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Statement

Notation (p. 1): $p_n$ is the $n$th prime, $\log_\nu$ is the $\nu$-fold iterated logarithm, and $G(X)=\sup_{p_n\le X}(p_{n+1}-p_n)$ is the largest gap between primes up to $X$. Rankin's bound, the paper's (1.1), is $G(X)\ge(c+o(1))(\log X)(\log_2X)(\log_4X)(\log_3X)^{-2}$ for $X$ sufficiently large; Rankin had $c=1/3$, and the best constant before this paper was Pintz's $c=2e^\gamma$.

Theorem 1 (p. 1).

$$\limsup_{n}\frac{p_{n+1}-p_n}{(\log p_n)(\log_2p_n)(\log_4p_n)(\log_3p_n)^{-2}}=\infty .$$

In words: for every constant $C>0$ there are infinitely many $n$ with $p_{n+1}-p_n>C(\log p_n)(\log_2p_n)(\log_4p_n)(\log_3p_n)^{-2}$.

The form the proof gives (abstract, pp. 2 and 4). The abstract states the result for every fixed $t$: there are consecutive primes below $x$ whose difference exceeds $t(1+o(1))(\log x)(\log\log x)(\log\log\log\log x) (\log\log\log x)^{-2}$. The reduction of Section 2 (p. 2) shows that if $[1,U]$ can be covered by classes $a_p\bmod p$ for the primes $p\le x$, with $x=(1-\epsilon)\log X$ and $U$ as in (2.1), then there is an interval in $[1,X]$ of length $(1-2\epsilon+o(1))C_U(\log X)(\log_2X)(\log_4X)(\log_3X)^{-2}$ containing no primes. The proof on p. 4 gives that covering for every fixed choice of $C_U$, so (1.1) holds with $c$ arbitrarily large. The paper notes (p. 1) that Ford, Green, Konyagin and Tao obtained the same result independently by a different method.

Remark (p. 1). The paper states that its method gives a quantitative improvement of (1.1), deferred to forthcoming work; no such bound is proved in the paper.

Source. J. Maynard, Large gaps between primes, Ann. of Math. (2) 183 (2016), no. 3, 915--933, doi:10.4007/annals.2016.183.3.3, read in the arXiv:1408.5110v2 preprint (28 October 2019) identified on the source card; the pages cited are the preprint's printed pages: Theorem 1 and the remark on p. 1, the reduction on p. 2, the proof of Theorem 1 from Proposition 5 on p. 4.

Read depth. Claims checked: the statement, the notation, the remark and the reduction of Section 2 were read clause by clause on the page images, and the proof of Theorem 1 assuming Proposition 5 (p. 4) was read through. The proof of Proposition 5 (pp. 4--17) was read for its structure but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pp. 1--4 and 17. The proof follows the Erdős--Rankin construction and changes only its last stage. If every integer in $[1,U]$ lies in a chosen class $a_p\bmod p$ for some prime $p\le x$, the Chinese remainder theorem gives a $U_0$ in $[x,x+\exp((1+o(1))x)]$ with $U_0+j$ composite for all $j\in[1,U]$; taking $x=(1-\epsilon)\log X$ gives the prime-free interval above, so it suffices to cover $[1,U]$ with $C_U$ arbitrarily large. With $y$, $z$, $U$ as in (2.1), the classes $a_p=1$ for $p\le y$ and $a_p=0$ for $y<p\le z$ leave the survivors $\mathcal R\cup\mathcal R'$ of (2.3)--(2.4): products $mp\le U$ with $p>z$ prime and $m$ $y$-smooth, and $y$-smooth $m\le U$, each with no common factor with $P_y$ after subtracting $1$. The set $\mathcal R'$ is small by Lemma 2 (p. 2), $|\mathcal R'|\ll x/(\log x)^{1+\epsilon}$. Splitting $\mathcal R$ by the even cofactor $m$ into the sets $\mathcal R_m$ of (2.5), Lemma 4 (p. 3) bounds $\sum\delta|\mathcal R_m|\log x$ over $m<Uz^{-1}(\log_2x)^{-2}$ by $O(\delta C_Ux)$, so for small $\delta$ disjoint intervals $\mathcal I_m\subseteq[x/2,x]$ of the length that Proposition 5 needs fit side by side, and that proposition covers each such $\mathcal R_m$ with the primes of $\mathcal I_m$. Lemmas 2, 3 and 4 leave $o(x/\log x)$ elements uncovered, and these are covered one at a time by primes in $[z,x/2]$ (p. 4). The proof of Proposition 5, completed at the top of p. 17, therefore completes the proof of Theorem 1.

Dependencies

Proposition 5 and Lemmas 2--4 of the same paper; Lemma 2 is quoted from Maier and Pomerance, Trans. Amer. Math. Soc. 322 (1990), Theorem 5.3 (the paper's reference [7]); Lemma 3 rests on a fundamental-lemma sieve and the Bombieri--Vinogradov theorem, citing Friedlander and Iwaniec, Opera de Cribro, Theorem 6.12 (reference [3]).

Bears on

• Problem 4: the problem asks whether for every $C>0$ infinitely many $n$ have $p_{n+1}-p_n>C\log n\log_2n\log_4n/(\log_3n)^2$. Since $\log p_n\sim\log n$, each iterated logarithm of $p_n$ is asymptotic to that of $n$, so Theorem 1 answers the question yes. The problem's claim page for this paper records the credit.
• Problem 1137: the problem asks whether $\max_{n<x}d_nd_{n-1}/(\max_{n<x}d_n)^2\to0$, with $d_n$ the $n$th prime gap. Theorem 1 is a lower bound for the largest single gap, the quantity squared in the denominator; it says nothing about products of two consecutive gaps and does not decide the question.