Large gaps in sets of primes and other sequences II. New bounds for large gaps between primes Random methods and weighted sieves

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1 Large gaps in sets of primes and other sequences II. New bounds for large gaps between primes Random methods and weighted sieves Kevin Ford University of Illinois at Urbana-Champaign May, 2018

2 Large gaps between primes Def: G(x) := max p n x (p n p n 1 ), p n is the n th prime. Theorem (F-Green-Konyagin-Maynard-Tao, 2018) G(x) log x log 2 x log 4 x. log 3 x Chains of gaps: G k (x) := max pn+k x min(p n+1 p n,..., p n+k p n+k 1 ) Theorem (F-Maynard-Tao, 2018+) For every k, G k (x) k log x log 2 x log 4 x log 3 x

3 Proving large gaps: Jacobsthal s function S T = {n Z : (n, Q T ) = 1}, Q T = p T p. Main goal: Find J(T ), the largest gap in S T. Lower bound (FGKMT, 2018). J(T ) T log T log 3 T log 2 T. Covering: J(T ) is the largest y so that there are a 2, a 3, a 5,... with {a p mod p : p T } [0, y]

4 Least prime in an arithmetic progression Let p(k, l) = min{p : p l (mod k)}, M(k) = max p(k, l). (l,k)=1 Upper bounds Linnik, M(k) k L. (Xylouris - L = 5.18). ERH: L = 2 + ε; Chowla conjecture: L = 1 + ε. Lower bounds Trivial: M(k) φ(k) log k. Prachar; Schinzel /62. For infinitely many k, M(k) φ(k) log k log 2 k log 4 k (log 3 k) 2. (1) Wagstaff (1978) - (1) holds for all prime k. Pomerance (1980) - (1) holds for almost all k, in fact all k with at most exp(log 2 k/ log 3 k) prime factors.

5 Least prime in an arithmetic progression, II Pomerance: M(k) φ(k) log k log 2 k log 4 k (log 3 k) 2 for almost all k. Lemma (Pomerance): Let j(m) be the maximal gap between numbers comprime to m. If 0 < m k/j(k) and (m, k) = 1 then M(k) > kj(m). Take m = p need a lower bound on j(m). p (1 δ) log k p k Corollary (FGKMT, 2018). If k has no prime factor log k, then M(k) φ(k) log k log 2 k log 4 k. (2) log 3 k Theorem (J. Li-K. Pratt-G. Shakan, 2017) Inequality (2) holds for all k with at most exp{(1/2 ε) log 2 k log 4 k log 3 k } prime factors.

6 Least Prime in an A.P. conjectures Conjecture (folklore): M(k) k log 2+ε k. Conjecture (Wagstaff, 1979): M(k) φ(k) log 2 k for most k Wagstaff s heuristic:given l < k(log k) 3, (l, k) = 1, the probability that l is prime is k/φ(k) log k. So P(l, l + k,..., l + m log k k all composite) Hence ( 1 k/φ(k) ) m log k log k e m/φ(k). P (M(k) mk log k) (1 e mk/φ(k)) φ(k) Threshhold value m φ(k) k log φ(k) exp{ φ(k)e mk/φ(k) }.

7 Least prime in AP: Refined conjectures Conjecture (Li-Pratt-Shakan, 2017) lim inf k M(k) φ(k) log 2 k = 1, M(k) lim sup k φ(k) log 2 k = Figure: Histogram for M(k)/φ(k) log(φ(k)) log k for k 10 6

8 Least prime in AP: Li-Pratt-Shakan conjecture Rough heuristic argument: coupon collectors problem p n - n-th prime, m k - a param, a Z/kZ E a - the event that p 1, p 2,..., p mk a (mod k) A k - the event {M(k) > p mk } = a E a. We have P(A k ) a P(E a) φ(k)e m/φ(k) If m = λφ(k) log φ(k), this is φ(k) 1 λ. (1) If λ 1, threshhold for being small. Justifies Wagstaff and lim inf. (2) When λ 2, threshhold for P(A k ) holding for infinitely many k (using Borel-Cantelli). Justifies lim sup.

9 New lower bounds on J(T ): outline 2 log 10 x z x y y = cx log x log 3 x log 2 x, z = x c log3 x log 2 x Want {a p mod p : p x} [0, y] 1 a p = 0 for p (z, x/4] [2, log 10 x]. Uncovered: z-smooth numbers and primes; 2 Random, uniform choice of a p, log 10 x < p z. 3 Strategic choice of a p, x/4 < p x/2 to cover many reminaing elements. 4 (trivial) Use single a p for each x/2 < p x to cover each remaining uncovered element.

