For $x, \lambda > 0$, define $$S_\lambda(x) := \#\{p_{n+1} \leq x : p_{n+1} - p_n \geq \lambda \log x\} ,$$ where $p_n$ is the $n$th prime number. It is known [1] that an uniform version of the Hardy-Littlewood prime k-tuples conjecture implies that for fixed $\lambda > 0$, $$S_\lambda(x) \sim e^{-\lambda} \frac{x}{\log x}$$ as $x \to +\infty$.

My question is: If we want only a lower bound of the form $$S_\lambda(x) \gg_\lambda \frac{x}{\log x}, \quad x > 2,$$ for every fixed $\lambda > 0$, has this been proved unconditionally?

Thank you for any reference or suggestion.

[1] Funkhouser,Goldston, Ledoan, Distribution of Large Gaps Between Primes, https://doi.org/10.1007/978-3-319-92777-0_3

  • $\begingroup$ Note that replacing formally $\lambda$ with $\log x$, the conjectured asymptotics provides a new heuristics for (a possibly slightly weaker version of) Cramer's conjecture. $\endgroup$ May 30 '19 at 22:23

The inequality $S_\lambda (x)\gg_{\lambda} \frac{x}{\log x}$ should follow from the case $m=0$ of the paper https://www.cambridge.org/core/journals/bulletin-of-the-australian-mathematical-society/article/positive-proportion-of-short-intervals-containing-a-prescribed-number-of-primes/BE20070F6B2864B2EFD65C268267BC5A

at least for small values of $\lambda$. More precisely, the result should be that there exists a large constant $C>0$ such that for any $\lambda<\frac{1}{C^4\log^2C}$, we have the desired lower bound for $S_{\lambda}(x)$.

Although the above cited paper could be more useful to analyze the case of general gaps between primes, for gaps of consecutive primes I would like to refer you to the following result https://www.ams.org/journals/tran/2010-362-05/S0002-9947-09-05009-0/

in which the requested inequality may be shown for values of $\lambda<\lambda_0$, for a certain $\lambda_0>1$ explicitly computed in that paper.

I doubt it has already been proven such lower bound unconditionally for any $\lambda>0$.

  • 4
    $\begingroup$ The authors of the second paper prove something slightly different; they show that for $\lambda < \lambda_0$, the interval $[x,x+\lambda \log x]$ is free of primes for a positive proportion of $x$. This almost implies that $p_{n+1}-p_n \geq \lambda \log p_n$ for a positive proportion of $n$, but one first has to eliminate the scenario that the prime gaps are all either less than $\lambda \log p_n$ or are really, really big (much bigger than $\log p_n$), in which case a positive proportion of $x$ might still be collapsed into a zero density set of $n$. $\endgroup$
    – Terry Tao
    May 30 '19 at 18:04
  • 2
    $\begingroup$ The only way I see how to eliminate this latter scenario is through unproven conjectures such as the two-point Hardy-Littlewood conjecture, which can be used with the second moment method to limit the proportion of time one spends in an extremely large prime gap. $\endgroup$
    – Terry Tao
    May 30 '19 at 18:06

I think I can establish the claim for all $\lambda < 1/4$. In principle one should be able to get up to $\lambda < 1.145\dots$ but as noted in other comments this would require upper bounds on the density of large intervals free of primes that do not appear to be known unconditionally.

For any $\lambda > 0$ and any $X \geq 1$, let $f_X(\lambda)$ be the proportion of $1 \leq x \leq X$ such that the interval $[x, x+\lambda \log X]$ is free of primes (one can replace $\log X$ here by $\log x$ if desired, it doesn't really change things). This is clearly a monotone decreasing function of $\lambda$ taking values in $[0,1]$. The problem is essentially equivalent to establishing a noticeable dip in the value of $f_X$. More precisely:

Proposition 1 Let $\lambda_0>0$. Then the following are equivalent:

  1. For all $\lambda < \lambda_0$, one has $S_\lambda(X) \gg_\lambda \frac{X}{\log X}$ for sufficiently large $X$.

  2. For all $\lambda < \lambda_0$, there exists $\Lambda>\lambda$ such that $f_X(\lambda) - f_X(\Lambda) \gg_\lambda 1$ for sufficiently large $X$.

