request is to use negative values of k_{param} = 6xy+x+y to identify composites of the form 6k_{orig}-1.
Let k_{param} < 0. Set k_{param} = -K_0 where K_0 > 0 (so K_0 = -(6xy+x+y)).
The number to test for compositeness is N = 6K_0-1.
From the framework, we know 6k_{param}+1 = (6x+1)(6y+1).
Substituting k_{param} = -K_0:
6(-K_0)+1 = (6x+1)(6y+1)
1-6K_0 = (6x+1)(6y+1)
Since K_0 > 0, 1-6K_0 must be negative. This implies that one of (6x+1) and (6y+1) is positive and the other is negative (which happens if x and y have opposite signs and are non-zero).
Then, N = 6K_0-1 = -(1-6K_0) = -((6x+1)(6y+1)) = |(6x+1)(6y+1)|.
This N is positive and \equiv -1 \pmod 6 if (6x+1)(6y+1) is \equiv 1 \pmod 6 and negative. (Actually, |(6x+1)(6y+1)| will be \equiv \pm 1 \pmod 6).
Since (6x+1)(6y+1) is \equiv 1 \pmod 6 when x,y are such that factors are \equiv \pm 1 \pmod 6, its negative is \equiv -1 \pmod 6.
So, N = |(6x+1)(6y+1)| is composite and \equiv -1 \pmod 6.
The k_{orig} for this number is K_0 = -k_{param} = -(6xy+x+y).
(e.g., x=1,y=-2 \implies k_{param} = -13. K_0 = 13. N=6(13)-1=77$. From the formula|(6(1)+1)(6(-2)+1)| = |(7)(-11)| = |-77| = 77`).

Thus, for k_{param} = 6xy+x+y (with x,y \in \mathbb{Z}\setminus\{0\}):

If k_{orig} = k_{param} and k_{param} > 0$, then6k_{orig}+1` is composite.
If k_{orig} = -k_{param} and k_{param} < 0 (so -k_{param}>0), then 6k_{orig}-1 is composite.

This means all k_{orig} values derived from 6xy+x+y (for x,y \in \mathbb{Z}\setminus\{0\}), when used appropriately for 6k+1 or 6k-1 forms, yield composites. +k correspond to 6k+1 numbers, and -k correspond to 6k-1 numbers. This ensures complete coverage of all composites and a method for sieving primes using just the linear algebraic form n=|6k+1| and its parameterization.

Part 2: Twin Prime Index Conjecture

Let:f(x,y)=∣6xy+x+y∣

where 𝑥,𝑦 ∈ 𝑍∖{0} (i.e., both are non-zero integers, so may be positive or negative).

Define the set:

𝐾composite = {𝑓(𝑥,𝑦): 𝑥≠0, 𝑦≠0}

Then: A positive integer 𝑘 is the index of a twin prime pair (6𝑘−1,6𝑘+1) if and only if:

𝑘∉𝐾composite

Therefore, the Twin Prime Conjecture is true if and only if:

𝑍+∖𝐾composite is infinite

In plain language:

There are infinitely many twin primes if and only if there are infinitely many positive integers 𝑘 that cannot be written in the form ∣6𝑥𝑦+𝑥+𝑦∣ for any non-zero integers 𝑥,𝑦.

The parameterization k = 6xy + x + y with x, y ∈ ℤ{0} achieves complete coverage of all composite indices through the symmetry of absolute values. Since {|6k-1|} = {|6k+1|}, we need only consider the single form |6k+1| to capture all composites in both 6k+1 and 6k-1 families.

When k > 0, the expression |6k+1| directly identifies composites of the form 6k+1.

When k < 0, the same expression |6k+1| = |6(-|k|)+1| = |-6|k|+1| = |-(6|k|-1)| = 6|k|-1 identifies composites of the form 6k-1.

The parameterization generates both positive and negative k values because x and y can independently be positive or negative (excluding zero).

This symmetry ensures that every composite number appearing as ∣6xy+x+y∣ in either 6k+1 or 6k-1 form corresponds to some k ∈ K_composite, making the characterization both necessary and sufficient: k is the index of a twin prime pair if and only if positive integer k ∉ K_composite = { |6xy + x + y| : x, y ∈ ℤ \ {0} }.

Part 3: Twin Prime Index Theorem.

The set
K_composite = { |6xy + x + y| : x, y ∈ ℤ \ {0} }
does not contain all positive integers. In fact, ℤ⁺ \ K_composite is infinite.

Proof.

