SMALL GAPS BETWEEN THE PIATETSKI-SHAPIRO PRIMES 6 HONGZE LI AND HAO PAN 1 0 2 Abstract. Suppose that 1 < c < 9/8. For any m 1, there exist infinitely r many n such that ≥ a c c c M [n ], [(n+1) ], ..., [(n+k0) ] { } contains at least m+1 primes, if k0 is sufficiently large (only depending on m 0 and c). 1 ] T N 1. Introduction . h at Letpk denotethek-thprime. Inviewoftheprimenumber theorem, theexpected m value of the prime gap pk+1 pk is near to logpk. In 1940, Erd˝os [3] showed that − [ p p liminf k+1 − k c0 5 k→∞ logpk ≤ v 5 for some constant 0 < c < 1. Later, the value of c was successively improved. In 0 0 0 2009, using a refinement of the Selberg sieve method, Goldston, Pintz and Yıldırım 9 [5] proved that 4 0 p p k+1 k . liminf − = 0. 1 k→∞ logpk 0 6 Furthermore, under the Elliott-Halberstam conjecture, they also showed that 1 : liminf(p p ) 16. v k→∞ k+1 − k ≤ i X In fact, if the twin prime conjecture is true, there are infinitely many k such that r p p = 2. In 2014, Zhang [18] for the first time proved unconditionally that a k+1 k − liminf(p p ) 7 107, k+1 k k→∞ − ≤ × i.e., the gap p p can be infinitely often bounded by a finite number. One k+1 k − ingredient ofZhang’sproofisanimprovement oftheBombieri-Vinogradovtheorem for the smooth moduli. Subsequently, the bound for p p was rapidly reduced. k+1 k − In 2015, with the help of a multi-dimensional sieve method, Maynard [10] gave a quite different proof of Zhang’s result and improved the bound to 600. Nowadays, 2010 Mathematics Subject Classification. Primary 11N05; Secondary 11L20,11N36, 11P32. Key words and phrases. Piatetski-Shapiroprime, Maynard-Taotheorem,Selberg sieve,Prime gaps. 1 2 HONGZELI ANDHAOPAN the best known bound is 246 [14]. Furthermore, using the Maynard’s sieve method, Maynard and Tao independently found that p p k+m k liminf − C m k→∞ logpk ≤ for any m 1, where C is a positive constant only depending on m. Now, basing m ≥ on the discussions of Maynard and Tao, the bounded gaps between the primes of some special forms are also investigated. For examples, the Maynard-Tao theorem has been extended to: the primes p with p+2 being analmost prime [8], the primes of the form [αn+β] [2], the primes having a given primitive root (under GRH) [13], the primes p = a2+b2 with a ǫ√p [16], etc.. In fact, the Maynard-Tao theorem is ≤ valid for any subset of primes with positive relative upper density satisfying some mean value theorem, we refer the reader to [9]. Let Nc = [nc] : n N , { ∈ } where [x] = max m x : m Z . That is, Nc the set of the integers of the form { ≤ ∈ } [nc]. In 1953, Piatetski-Shapiro [12] showed that there are infinitely many primes lying in Nc provided 1 < c < 12/11. In fact, he got x1/c 1 = (1+o(1)) . logx p≤x,p∈Nc X The primes lying in Nc are usually called Piatetski-Shapiro primes, so that the Piatetski-Shapiro primes form a thin set of primes, and the average gap between the primes in [1,x] Nc is about x1−1/clogx. The upper bound for c satisfying ∩ x1/c 1 ≫ logx p≤x,p∈Nc X has been improved many times during the past six decades. Now the best admis- sible range of c is (1,243/205) by Rivat and Wu [15]. It is natural to ask how small the gaps between the primes in Nc can be. Let p(c) be the k-th