Proof Suppose 1 holds and $\lambda < \lambda_0$. From the PNT and Markov's inequality one has $S_\Lambda(X) \ll \frac{1}{\Lambda} \frac{X}{\log X}$, thus for $\Lambda = O_\lambda(1)$ large enough one has $S_\lambda(X) - S_\Lambda(X) \gg_\lambda \frac{X}{\log X}$. Thus there are $\gg_\lambda \frac{X}{\log X}$ prime gaps in $[1,x]$ of size between $(\lambda-o(1)) \log X$ and $(\Lambda+o(1)) \log X$ (since $\log p_n = (1+o(1)) \log X$ for almost all primes $p_n \leq x$). This easily implies that $f_X(\lambda) - f_X(\Lambda) \gg_\lambda 1$ for sufficiently large $X$ (possibly after tweaking $\lambda$ and $\Lambda$ by $o(1)$ first).

Conversely, if 2 holds, then for a proportion $\gg_\lambda 1$ of $x \in [1,X]$, the interval $[x,x+\lambda \log X]$ is free of primes while the interval $[x,x+\Lambda \log X]$ has a prime. Let $A$ be large. The proportion of $x$ for which $[x,x+\Lambda \log X]$ has a prime but $[x-A\Lambda \log X, x]$ does not is $O(1/A)$, so by taking $A = O_\lambda(1)$ large enough, we see that for a proportion $\gg_\lambda 1$ of $x \in [1,X]$, the intervals $[x-A \Lambda \log X, x]$ and $[x+\lambda \log X,x+\Lambda \log X]$ contain primes but $[x,x+\lambda \log X]$ does not, thus $x$ lies in a prime gap of size between $\lambda \log X$ and $(A+1)\Lambda \log X$. From this one can establish $S_\lambda(X) \gg_\lambda \frac{X}{\log X}$ without difficulty. $\Box$

The claim now follows from this proposition and

Proposition 2 For any $\lambda > 0$, one has $$ 1 - \lambda+o(1) \leq f_X(\lambda) \leq 1 - \frac{\lambda}{1+4\lambda}+o(1)$$.

Proof Let $N_\lambda$ be the number of primes in $[x,x+\lambda \log X]$ where $x$ is drawn uniformly at random from $[1,X]$, so $f_X(\lambda)$ is the probability that $N_\lambda=0$. From the prime number theorem and the first moment method one has $$ {\bf E} N_\lambda = \lambda + o(1)$$ which by Markov's inequality gives that ${\bf P}(N_\lambda \geq 1) \leq \lambda + o(1)$, which gives the lower bound on $f_X(\lambda)$. For the upper bound we need to control the second moment $$ {\bf E} \binom{N_\lambda}{2}.$$ Gallagher's calculation assuming Hardy-Littlewood predicts this quantity to be $\lambda^2/2 + o(1)$ (and more generally $N_\lambda$ should be Poisson distributed with parameter $\lambda$). This is not known unconditionally, but if one uses the standard Selberg sieve estimate coming from Bombieri-Vinogradov (e.g. Theorem 7.16 of Opera del Cribro), which is an upper bound that is worse than Hardy-Littlewood by a factor of $4$, one gets an upper bound of $2\lambda^2 + o(1)$. In particular $$ {\bf E} N_\lambda^2 \leq \lambda + 4 \lambda^2 + o(1).$$ But from Cauchy-Schwarz one has $$ {\bf E} N_\lambda^2 \geq ({\bf E} N_\lambda)^2 / (1-f_X(\lambda))$$ and if one puts together all these inequalities one obtains the claim. $\Box$

If one could prove an upper bound on $f_X(\Lambda)$ that went to zero as $\Lambda \to \infty$ then we could use the above arguments to show that $S_\lambda(x) \gg_\lambda \frac{x}{\log x}$ for all $\lambda < 1$. The paper https://www.ams.org/journals/tran/2010-362-05/S0002-9947-09-05009-0/ linked to in the previous answer would extend this to $\lambda < 1.145...$ due to slightly improved lower bounds on $f_X(\lambda)$ for $\lambda$ near $1$.

