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Statement

With $Y(x)$ as in Definition 1 (the largest $y$ such that residue classes $a_p\bmod p$, one for each prime $p\le x$, cover $[y]=\{1,\ldots,\lfloor y\rfloor\}$) and $P(x)$ the product of the primes less than or equal to $x$:

Lemma 1.1 (p. 3). $G(P(x)+Y(x)+x)\ge Y(x)$, where $G(X)$ is the largest gap between consecutive primes less than $X$.

Display (1.3) (p. 4). If $n$ is a positive integer, Jacobsthal's function $j(n)$ is "the maximal gap between integers coprime to $n$. In particular $j(P(x))$ is the maximal gap between numbers free of prime factors $\le x$, or equivalently $1$ plus the longest string of consecutive integers, each divisible by some prime $p\le x$." The construction in the proof of Lemma 1.1 "in fact proves that"

$$Y(x)=j(P(x))-1.$$

Source. K. Ford, B. Green, S. Konyagin, J. Maynard and T. Tao, Long gaps between primes, arXiv:1412.5029v3 (14 July 2016, 40 pp.); Lemma 1.1 with its proof on p. 3 and (1.3) on p. 4, read on the page images and in the text layer. Published in J. Amer. Math. Soc. 31 (2018), no. 1, 65--105, DOI 10.1090/jams/876; the journal text was not compared.

Read depth. Claims checked for both statements; the proof of Lemma 1.1 was read in full (below) and its steps were followed here. Display (1.3) carries no separate proof in the paper; the two inequalities it asserts were checked here from the definitions: a covering of $[y]$ yields, through the $m$ of the proof, the $y$ consecutive integers $m+1,\ldots,m+y$ each divisible by a prime $p\le x$, so $j(P(x))\ge y+1$; conversely a run of $j(P(x))-1$ consecutive integers $m+1,\ldots,m+j-1$ each sharing a factor with $P(x)$ gives the covering $a_p:=-m\bmod p$ of $[j-1]$, so $Y(x)\ge j(P(x))-1$. This check is an authored remark, not the paper's text.

Proof pointer

The paper's proof (p. 3) is a Chinese remainder construction. Fix a covering of $[y]$, $y=Y(x)$, by one class $a_p\bmod p$ for each prime $p\le x$, and take $m\in(x,x+P(x)]$ with $m\equiv-a_p\pmod p$ for every such $p$. Each $t\in[y]$ lies in some class $a_p\bmod p$, so $p$ divides $m+t$, and $m+t>x\ge p$ makes $m+t$ composite. The $y$ consecutive integers $m+1,\ldots,m+y$ are therefore all composite, and as $m+y\le P(x)+Y(x)+x$ this gives $G(P(x)+Y(x)+x)\ge Y(x)$.

Dependencies

The Chinese remainder theorem only.

Bears on

• Problem 687: the transfer from $Y(x)$ to prime gaps and the identity with Jacobsthal's function that the site's commentary alludes to ("associated with Jacobsthal").
• Problem 970: (1.3) is the identity through which the paper's bound (1.2) becomes a lower bound for the Jacobsthal function of a number with $\pi(x)$ prime factors.
• Problem 929: the same Chinese remainder construction shows that $S(k)$ is the least $x$ with $Y(x)\ge k$ (made explicit on the problem page).