GAPS BETWEEN PRIMES IN BEATTY SEQUENCES arXiv:1411.2989v2 [math.NT] 23 Sep 2015 ROGER C. BAKER AND LIANGYI ZHAO Abstract. In this paper, we study the gaps between primes in Beatty sequences following the methods in the recent breakthrough of [9]. 2010 Mathematics Subject Classficiation: 11B05, 11L20, 11N35 1. Introduction Let pn denote the n-th prime and t a natural number with t ≥ 2. It has long been conjectured that lim inf (pn+t−1 − pn ) < ∞. n→∞ This was established recently for t = 2 by Y. Zhang [12] and shortly after for all t by J. Maynard [9]. Maynard showed that for N > C(t), the interval [N, 2N ) contains a set S of t primes of diameter D(S) ≪ t3 exp(4t), where D(S) := max{n : n ∈ S} − min{n : n ∈ S}. In the present paper, we adapt Maynard’s method to prove a similar result where S is contained in a prescribed set A (see Theorem 1). We then work out applications (Theorems 2 and 3) to a section of a Beatty sequence, so that A = {[αm + β] : m ≥ 1} ∩ [N, 2N ). The number α is assumed to be irrational with α > 1, while β is a given real number. We require an auxiliary result (Theorem 4) for the estimation of errors of the form X (N − N ′ )|I| , Λ(n) − ϕ(q) ′ N ≤n<N γn∈I mod 1 n≡a mod q where I is an interval of length |I| < 1 and γ = α−1 . Theorem 4 is of “Bombieri-Vinogradov type”; for completeness, we include a result of Barban-Davenport-Halberstam type for these errors (Theorem 5). We note that Chua, Park and Smith [5] have already used Maynard’s method to prove the existence of infinitely many sets of k primes of diameter at most C = C(α, k) in a Beatty sequence [αn], where α is irrational and of finite type. However, no explicit bound for C is given. In this paragraph, we introduce some notations to be used throughout this paper. We suppose that t ∈ N, N ≥ C(t) and write L = log N , log L D0 = . log log L Moreover, (d, e) and [d, e] stand for the great common divisor and the least common multiple of d and e, respectively. τ (q) and τk (q) are the usual divisor functions. kxk is the distance of between x ∈ R and the Date: July 12, 2018. 1 2 ROGER C. BAKER AND LIANGYI ZHAO nearest integer. Set P (z) = Y p<z p with z ≥ 2 and ψ(n, z) =  1 if (n, P (z)) = 1, 0 otherwise. X(E; n) stands for the indicator function of a set E and P for the set of primes. Let ε be a positive constant, sufficiently small in terms of t. The implied constant “≪”, when it appears, may depend on ε and on A (if A appears in the statement of the result). “F ≍ G” means both F ≪ G and G ≪ F hold. As usual, e(y) = exp(2πiy), and o(1) indicates a quantity tending to 0 as N tends to infinity. Furthermore, X X⋆ X′ , , χ mod q χ mod q χ mod q denote, respectively, a sums over all Dirichlet characters modulo q, a sum over nonprincipal characters modulo q and a sum restricted to primitive characters, other than χ = 1, modulo q. We write χ̂ for the primitive character that induces χ. A set H = {h1 , · · · , hk } of distinct non-negative integers is admissible if for every prime p, there is an integer ap such that ap 6≡ h (mod p) for all h ∈ H. In Sections 1 and 2, let θ be a positive constant. Let A be a subset of [N, 2N )∩N. Suppose that Y > 0 and Y /q0 is an approximation to the cardinality of A, #A. Let q0 , q1 be given natural numbers not exceeding N with (q1 , q0 P (D0 )) = 1 and ϕ(q1 ) = q1 (1 + o(1)). Suppose that n ≡ a0 (mod q0 ) for all n ∈ A with (a0 , q0 ) = 1. An admissible set H is given with h≡0 (mod q0 ) (h ∈ H) and p|h − h′ , with h, h′ ∈ H, h 6= h′ , p > D0 implies p|q0 . (1.1) We now state “regularity conditions” on A. (I) We have X (1.2) µ2 (q)τ3k (q) X n≡aq mod qq0 q≤N θ (q,q0 q1 )=1 X(A; n) − Y Y ≪ qq0 q0 Lk+ε (any aq ≡ a0 (mod q0 )). (II) There are nonnegative functions ̺1 , · · · , ̺s defined on [N, 2N ) (with s a constant, 0 < a ≤ s) such that (1.3) X (P; n) ≥ ̺1 (n) + · · · + ̺a (n) − (̺a+1 (n) + · · · + ̺s (n)) for n ∈ [N, 2N ). There are positive Yg,m (g = 1, · · · , s and m = 1, · · · , k) with Yg,m = Y (bg,m + o(1)) L−1 . where the positive constants bg,m satisfy (1.4) b1,m + · · · + ba,m − (ba+1,m + · · · + bs,m ) ≥ b > 0, for m = 1, · · · , k. Moreover, for m ≤ k, g ≤ s and any aq ≡ a0 (mod q0 ) with (aq , q) = 1 defined for q ≤ xθ , (q, q0 q1 ) = 1, we have (1.5) X q≤N θ (q,q0 q1 )=1 µ2 (q)τ3k (q) X n≡aq mod qq0 ̺g (n)X ((A + hm ) ∩ A; n) − Yg,m Y . ≪ ϕ(q0 q) ϕ(q0 )Lk+ε Finally, ̺g (n) = 0 unless (n, P (N θ/2 )) = 1. Theorem 1. Under the above hypotheses on H and A, there is a set S of t primes in A with diameter not exceeding D(H), provided that k ≥ k0 (t, b, θ) (k0 is defined at the end of this section). GAPS BETWEEN PRIMES IN BEATTY SEQUENCES 3 In proving Theorem 2, we shall take s = a = 1, q0 = q1 = 1, ρ1 (n) = X(P; n). A more complicated example with s = 5, of the inequality (1.3), occurs in proving Theorem 3, but again, q0 = q1 = 1. We shall consider elsewhere a result in which q0 , q1 are large. Maynard’s Theorem 3.1 in [10] overlaps with our Theorem 1, but neither subsumes the other. Theorem 2. Let α > 1, γ = α−1 and β ∈ R. Suppose that kγrk ≫ r−3 (1.6) for all r ∈ N. Then for any N > c1 (t, α, β), there is a set of t primes of the form [αm + β] in [N, 2N ) having diameter < C2 α(log α + t) exp(8t), where C2 is an absolute constant. Theorem 3. Let α be irrational with α > 1 and β ∈ R. Let r ≥ C3 (α, β) and 1 b 1 − < 2 , b ∈ N, (b, r) = 1. α r r Let N = r2 . There is a set of t primes of the form [αn + β] in [N, 2N ) having diameter < C4 α (log α + t) exp(7.743t), where C4 is an absolute constant. Theorem 3 improves Theorem 2 in that α can be any irrational number in (1, ∞) and 7.743 < 8, but we lose the arbitrary placement of N . Turning our attention to our theorem of Bombieri-Vinogradov type, we write E(N, N ′ , γ, q, a) = sup I X N ≤n<N ′ γn∈I mod 1 n≡a mod q Λ(n) − (N ′ − N )|I| . ϕ(q) Here, I runs over intervals of length |I| < 1. Theorem 4. Let A > 0, γ be a real number and b/r a rational approximation to γ, (1.7) γ− b 1 ≤ , N ε ≤ r ≤ N 3/4 , (b, r) = 1. r rN 3/4 Then for N < N ′ ≤ 2N and any A > 0, we have X max E(N, N ′ , γ, q, a) ≪ N L−A . (1.8) q≤min(r,N 1/4 )N −ε (a,q)=1 Our Barban-Davenport-Halberstam type result is the following. Theorem 5. Let A > 0 and γ be an irrational number. Suppose that for each η > 0 and sufficiently large r ∈ N, we have kγrk > exp(−rη ). (1.9) Let N L−A ≤ R ≤ N . Then for N < N ′ ≤ 2N , (1.10) q X X q≤R a=1 (a,q)=1 E(N, N ′ , γ, q, a)2 ≪ N RL(log L)2 . 