## Large gaps between consecutive prime numbers.(English)Zbl 1338.11083

Let $$G(X)$$ be the length of the largest interval between primes up to $$X$$. Then it is shown that $\frac{G(X)}{(\log X)(\log\log X)(\log\log\log X)^{-2}(\log\log\log\log X)} \rightarrow\infty$ as $$X\rightarrow\infty$$. It was proved by R. A. Rankin [J. Lond. Math. Soc. 13, 242–247 (1938; Zbl 0019.39403)] that the above expression has a positive lower bound, and for many years it has been a well-known open problem to do better.
An alternative proof of the above result has simultaneously been obtained by J. Maynard [Ann. Math. (2) 183, No. 3, 915–933 (2016; Zbl 1353.11099)], by a different method, and the authors have joined forces to prove a yet stronger result, in work to appear.
The argument shows something slightly stronger. Let $$Y(x)$$ be maximal such that for every prime $$p\leq x$$ one can choose a residue class $$a_p$$ so that every integer $$n\in[1,Y(x)]$$ lies in at least one residue class $$a_p\pmod {p}$$. Then one has $\frac{Y(x)}{x(\log x)(\log\log x)^{-2}(\log\log\log x)} \rightarrow\infty$ as $$x\rightarrow\infty$$.
As with previous work on this problem, one chooses $$a_p=0$$ for a middle range of primes, $$z<p\leq cx$$, say (for some small positive constant $$c$$). For the primes $$p\leq z$$ one makes a random choice for $$a_p$$, which leaves a set $$V$$, say, consisting of $$O(Y(x)(\log\log x)(\log x)^{-1}(\log z)^{-1})$$ unsieved elements. These have to be handled using primes from the remaining range $$cx<p\leq x$$. Certainly one can remove at least one element of $$V$$ for each prime $$p$$, and J. Pintz [J. Number Theory 63, No. 2, 286–301, Art. No. NT972081 (1997; Zbl 0870.11056)], who held the previous record for this problem, showed that one can arrange to remove two elements in general.
The present paper uses results from the work of B. Green and T. Tao [Ann. Math. (2) 171, No. 3, 1753–1850 (2010; Zbl 1242.11071); Ann. Math. (2) 175, No. 2, 465–540 (2012; Zbl 1251.37012); erratum ibid. 179, No. 3, 1175–1183 (2014); B. Green et al., Ann. Math. (2) 176, No. 2, 1231–1372 (2012; Zbl 1282.11007)], concerning progressions $q, q+r!p, q+2r!p,\dots, q+(r-1)r!p$ of primes. The choice $$a_p=q$$ allows one to eliminate $$r$$ primes from $$V$$, but one has to show that many of these progressions are (nearly) disjoint. It is at this stage that one makes use of the fact that the $$a_p$$ for small primes $$p$$ were chosen randomly.

### MSC:

 11N05 Distribution of primes 11N13 Primes in congruence classes 11N36 Applications of sieve methods 11B30 Arithmetic combinatorics; higher degree uniformity

### Keywords:

primes; long gaps; sieve; Rankin; progression
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### References:

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