prime in Nc. Of course, we can’t expect that the p(c) p(c) can be k k+1− k bounded by a finite number, since (n+1)c nc cnc−1 implies − ≈ (c) (c) (c) 1−1/c p p c+o(1) p . k+1 − k ≥ · k However, in this paper, we shall show(cid:0) that if (cid:1)1 <(cid:0)c <(cid:1)9/8, then for any m 1, ≥ (c) (c) p p liminf k+m − k < C , k→∞ p(c) 1−γ m k where γ = 1/c and Cm > 0 is a const(cid:0)ant(cid:1)only depending on m and c. That is, we have (c) (c) (c) 1−γ p p C p (1.1) k+m − k ≤ m k (cid:0) (cid:1) SMALL GAPS BETWEEN THE PIATETSKI-SHAPIRO PRIMES 3 for infinitely many k. Our main result is that Theorem 1.1. Suppose that 1 < c < 9/8 and m 1. If ≥ k eCm 0 ≥ where C > 0 is a constant only depending on c, then there are infinitely many n such that the set [nc], [(n+1)c], ..., [(n+k )c] 0 { } contains at least m+1 primes. Althoughtherealsoexists ameanvaluetheoremforthePiatetski-Shapiro primes [11], the main difficulty in the proof of Theorem 1.1 is that the Piatetski-Shapiro primes are too sparse. According to the Maynard sieve method, we have to use the weight 2 λ d0,...,dk (cid:18) di|[X(n+i)c] (cid:19) for0≤i≤k0 rather than 2 λ d0,...,dk (cid:18) diX|n+hi (cid:19) for0≤i≤k0 which is applicable to those subsets of primes with positive relative upper density. Then our problem can be reduced to consider ̟([(n+i )c]) 0 X≤n≤2X, n≡b (mod W) X di|[(n+i)c]for0≤i≤k0 for some 0 i k , where W is the product of the primes less than logloglogX 0 0 ≤ ≤ and ̟(n) = logn or 0 according to whether n is prime or not. However, the above sum is not easy to estimate. Our strategy is to construct a suitable smooth function χ with 0 χ(n) 1. Then it becomes possible to estimate ≤ ≤ χ([(n+i )c]) ̟([(n+i )c]). 0 0 · X≤n≤2X, n≡b (mod W) X di|[(n+i)c]for0≤i≤k0 In fact, when χ([(n+i )c]) > 0, [nc], [(n+1)c], ..., [(n+k )c] form an arithmetic 0 0 progression. Furthermore, as we shall see later, we only need to use the Siegel- Walfisz theorem, rather than any mean value theorem. In the next section, we shall extend the Maynard sieve method to the Piatetski- Shapiroprimes. TheproofofTheorem1.1willbeconcludedinSection3. Through- out this paper, f(x) g(x) means f(x) = O g(x) as x tends to . And ǫ ≪ ∞ ≪ means the implied constant in only depends on ǫ. Furthermore, as usual, we ≪ (cid:0) (cid:1) define e(x) = exp(2π√ 1x) for x R and x = min x m . m∈Z − ∈ k k | − | 4 HONGZELI ANDHAOPAN 2. Maynard’s sieve method for the Piatetski-Shapiro primes Let 1 σ = min c 1, 9 8c , 0 200 · { − − } and let k be a large integer to be chosen later. Suppose that f(t ,t ,...,t ) is a 0 0 1 k0 symmetric smooth function whose support lies on the area (t ,...,t ) : t ,...,t 0, t + +t 1 . { 0 k0 0 k0 ≥ 0 ··· k0 ≤ } Let R = Xσ0 and define logd logd logd k0 λ = f 0, 1,..., k0 µ(d ). d0,d1,...,dk0 logR logR logR j (cid:18) (cid:19)j=0 Y Clearly λ = 0 provided d d d > R. d0,...,dk0 0 1··· k0 Now suppose that X is sufficiently large and let W = p. p≤logloglogX Y For convenience, below we write n X for X n 2X. The following lemma is ∼ ≤ ≤ one of the key ingredients of Maynard’s sieve method. Lemma 2.1. λ λ d0,...,dk0 e0,...,ek0 [d ,e ] [d ,e ] d0,...,dkX0,e0,...,ek0 0 0 ··· k0 k0 W,[d0,e0],...,[dk0,ek0] coprime 1+o(1) Wk0+1 ∂k0+1f(t ,...,t ) 2 = 0 k0 dt dt