4 ROGER C. BAKER AND LIANGYI ZHAO There are weaker results overlapping with Theorems 4 and 5 by W. D. Banks and I. E. Shparlinski [4]. Let γ be irrational, η > 0 and suppose that kγrk ≤ exp (−rη ) for infinitely many r ∈ N. Then (1.10) fails (so Theorem 5 is optimal in this sense). To see this, take N = exp(rη/2 ), N ′ = 2N , R = N L−8/η . We have, for some u ∈ Z, γn − un 1 ≤ 2N r−1 exp(−rη ) < , (n ≤ 2N ). r 4r From this, we infer that γn 6∈  1 3 , 4r 4r  (mod 1), (n ≤ 2N ). So N2 E(N, 2N, γ, q, a)2 ≥ 4r2 ϕ(q) , (q ≤ R, (a, q) = 1). Therefore, q X X q≤R a=1 (a,q)=1 E(N, 2N, γ, q, a)2 ≥ N2 N2 X 1 = N RL4/η . > 4r2 ϕ(q) r2 q≤R We now turn to the definition of k0 (t, b, θ). For a smooth function F supported on ) ( k X k xi ≤ 1 , Rk = (x1 , · · · , xk ) ∈ [0, 1] : i=1 set Ik (F ) = Z1 0 ··· Z1 0 F (t1 , · · · , tk )2 dt1 · · · dtk , and (m) Jk (F ) = Z1 0 for m = 1, · · · , k. Let ··· Z1 0   Z1 0 2 F (t1 , · · · , tk )dtm  dt1 · · · dtm−1 dtm+1 · · · dtk Mk = sup F k X (m) Jk (F ) m=1 Ik (F ) , where the sup is taken over all functions F specified above and subject to the conditions Ik (F ) 6= 0 and (m) Jk (F ) 6= 0 for each m. Sharpening a result of Maynard [9], D. H. J. Polymath [11] gives the lower bound (1.11) Mk ≥ log k + O(1). Now let k0 (t, b, θ) be the least integer k for which (1.12) Mk > 2t − 2 . bθ 5 GAPS BETWEEN PRIMES IN BEATTY SEQUENCES 2. Deduction of Theorem 1 from Two Propositions We first write down some lemmas that we shall need later. Lemma 1. Let κ, A1 , A2 , L > 0. Suppose that γ is a multiplicative function satisfying 0≤ γ(p) ≤ 1 − A1 p for all prime p and −L ≤ X γ(p) log p z − κ log ≤ A2 p w w≤p≤z for any w and z with 2 ≤ w ≤ z. Let g be the totally multiplicative function defined by g(p) = γ(p) . p − γ(p) Suppose that G : [0, 1] → R is a piecewise differentiable function with |G(y)| + |G′ (y)| ≤ B for 0 ≤ y ≤ 1 and (2.1) S= −1  κ Y γ(p) 1 . 1− 1− p p p Then for z > 1, we have   Z X
S(log z)κ 1 κ−1 log d 2 = t G(t)dt + O SLB(log z)κ−1 . µ(d) g(d)G log z Γ (κ) 0 d<z The implied constant above depends on A1 , A2 , κ, but is independent of L. Proof. This is [7, Lemma 4].  Throughout this section, we assume that the hypotheses of Theorem 1 hold. Moreover, we write Y Y p, W2 = W1 = p, R = N θ/2−ε . p≤D0 p≤D0 or p|q0 q1 p∤q0 Recalling the definition of admissible set, we pick a natural number ν0 with (ν0 + hm , W2 ) = 1 (m = 1, · · · , k). Lemma 2. Suppose that γ(p) = 1 + O(p−1 ) if p ∤ W1 and γ(p) = 0 if p|W1 . Let κ = 1 and S as defined in (2.1). We have
ϕ(W1 ) S= 1 + O(D0−1 ) . W1 Proof. We have S= Y  p|W1 1 1− p  Y    −1  1 ϕ(W1 ) Y 1 1 1− +O = 1 − p p2 p W1 p>D0 p∤q0 q1 p∤W1
1 + O(p−2 ) , from which the statement of the lemma can be readily obtained.  Lemma 3. Let H > 1, T1 = X d≤R (d,W1 )=1 µ2 (d) X 4ω(a) and T2 = d a a|d X H<d≤R µ2 (d) X −1/2 a . d2 a|d 6 ROGER C. BAKER AND LIANGYI ZHAO Then, we have (2.2) ϕ(W1 ) L W1 T1 ≪ and T2 ≪ H −1 . (2.3) Proof. Let γ(p) = 0 if p|W1 and p2 + 4p p2 + p + 4 if p ∤ W1 . Then g(p), as defined in the statement of Lemma 1, is γ(p) = g(p) = 4 1 + 2 p p if p ∤ W1 . Therefore, if d is square-free and (d, W1 ) = 1,   1 X 4ω(a) 4 1Y 1+ = g(d). = d a d p a|d p|d Otherwise, if (d, W1 ) 6= 1, then g(d) = 0. Using Lemma 1 with G(y) = 1 and Lemma 2, we have     X
ϕ(W1 ) log d ϕ(W1 ) = T1 = µ2 (d)g(d)G L , 1 + O(D0−1 ) log R + O log R W1 W1 d≤R where we can take L= X log p ≪ log D0 + log ω(q0 ) ≪ log L. p p|W1 Combining everything, we get (2.2). To prove (2.3), we interchange the summations and get X X X T2 ≤ a−5/2 k −2 ≪ a−3/2 H −1 ≪ H −1 , a≤R Ha−1 <k≤Ra−1 a≤R completing the proof of the lemma.  Lemma 4. Let f0 , f1 be multiplicative functions with f0 (p) = f1 (p) + 1. Then for squarefree d, e, X 1 1 f1 (k). = f0 ([d, e]) f0 (d)f0 (e) k|d,e Proof. We have Y X Y Y 1 1 1 (f0 (p))−1 . f1 (k) = (1 + f1 (p)) = f0 (p) = f0 (d)f0 (e) f0 (d)f0 (e) f0 (d)f0 (e) k|d,e p|(d,e) p|(d,e) The lemma follows from this. p|[d,e]  We now prove two propositions that readily yield Theorem 1 when combined. To state them, we define weights yr and λr for tuples r = (r1 , · · · , rk ) ∈ Nk having the properties (2.4) k Y i=1 ri , W1 ! = 1, µ2 k Y i=1 ri ! = 1. 7 GAPS BETWEEN PRIMES IN BEATTY SEQUENCES We set yr = λr = 0 for all other tuples. Let F be a smooth function with |F | ≤ 1 and the properties given at the end of Section 1. Let   log r1 log rk , (2.5) yr = F ,··· , log R log R and (2.6) λd = k Y yr X µ(di )di r di |ri ∀i i=1 We have Qk i=1 ϕ(ri ) . λr ≪ Lk (2.7) (see (5.9) of [9]). For n ≡ ν0 (mod W2 ), let  X wn =  di |n+hi ∀i and wn = 0 for all other natural numbers n. 2 λd  , Proposition 1. Let X S1 = wn X(A; n). N ≤n<2N Then (1 + o(1))ϕ(W1 )k Y (log R)k Ik (F ) . q0 W1k W2 S1 = Proposition 2. Let S2 (g, m) = X wn ̺g (n + hm ). N ≤n<2N n∈A∩(A−hm ) Then for 1 ≤ g ≤ s and 1 ≤ m ≤ k, (m) bg,m (1 + o(1))ϕ(W1 )k+1 Y (log R)k+1 Jk S2 (g, m) = ϕ(q0 )ϕ(W2 )W1k+1 L (F ) . Before proving the above propositions, we shall deduce Theorem 1 from them. Proof of Theorem 1. Let Z= Y ϕ(W1 )k (log R)k q0 W1k W2 wn k X and S(N ) = X N ≤n<2N n∈A Since wn ≥ 0, (1.3) gives that S(N ) ≥ k X m=1 a X g=1 m=1 ! X (P ∩ A; n + hm ) − (t − 1) . S2 (g, m) − s X g=a+1 ! S2 (g, m) − (t − 1)S1 . Using Propositions 1 and 2, the right-hand side of the above is ! !   a s k X X X θ (m) (1 + o(1)) Z bg,m − bg,m Jk (F ) − ε − (t − 1)Ik (F ) . 