Dickson's conjecture, in simple terms, says that for any choice of $a_1,b_1,a_2,b_2,...,a_k,b_k\in\Bbb N$ we have, for infinitely many $n\in\Bbb N$, that all of $a_1+nb_1,...,a_k+nb_k$ are prime, unless there is a trivial condition why it's not so. What I'm interested in is a slight extension of this theorem, which namely also has some constraints saying which numbers are supposed to be composite. For example, I might be interested in sexy primes not forming prime triplets, so we would have constraints $n,n+6$ prime and $n+2,n+4$ composite.
My question is: is this conjecture even stronger than Dickson's conjecture? Answer isn't obviously "yes", because we may be able to encode some information using the form of primes, so, for example above, if we take into Dickson's conjecture $30n+1,30n+7$ we get that it's a sexy prime pair, and inbetween numbers, $30n+3,30n+5$ are composite (for $n>1$).
Same can be applied to Bunyakowsky's conjecture - do compositeness contraints strengthen the conjecture?
Thanks in advance.
This conjecture is equivalent to Dickson's (and Bunyakowsky's, respectively), because you can transform your instance into another which forces the compositeness you want.
If you want $n+2, n+4$ to be composite, just pick some big primes (let's do 101 and 103) and constrain $n+2$ to be divisible by 101 and $n+3$ to be divisible by 103. Then use the CRT to find $n$ such that $n\equiv-2\pmod{101}$ and $n\equiv-4\pmod{103}$, which in this case is $n\equiv99\pmod{10403}$. So your revised problem is to find $n$ such that $10403n+99$ and $10403n+105$ are prime, which guarantees that $10403n+101$ and $10403n+103$ are composite.
It's not hard to show that you can pick primes big enough that they won't interfere with anything else, basically just make them larger than any of the numbers appearing elsewhere in the problem. Of course if you want to optimize you can choose them smaller.
Edit: Note that this approach does not work directly with the density version, first formulated (as far as I know) by Bateman, Horn, & Stemmler. I think that with a bit more work you can show that they appear with the same density as without the compositeness condition (!). You'll need to use a limiting argument which is somewhat delicate.