We are all familiar with the Mersenne primes $$M_n = 2^n-1$$ and we indeed know that there are some $M_n$ that are prime. However, it is still open whether there are infinitly many $M_n$ that are prime. Now consider $$ F_n(k) = 2^n - k $$ with $k \geq 3$ and $k$ odd. I would like to know whether one can answer following questions for some $k$:
(1) Is $F_n(k)$ prime for some $n$?
(2) Is $F_n(k)$ prime for infinitely many $n$?
(3) (If not) For how many $n$ is $F_n(k)$ prime?
The first prime values of $F_n(3) = 2^n - 3$ are given in A050415 and the first prime values of $F_n(5) = 2^n - 5$ are given in A156560.
My hope is that it is easier to show (1), (2) and (3) for the form $2^n-k$ instead of $2^n-1$. In case someone is aware of papers, etc. concerning this 'generalization' I would be glad about the reference.
Edit: Another example: For $k=5$ the first prime value is $F_{39}(5)= 549755813881$.
Here's a partial answer (an answer to (1) and more of a comment about (2) and (3)). A lot of this can be found in this OEIS entry.
1) Not necessarily. If $k=509203$ (a Riesel Number), all of the terms are composite.
2) This is a hard problem, and I would be very surprised if anything of this form could be proven (problems of this form tend to be hard: as far as I know, close to the best known result like this is Dirichlet's theorem on arithmetic progressions, and it is not even known whether there exists infinitely many primes in the sequence $\{n^2+1\}$).
3) Again, this is hard in the general case. Due to the existence of Riesel numbers and their covering sets, I wouldn't be surprised if there exists one sufficiently close to a power of $2$ so that $2^n-k$ is indeed a prime, but no other terms are (or something like that). However, I don't think this problem is anything close to tractable in the general case.
Lemma: If there exists a set $P$ of primes such that, for all $n$, there exists a prime $p\in P$ such that $k2^n\equiv 1\bmod p$, then, for all $n$, there exists a prime $p\in P$ such that $2^n\equiv k\bmod p$.
Proof sketch:
Let the smallest $p\in P$ such that $k2^n\equiv 1\bmod p$ for a given $n$ be called $f(n)$, and let the smallest $p\in P$ such that $2^n\equiv k\bmod p$ be $g(n)$.
Step 1: Prove that $f(n)$ is periodic.
Step 2: Prove that one can "invert" the period of $f$ to obtain $g$.