2 g=a+1 m=1 g=1 8 ROGER C. BAKER AND LIANGYI ZHAO Here we have used ϕ(q0 )ϕ(q1 )ϕ(W2 ) W1 ϕ(q1 ) = 1 + o(1). = 1 and q0 q1 W2 ϕ(W1 ) q1 Therefore, using (1.4), we get !   k X θ (m) Jk (F ) S(N ) ≥ (1 + o(1)) Z b − ε − (t − 1)Ik (F ) > 0, 2 m=1 for a suitable choice of F . The positivity of the above expression is a consequence of (1.12). Therefore, there must be at least one n ∈ A for which k X m=1 X (P ∩ A; n + hm ) > t − 1. For this n, there is a set of t primes n + hm1 , · · · , n + hmt in A.  3. Proof of Propositions 1 and 2 This section is devoted to the proofs of the two propositions. Proof of Proposition 1. We first show that   Y X Y ϕ(W1 )k Lk yr2 . (3.1) S1 = +O Qk q0 W2 r q0 W2 W1k D0 i=1 ϕ(ri ) From the definition of wn , we get (3.2) S1 = X d, e X λd λe X (A; n) . N ≤n<2N n≡ν0 mod W2 [di ,ei ]|n+hi ∀i Recall that n ≡ a0 (mod q0 ) for all n ∈ A. The inner sum of the above takes the form X X (A; n) , N ≤n<2N n≡aq mod qq0 where q = W2 k Y [di , ei ], i=1 provided that W2 , [d1 , e1 ], · · · , [dk , ek ] are pairwise coprime. The latter restriction reduces to (3.3) (di , ej ) = 1 for all i 6= j, and we exhibit this condition on the summation by writing X′ . d, e Outside of conditions P′ d, e , the inner sum is empty. To see this, suppose that p|di , p|ej with i 6= j, then the [di , ei ]|n + hi , and [dj , ej ]|n + hj imply that p|hi − hj . This means that either p ≤ D0 or p|q0 , both contrary to p|di . Counting the number of times a given q can arise, we get  2 X Y X′ λd λe (3.4) S1 − ≪ max |λd | µ2 (q)τ3k (q) Qk d q0 W2 [d , e ] i=1 i i d, e q≤R2 W 2 (q,q0 )=1 X n≡aq mod qq0 X (A; n) − Y . qq0 9 GAPS BETWEEN PRIMES IN BEATTY SEQUENCES Since R2 W2 ≤ N θ , we can appeal to (1.2) and (2.7) to majorize the right-hand side of (3.4) by ≪ Y 2k−(k+ε) ϕ(W1 )k Y Lk . L ≪ q0 q0 W2 W1k D0 Applying Lemma 4 with f1 = ϕ, we see that k X′ Y XY S1 = ϕ(ui ) q0 W2 u i=1 d, e ui |di ,ei ∀i λd λe Qk i=1 di ei +O  ϕ(W1 )k Y Lk q0 W2 W1k D0  . Now we follow [9] verbatim to transform this equation into k Y XY ϕ(ui ) S1 = q0 W2 u i=1 s (3.5) X∗ 1,2 ,··· ,sk,k−1 Here (3.6) P∗ Y µ (si,j ) 1≤i,j≤k i6=j λd λe X d, e ui |di ,ei ∀i si,j |di ,ej ∀i6=j Qk i=1 di ei +O  ϕ(W1 )k Y Lk q0 W2 W1k D0  . indicates that (si,j , ui uj ) = 1 and (si,j , si,c ) = 1 = (si,j , sd,j ), for c 6= j, d 6= i. Now define Y Y aj = u j sj,i , bj = uj si,j . i6=j i6=j Y X As in [9], we recast (3.5) as (3.7) k Y X Y µ(ui )2 S1 = q0 W2 u i=1 ϕ(ui )s X∗ 1,2 ,··· ,sk,k−1 µ (si,j ) 1≤i,j≤k i6=j d, e ui |di ,ei ∀i sij |di ,ej ∀i6=j µ(si,j ) ya yb + O ϕ(si,j )2  ϕ(W1 )k Y Lk q0 W2 W1k D0  . For the non-zero terms on the right-hand side of (3.7), either si,j = 1 or si,j > D0 . The terms of the latter kind (for given i, j, i 6= j) contribute k    k2 −k−1 X µ(si,j )2 X µ(s) X µ(u)2  Y  Y      (3.8) ≪ = U1 U2 U3 ,   2 2 q0 W2 ϕ(u) ϕ(si,j ) ϕ(s) q0 W2 si,j >D0 u<R (u,W1 )=1 s≥1 say. Clearly, U3 ≪ 1. Now if u is squarefree, we have  −1 1 1X1 1 1Y 1− ≪ = ϕ(u) u p u a p|u a|u and   1 1 X 2ω(a) 2 1 Y 1 X −1/2 = 1 + a . ≪ ≪ 2 2 2 2 ϕ(u) u p u a u p|u a|u a|u So (2.2) and (2.3) give, respectively, U1 ≪  k ϕ(W1 ) L W1 and U2 ≪ Hence, the right-hand side of (3.8) is ≪ and we have (3.1). ϕ(W1 )k Y Lk q0 W2 W1k D0 1 . D0 10 ROGER C. BAKER AND LIANGYI ZHAO Now, we shall deduce Proposition 1 from (3.1). Mindful of (2.6), we have 2    k Y X µ(ui )2 log uk ϕ(W1 )k Y Lk Y log u1 +O F ,··· , S1 = . q0 W2 ϕ(ui ) log R log R q0 W2 W1k D0 u i=1 (ul ,uj )=1 ∀l6=j (ul ,W1 )=1 ∀l Note that the common prime factors of two integers both coprime to W1 are strictly greater than D0 . Thus, we may drop the condition (ul , uj ) = 1 in the above expression at the cost of an error of size k  ≪ X Y q0 W2 p>D0 X k Y µ(ui )2 ϕ(ui ) u1 ···uk <R i=1 p|ul ,uj (ul ,W1 )=1 ∀l ≪ X  1 Y  2 q0 W2 (p − 1)  p>D0 X u<R (u,W1 )=1 k k µ(u)2   ≪ ϕ(W1 ) Y L , ϕ(u)  q0 W2 W1k D0 by virtue of (2.2). It remains to evaluate the sum k Y µ(ui )2 X (3.9) u i=1 (ul ,W1 )=1 ∀l ϕ(ui ) F  log uk log u1 ,··· , log R log R 2 . This requires applying Lemma 1 k times with γ(p) =  0 1 p|W1 , p ∤ W1 . We take A1 and A2 to be suitable constants and L≪1+ X log p ≪ log L p p|W1 as noted earlier. In the j-th application, we replace the summation over uj by the integeral over [0, 1]. Ultimately, we express the sum in (3.9) in the form   ϕ(W1 )(log L)Lk−1 ϕ(W1 )k k (log R) Ik (F ) + O W1k W1k and Proposition 1 follows at once.  We shall need the following lemma in the proof of Proposition 2. Lemma 5. Let 1 ≤ m ≤ k and suppose that rm = 1. Let yr(m) = k Y µ(ri )g(ri ) i=1 X λd d ri |di ∀i dm =1 Qk i=1 ϕ(di ) . Then yr(m) = X yr1 ,··· ,rm−1 ,am ,rm+1 ,··· ,rk am ϕ(am ) +O  ϕ(W1 )L W1 D0  Proof. Following [9] verbatim, we have (3.10) yr(m) = k Y i=1 µ(ri )g(ri ) X a ri |ai ∀i ya Y µ(ai )ri . ϕ(ai ) i=1 ϕ(ai ) i6=m Qk . GAPS BETWEEN PRIMES IN BEATTY SEQUENCES 11 Fix j, 1 ≤ j ≤ k. In (3.10), the nonzero terms will have either aj = rj or aj > D0 rj . The contribution from the terms with aj 6= rj is    (3.11) ≪ k Y i=1  X g(ri )ri   aj >D0 rj rj |aj  µ(aj )2     2 ϕ(aj ) am <R (am ,W1 )=1 Now, as before, from (2.2) and (2.3), X µ(aj )2 1 ≪ , ϕ(aj )2 D0 ϕ(rj )2 aj >D0 rj rj |aj and X Y X µ(ai )2 µ(am )2   .  ϕ(am ) ϕ(ai )2 X am <R (am ,W1 )=1 1≤i≤k ri |ai i6=j,m ϕ(W1 ) µ(am )2 L ≪ ϕ(am ) W1 X µ(ai )2 µ(ri )2 X µ(k) 1 ≤ ≪ , 2 2 2 ϕ(ai ) ϕ(ri ) ϕ(k) ϕ(ri )2 k ri |ai majorizing (3.11) by ≪ Hence (3.10) becomes yr(m) = k Y ϕ(W1 )L g(ri )ri ϕ(W1 ) L≪ . 2 ϕ(ri ) W1 D0 W1 D0 i=1   k Y g(ri )ri X yr1 ,··· ,rm−1 ,am ,rm+1 ,··· ,rk ϕ(W1 )L , + O ϕ(ri )2 a ϕ(am ) W1 D0 i=1 m and the proof is completed by applying Lemma 2.  