Step 1: Algebraic Characterization.

Let k₀ = 6xy + x + y.
Let X = 6x + 1 and Y = 6y + 1.
Then (6x + 1)(6y + 1) = 36xy + 6x + 6y + 1 = 6(6xy + x + y) + 1 = 6k₀ + 1.
So, k₀ = (XY – 1)/6.

The set K_composite is defined as { |k₀| : x, y ∈ ℤ \ {0} }.
Thus, for any k ∈ K_composite,
k = |(XY – 1)/6|.

Since x, y ∈ ℤ \ {0} (non-zero integers), X and Y must satisfy:
X ≡ 1 (mod 6) and X ≠ 1,
Y ≡ 1 (mod 6) and Y ≠ 1.
Let S₁ = { m ∈ ℤ : m ≡ 1 (mod 6) and m ≠ 1 }.
Then X, Y ∈ S₁.

The characterization is an equality:
K_composite = { |(XY – 1)/6| : X, Y ∈ S₁ }
because for any X, Y ∈ S₁, we can recover x = (X – 1)/6 and y = (Y – 1)/6, which are non-zero integers. Also, since X,Y ∈ S₁ implies XY ≠ 1 (as X,Y ≠ ±1 simultaneously in a way that XY=1, e.g. -1 ∉ S₁), (XY-1)/6 ≠ 0, so k > 0. Thus K_composite ⊂ ℤ⁺.

Step 2: Density of S₁.

The set S₁ consists of integers of the form 6j + 1 where j ∈ ℤ and j ≠ 0.
Examples: S₁ = { …, -17, -11, -5, 7, 13, 19, … }.
The asymptotic density of S₁ within ℤ (or of its positive/negative parts within ℤ⁺/ℤ⁻ respectively) is 1/6.

Step 3: Sparseness of Products P = { XY : X, Y ∈ S₁ }.

Let P = { XY : X, Y ∈ S₁ }. All elements p ∈ P satisfy p ≡ 1 (mod 6).
However, not every integer congruent to 1 (mod 6) is in P. For instance, 31 ≡ 1 (mod 6), but 31 is prime. Since 1 ∉ S₁ and 31 ∉ S₁, 31 cannot be expressed as a product XY with X, Y ∈ S₁.

It is a known result in multiplicative number theory (e.g., from the study of the “multiplication table problem” and related work by Erdős or Ford) that the number of distinct products formed from elements of a sparse arithmetic set (like S₁) grows sublinearly. That is,
#{ p ∈ P : |p| ≤ M } = o(M)
(this notation means the count divided by M tends to 0 as M → ∞).
Consequently, the set P has asymptotic density 0 in ℤ.

Step 4: Sparseness of K_composite.

Let g₀(p) = (p – 1)/6. Then K_composite = { |g₀(p)| : p ∈ P }.
Consider the number of elements in K_composite up to a positive integer M:
#{ k ∈ K_composite : k ≤ M }.
If k ≤ M, then |(p – 1)/6| ≤ M, which implies |p – 1| ≤ 6M, so |p| is approximately bounded by 6M (e.g., |p| ≤ 6M + 1).

The number of such k is bounded by the number of distinct values of p ∈ P such that |p| ≤ 6M + 1:
#{ k ∈ K_composite : k ≤ M } ≤ #{ p ∈ P : |p| ≤ 6M + 1 }.
Since #{ p ∈ P : |p| ≤ 6M + 1 } = o(6M + 1) = o(M), it follows that
#{ k ∈ K_composite : k ≤ M } = o(M).

Therefore, the asymptotic density of K_composite in ℤ⁺ is 0:
lim (M→∞) [ #{ k ∈ K_composite : k ≤ M } / M ] = 0.

Step 5: Conclusion.

A subset of ℤ⁺ with zero asymptotic density cannot be cofinite (i.e., its complement cannot be finite).
Therefore, the complement ℤ⁺ \ K_composite must be infinite.

This demonstrates that infinitely many positive integers are not of the form |6xy + x + y| for any non-zero integers x, y.

(Q.E.D.)

Reference Note:

The sparseness of the product set P (Step 3) can be more formally justified by citing classical results in number theory:

P. Erdős (1960), “An asymptotic inequality in the theory of numbers,” Vestnik Leningrad. Univ. 13, 41-49.
K. Ford (2008), “The distribution of integers with a divisor in a given interval],” Annals of Mathematics (2), 168(2), 367–433.

These works address the number of distinct entries in multiplication tables and related problems concerning the density of product sets.