Given a natural $n \ge 2$, we can construct a set of all powers of two from $2^n$ to $2^{4n}$:
$$\{2^n, 2^{n+1}, 2^{n+2}, \dots, 2^{4n}\}$$
How close does one of these numbers come to a prime in the worst case?
AN EXAMPLE
For example, if $n=2$, the set is:
$$\{2^2, 2^3, 2^4, 2^5, 2^6, 2^7, 2^8\}$$
$2^2=4$ is $1$ away from 5, and we can't get any closer for any power of two, so the answer is $1$.
WHAT I'M TRYING TO FIND
I'm looking for an upper bounds for the distance we can get from elements of the set (any set that fits the description above) to a prime. In other words, for a set that fits the description, and in a really bad scenario, where all the primes are far from the set, how close will they be to the set?
Opperman's conjecture would imply that there is at least one prime between $2^n$ and $2^n + 2^{n/2}$ [EDIT] if $n$ is even.