Full set of congruences uniquely defining all twin primes

Question

I am looking at the sequence A002822 which essentially is equivalent to the twin prime sequence. Based on initial investigations, it has the following peculiar congruential structure.

Pattern A

For each prime $p\geq 5$, there exists a residue $a_p$ with $0\leq a_p <p$ such that if $n \bmod p = \pm a_p$ then $n$ does not belong to the sequence, except possibly if $n=a_p$. Otherwise, $n$ belongs to the sequence. The values of $a_p$, respectively for $p = 5, 7, 11, 13, 17, 19, 23, 29, 31$, are $a_p = 1, 1, 2, 2, 3, 3, 4, 5, 5$.

Pattern B

Besides these congruential exceptions, all values in A002822 are equidistributed modulo $p$, regardless of $p$.

Question: the original question was: can you prove pattern A? It has been positively answered, see the answer below.

Update: pattern C and new question

I found a new pattern and I am wondering if I should create a separate question about it. I am very interested in the answer.

Pattern C is simply the fact that $a_p=\lfloor \frac{p}{6}+\frac{1}{2}\rfloor$ (assuming $p\geq 5$ is prime). At this point, this is just an empirical observation, verified up to $p=181$.

Motivation

Assuming patterns A and B are correct, we then get the following, unless I am wrong. Let $f(x)$ be the number of elements less than or equal to $x$ in the sequence in question. Then

$$f(x)\sim x\cdot \prod_{5\leq p \leq x}\frac{p-2}{p}\sim C\cdot\frac{x}{(\log x)^2} \mbox{ as } x\rightarrow\infty.$$

Note that we ignored the fact that some of the elements of the sequence are unaccounted for in the above formula: these are the elements for which $a_p$ is itself part of the sequence. Not a big deal, since the interest is to prove that the sequence has infinitely many elements (that is, there are infinitely many twin primes). Also, I am wondering if my asymptotic formula is compatible with the first Hardy-Littlewood conjecture. It is, except maybe for the constant $C$, unless I am mistaken.

Finally, could this lead to a non-probabilistic, non-heuristic argumentation to help prove the twin prime conjecture? Or is what I discovered already well-known, or wrong, or non amenable to anything useful?

Answer 1

This answer addresses only the question of the correctness of the congruential structure observed by OP. The pattern he reports is provably correct.

Consider composite numbers of the form $n=36m^2-1$. $n \text{ is a semiprime } \iff (6m-1,6m+1) \in \mathbb P \iff 6m-1,6m+1 \text { are twin primes} \\ n \text{ is not a semiprime } \Rightarrow n \text{ has at least one prime factor } p_i<6m-1 $

Theorem: If $p_i=6a \pm 1 \le 6m-1 \land p_i \mid 36m^2-1$, then $m \equiv \pm a \bmod p_i$.

Case 1: $$p_i=6a-1 \Rightarrow s_j=\frac{36m^2-1}{p_i}=6(a+k)+1 \\ 36m^2-1=36a^2+36ak+6a-6a-6k-1 \\ 36m^2=36a^2+36ak-6k$$ Since $36$ divides the LHS, $36$ must divide the RHS, so $k=6n,\ n \ge 0$. Substituting $$m^2=a^2+6an-n \\ m^2-a^2=(m-a)(m+a)=n(6a-1)=n\cdot p_i \\ m \equiv \pm a \bmod p_i$$

Case 2: $$p_i=6a+1 \Rightarrow s_j=\frac{36m^2-1}{s_i}=6(a+k)-1 \\ 36m^2-1=36a^2+36ak-6a+6a+6k-1 \\ 36m^2=36a^2+36ak+6k$$ Since $36$ divides the LHS, $36$ must divide the RHS, so $k=6n,\ n > 0$. Substituting $$m^2=a^2+6an+n \\ m^2-a^2=(m-a)(m+a)=n(6a+1)=n\cdot p_i \\ m \equiv \pm a \bmod p_i$$

Corollary: $6m-1,6m+1 \in \mathbb P \iff \forall a<m,\ m\not \equiv \pm a \bmod p_i$. Note that this corollary excludes the case $a=m$, for which $m\equiv a$ with respect to any modulus. If $m=a$ (whence $n=0$ and $b=a$) is the only instance where $m \equiv \pm a \bmod p_i$, then $6m-1,6m+1$ are twin primes.