# Full set of congruences uniquely defining all twin primes

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 $$0\leq a_p such that if $$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?

• I suspect that what you have found isn't wrong, but the difficulty arises in proving it... Jul 19, 2021 at 14:19
• The fact that none of the elements of the sequence is congruent to $1$ or $4$ modulo $5$ is listed in the OEIS entry (and probably a trivial fact), but that's the only congruence being mentioned. Jul 19, 2021 at 14:25
• Perhaps I am missing something. Surely the point is just that if $m$ is such that $6m\equiv \pm 1 \pmod p$ then of course $p$ divides one of $6m\pm 1$. For instance, if $p=5$, them we must avoid $m\equiv \pm 1 \pmod 5$. Is this the phenomenon you are speaking of?
– lulu
Jul 19, 2021 at 14:30
• Patterns won't solve the twin prime conjecture. No pattern that has yet been found, holds for larger twin prime pairs. This is the reason that the record size for a twin prime pair is "just" about $388\ 000$ digits (see here) , whereas the largest prime has more than $24.8$ million digits. Jul 19, 2021 at 15:20
• Remarkable that it has been shown that infinite many prime gaps do not exceed $246$. This is the best result that could be unconditionally proven. Not a single even prime gap is known to occur infinite many often, although almost surely , every even prime gap occurs infinite many often. A proof seems to be out of reach. Jul 19, 2021 at 15:24

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: $$6m-1,6m+1 \in \mathbb P \iff \forall a. Note that this corollary excludes the case $$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.