# A question about distribution of prime numbers

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.

• en.wikipedia.org/wiki/Prime_gap and for first occurrence of each gap oeis.org/A000230 – Will Jagy Jun 15 '16 at 18:26
• for part B, the maximal gaps are tiny, as high as this has been computed (up to $4 \cdot10^{18},$ with prime $p$ and next prime $p+g,$ for $p \geq 11,$ we find $q < \log^2 p.$ No proof of this expected for the next 1000 years or so. What can be proved is quite modest. math.stackexchange.com/questions/1815044/… – Will Jagy Jun 15 '16 at 18:40
• always something; gap should say $g$ in both places, not $q$ – Will Jagy Jun 15 '16 at 18:46