It is wellknown that using the sieve of Eratosthenes, we can generate the sequence of primes $2,3,5,7,11,13,17,19,\ldots\), which is known to be infinite. It is a longstanding problem to prove if there are (or there are not) infinitely many twin primes \((3,5),~(5,7),~(11,13),~(17,19),~(29,31),\ldots\).
There exists an efficient method[^6913] to sieve all twin primes of the form \((6k1,6k+1)\), \(k=1,2,\ldots\).
Note that these are all twin primes except the twin primes \((3,5)\). This is because even almost all primes (and not only twin primes) can be written in this form^{1}.
The following figure visualizes this sieve method:
With this respect, the sequence \(k=1, 2, 3, 5, 7, 10, 12, 17, 18, 23, 25, 30, 32, 33,\ldots\) can be called the twin prime sequence.
The following lemma formalizes this result.
Let $n\ge 1$ be an integer. The integer $a_n:=36n^21$ is the product of two
twin primes $p=6n1$ and $q=6n+1$ if and only if $n\not\equiv\pm f_s\mod s$ for
all primes $s$ with $5\le s\le\sqrt q,$ where, using the floor function,
the residue classes $f_s$ are defined by $$f_s:=\left\lfloor\frac {s+1}6\right\rfloor.$$
Table of Contents
Proofs: 1
References
Bibliography
 Piotrowski, Andreas: "Anmerkungen zur Verteilung der Primzahlzwillinge", Master’s thesis, Frankfurt am Main, 1999
Footnotes

This is because for all remaining natural numbers \(n\), there is a \(k\ge 1\) such
that \(n=6k2\), or \(n=6k+2\), or \(n=6k3\), or \(n=6k+3\), and all these numbers
are divisible by \(2\) and \(3\). Since they are composite (i.e. not prime), all the
remaining primes (and twin primes) must be of the form \(n=6k1\) or \(n=6k+1\). ↩