So as I understand it, the current state of the twin primes problem is that an unconditional proof of infinitely many pairs of primes separated by 246 exists, and I think a conditional proof of pairs separated by 6. I think this also can be said for any fixed gap larger than 246.
My understanding was also that if you examine the distribution of primes using the subset of some $ax+b$ to filter the integers, you still end up with properties that for the most part approach identical to what you see with the full set of integers.
For instance, $257x+k$ (for at least one $k\leq 256$) should have infinitely many pairs of primes separated by $2$, yes? I understand that's not proof of anything, but isn't it overwhelmingly strong evidence in favor of twin primes being true? (Though yeah, I realize we have no shortage of strong circumstantial evidence.)
An example of what I'm referring to: a virtual twin prime pair for $257x+1$ is found at $\{74017,74531\}$. So, if you run $x$ over $\mathbb N$, I mean primes separated by a difference of 2 in that $x$, or in absolute terms, prime pairs which are $\equiv 1 \pmod{257}$ with a difference of 514. It need not be that specific example, of course, but presumably it's not hard to show that some suitable residue class would have this property.
$$a=257x+k\land k=2j+1\implies x=2m\lor 2\mid a$$ $$a=257x+k\land k=2j\implies x=2m+1\lor 2\mid a$$ $$a=257x+k\land k=3j+2\implies x\neq 3m+2\lor 3\mid a$$ $$a=257x+k\land k=3j+1\implies x\neq 3m+1\lor 3\mid a$$ $$a=257x+k\land k=3j\implies x\neq 3m\lor 3\mid a$$ $$a=257x+k\land k=5j+4\implies x\neq 5m+3\lor 5\mid a$$ $$a=257x+k\land k=5j+3\implies x\neq 5m+1\lor 5\mid a$$ $$a=257x+k\land k=5j+2\implies x\neq 5m+4\lor 5\mid a$$ $$a=257x+k\land k=5j+1\implies x\neq 5m+2\lor 5\mid a$$ $$a=257x+k\land k=5j\implies x\neq 5m\lor 5\mid a$$ $$a=257x+k\land k=7j+6\implies x\neq 7m+3\lor 7\mid a$$ $$a=257x+k\land k=7j+5\implies x\neq 7m+6\lor 7\mid a$$ $$a=257x+k\land k=7j+4\implies x\neq 7m+2\lor 7\mid a$$ $$a=257x+k\land k=7j+3\implies x\neq 7m+5\lor 7\mid a$$ $$a=257x+k\land k=7j+2\implies x\neq 7m+1\lor 7\mid a$$ $$a=257x+k\land k=7j+1\implies x\neq 7m+4\lor 7\mid a$$ $$a=257x+k\land k=7j\implies x\neq 7m\lor 7\mid a$$
Forgetting for a moment that $j$ and $m$ aren't indexed (as they should technically be). You can use set difference to sieve massive amounts of $x$ values potentially. $k=210j+1$ ( so 1 or 211 in your range) eliminates all but 48 of the remainders mod 210 for $x$ values. Of those only 30 come in pairs. Here's the results
[3, 5, 9, 15, 21, 23, 29, 33, 35, 41, 45, 51, 59, 63, 65, 69, 71, 75, 83, 89, 93, 99, 101, 105, 111, 113, 119, 125, 129, 131, 135, 141, 143, 149, 153, 155, 159, 161, 171, 173, 183, 185, 189, 191, 195, 201, 203, 209]only another 104 to check if you're smart because of symmetry. I'll leave that up to you and CRT like methods.