A slight generalization of the Sieve of Sundaram that might shed light on the $6n \pm 1$ phenomenon of sequence A002822.

Question

There's the $n$ such that $6n \pm 1$ is a twin prime pair sequence: https://oeis.org/A002822 It contains all twin prime averages (divided by $6$) other than $4$.

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Notice this sequence: Positive numbers $n = 6xy + x + y$ such that $x,y \in \Bbb{Z}$. That sequence is the complement (all entries found in $\Bbb{N}$ that are not in A002822, by its very definition).

But notice the $6xy + x + y$ part. That reminded me of the Sieve of Sundaram's $2ij + i + j$ form. Thus:

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Start with a list of integers from $81$ to $N$, and cross out all elements of the form $6ij + ki + lj$ such that $k, l \in \{1,3,5\}$ and $i,j \geq 1$. Then taking the remaining elements, multiplying them by $6$ and adding respectively $lk = 1,3,5,9,15, $ or $25$; and of course we union the result of each case / cross-out form.

We have the following possibilities:

$$ 6(6ij + i + j) + 1 = (6i + 1)(6j + 1) \\ 6(6i + i + 3j) + 3 = (6i + 3)(6j + 1) \\ 6(6ij + i + 5j) + 5 = (6i + 5)(6j + 1) \\ 6(6ij + 3i +3j) + 9 = (6i + 3)(6j + 3) \\ 6(6ij + 3i + 5j) + 15 = (6i+5)(6j+3) \\ 6(6ij + 5i +5j) + 25 = (6i + 5)(6j + 5) \\ $$

The above will cover all odd numbers $\geq 81$ which we find by setting $i=j = 1$ and computing the smallest formula above which is $(6i + 1)(6j + 1) = 81$.

Thus we've filtered out all and only composites $\geq 81$, clearly, and via inspection of the above factorizations.

See: Sieve of Sundaram.

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However, all primes $\gt 3$ are of the form $6n + 1$ or $6n + 5$. Thus we can cross out lines $2,3,5$ in the above 6 formulas, or namely the lines with added $3,9,15$. But $6n + 25 = 1 \pmod 6$ so it is left and the list of cross-out forms in the generalized Sundaram sieve becomes:

$$ 6ij + i + j \\ 6ij + i + 5j \\ 6ij + 5i + 5j \\ i,j \geq 1 $$

Thus, that is the Sundaram sieve method genralized to a multiple of $6$ instead of just $2$. You could probably do similarly, with other multiples.

Question. Can we remove even more unecessary cross-out forms so that there would be less than $3$?

Answer 1

No, you can't remove more, $3$ is the minimum. In fact, there are $4$, but one of them is superfluous because of symmetry. Before getting to the point, a first observation: the $6xy+x+y$ form you mention at the top of your question is equivalent to, and better known as $6xy \pm x \pm y$, and if you take your $3$ final forms modulo 6, you'll realize they're in fact identical to that form.

And there's a good reason for that: the proof that positive integers $n$ that can't be written in the $6xy \pm x \pm y$ form are bijectively related to twin primes is mostly just a simple change of coordinates for an elliptic curve equation. The derivation should prove a fun exercise. And essentially, you'll arrive at $$ 6xy \pm x \pm y = n \quad \text{has no solutions for $(x,y) \in \mathbb{Z}^2$.}\\ \big\Updownarrow \\ X^2-Y^2=6n \pm 1 \quad \text{both have only one solution.} $$ So yeah, I'm afraid you do need all $3$ of them.

Answer 2

First note that $6(0)+5$ is prime, so you would need it in $6(1)-1$ form for the last inequality to hold. But yes, the Sieve of Sundaram has generalizations to all multiplicative groups. It can be reduced to $6ij\pm i\pm j$ if you need to save space but might want to save that for mod 30 with 36 cases.

In Addition, it can be generalized to find semiprimes with lower bounds on factors etc. Just let $i,j$ not be also representable using some other values. Then it's a product of primes. If exactly 1 is of the form of a semiprime has 3 prime factors ... But 3 cases exist with 2 possibilities each.