10 Stage 2: random, uniform choice of a p Q 1 - the set of uncovered elements after stage 1 (mainly primes). Q 2 (a) := the set of uncovered elements after stage 2 = Q 1 \ (a p mod p), p P where P is the set of primes in (log 10 x, z]. Lemma w.h.p., Q 2 (a) σπ(y), σ := p P (1 1/p) Proof. Recall Q 1 π(y). We calculate 1st, 2nd moments: E Q 2 (a) = P(n Q 2 (a)) n Q 1 = P(n a p (mod p)) = σ Q 1. n Q 1 p P

11 The Lemma follows from the 1st, 2nd moment bounds plus Chebyshev s inequality. Lemma w.h.p., Q 2 (a) σπ(y), σ := p P (1 1/p) Proof (continued). For the 2nd moment, E Q 2 (a) 2 = P(n 1, n 2 Q 2 (a)) n 1,n 2 Q 1 = E Q 2 (a) + n 1,n 2 Q 1 n 1 n 2 P(n i a p (mod p); i = 1, 2). p P Now P(n i a p (mod p); i = 1, 2) = 1 2/p unless p n 1 n 2, which occurs for O(log x) primes p. Get E Q 2 (a) 2 = σ 2 ( 1 + O((log x) 9 ) ) n 1,n 2 Q 1 = (σ Q 1 ) 2 ( 1 + O((log x) 9 ) ).

12 Random residues: higher correlations Define the random sifted set S(a) = Z \ (a p mod p). p P In particular, Q 2 (a) = Q 1 S(a). Lemma (S(a) correlations) Let n 1,..., n t be distinct integers in [ y, y], with t log x. Then P(n 1,..., n t S(a)) = σ t ( 1 + O(t 2 / log 9 x) ).

13 Stage 3: Strategic choices We choose a p, x/4 < p x/2 to have two properties: (a) the sets e p := (a p x/4 < p x/2; mod p) Q 2 (a) are large (on average) for (b) the collection of sets {e p : x/4 < p x/2} covers most of Q 2 (a) efficiently (little overlap). Item (a) is accomplished using a weighted, prime detecting sieve. Recall that Q 1, and hence Q 2 (a) consists mainly of primes. The average of e p, over all choices of a p is Q 2 (a) p Q 2(a) x σy log x = o(1). So a random (uniform) choice for a p is very inefficient! Item (b) is accomplished using hypergraph covering methods.

14 Primes in sparse A.P. s: weighted sieves Admissible k-tuple h 1,..., h k Prime-detecting weight fcn. w(n) = w(n; h) (GPY-Maynard-Tao) Goal: Find w(n) which is large when many of the numbers n + h i are prime, and small otherwise, and such that the sums T 1 (N) = n N w(n), T 2 (N) = n N can both be evaluated asymptoticlaly. If k 1(n + h j prime)w(n) j=1 T 2 (N) rt 1 (N), (w) then there are some values of n N such that the set {n + h 1,..., n + h k } contains at least r primes. Theorem (Maynard, 2016) For k (log N) 1/5, h i x c, weights s.t. (w) holds with r log k.

15 Sieve weights Fix an admissible k-tuple 1 h 1 < < h k k 2, k (log x) 1/5. Let x/4 < p x/2. Then h p := (h 1 p,..., h k p) is admissible. Define the weight w(p, n) by w(p, n) = w(n; h p ); (0 n y). Two crucial estimates (after suitable notmalization) Theorem (FGKMT) (a) (b) 1 π(y) w(p, n) 1 n y 1 P p P (p P); k w(p, q h i p) log 2 x (x < q y] i=1 (a) is a T 1 sum; (b) is a T 2 -type sum (with a different k-tuple).

16 Weighted choice of a p for x/4 < p x/2 Select a random number in n p [0, y] with probability proportional to w(p, n); that is P(n p = n) := w(p, n) l w(p, l) (0 n y). Bigger weight when many of n + h i p are prime. For each p P and fixed (non-random) vector a, let X p ( a) := P(n p + h i p S(a) for all i = 1,..., k). Lemma (Q 2 (a) correlations) + Chebyshev X p (a) σ k w.h.p. Spse a = a. Define r.v. m p by P(m p = m a = a) := Z p( a; m) X p ( a), { P(n p = m) if m + h j p S( a) for j = 1,..., k Z p ( a; m) = 0 otherwise,

17 weights, II P(m p = m a = a) := Z p( a; m) X p ( a), { P(n p = m) if m + h j p S( a) for j = 1,..., k Z p ( a; m) = 0 otherwise, Let a p m p (mod p), x/4 < p x/2. Then (1) m p + h i p S(a) for all i; (2) (on avg) Q 2 (a) (a p mod p) log k log 2 x (the k-tuple contains many primes) If the sets e p = Q 2 (a) (a p mod p) have little overlap (efficiently chosen), they cover about x log 2 x/ log x elements. Good if σ y log x cx log 2 x log x.