Building on the previous probabilistic approach to the Hardy-Littlewood twin prime conjecture, today we analyze the independence and distribution of primes in the form 6k±1. We will use a combination of the Prime Number Theorem, probabilistic reasoning, and the Chinese Remainder Theorem (CRT). Let’s break down the steps:

*Revised Proof of Independence of Events A_k and B_k*

**1. Probability Space:**

- Let Ω be the set of all positive integers.
- Define P as the asymptotic density of a set of integers. For a set A ⊆ Ω, P(A) = lim (n → ∞) [ |A ∩ {1, 2, …, n}| / n ], if the limit exists.

**2. Event Definitions:**

- Let A_k be the event that 6k – 1 is prime.
- Let B_k be the event that 6k + 1 is prime.

**3. Prime Number Theorem (PNT):**

- By the PNT, the asymptotic density of primes is zero, and for large x, P(x is prime) ≈ 1/ln(x)

**4. Chinese Remainder Theorem (CRT) Formalization:**

- For a fixed k and a finite set of primes S = {p_1, p_2, …, p_r}, define: M_S = ∏_{i=1}^r p_i (product of primes in S)
- By the CRT, there exists a bijection between:
- Residue classes of 6k-1 modulo M_STuples of residue classes (a_1 mod p_1, a_2 mod p_2, …, a_r mod p_r)

- By the CRT, there exists a bijection between:
- Similarly for 6k+1

**5. Conditional Events:**

- Define E_S(A_k) as the event that 6k-1 is not divisible by any prime in S
- Define E_S(B_k) as the event that 6k+1 is not divisible by any prime in S

**6. Independence from CRT:**

- For any prime p_i in S:
- 6k-1 ≢ 0 (mod p_i) corresponds to p_i – 1 allowed residue classes

- 6k+1 ≢ 0 (mod p_i) corresponds to p_i – 1 different allowed residue classes

- By the CRT bijection, the choices of residue classes for 6k-1 and 6k+1 are independent across different primes

- Therefore, P(E_S(A_k) ∩ E_S(B_k)) = P(E_S(A_k)) · P(E_S(B_k))

**7a. Limit Argument:**

- As S approaches the set of all primes: lim_{S → all primes} P(E_S(A_k)) = P(A_k) lim_{S → all primes} P(E_S(B_k)) = P(B_k)
- The error in this approximation goes to zero because:
- The density of primes is zero (by PNT)

- For any ε > 0, there exists a finite set of primes S such that |P(A_k) – P(E_S(A_k))| < ε and |P(B_k) – P(E_S(B_k))| < ε

- The error in this approximation goes to zero because:

** 7b. Explicit Error Bound:**

- Let π(x) be the prime counting function (number of primes ≤ x).
- By Chebyshev’s Theorem, there exist positive constants c1 and c2 such that:
- c1 · x/ln(x) ≤ π(x) ≤ c2 · x/ln(x) for x > 1

- Let p_S be the largest prime not in S. Then the error in our approximation is bounded by:
- |P(A_k) – P(E_S(A_k))| ≤ ∑_{p > p_S} 1/pUsing the integral comparison test and Chebyshev’s upper bound:
- ∑_{p > p_S} 1/p ≤ ∫_{p_S}^∞ 1/(x ln(x)) dx + O(1/ln(p_S))
- = ln(ln(x))|_{p_S}^∞ + O(1/ln(p_S))
- = -ln(ln(p_S)) + O(1/ln(p_S))
- Therefore, |P(A_k) – P(E_S(A_k))| ≤ -ln(ln(p_S)) + O(1/ln(p_S))

- As p_S → ∞ (i.e., as S approaches the set of all primes), this error bound approaches 0.

**8. Combining PNT and CRT:**

- By the PNT, for large k: P(A_k) ≈ 1/ln(6k-1) and P(B_k) ≈ 1/ln(6k+1)
- From steps 6 and 7: P(A_k ∩ B_k) = lim_{S → all primes} P(E_S(A_k) ∩ E_S(B_k)) = lim_{S → all primes} [P(E_S(A_k)) · P(E_S(B_k))] = P(A_k) · P(B_k)

**9. Conclusion: **

We have shown that P(A_k ∩ B_k) = P(A_k) · P(B_k) for large k, demonstrating the asymptotic independence of the events A_k and B_k.

**Revised Symmetry-Focused Proof of Asymptotic Independence of Events A_k and B_k**

**Definitions and Asymptotic Density:**- Define the asymptotic density d(X) of a set X of integers as: d(X) = lim_{n→∞} |{k ∈ X : |k| ≤ n}| / (2n + 1), if the limit exists.
- Let d(A_k) be the asymptotic density of integers k such that |6k-1| is prime.
- Let d(B_k) be the asymptotic density of integers k such that |6k+1| is prime.
- Note: Asymptotic density is not a probability measure, as it’s not countably additive.