Now we proceed to the proof of Proposition 2. Proof of Proposition 2. Let (m) = max yr(m) , ymax r where (3.12) (m) yr is defined in Lemma 5. We shall first show that    2  2 (m) (m) k−2 k−1 y Y L ϕ (W ) y k−ε X max u 1 Yg,m YL   S2 (g, m) = +O + . Qk k−1 ϕ(q0 )ϕ(W2 ) u ϕ(q ) ϕ(q0 )ϕ(W2 )W1 D0 0 i=1 g(ui ) From the definition of wn , we have (3.13) S2 (g, m) = X d,e λd λe X ̺g (n + hm ). n∈A∩(A−hm ) N ≤n<2N, n≡ν0 mod W2 [di ,ei ]|n+hi ∀i P′ P As in the proof of Proposition 1, d,e reduces to d,e . Let n′ = n + hm . Since n + hm ≡ a0 (mod q0 ) for n ∈ A, the inner sum of (3.13) reduces to X X (A ∩ (A + hm ), n′ ) ̺g (n′ ). T (d, e) := n′ ≡ν0 +hm mod W2 n′ ≡a0 mod q0 n′ ≡hm −hi mod [di ,ei ]∀i Recall that ̺g (n′ ) = 0 if n′ is divisible by a prime divisor of [di , ei ]. Since one condition of the summation is [dm , em ]|n′ , we have T (d, e) = 0 unless dm = em = 1. When dm = em = 1, X T (d, e) = X (A ∩ (A + hm ), n) ̺g (n). n≡aq mod qq0 12 ROGER C. BAKER AND LIANGYI ZHAO Here we have q = W2 k Y [di , ei ], (aq , q) = 1, aq ≡ a0 (mod q0 ). i=1 For (aq , q) = 1, we need (hm − hi , [di , ei ]) = 1 whenever m 6= i, which was noted earlier. Arguing as in the proof of Proposition 1, (1.5) now gives Yg,m S2 (g, m) = ϕ(q0 )ϕ(W2 ) λd λe X′ Qk ϕ([di , ei ]) X∗ Y d,e dm =em =1 With aj and bj as in (3.6), we follow [9] to obtain k (3.14) X Y µ(ui )2 Yg,m S2 (g, m) = ϕ(q0 )ϕ(W2 ) u i=1 g(ui ) s i=1 1,2 ,··· ,sk,k−1 1≤i,j≤k i6=j +O  Y Lk−ε ϕ(q0 )  . µ(si,j (m) (m) y y +O g 2 (si,j ) a b  Y Lk−ε ϕ(q0 )  . Here q is the totally multiplicative function with g(p) = p−2 for all p and we have used Lemma 4 with f1 = g. The contribution to the sum in (3.14) from si,j 6= 1 (for given i, j) is k−1  2   !k(k−1)−1  (m) Y ymax X µ(si,j )2 X µ(s)2  X µ(u)2      ≪ 2 2 ϕ(q0 )ϕ(W2 )L  g(u)  g(s) g(s ) i,j s u<R si,j >D0 (3.15) (u,W1 )=1  2 (m) Y ymax V1 V2 V3 , = ϕ(q0 )ϕ(W2 )L say. Clearly, V2 ≪ 1. Using (2.2) while mindful of the estimate 1 1 X 2ω(a) ≪ g(s) s a a|s yields that V1 ≪  k−1 ϕ(W1 ) L . W1 From (2.3) and the observation that, for s squarefree, 1 1 X 4ω(a) 1 X −1/2 a , ≪ ≪ g 2 (s) s2 a s2 a|s a|s we get that V3 ≪ D0−1 . Note the bound in (3.15) is  2 (m) Y ymax Lk−2  ϕ(W ) k−1 1 1 , ≪ ϕ(q0 )ϕ(W2 ) W1 D0 and we have established (3.12). 13 GAPS BETWEEN PRIMES IN BEATTY SEQUENCES Now we use Lemma 5 in (3.12), recalling (2.5). When rm = 1,   X µ(u)2 log rm−1 log u log rm+1 log rk log r1 yr(m) = F ,··· , , , ,··· , ϕ(u) log R log R log R log R log R Q (u,W1 ki=1 ri )=1 (3.16)   ϕ(W1 )L . +O W1 D0 From this, we find that (m) ymax ≪ ϕ(W1 ) L. W1 We shall apply Lemma 1 to (3.16) with κ = 1,  Qk 1, p ∤ W1 i=1 ri γ(p) = 0, otherwise, A1 , A2 suitably chosen and L ≪ log L (similar to the proof of (2.2)). Define Fr(m) = Z1 F 0  log r1 log rm−1 log rm+1 log rk ,··· , , tm , ,··· , log R log R log R log R We obtain that yr(m) k Y ϕ(ri ) ri i=1 ϕ(W1 ) = log R W1 ! Fr(m) +O  ϕ(W1 )L W1 D0   dtm . . Inserted into (3.12), the above produces a main term (log R)2 Yg,m ϕ(W1 )2 ϕ(q0 )ϕ(W2 )W12 (3.17) X k Y ϕ(ri )µ(ri )2  g(ri )ri2 r i=1 (ri ,W1 )=1∀i (ri ,rj )=1∀i6=j rm =1 Fr(m) 2 and an error term of size ≪  X  X Yg,m ϕ(W1 )2 L2 Y ϕ(W1 )2 L2  ≪ Q 2 ϕ(q0 )ϕ(W2 ) r W12 D0 ki=1 g(ri ) ϕ(q0 )ϕ(W2 )W1 D0  r<R rm =1 (r,W1 )=1 −1 Recall that Yg,m ≪ Y L an error of size ≪ k−1 1   g(r)  ≪ Y ϕ(W1 )k+1 Lk . ϕ(q0 )ϕ(W2 )W1k+1 D0 . Now we remove the condition (ri , rj ) = 1 from (3.17). As before, this introduces    X L2 Y ϕ(W1 )2  X ϕ(p)2    2  2 2 ϕ(q0 )ϕ(W2 )W1 g(p) p p>D0 r<R (r,W1 )=1 k−1 µ(r)2 ϕ(r)   g(r)r  ≪ Y Lk ϕ(W1 )k+1 ϕ(q0 )ϕ(W2 )W1k+1 D0 by an application of Lemma 3. Combining all our results, we get   k X Y ϕ(ri )2 µ(ri )2  (m) 2 (log R)2 Yg,m ϕ(W1 )2 Y ϕ(W1 )k+1 Lk +O S2 (g, m) = . Fr ϕ(q0 )ϕ(W2 )W12 g(ri )ri2 ϕ(q0 )ϕ(W2 )W1k+1 D0 r i=1 (ri ,W1 )=1∀i rm =1 The last sum is evaluated by applying Lemma 1 to each summation variable in turn, taking ( p3 −2p2 +p p3 −p2 −2p+1 , p ∤ W1 γ(p) = 0, p|W1 14 ROGER C. BAKER AND LIANGYI ZHAO to produce the right value of γ(p)/(p − γ(p)). Of course
ϕ(W1 ) 1 + O(D0−1 ) W1 by Lemma 2, while L ≪ log L. Our final conclusion is that S= (m) S2 (g, m) = (log R)k+1 Yg,m ϕ(W1 )k+1 Jk ϕ(q0 )ϕ(W2 )W1k+1 (1 + o(1)) completing the proof.  4. Further Lemmas Let γ = α−1 . As noted in [4], the set of [αm + β] in [N, 2N ) may be written as {n ∈ [N, 2N ) : γn ∈ (γβ − γ, βγ] (mod 1)}. Lemma 6. Let I = (a, b) be an interval of length l with 0 < l < 1 and let h be a natural number satisfying 0 < −hγ < 2ε (mod 1), where 2ε < l. Let A = {n ∈ [N, 2N ) : γn ∈ I Then (mod 1)}. A ∩ (A + h) = {n ∈ [N + h, 2N ) : γn ∈ J where J is an interval of length l′ with l − 2ε < l′ < l. (mod 1)} Proof. Let t ≡ −hγ (mod 1), 0 < t < 2ε. Clearly A ∩ (A + h) consists of the integers in [N + h, 2N ) for which γn ∈ (a, b) (mod 1), γn + t ∈ (a, b) (mod 1). The lemma follows with J = (a, b − t).  Lemma 7. Let I be an interval of length l, 0 < l < 1. Let x1 , · · · , xN be real. Then (i) There exists z such that # {j ≤ N : xj ∈ z + I (ii) We have (for aj ≥ 0, j = 1, · · · , N and L ≥ 1) N X aj − l j=1 xj ∈I mod 1 N X j=1 aj ≪ L−1 N X (mod 1)} ≥ N l. aj + j=1 L X h−1 h=1 N X aj e(hxj ) . j=1 Proof. We leave (1) as an exercise; (2) is a slight variant of [1, Theorem 2.1].  Lemma 8. Let 1 ≤ Q ≤ N and F a nonnegative function defined on Dirichlet characters. Then for some Q1 , 1 ≤ Q1 ≤ Q, X X′ X X⋆ LQ F (χ̂) ≪ F (ψ). Q1 q≤Q χ mod q Q1 ≤q1 <2Q1 ψ mod q1 Proof. We recall that χ̂ is the primitive character that induces χ, so that F (χ̂) may be quite different from F (χ). The left-hand side of the claimed inequality is X X X⋆ F (ψ) q1 ≤Q ψ mod q1 χ mod q q≤Q, q1 |q ψ induces χ 1≤ X X⋆ q1 ≤Q ψ mod q1 The lemma follows on applying a splitting-up argument to q1 . F (ψ) Q . q1  15 GAPS BETWEEN PRIMES IN BEATTY SEQUENCES Lemma 9. Let f (j) (j ≥ 1) be a periodic function with period q,   n X nj f (j)e − , S(f, n) = q j=1 F > 0, and R ≥ 1. Let H(y) be a real function with H ′ (y) monotonic and |H ′ (y))| ≤ F y −1 for R ≤ y ≤ 2R. Then for J = [R, R′ ] with R < R′ ≤ 2R,  Z  X X X |S(f, n)| ny R|S(f, 0)| −1 f (m)H(m) − q S(f, n) e + H(y) dy ≪ + , q qF n ′ −1 m∈J 1≤|n|≤2F qR |n|∈J J where J ′ = [min{2F qR−1 , q/2}, max{2F qR−1 , q} + q]. Proof. This is [2, Theorem 8].  