For prime numbers $p$, let $f(p) = \min\{k: p+k \text{ is prime}\}$.
(BTW, is there a standard notation for this function?)
For $p=113$ we have $f(p) = f(113) = 14$, i.e. the next prime number above $113$ is fully $14$ units above it.
The prime $113$ is both
- The smallest prime $p$ for which $f(p)$ is so big; and
- The largest prime $p$ for which $f(p)/p$ is so big.
(I have seen the second fact somewhat credibly stated with an argument, but I don't remember the argument.)
The same can be said of $p=2$ and of $p=3$, if I surmise correctly.
So consider the sets \begin{align} A & = \left\{ p : \text{$p$ is the smallest prime for which $f(p)$ is as big as it is.}\right\} \\[10pt] B & = \left\{p : \text{$p$ is the largest prime for which $f(p)/p$ is as big as it is.} \right\}. \end{align} Are there any interesting results about these? (As opposed to results about just one of them?)
At one extreme one might have $A=B$; at the opposite extreme $A\cap B$ is finite, and maybe even a very small finite number.