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.

  • 2
    $\begingroup$ en.wikipedia.org/wiki/Prime_gap and for first occurrence of each gap oeis.org/A000230 $\endgroup$ – Will Jagy Jun 15 '16 at 18:26
  • 2
    $\begingroup$ 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/… $\endgroup$ – Will Jagy Jun 15 '16 at 18:40
  • $\begingroup$ always something; gap should say $g$ in both places, not $q$ $\endgroup$ – Will Jagy Jun 15 '16 at 18:46

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