**Symmetry:**- Observe that |6k-1| = |6k+1| for all integers k, establishing a fundamental symmetry.

**Prime Number Theorem (PNT) Application:**- By the PNT, for x > 2, π(x) = x/ln(x) + O(x/ln^2(x)), where π(x) is the prime counting function.
- Thus, for large |x|, d({k : |x| is prime}) = 1/ln(|x|) + O(1/ln^2(|x|))

**Chinese Remainder Theorem (CRT) and Mirror Image Formalization:**- For a fixed k and a finite set of primes S = {p_1, p_2, …, p_r}, define: M_S = ∏_{i=1}^r p_i
- By the CRT, there exists a bijection φ between:
- Residue classes of |6k-1| modulo M_S
- Tuples of residue classes (a_1 mod p_1, a_2 mod p_2, …, a_r mod p_r)

- For primes p > 2, define a “mirror image” function μ on residue classes modulo p: μ(a mod p) = (-a mod p)
- Extend μ to tuples: μ(a_1, …, a_r) = (μ(a_1), …, μ(a_r)) for p_i > 2
- Key Property: For any prime p > 3, if |6k-1| ≡ a (mod p), then |6k+1| ≡ μ(a) (mod p)

**Conditional Sets:**- Define E_S(A_k) = {k : |6k-1| is not divisible by any prime in S}
- Define E_S(B_k) = {k : |6k+1| is not divisible by any prime in S}

**Independence and CRT:**- For any prime p_i > 3 in S:
- |6k-1| ≢ 0 (mod p_i) corresponds to p_i – 1 allowed residue classes
- |6k+1| ≢ 0 (mod p_i) corresponds to the μ-images of these p_i – 1 classes

- For p = 2, both |6k-1| and |6k+1| are odd, so this case is trivial and disjoint
- By the CRT bijection φ and the mirror image property: d(E_S(A_k) ∩ E_S(B_k)) = ∏_{p_i ∈ S, p_i > 2} [(p_i – 1)/p_i]^2 · (1/2)
- This factorization demonstrates independence across different primes

- For any prime p_i > 3 in S:
**Error Analysis:**- Let ε_S(A_k) = |d(A_k) – d(E_S(A_k))|
- Using Mertens’ third theorem and partial summation: ε_S(A_k) = O(1/ln(p_S)), where p_S is the largest prime not in S
- As S approaches the set of all primes, p_S → ∞, so ε_S(A_k) → 0
- The same argument applies to ε_S(B_k)

**Asymptotic Independence:**- By the PNT and symmetry, for large |k|: d(A_k) = 1/ln(|6k-1|) + O(1/ln^2(|6k-1|)) d(B_k) = 1/ln(|6k+1|) + O(1/ln^2(|6k+1|))
- Combining the results from steps 6 and 7: |d(A_k ∩ B_k) – d(A_k) · d(B_k)| ≤ ε_S(A_k) + ε_S(B_k) + ε_S(A_k)ε_S(B_k) → 0 as |k| → ∞

**Conclusion: **We have shown that the difference between the joint asymptotic density of A_k and B_k and the product of their individual asymptotic densities tends to zero as |k| → ∞. This demonstrates the asymptotic independence of A_k and B_k in terms of their asymptotic densities.

**Illustrative Examples:**

- For p = 5: The residue classes for |6k-1| not divisible by 5 are {1, 2, 3, 4}. The corresponding residue classes for |6k+1| are {1, 2, 3, 4}. The mirror image function μ maps these as: μ(1) = 4, μ(2) = 3, μ(3) = 2, μ(4) = 1
- For p = 11: The residue classes for |6k-1| not divisible by 11 are {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. The corresponding residue classes for |6k+1| are {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. The mirror image function μ maps these as: μ(1) = 10, μ(2) = 9, μ(3) = 8, μ(4) = 7, μ(5) = 6, μ(6) = 5, μ(7) = 4, μ(8) = 3, μ(9) = 2, μ(10) = 1

(user thoughts:I used Claude for much of this. Claude seems very good at math and the text formatting is natively neater out of Claude. ChatGPT looks great on screen, is great at math, and does some fantastic stuff with code and code execution; but the LaTeX is a pain in the ass for formatting in other media. Gemini also does great pasting but can be aggravating with some math approaches. Many times, I will take a solution worked first in ChatGPT back to Gemini and then ideally post the revised Gemini output into the blog in order to ensure that the models agree and to reduce the formatting overhead. I've been incorporating Claude more in this process. Overall the other AI seem to "like" the solutions produced by Claude which may reduce a "peer review cycle" in AI; and combined with the formatting aspects makes it pleasant to work with so far. Perplexity.ai also has its place and does a great job at validating some math proofs and finding relevant side references for expanding this kind of mathematical inquiry in the "peer review cycle".)