Maynard 2016 large gaps between primes

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proposition_5: Maynard's key proposition: for each even m below U z^{-1} (log_2 x)^{-2}, the primes of the leftover set R_m can all be put in residue classes, one for each prime of any interval in [x/2, x] of length at least δ|R_m| log x, once x is large in terms of δ and C_U.

theorem_1: Maynard's theorem that (p_{n+1} - p_n) divided by (log p_n)(log_2 p_n)(log_4 p_n)(log_3 p_n)^{-2} has infinite limit superior, so Rankin's lower bound for the largest prime gap holds with every constant; it answers Problem 4 yes.

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Maynard, James, Large gaps between primes. Ann. of Math. (2) 183 (2016), no. 3, 915-933, DOI 10.4007/annals.2016.183.3.3. arXiv:1408.5110. The copy read for this card is the arXiv:1408.5110v2 preprint (28 October 2019), not the Annals version; the theorem, lemma and equation labels and the page numbers below are read from that preprint. arXiv:1408.5110 is Maynard's single-author paper, not the five-author Ford-Green-Konyagin-Maynard-Tao paper. The arXiv record names arXiv's non-exclusive distribution license (arXiv:1408.5110), every other right reserved.

Theorem 1 (p. 1) states that limsup_n (p_{n+1} - p_n) / ((log p_n)(log_2 p_n)(log_4 p_n)(log_3 p_n)^{-2}) = infinity; the argument shows that Rankin's 1938 lower bound (1.1) for G(X) = sup_{p_n <= X}(p_{n+1} - p_n) holds with the constant c arbitrarily large, instead of the best previous value c = 2e^gamma due to Pintz. The proof follows the Erdos-Rankin construction, choosing residue classes a_p = 1 for primes 1 < p <= y and a_p = 0 for y < p <= z with y = exp((1-eps) log x log_3 x / log_2 x), z = x / log_2 x and U = C_U x log y / log_2 x (2.1), and modifies only the final stage. The survivors (2.3) are the union of the sets R and R' of (2.4): R is the set of products mp <= U with p > z prime, m y-smooth and (mp-1, P_y) = 1, and R' the set of y-smooth m <= U with (m-1, P_y) = 1. For each even m below U z^{-1} (log_2 x)^{-2}, the primes of the set R_m of (2.5) are covered by classes for the primes of a short interval in [x/2, x] (Proposition 5, p. 3), chosen at random from weights taken from the recent work on small gaps between primes, which is what allows C_U to be taken arbitrarily large. By Lemmas 2, 3 and 4 this covers all but o(x/log x) elements of R and R' using primes in [x/2, x], and the elements left over are covered one at a time by primes in [z, x/2] (p. 4). Lemma 2 (p. 2) bounds |R'| << x/(log x)^{1+eps}, quoting Maier and Pomerance. The paper notes that Ford, Green, Konyagin and Tao independently obtained the same result by a different route through linear equations in the primes, and a remark states that the method also yields a quantitative improvement to Rankin's bound, deferred to later work.

Source: https://arxiv.org/abs/1408.5110.

Read status. Claims checked: Theorem 1, the remark, the construction of Section 2, Lemma 2 and Proposition 5 with its equivalent form (pp. 1--4) were read clause by clause on the page images, and the proof of Theorem 1 from 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.

Bears on. #4: Theorem 1 answers the question yes, since log p_n ~ log n makes the problem's scale in n asymptotic to the theorem's scale in p_n; Proposition 5 is the step that lets the constant grow. #1137: Theorem 1 bounds the largest single gap from below, the quantity squared in the problem's denominator; it says nothing about the products d_n d_{n-1} of two consecutive gaps and does not decide the question. #689: the problem's page cites the paper as one cited in its discussion thread; the paper's covering, one class for each prime p <= x, puts each integer of [1, U] in at least one chosen class, and it proves nothing about covering each integer twice, so it does not decide the question.

Results.

• Theorem 1 (p. 1): limsup_n (p_{n+1} - p_n)/((log p_n)(log_2 p_n)(log_4 p_n)(log_3 p_n)^{-2}) = infinity, so Rankin's bound holds with arbitrarily large constant.
• Proposition 5 (p. 3): for fixed delta > 0, even m < U z^{-1} (log_2 x)^{-2} and an interval I_m in [x/2, x] of length at least delta |R_m| log x, once x > x_0(delta, C_U) one residue class for each prime of I_m covers every prime of R_m.
• Lemma 2 (p. 2), quoted from Maier and Pomerance and given no page here: the set R' of y-smooth m <= U with (m-1, P_y) = 1 satisfies |R'| << x/(log x)^{1+eps}.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.