On the difference between consecutive primes

Question

Let $(p_n)$ be the sequence of prime numbers and $g_n = p_{n+1} - p_n$

Question: Is it known that $g_n \le n$?

Remark: it's known that $g_n < p_n^{\theta}$ with $\theta = 0.525$ for $n$ sufficiently large (see here), and that $p_n < n(\ln n + \ln\ln n )$ for $n \ge 6$ (see here).
It follows that $g_n < (n(\ln n + \ln\ln n))^{\theta}$ for $n$ sufficiently large.
But $(n(\ln n + \ln\ln n))^{\theta}<n$, for $n \ge 2$.
Conclusion: it's known that $g_n \le n$ for $n$ sufficiently large. Is it known for all $n$?

Note that $2=g_2<2$ is not correct. Also $4=g_4<4$ does not hold. — Dietrich Burde, Oct 19 '14 at 18:07
Thus a suitable conjecture is that $g_n\le n$ for all $n$, while it is known (from the results quoted by the OP) only to hold for almost all $n$. — Hagen von Eitzen, Oct 19 '14 at 18:12
Checking for the set of primes under $10^8$, the inequality holds. BTW, only for n=1,2,4 we get $g_n=n$. — Galc127, Oct 19 '14 at 18:35
What a wonderful conjecture: simple, non trivial, obviously true and proven for almost every natural number! A great example of complications arising from adding and subtracting primes. — Lehs, Oct 20 '14 at 07:41
I'm probably missing something but dont all twin primes refute this claim? — KBusc, Oct 20 '14 at 13:45
@KBusc. For twin primes, $g_n=2<n$ is fine for all $n>4$, right ? — Dietrich Burde, Oct 20 '14 at 13:49
ahh had gn and n switched around in my head. thanks for clearing that up — KBusc, Oct 20 '14 at 13:50

score 5 · Accepted Answer · answered Oct 21 '14 at 01:48

This should be really easy to answer using Pierre Dusart's explicit estimates on prime-related functions (and probably the older Rosser-Schoenfeld inequalities). For instance, Proposition 6.8 in "Estimates of some functions on primes without R.H." states that for $x \ge 396738$ there is always a prime in the interval $(x, x + x/(25\ln^2 x)]$.

Since that gap is significantly smaller than $x/\ln x$ and quite explicit, this is certainly less than $\pi(x)$, establishing the inequality for large $n$. Combined with Galc127's comment about verifying small $n$, that should cover all cases.

From the inequality $n(\ln n + \ln\ln n - 1)<p_n < n(\ln n + \ln\ln n)$ (coming from a previous paper of Pierre Dusart) we deduce that $g_n < n+2\ln(n)$. But you're right, this more recent paper gives a proof for $g_n \le n$. — Sebastien Palcoux, Oct 21 '14 at 20:21

Dietrich Burde · Answer 2 · 2014-10-21T09:04:01.413

2

I am not an expert, but searching the literature gave me the impression that the result is not explicitly stated. There are several asymptotic results of the kind $g_n \ll p_n^{\epsilon}$, e.g., the result by Baker, Harman and Pintz from $2001$, which showed that $g_n ≪ p_n^{\frac{21}{40}}$. Assuming RH one can improve this further.
The question is whether or not any of these results can be made explicit, so that we can find an explicit constant, say $c<10^8$ with $g_n<p_n^{\epsilon}$ for all $n\ge c$, and some reasonable small fixed $\epsilon$ with $\frac{1}{2}< \epsilon <1$. If it is possible, we are done, and we have that $g_n<n$ for all $n>4$.

Edit: The explicit estimates of Dusart (see the answer of Erick Wong) should give us the result. There is also a reference to a paper of Schoenfeld where he shows that $g_n\le 652$ for all $p_n\le 2.685\cdot 10^{12}$.

edited Oct 21 '14 at 09:04

answered Oct 19 '14 at 18:21

Dietrich Burde

140,055

The current record is $g_n \le 1476$ for all $p_n \le 1425172824437699411 \simeq 1.425 \cdot 10^{18}$ (see here), with $\pi(10^{18}) \simeq 2.474 \cdot 10^{16} $ – Sebastien Palcoux Oct 21 '14 at 11:44
Very good. So now you can try to cover all cases. It should be really easy, as Erick Wong says. However, I have not seen this somewhere yet. – Dietrich Burde Oct 21 '14 at 11:51
2

$g_n \le \frac{p_n}{25ln^2(p_n)} \le \frac{n(ln(n) + ln(ln(n)))}{25ln^2(n(ln(n) + ln(ln(n))))} \le n$, for $p_n>396738$ (i.e. $n>33608$). It follows that $\forall n \ge 1, \ g_n \le n$, by the comment of Galc127. – Sebastien Palcoux Oct 21 '14 at 12:07
1

Should the log square in the denominator be n(\ln n + \ln\ln n - 1)? Small note: Because $n$ is on both side and a factor of the numerator we have $\frac{(ln(n) + ln(ln(n)))}{25ln^2(n(ln(n) + ln(ln(n))))} \le 1$ or with my concern $\frac{(ln(n) + ln(ln(n)))}{25ln^2(n(ln(n) + ln(ln(n)) -1))} \le 1$. – John Nicholson Apr 01 '15 at 19:15

score 1 · Answer 3 · answered Jul 14 '16 at 23:50

I was thinking about this problem, the other day, and came to this conclusion ...

Let's define the sequence $\epsilon_n$ such that $$p_{n+1}-p_n=p_n^{\epsilon_n}$$ According to Bertrand's postulate $$p_{n+1}-p_n=p_n^{\epsilon_n}<p_n$$ thus $$0<{\epsilon_n} <1$$ So that $$\frac{p_{n+1}-p_n}{n}=\frac{p_n^{\epsilon_n}}{n}=\frac{1}{p_n^{1-\epsilon_n}}\cdot \frac{p_n}{n\ln{n}}\cdot \ln{n}<(1+\epsilon)\cdot \frac{\ln{n}}{p_n^{1-\epsilon_n}}<(1+\epsilon)\cdot \frac{\ln{n}}{n^{1-\epsilon_n}}$$ Just because $p_n>n$ and $$\lim_{n \to \infty }\frac{p_n}{n\ln{n}}=1$$ In fact $\epsilon_n< \theta$ which means $1-\epsilon_n > 1- \theta>0$ (from some n, obviously) thus $$0<\frac{p_{n+1}-p_n}{n}<(1+\epsilon)\cdot \frac{\ln{n}}{n^{1-\theta}}\underset{n \to \infty }{\rightarrow}0$$

On the difference between consecutive primes

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