# Why is “jump” between two primes (almost) always prime or 1 up to 1000?

While looking at prime numbers between 1 to 1000 I noticed that the number of non-primes between prime numbers are almost always also prime or 1. In other words, if we take the prime gap, $$g_n=p_{n+1}-p_n,$$ for prime numbers larger than 2, and subtract 1 from it, we almost always get a prime number or 1. In the cases (between 1 to 1000) where $$g_n-1$$ is not equal to a prime number or 1 it is instead equal to 9. Below is a figure I made to better explain:

What is the reason that the "jump length" (i.e. numbers in red or blue above) between primes are either 1, 9 or also prime when looking at 1 to 1000?

• Because the gap between consecutive primes is always an even number, except for $3-2=1$ – rtybase Sep 25 '18 at 21:21
• It does, $p_{282}=1831, p_{283}=1847$ and then again almost immediately, $p_{295}=1933,p_{296}=1949$. – lulu Sep 25 '18 at 21:34
• General note: it's tempting, but risky, to read too much into numerical computations for such small numbers. In this case, there are many small primes, so any random-y list of small odd numbers is going to contain lots of primes. To study anything like this, I'd want to look at much larger primes. – lulu Sep 25 '18 at 21:38
• You can assume Cramer's conjecture: $g_n \approx O\left((\log{p_n})^2\right)$ and draw one or two conclusions, like $\left((\log(1000))^2\right)\approx 48$, consider only even gaps, that's $24$ and there are $14$ primes (or "jumps") between $1$ and $48$ (excluding $2$), thus the chance is $\frac{14}{24}=\frac{7}{12}>\frac{1}{2}$. – rtybase Sep 25 '18 at 21:58
• Absolutely agree with @lulu on this one. "Law of small numbers." A cursory look at oeis.org/A046933 seems to confirm your hypothesis. But look at the "b-file": it's still mostly primes, but 9, 15, 25 make more appearances the further along you go. – Robert Soupe Sep 26 '18 at 4:13

This phenomenon was studied by Odlyzko, Rubinstein and Wolf (with whom I have written a paper on primes).

Definition: An integer $$d$$ is called a jumping champion for a given $$x$$ if d is the most common gap between consecutive primes up to $$x$$.

They postulated that most common gap between primes or the jumping champions are the product of prime numbers i.e. the first jumping champion is $$2$$. The after some time, $$2*3 = 6$$ takes over as the next jumping champion. At around $$1.74*10^{35}$$, the number $$2*3*5 = 30$$ takes over as the next jumping champion. At around $$10^{425}$$, the number $$2*3*5*7 = 210$$ takes over as the next jumping champion and so on.

This means that per your definitions, if we subtract $$1$$ from the gap between primes then the most common results would be $$1$$ initially. This would be overtaken by $$5$$ which would in turn be overtaken by $$29$$ and this would be overtaken by $$209$$ as you go higher up the number line.

So it is a co-incidence that $$5, 29$$ are prime but $$209 = 11*19$$ is not. This means if you looked large enough to reach the range of $$209$$, you would actually have the opposite conclusion based on experimental evidence that gap between prime minus $$1$$ is often a non prime.

Equivalent question: With the above arguments, your question is equivalent to

How often is the primorial number $$2.3.5.7\cdots p - 1$$ a prime?

Reference:

1. Odlyzko, A.; Rubinstein, M.; and Wolf, M. "Jumping Champions." Experiment. Math. 8, 107-118, 1999.
• Nice answer! Do you know whether any result of this form (classifying jumping champions) has been proven, or if it's all just conjecture so far? – Carl Schildkraut Sep 26 '18 at 6:47
• As with most problems with primes, we have way more conjectures on jumping champions than theorems. However you can find a few theorems on them by Goldston in this paper. arxiv.org/pdf/1102.4879.pdf – Nilotpal Kanti Sinha Sep 26 '18 at 6:50