For a finite sequence {ak : K ≤ k < K ′ }, set  kak2 =  X K≤k<K ′ 1/2 |ak |2  . Lemma 10. Let R ≥ 1, M ≥ 1, H ≥ 1. Let β be real and H u1 ≤ 2 (4.1) β− r1 r1 where r1 ≥ H and (u1 , r1 ) = 1. Then for M1 ∈ N,     MX 1 +M 1 HM (4.2) min R, ≪ + 1 (R + r1 log r1 ) . kmβk r1 m=M1 +1 If M < r1 and M β− u1 1 ≤ , r1 2r1 then M X 1 ≪ r1 log 2r1 . kmβk m=1 (4.3) Proof. For (4.2), it suffices to show that a block of [r1 /H] consecutive m’s contribute r1 X r1 ≪R+ . l l=1 Writing m = m0 + j, 1 ≤ j ≤ [r1 /H], (m0 + j)β − m0 β − jH 1 ju1 ≤ 2 ≤ , r1 r1 r1 so there are O(1) values of j for which the bound k(m0 + j)βk ≥ 1 ju1 m0 β + 2 r1 fails. Our block estimate follows immediately. The argument for (4.3) is similar. In this case, mβ − 1 mu1 ≤ , r1 2r1 16 ROGER C. BAKER AND LIANGYI ZHAO if 1 ≤ m ≤ M . Therefore, the left-hand side of (4.3) can be estimated by ′ Pr1 l=1 r1 /l.  Lemma 11. Let N < N ≤ 2N , M K ≍ N , N ≥ K ≥ M ≥ 1. Suppose that H u ≤ 2 , (u, r) = 1, H ≤ r ≤ N. (4.4) γ− r r Let (am )M≤m<2M , (bk )K≤k<2K be two sequences of complex numbers. Then (4.5) S := X X X X am bk χ(mk)e(γmk) Q≤q<2Q χ mod q M≤m<2M K≤k<2K N ≤mk<N ′ satisfies the bound 3/2 S ≪ kak2 kbk2 L D 1/2 where   Q3/2 H 1/2 N 1/2 2 1/2 3/2 1/2 1/2 3/2 1/2 Q M + , +Q H K +Q r r1/2 D = max #{q ∈ [Q, 2Q) : n = lq}. n<N ′ Proof. Let S be the sum obtained from S by removing the condition N ≤ mk < N ′ . It suffices to prove the same bound, with L1/2 in place of L3/2 , for S ′ , since the condition can be restored at the cost of a factor of L. See [8, Section 3.2]. We have S′ ≤ X X X Q≤q<2Q χ mod q M≤m<2M |am | X bk χ(k)e(γmk) = X Sq , q K≤k<2K say. We may also assume that bk = 0 if (k, q) > 1. By Cauchy’s inequality, and with summations subject to the obvious restrictions on m, k1 and k2 , X XXX Sq2 ≤ ϕ(q)kak22 bk1 bk2 χ(k1 )χ(k2 )e(γm(k1 − k2 )). χ mod q m k1 k2 Bringing the sum over χ inside we see that the right-hand side of the above is ϕ(q)2 kak22 X k1 ,k2 k1 ≡k2 mod q bk1 bk2 X m e(γm(k1 − k2 )) ≤ ϕ(q)2 kak22 X k1 |bk1 |2 X k1 ≡k2 mod q X m e(γm(k1 − k2 )) upon using the parallelogram rule  1 2 2 |bk1 | + |bk2 | . 2 Now summing the geometric sum over m and then summing over q, we see that  X X X 2 3 2 2 2 2 2 min M, (4.6) Sq ≪ Q kak2 kbk2 M + Q kak2 kbk2 |bk1 bk2 | ≤ Q≤q<2Q Q≤q<2Q 1≤l<K/q 1 kγlqk  . Now we combine the variables l and q and then apply (4.2), leading to   X HK 2 3 2 2 2 2 2 Sq ≪ Q kak2 kbk2 M + Q kak2 kbk2 D + 1 (M + r log r) r Q≤q<2Q    HN 3 2 2 2 ≪ kak2 kbk2 Q M + LQ D + HK + M + r . r The desired bound for S ′ follows by another application of Cauchy’s inequality.  Lemma 12. Under the hypotheses of Lemma 11, suppose that 4M Q < N , bk = 1 for K ≤ k < 2K and |am | ≤ 1 for M ≤ m < 2M . Define D as in Lemma 11. Then 17 GAPS BETWEEN PRIMES IN BEATTY SEQUENCES (i) We have S ≪ Q3/2 LD (ii) If 4M Q < r and  QM H +1 r 4M Q γ −   K +r . Q u 1 ≤ , r 2r then S ≪ LDQ3/2 r. Proof. Let Im (here and after) denote a subinterval of [N/m, N ′ /m). We have S ≤ QS ∗ + S ∗∗ , where, for a suitably chosen nonprincipal χq (mod q), S∗ = X X X χq (k)e(γmk) X X X χ0 (k)e(γmk) . Q≤q<2Q M≤m<2M k∈Im and S ∗∗ = Q≤q<2Q M≤m<2M k∈Im To prove part (i), it suffices to show that S ∗ ≪ Q1/2 LD and  QM H +1 r  QM H +1 S ≪ QLD r We give the proof for S ∗ ; the proof for S ∗∗ is similar. ∗∗   K +r Q   K +1 . Q Given q and m, Lemma 9 together with, using the notation from Lemma 9, √ |S(χ, q)| ≤ q (see Chapter 9 of [6]) gives X 1 χq (k)e(γmk) − q k∈Im X S(χq , n) 1≤|n|≪Mq Z e Im    n + γm y dy q ≪ q −1/2 M −1 + q 1/2 Therefore X k∈Im χq (k)e(γmk) ≪ q 1/2 L + q −1/2 1≤|n|≪Mq Summing over m and q, S ∗ ≪ M Q3/2 L + Q1/2 X X X  X 1≤n≪Mq min K, X min Q≤q<2Q M≤m<2M 1≤|n|≪Mq  n−1 ≪ q 1/2 L. 1 γm − n q  . K 1 , Q |γmq − n| The contribution to the right-hand side of the above from n’s with |n − γmq| > 1/2 is Now combining the variables m, q, (4.7) ≪ M Q3/2 L. S ∗ ≪ M Q3/2 L + Q1/2 D X MQ≤m′ <4MQ min  K 1 , Q kγm′ k  .  . 18 ROGER C. BAKER AND LIANGYI ZHAO We can now deduce the desired bound for S ∗ by applying (4.2). Now for part (ii), we note that (4.3) is applicable to the reciprocal sum in (4.7) with 4M Q and γ in place of M and β. Hence S ∗ ≪ M Q3/2 L + Q1/2 Dr log 2r ≪ DLQ1/2 r since 4M Q < r. Similarly S ∗∗ ≪ DLQr, and part (ii) follows.  Lemma 13. Suppose that γ− u LA+1 ≤ r r2 with (u, r) = 1 and that r2 ≤ N ≤ r2 L2A+2 . Then (i) For Q < N 2/7−ε , N 4/7 ≪ K ≪ N 5/7 and any am , bk with |am | ≤ τ (m)B , |bk | ≤ τ (k)B , where B is an absolute constant, the sum S in (4.5) satisfies the bound S ≪ QN 1−ε/4 . (4.8) (ii) For Q ≤ N 2/7−ε , M ≪ N 4/7 and bk = 1 for K ≤ k < 2K, |am | ≤ 1 for M ≤ m < 2M , the sum S in (4.5) satisfies (4.8). Proof. In order to prove (i), we use Lemma 11. As D ≪ N ε/15 ,     SQ−1 N −1+ε/4 ≪ Q−1 N −1/2+ε/3 Q2 N 3/14 + Q3/2 N 5/14 ≪ N −1/2+ε/2 QN 3/14 + Q1/2 N 5/14 ≪ 1. To prove (ii), we break the situation into two cases. If K < N 1−ε , then by (i) of Lemma 12   N 1−ε 1/2 −1 −1+ε/4 1/2 −1+ε/2 N + MQ + SQ N ≪Q N ≪ N 1/7−1/2+ε + N 3/7+4/7−1−ε + N −ε/2 ≪ 1. Q If K ≥ N 1−ε , then M ≪ N ε and (ii) of Lemma 12 is applicable since u ≪ N −1+2/7+ε . 4M Q γ − r Hence SQ−1 N −1+ε/4 ≪ Q1/2 N −1/2+ε ≪ 1, giving the desired majorant.  Lemma 14. Let f be an arbitrary complex function on [N, 2N ). Let N < N ′ ≤ 2N . The sum X S= Λ(n)f (n) N ≤n<N ′ can be decomposed into O(L2 ) sums of the form X M<m≤2M am X f (mk) or K≤k<2K N ≤mk<N ′ ZN ′ X am N M≤m<2M X f (mk) k≥w K≤k<2K N ≤mk<N ′ dw w with M ≤ N 1/4 and |am | ≤ 1, together with O(L) sums of the form X X am bk f (mk) M<m≤2M K≤k<2K N ≤mk<N ′ with N 1/2 ≤ K ≪ N 3/4 and kak2 kbk2 ≪ N 1/2 L2 . Proof. This follows from the arguments in [6, Chapter 24] by taking U = V = N 1/4 . We record a special case of [3, Lemma 14]. For more background on the “Harman sieve”, see [8].  GAPS BETWEEN PRIMES IN BEATTY SEQUENCES 19 Lemma 15. Let W (n) be a complex function with support in (N, 2N ] ∩ Z, |W (n)| ≤ N 1/ε . For r ∈ N, z ≥ 2, let X W (rn). (4.9) S ∗ (r, z) = (n,P (z))=1 Suppose that for some constant c > 0, 0 ≤ d ≤ 1/2, and for some Y > 0, we have, for any coefficients am , bk with |am | ≤ 1, |bk | ≤ τ (k), X X (4.10) am W (mk) ≪ Y, m≤2N c k and X (4.11) am N c ≤m≤2N c+d X k bk W (mk) ≪ Y. Let ur (r ≤ N c ) be complex numbers with |ur | ≤ 1 and ur = 0 for (r, P (N ε )) > 1. Then X
