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There are no prime numbers between the two primes $113$ and $127$. That gap seems quite large by comparison to the sizes of the numbers in it. $$ \frac{\text{size of gap}}{\text{prime just below the gap}} = \frac{14}{113} = 0.12389\ldots $$ Is that the biggest that this particular statistic ever gets?

Is that the last time this particular statistic ever gets that big?

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Are there restrictions on how big the gap has to be? i.e. $3$ and $5$ gives $2/3=.666...$ – MITjanitor Mar 30 '13 at 22:01
@GitGud, even if he does, it goes to $0$. (Since $p_{k+1}/p_k \to 1$ as $k \to \infty$, the limit of $p_{k+1}/p_k - 1 \to 0$.) – George V. Williams Mar 30 '13 at 22:04
@MITjanitor biggest would probably gap between 2,3 which gives 1/2 = .5 – DanZimm Mar 30 '13 at 22:07
I can't speak to the answer, but for $n$ sufficiently large $g_n < p_n^\theta$ for some $\theta < 1$ (Hoheisel 1930). So the statistic in question is bounded above by $p_n^\theta / p_n$ which $ p_n^{\theta-1} $, which goes to $0$ as $n$ goes to $\infty$. This implies that there is a maximum. – George V. Williams Mar 30 '13 at 22:21
Assuming this page is reliable, the results in section 4 seem to indicate that it's the last time modulo a few that you can check by hand. See also this – Martin Mar 30 '13 at 22:42

Indeed, the answer is positive. We have:

$$R_n := \frac{\text{size of gap}}{\text{prime just below the gap}} = \frac {p_{n+1} - p_n} {p_n} = \frac {p_{n+1}} {p_n} - 1$$

Using some well known approximations (Rosser's theorem) on $p_n$, we have

$$p_{n+1} \le (n+1) \log(n+1) + (n+1) \log \log(n+1)$$ $$p_n \ge n \log n + n \log \log n - n$$


$$R_n + 1 \le \frac {n+1} n \frac {\log (n+1) + \log \log (n+1)} {\log n + \log \log n - 1}$$

RHS is a decreasing function, and so we have $R_n < 0.12389\ldots$ for all $n \ge 1296$. The remaining cases $31 \le n \le 1295$ may be checked manually.

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Rohrbach and Weis show that (as Martin notes in the comments):

$$ g_n < \frac{p_n}{13} \quad \quad n > 118$$

From this we have that your statistic is bounded by: $ \frac{1}{13} \approx 0.0769231 $ for $n > 118$. The 118th prime is $647$, and a quick numerical analysis verifies this.

Mathematica code to verify this (which should print nothing):

Do[If[Prime[k + 1]/Prime[k] - 1 >= 14/113, Print[k]], {k, 31, 119}]
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