One can probably improve the upper bound on $f_X(\lambda)$ a little bit by using the inequality $f_X(\lambda) \binom{-k}{2} \leq {\bf E} \binom{N_\lambda - k}{2}$ for any natural number $k$, then optimising in $k$, but I don't think this actually improves the $1/4$ threshold. The Elliott-Halberstam conjecture would let one replace $1/4$ with $1/2$ though. Meanwhile, the Maynard sieve should be able to slightly improve upon the Markov inequality lower bound for $f_X(\lambda)$ for any value of $\lambda>0$ and thus lead to modest improvement in the $1/4$ threshold (this may possibly already follow from the paper linked to above). On the other hand getting the upper bound for $f_X(\Lambda)$ below $1/2$ seems to require getting around the parity problem (in particular eliminating the scenario in which the integers split into long intervals, with the Liouville function on almost primes biased to be +1 on half of these long intervals (as measured by density), and biased to be -1 on the other half).

  • 1
    $\begingroup$ I would like to point out that the above upper bound estimate for $f_X(\lambda)$ corresponds to one of the main results in the unpublished Sardari's paper: arxiv.org/abs/1802.08524 (the technique to deduce such upper bound is basically the same, the only difference being that here the result is presented in $\lim_{X\longrightarrow \infty}$, whereas in Sardari's paper in $\liminf_{X\longrightarrow \infty}$). $\endgroup$ Jan 31 '20 at 16:38
  • $\begingroup$ Also, I would like to notice that the above reference (Theorem 7.16 of Opera de Cribro) only gives an upper bound that is worse than Hardy-Littlewood by a factor of 8. However, the Selberg sieve estimate coming from Bombieri-Vinogradov does give the factor of 4 mentioned above. $\endgroup$ Feb 4 '20 at 12:34

The following paper of Goldston, Pintz, and Yildirim seem to completely answer your question. See their Theorem 1.

D. Goldston, J. Pintz, C. Yildirim, Primes in tuples IV: Density of small gaps between consecutive primes, Acta Arithmetica 160 (2013), 37-53.

  • $\begingroup$ I checked the paper and it actually deals with the opposite problem, that is, $p_{n+1} - p_n \leq \lambda \log x$. My question is about $p_{n+1} - p_n \geq \lambda \log x$. $\endgroup$
    – Kello
    May 26 '19 at 15:56
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    $\begingroup$ @Kello then I don't understand your question. You want to ask for the behaviour of $S_{\lambda}(x)$ when $\lambda$ is small... but it is clear from just the prime number theorem that $S_1(x) \sim \pi(x)$, so for any $\lambda < 1$ the lower bound you want will hold. $\endgroup$ May 26 '19 at 16:11
  • $\begingroup$ How $S_1(x) \sim \pi(x)$ is clear from the prime number theorem? Actually, the conditional result of [1] implies that $S_1(x) \sim e^{-1} \pi(x)$. $\endgroup$
    – Kello
    May 26 '19 at 17:56
  • $\begingroup$ @Kello you are right that I was too cavalier to assert the asymptotic formula. However the prime number theorem pretty quickly gives that the average gap between consecutive primes $p, p^\prime$ is $\sim \log p$, so in particular, it immediately implies that a positive proportion of gaps must exceed this average. Thus it is clear that $S_1(x) \gg \pi(x)$. $\endgroup$ May 26 '19 at 18:00
  • 4
    $\begingroup$ "the average gap between consecutive primes $p,p′$ is ∼$logp$, so in particular, it immediately implies that a positive proportion of gaps must exceed this average". Really? For all I can tell, the PNT allows the primes to go in long blocks of consecutive integers with moderately big gaps in between. $\endgroup$
    – fedja
    May 26 '19 at 18:24

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