ur S ∗ r, (2N )d ≪ Y L3 . r≤(2N )c The following application of Lemma 15 will be used in the proof of Theorem 3. We take X X (4.12) W (n) = ηχ χ(n)e(γn) Q≤q<2Q χ mod q ′ for N ≤ n < N ; otherwise, W (n) = 0. Here ηχ is arbitrary with |ηχ | ≤ 1. Lemma 16. Suppose that LA+1 u , (u, r) = 1, N = r2 , 1 ≤ Q ≤ N 2/7−ε . ≤ r r2 Define S ∗ (r, z) as above with W defined in (4.12). Then   X ur S ∗ r, (2N )1/7 ≪ N L−A γ− r≤(2N )4/7 for every A > 0, provided that |ur | ≤ 1, ur = 0 for (r, P (N ε )) > 1. Proof. We need to verify (4.10) and (4.11) with c = 4/7, d = 1/7 and Y = N L−A−3 . This is an application of Lemma 13.  We now introduce some subsets of Rj needed in the proof of Theorem 3. Write Ej for the set of j-tuples αj = (α1 , · · · , αj ) satisfying 1 1 ≤ αj < αj−1 < · · · < α1 ≤ and α1 + α2 + · · · + αj−1 + 2αj ≤ 1. 7 2 A tuple αj is said to be good if some subsum of α1 + · · · + αj is in [2/7, 3/7] ∪ [4/7, 5/7] and bad otherwise. We use the notation pj = (2N )αj . For instance, the sum X ψ(n3 , p2 ) p1 p2 n3 =k (2N )1/7 ≤p2 <p1 <(2N )1/2 will be written as X p1 p2 n3 =k α2 ∈E2 ψ(n3 , p2 ). 20 ROGER C. BAKER AND LIANGYI ZHAO Lemma 17. Let γ, u/r, N , Q be as in Lemma 16 and E be a subset of Ej defined by a bounded number of inequalities of the form (4.13) c1 α1 + · · · + cj αj < cj+1 (or ≤ cj+1 ). Suppose that all points in E are good and that throughout E, zj is either the function zj = (2N )αj or the constant zj = (2N )1/7 . Then for arbitrary ηχ with |ηχ | ≤ 1, X X X ηχ χ(p1 · · · pj nj+1 )e(γp1 · · · pj nj+1 )ψ(nj+1 , zj ) ≪ N L−A , Q≤q<2Q χ mod q N ≤p1 ···pj nj+1 <N ′ αj ∈E for every A > 0. Proof.QThis is a consequence of (i) of Lemma 13. On grouping a subset of the variables as a product m = i∈S pi , with S ⊂ {1, · · · , j}, we obtain a sum S of the form appearing in (i) of Lemma 13, except that a bounded number of inequalities of the form (4.13) are present. These inequalities may be removed at the cost of a log power, by the mechanism noted earlier. See page 184 of [3] for a few more details of a similar argument. The lemma follows at once.  Lemma 18. Let D = {(α1 , α2 ) ∈ E2 : (α1 , α2 ) is bad, α1 + 2α2 > 5/7}. Then X X (P; n) − ψ(n3 , p2 ) = ̺1 (n) + ̺2 (n) + ̺3 (n) − ̺4 (n) − ̺5 (n). p1 p2 n3 =n α2 ∈D Here ̺1 (n) = ψ(n, (2N )1/7 ), ̺4 (n) = X ψ(n2 , (2N )1/7 ), ̺2 (n) = p1 n2 =n α1 ∈E1 ̺5 (n) = X X ψ(n3 , (2N )1/7 ), p1 p2 n3 =n α2 ∈E2 \D X ψ(n4 , (2N )1/7 ) and ̺3 (n) = p1 p2 p3 n4 =n α3 ∈E3 (α1 ,α2 )∈E2 \D ψ(n5 , p4 ). p1 p2 p3 p4 n5 =n α4 ∈E4 (α1 ,α2 )∈E2 \D Proof. We repeatedly use Buchstab’s identity in the form X ψ(m, z) = ψ(m, w) − ψ(h, p) (2 ≤ w < z). ph=m w≤p<z Thus X(P; n) = ψ(n, (2N )1/2 ) = ψ(n, (2N )1/7 ) − (4.14) = ̺1 (n) − ̺4 (n) + X(P; n) − X p1 p2 n3 =n α2 ∈D X p1 p2 n3 =n α2 ∈E2 \D ψ(n2 , p1 ) ≤p1 <(2N ) p1 n2 =n 1/2 ψ(n3 , p2 ), p1 p2 n3 =n α2 ∈E2 ψ(n3 , p2 ) = ̺1 (n) − ̺4 (n) + Continuing the decomposition of the last sum, X X ψ(n3 , (2N )1/7 ) − ψ(n3 , p2 ) = (4.15) p1 p2 n3 =n α2 ∈E2 \D (2N ) 1/7 X X X ψ(n3 , p2 ). p1 p2 n3 =n α2 ∈E2 \D ψ(n4 , (2N )1/7 ) + p1 p2 p3 n4 =n α3 ∈E3 (α1 ,α2 )∈E2 \D Combining (4.14) and (4.15), we complete the proof of the lemma. X ψ(n5 , p4 ). p1 p2 p3 p4 n5 =n α4 ∈E4 (α1 ,α2 )∈E2 \D  21 GAPS BETWEEN PRIMES IN BEATTY SEQUENCES Lemma 19. Let r, u/r, N and Q be as in Lemma 16 with ̺1 , · · · , ̺5 as in Lemma 18; we have X X X ηχ ̺j (n)χ(n)e(γn) ≪ QN L−A Q≤q<2Q χ mod q N ≤n<N for arbitrary ηχ with |ηχ | ≤ 1 and any A > 0. Proof. This follows from Lemmas 16 and 17 for j = 1, 2, 4, 5 on noting that α1 + α2 + α3 ≤ α1 + 2α2 ≤ 5/7 for j = 5, so that either α3 is good or α1 + α2 + α3 < 4/7 (similarly for j = 2). For j = 3, we need to show that each α4 counted is good. Suppose that some α4 is bad. We have α1 + α2 + α3 + 2α4 ≤ 1. Hence α1 + α2 + α3 ≤ 5/7 from which we infer that α1 + α2 + α3 < 4/7. Therefore, α1 + α2 < 3/7. But we know that α1 + α2 > 2/7. This makes α4 good, a contradiction.  5. Proof of Theorems 4 and 5 Proof of Theorem 4. With a suitable choice of aq , (aq , q) = 1, we have max E(N, N ′ , γ, q, a) ≤ sup (a,q)=1 I X N ≤n<N ′ γn∈I mod 1 n≡aq mod q Λ(n) − |I| X Λ(n) + X N ≤n<N ′ n≡aq mod q N ≤n<N ′ n≡aq mod q Λ(n) − N′ − N ϕ(q) = T1 (q) + T2 (q), say. In view of the Bombieri-Vinogradov theorem, we need only bound ≪ X q≤N 1/4−ε L−A−1 X X X Λ(n) + P q≤min(r,N 1/4 )N −ε h≤LA+1 N ≤n<N ′ n≡aq mod q q 1 h T1 (q), which is, applying Lemma 7, X Λ(n)e(γnh) . N ≤n<N ′ n≡aq mod q Let H = LA+1 . Mindful of the Brun-Titchmarsh inequality, it remains to show that for 1 ≤ h ≤ H, X q≤min(N 1/4 ,r)N −ε X N ≤n<N ′ n≡aq mod q Λ(n)e(γnh) ≪ N L−A−1 . Reducing hu/r into lowest terms, we need only show that X X ηq Λ(n)e(γn) ≪ N L−A−1 q≤min(N 1/4 ,r)N −ε/2 N ≤n<N ′ n≡aq mod q under the modified hypothesis (4.4) on γ (with H = LA+1 ), whenever |ηq | ≤ 1. Using Lemma 14, it suffices to show that X X ηq (5.1) q≤min(N 1/4 ,r)N −ε/2 X M≤m<2M K≤k<2K N ≤mk<N ′ mk≡aq mod q am bk e(γmk) ≪ N L−A−3 under either of the following sets of conditions. (a) kak2 kbk2 ≪ N 1/2 L2 , N 1/2 ≤ K ≤ N 3/4 ; (b) |am | ≤ 1, bk = 1 for k ∈ Im ⊂ [K, 2K), bk = 0 otherwise, M ≤ N 1/4 . We use Dirichlet characters to detect the congruence relation in (5.1) and we require the estimate X X X X ηq χ(aq ) am bk χ(mk)e(γmk) ≪ N L−A−4 . ϕ(q) 1/4 −ε/2 q≤min(N ,r)N χ mod q M≤m<2M K≤k<2K N ≤mk<N ′ 22 ROGER C. BAKER AND LIANGYI ZHAO It suffices to show that (5.2) S := X X X X Q≤q<2Q χ mod q M≤m<2M K≤k<2K N ≤mk<N ′ am bk χ(mk)e(γmk) ≪ QN L−A−6 for Q ≤ min(N 1/4 , r)N −ε/2 . In case (a), we apply Lemma 11, which gives   Q3/2 N 1/2 3/2 1/2 3/2 1/2 S ≪ N 1/2+ε/6 Q2 M 1/2 + + Q K + Q r r1/2 N 1+ε/6 Q3/2 + Q3/2 N 7/8+ε/6 . r1/2 Each one of these three terms is ≪ QN L−A−6 as ≪ N 3/4+ε/6 Q2 + N 3/4+ε/6 Q2 (QN L−A−6 )−1 ≪ QN −1/4+ε/5 ≪ 1, since Q ≤ rN −ε/2 , and N 1+ε/6 Q3/2 r−1/2 (QN L−A−6 )−1 ≪ Q1/2 N ε/4 r−1/2 ≪ 1, N 7/8+ε/6 Q3/2 (QN L−A−6 )−1 ≪ N −1/8+ε/5 Q1/2 ≪ 1. In case (b), we use Lemma 12. Suppose that K < N 1−ε/4 ; (i) of Lemma 12 gives   K N 3/2 ε/6 + QM + +r . S≪Q N r Q Each of the above four terms is ≪ QN L−A−6 , since Q3/2 N 1+ε/6 (QN L−A−6 )−1 ≪ Q1/2 r−1 N ε/5 ≪ 1, r Q5/2 N ε/6 M (QN L−A−6 )−1 ≪ Q3/2 N −3/4+ε/5 ≪ 1, Q1/2 N ε/6 K(QN L−A−6 )−1 ≪ KN −1+ε/4 ≪ 1 and Q3/2 N ε/6 r(QN L−A−6 )−1 ≪ Q1/2 N −1/4+ε/5 ≪ 1. Now suppose that K ≥ N 1−ε/4 . Then 4M Q ≪ QN ε/4 , thus 4M Q < r and u u 1 ≪ M QN −3/4 , hence 4M Qr γ − ≤ . r r 2 So (ii) of Lemma 12 gives comfortably: 4M Qr γ − S ≪ N ε Q3/2 r ≪ QN L−A−6 , completing the proof.  Proof of Theorem 5. We first show that the contribution to the sum in (1.10) from q ≤ LA+1 is ≪ N 2 L−A ≪ N R. Since, for some Q ≤ LA+1 , X q≤LA+1 q X a=1 (a,q)=1 E2 ≪ N X q≤LA+1 q X NL 1 E(N, N ′ , γ, q, a) ≪ ϕ(q) a=1 Q (a,q)=1 X Q≤q<2Q max E(N, N ′ , γ, q, a), (a,q)=1 23 GAPS BETWEEN PRIMES IN BEATTY SEQUENCES it suffices to show for this Q that X (5.3) Q≤q<2Q max E(N, N ′ , γ, q, a) ≪ QN L−A−1 . (a,q)=1 We may suppose that A is large. Arguing as in the proof of Theorem 4, we need only show that (5.2) follows from either (a) or (b). By Dirichlet’s theorem, there is a rational approximation b/r to γ satisfying (1.7). For any η > 0, N −3/4 ≥ kγrk ≫ exp(−rη ), 5A hence r ≫ L . Now we apply Lemma 11 to prove the desired bound under (a). Since D ≤ Q ≤ LA+1 , the term kak2 kbk2 L2 D1/2 Q3/2 H 1/2 N 1/2 r−1/2 presents no difficulty; the other terms are clearly all small enough. For the bound under (b), a similar remark applies to Lemma 12 and the terms Q3/2 LDN Hr−1 if K < N 1−ε/4 and LDQ3/2 r if K ≥ N 1−ε/4 . This establishes (5.3). It remains to examine the contribution to the sum in (1.10) from q ∈ [Q, 2Q) with LA+1 ≤ Q ≤ R. We have 2 X Q≤q<2Q q X a=1 (a,q)=1 E(N, N ′ , γ, q, a)2 ≪ XX q X sup I a N <n≤N ′ {γn}∈I n≡a mod q + Λ(n) − |I|  XX   q a X Λ(n) N <n≤N ′ n≡a mod q X N <n≤N ′ n≡a mod q Λ(n) − 2 N′ − N   = T1 (Q) + T2 (Q), ϕ(q)  say. Since T2 (Q) is covered by a slight variant of the discussion in [6, Chapter 29], we focus our attention on T1 (Q). By Lemma 7, 2   2 T1 (Q) ≪ X Q≤q<2Q q X a=1 (a,q)=1  L−2A   X N <n≤N ′ n≡a mod q = T3 (Q) + T4 (Q),  Λ(n)  + X Q≤q<2Q q X X 1   h A a=1 (a,q)=1 h≤L X N <n≤N ′ n≡a mod q  Λ(n)e(γnh)   say. The Brun-Titchmarsh Theorem gives a satisfactory bound for T3 (Q). Applying Cauchy’s inequality to T4 (Q), we get 2  X 1 X 1  T4 (Q) ≤  h h A A  h≤L h≤L 2 ≪ (log L) X Q≤q<2Q X Q≤q<2Q 1 ϕ(q) q X X Λ(n)e(γnh) a=1 N <n≤N ′ (a,q)=1 n≡a mod q 2 X X Λ(n)χ(n)e(γnh) , χ mod q N <n≤N ′ for some h ≤ LA . From this point, we can conclude the proof by following, with slight changes, the argument in [6, pp. 170-171].  24 ROGER C. BAKER AND LIANGYI ZHAO 6. Proof of Theorems 2 and 3 −1 Proof of Theorem 2. Let γ = α and N ≥ C1 (α, t), 0 < ε < C2 (α, t). By Dirichlet’s theorem, there is a reduced fraction b/r satisfying (1.7). Our hypothesis on α implies that N −3/4 ≥ kγrk ≫ r−3 , r ≫ N 1/4 . Let h′′1 , · · · , h′′l be the first l primes in (l, ∞). Any translate H = {h′1 , · · · , h′k } + h, h ∈ N with {h′1 , · · · , h′k } ⊂ {h′′1 , · · · , h′′l }, is an admissible set. Using (i) of Lemma 7, we choose h′1 , · · · , h′k so that (6.1) k ≥ εγl and for some real η, −γh′m ∈ (η, η + εγ) (mod 1) for every m = 1, · · · , k. Now choose h ∈ N, h ≪γ 1 so that hγ ∈ (η − εγ, η) (mod 1). Thus, writing hm = h′m + h, we have −γhm = −γh′m − γh ∈ (0, 2εγ) (mod 1). We apply Theorem 1 to the set A = {n ∈ [N, 2N ) : γm ∈ I (mod 1)} where I = (γβ − γ, γβ), taking q0 = q1 = 1, s = 1, ̺(n) = X(P; n), θ = 1/4 − ε, b = 1 − 2ε, Y = γN, Y1,m = lm Z2N N 1 lm Y dt = (1 + o(1)) . log t Lγ Here Jm , lm are the interval J and its length l in Lemma 6 (with εγ in place of ε), so that γ > lm > γ(1 − 2ε). Since (1.2) can be proved in a similar (but simpler) fashion to (1.5), we only show that (1.5) holds. We can rewrite this in the form (6.2) X 2 µ (q)τ3k (q) X N +hm ≤p<2N p≡aq mod q γp∈Jm mod 1 q≤x1/4−ε lm 1− ϕ(q) Z2N dt ≪ N L−k−ε . log t N The function E(N, N ′ , γ, q, a) appearing in Theorem 4 is not quite in the form that we need. However, discarding prime powers and using partial summation in the standard way, we readily deduce a variant of (6.2) from Theorem 4, in which N L−A appears in place of N L−k−ε , and the weight µ2 (q)τ3k (q) is absent. We then obtain (6.2) by using Cauchy’s inequality; see [9, (5.20)] for a very similar computation. We are now in a position to use Theorem 1, obtaining a set of S of t primes in A ∩ [N, 2N ), which of course have the form [αn + β], with D(S) ≤ hk − h1 ≤ h′′l provided that 2t − 2 . (6.3) Mk > (1 − 2ε)(1/4 − ε) We take l to be the least integer with log(εγl) ≥ 2t − 2 +C (1 − 2ε)(1/4 − ε) GAPS BETWEEN PRIMES IN BEATTY SEQUENCES 25 for a suitable absolute constant C, so that (6.3) follows from (6.1) and (1.11). Therefore, γl ≪ exp(8t), l ≪ α exp(8t), D(S) ≪ l log l ≪ α(t + log α) exp(8t), completing the proof.  In the proof of Theorem 3, we shall need the following. Lemma 20. Let D be as in Lemma 18 and let ω0 (t) denote Buchstab’s function. (i) The points of D lie in two triangles A1 , A2 , where A1 has vertices       2 3 2 2 5 5 , , , , , 21 21 7 14 7 7 and A2 has vertices  (ii) For j = 1, 2, let Ij = Z Aj 1 3 , 2 14      3 2 1 1 , , . , , 7 7 2 4 1 ω0 α1 α22  1 − α1 − α2 α2  dα1 dα2 . Then I1 < 0.03925889 and I2 < 0.0566295. Proof. Let (α1 , α2 ) ∈ D. If α1 + α2 > 5/7, then we have 1 5 , α1 + 2α2 ≤ 1, α1 ≤ . 7 2 This defines a triangle which is easily verified to be A2 . If α1 + α2 ≤ 5/7, then as α2 is bad, we have in turn α1 + α2 > α1 + α2 < 4 3 2 , α1 < , α1 < . 7 7 7 Altogether, we have 2 5 , α1 < , α2 < α1 . 7 7 This defines a triangle which we can verify to be A1 . This proves (i). Now (ii) requires a computer calculation, which was kindly carried out by Andreas Weingartner.  α1 + 2α2 > Proof of Theorem 3. With a different value of l, we choose h′′1 , · · · , h′′l and h1 , · · · , hk exactly as in the proof of Theorem 2. In applying Theorem 1, we also take I, A, q0 , q1 , Y , Jm , lm as in that proof, but now θ = 2/7 − ε, s = 5, a = 3; the functions ̺1 (n), · · · , ̺5 (n) are given in Lemma 18. There is little difficulty in verifying (1.2) by a similar but simpler version of the proof of (1.5). So we concentrate on (1.5). We recall that this can be rewritten as (6.4) X q≤xθ µ2 (q)τ3k (q) X n≡aq mod q γn∈Jm mod 1 N +hm ≤n<2N ̺g (n) − Yg,m ≪ N L−k−ε . ϕ(q) We define Yg,m by Yg,m = lm X ̺g (n). N ≤n<2N It is well known that (6.5) Yg,m = l m cg N (1 + o(1)) , L 26 ROGER C. BAKER AND LIANGYI ZHAO where cg is given by a multiple integral. In fact, we have   Z 1 1 − α1 − α2 c1 + c2 + c3 − c4 − c5 = 1 − dα1 dα2 . ω0 α1 α22 α2 α2 ∈D Similar calculations are found in [8, Chapter 1]. Fix m and g. By analogy with the proof of Theorem 4, we can obtain (1.5) by showing X (6.6) q≤N 2/7−ε X ′ N ≤n<N n≡aq mod q ̺g (n) − 1 ϕ(q) X N ≤n<N ′ ̺g (n) ≪ N L−A for every A > 0 and X (6.7) q≤N 2/7−ε X N ≤n<N ′ n≡aq mod q ̺g (n)e(γnh) ≪ N L−A for 1 ≤ h ≤ LA+1 and for every A > 0. Again adapting the argument of Theorem 4, we see that (6.7) is a consequence of Lemma 19. For (6.6), it suffices to show, recalling Lemma 8, that for arbitrary ηχ ≪ 1 and Q ≤ N 2/7−ε , X X X⋆ ηχ ̺g (n)χ(n) ≪ QN L−A (6.8) Q≤q<2Q χ mod q N ≤n<N ′ for every A > 0. This can be readily deduced from the Siegel-Walfisz theorem for Q ≤ L2A , so we assume that Q > L2A . We apply Lemma 15 with W (n) = X⋆ X ηχ χ(n) Q≤q<2Q χ mod q if N ≤ n < N ′ and W (n) = 0 otherwise. For example, when g = 3, the left-hand side of (6.8) is X X W (p1 p2 p3 n4 ) = S ∗ (p1 p2 p3 , (2N )1/7 ). α3 ∈E3 (α1 ,α2 )∈E2 \D ′ N ≤p1 p2 p3 n4 <N (n4 ,P ((2N )1/7 )=1 α3 ∈E3 (α1 ,α2 )∈E2 \D We shall show that (4.10) and (4.11) hold with Y = QN L−A−3 , c = 4/7 and d = 1/7. (We could reduce the constraints on c and d, but that would not be useful in the present context.) Once we have done this, we can follow the proof of Lemma 19 to prove (6.8). To prove (4.10), we use the Polya-Vinogradov bound for character sums to obtain X X X⋆ X X am W (nk) = ηχ χ(mk) m≤2N 4/7 k m≤2N 4/7 ≪L X Q≤q<2Q χ mod q N ≤mk<N ′ X m≤2N 4/7 Q≤q<2Q q 1/2 ≪ LQ3/2 N 4/7−ε ≪ QN L−A−3 . 27 GAPS BETWEEN PRIMES IN BEATTY SEQUENCES Now to prove (4.11), we note that by the method of [8, Section 3.2] mentioned earlier, it suffices to show that X X am bk W (mk) ≪ QN L−A M≤m<2M K≤k<2K whenever |am | ≤ 1 and |bk | ≤ τ (k), N 4/7 ≪ M ≪ N 5/7 , M K ≍ N . That is, it suffices to show that (6.9) X X⋆ X X am χ(m) K≤k<2K Q≤q<2Q χ mod q M≤m<2M bk χ(k) ≪ QN L−A . Following the proof of (6) in [6, Chapter 28], the left-hand side of (6.9) is   ≪ L(M + Q2 )1/2 (K + Q2 )1/2 kak2 kbk2 ≪ L3 N 1/2 + M 1/2 Q + Q2 N 1/2 ≪ QN L−A , since L3 Q−1 N ≪ L3−A N , L3 M 1/2 N 1/2 ≪ L3 N 6/7 ≪ N L−A and L3 QN 1/2 ≪ L3 N 11/14 ≪ N L−A . This proves (1.5) with the present choice of A, Yg,m , etc. Applying Theorem 1, we find that there is a set S of t primes in A (and thus of the form [αm + β]) having diameter D(S) ≤ hk − h1 ≪ l log l provided that Mk > 2t − 2 . b(2/7 − ε) Here b must have the property b1,m + b2,m + b3,m − b4,m − b5,m ≥ b > 0; that is, lm (c1 + c2 + c3 − c4 − c5 ) ≥ bγ > 0. We can choose  b = (1 − 2ε) 1 − 1 ω0 α1 α22 Z α2 ∈D  1 − α1 − α2 α2  Using Lemma 20, we see that  dα1 dα2  . b > 0.90411. Now we proceed just as the proof of Theorem 2. We may choose any l for which log(εγl) ≥ 2t − 2 +C 0.90411(2/7 − ε) for a suitable constant C, and now it is a simple matter to deduce that D(S) < C4 α(log α + t) exp(7.743t), where C4 is an absolute constant.  Acknowledgments. This work was done while L. Z. held a visiting position at the Department of Mathematics of Brigham Young University (BYU). He wishes to thank the warm hospitality of BYU during his thoroughly enjoyable stay in Provo. 28 ROGER C. BAKER AND LIANGYI ZHAO References [1] R. C. Baker, Diophantine Inequalities, London Mathematical Society Monographs New Series, vol. 1, Oxford University Press, Oxford, 1986. [2] R. C. Baker and W. D. Banks, Character sums with Piatetski-Shapiro sequences, Q. J. Math. 66 (2015), no. 2, 393–416. [3] R. C. Baker and A. Weingartner, A ternary Diophantine inequality over primes, Acta Arith. 162 (2014), no. 2, 159–196. [4] W. D. Banks and I. E. Shparlinski, Prime numbers with Beatty sequences, Colloq. Math. 115 (2009), no. 2, 147–157. [5] L. Chua, S. Park, and G. D. Smith, Bounded gaps between primes in special sequences, Proc. Amer. Math. Soc. 143 (2015), no. 11, 4597–4611. [6] H. Davenport, Multiplicative Number Theory, Third Edition, Graduate Texts in Mathematics, Springer-Verlag, New York, 2000. [7] D. A. Goldston, S. W. Graham, J. Pintz, and C. Y. Yıldırım, Small gaps between products of two primes, Proc. London Math. Soc. (3) 98 (2009), no. 3, 741–774. [8] G. Harman, Prime Detecting Sieves, London Mathematical Society Monographs Series, vol. 33, Princeton University Press, Princeton, 2007. [9] J. Maynard, Small gaps between primes, Ann. of Math. (2) 181 (2015), no. 1, 383–413. [10] , Dense clusters of primes in subsets, Preprint (2016). arXiv: 1407.4897. [11] D. H. J. Polymath, Variants of the Selberg sieve, and bounded intervals containing many primes, Res. Math. Sci. (2014), 1:12. [12] Y. Zhang, Bounded gaps between primes, Ann. of Math. (2) 179 (2014), no. 3, 1121–1174. Roger C. Baker Department of Mathematics Brigham Young University Provo, UT 84602, U. S. A. Email: baker@math.byu.edu Liangyi Zhao School of Mathematics and Statistics University of New South Wales Sydney, NSW 2052 Australia Email: l.zhao@unsw.edu.au