dt . (logR)k0+1 · φ(W)k0+1 ZRk0+1(cid:18) ∂t0···∂tk0 (cid:19) 0 1··· k0 (cid:3) Proof. See [14, Lemma 30]. Set 1 γ = . c The conventional way to capture the Piatetski-Shapiro primes is to use the Fourier expansion of x and the fact { } 1, if n Nc, [(n+1)γ] [nγ] = ∈ − 0 otherwise, ( when nγ > 0. However, notice that [(n+1)γ] [nγ] = 1 if and only if { } − nγ > 1 ((n+1)γ nγ) = 1 γnγ−1 +O(nγ−2). { } − − − So if nγ lies in the short interval [1 δnγ−1,1) for some constant 0 < δ < γ, { } − then n Nc. The following lemma is a classical result in number theory, and is ∈ frequently used for the problems of Diophantine approximation. SMALL GAPS BETWEEN THE PIATETSKI-SHAPIRO PRIMES 5 Lemma 2.2. Suppose that 0 α < β 1 and ∆ > 0 with 2∆ < β α. For any ≤ ≤ − r 1, there exists a smooth function ψ with the period 1 satisfying that ≥ (i) ψ(x) = 1 if α + ∆ x β ∆, ψ(x) = 0 if x α or x β, and ≤ { } ≤ − { } ≤ { } ≥ ψ(x) [0,1] otherwise; ∈ (ii) ψ(x) = (β α ∆)+ a e(jx), j − − |j|≥1 X where 1 1 a min , β α ∆, . j ≪r j − − ∆r j r+1 (cid:26)| | | | (cid:27) (cid:3) Proof. This is [17, Lemma 12 of Chapter I]. Lemma 2.3. Let f (x) = jxγ +C x+C x1−γ, j 1 2 where C ,C are constants and C = o(X2γ−1). Suppose that σ > 0 and 9(1 1 2 2 | | − γ) + 12σ < 1. Then there exists a sufficiently small ǫ > 0 (only depending on c and σ), such that X1−γ min 1, Λ(n)e f (n) X1−σ−ǫ, (2.1) j ǫ H · ≪ (cid:26) (cid:27) j∼H(cid:12)n∼X (cid:12) X(cid:12)X (cid:0) (cid:1)(cid:12) (cid:12) (cid:12) where 1 H X1−γ+σ+ǫ. (cid:12) (cid:12) ≤ ≤ Proof. This is just (2.10) of [1], although Balog and Friedlander only considered f (x) = jxγ +C x. In fact, in their proof, only the fact f′′(x) γ(γ 1) jxγ−2 is j 1 j ≈ − · used. So the same discussions are also valid for f (x) = jxγ +C x+C x1−γ. (cid:3) j 1 2 Suppose that ǫ is the ǫ corresponding to c and σ = 8σ in Lemma 2.3. Let 0 0 c cδ 0 δ = , η = . 0 0 9 16k 0 Suppose that χ is the smooth function described in Lemma 2.2 with β α α = 1 2δ Xγ−1, β = 1 δ Xγ−1, ∆ = − , r = [100ǫ−1]. − 0 − 0 4 0 And let ψ be the smooth function described in Lemma 2.2 with β α α = η , β = 2η , ∆ = − , r = [100σ−1]. 0 0 4 0 Define logn, if n is prime, ̟(n) = 0, otherwise. ( 6 HONGZELI ANDHAOPAN For n Nc and h Z, let ∈ ∈ s (n) = [([nγ]+h+1)c]. h Proposition 2.1. k0 2 ̟(s (n))χ(s (n)γ)ψ(cs (n)1−γ) λ h h h · d0,d1,...,dk0 n∼XX, n∈Nc (cid:18)Xh=0 (cid:19) (cid:18)diX|si(n) (cid:19) sj(n)≡1 (mod W) 0≤i≤k0 for 0≤j≤k0 1+o(1) Wk0−1 9δ η (k +1)Xγ ∂k0f(0,t ,...,t ) 2 = 0 0 0 1 k0 dt dt . (logR)k0 · φ(W)k0+1 · 16 ZRk0 (cid:18) ∂t1···∂tk0 (cid:19) 1··· k0 (2.2) Assume that n X and ∼ δ 1 0Xγ−1 > nγ > 1 4δ Xγ−1. 0 − 4 { } − Then [(n+1)γ] = [nγ +γnγ−1 +O(nγ−2)] > [nγ]. So n Nc. Furthermore, if h is an integer with h k , we also have 0 ∈ | | ≤ ([nγ]+h+1)c = (nγ nγ +h+1)c −{ } =n+chn1−γ +c(1 nγ )n1−γ +O(n1−2γ) −{ } =n+h [cn1−γ]+h cn1−γ +c(1 nγ )n1−γ +O(n1−2γ). · ·{ } −{ } Clearly c(1 nγ )n1−γ 4cδ Xγ−1 (2X)1−γ 23−γcδ 0 0 −{ } ≤ · ≤ and cδ cδ c(1 nγ )n1−γ 0Xγ−1 X1−γ = 0. −{ } ≥ 4 · 4 On the other hand, if η < cn1−γ < 2η , 0 0 { } then we have cδ h cn1−γ = h cn1−γ 2η k < 0 0 0 { · } ·{ } ≤ 7 for those 0 h k , i.e., 0 ≤ ≤ cδ δ 0 h cn1−γ +c(1 nγ )n1−γ 1 0. 4 ≤ ·{ } −{ } ≤ − 2 Similarly, if k h < 0, then 0 − ≤ cδ η h cn1−γ 2η k 0 0 0 0 − ≥ ·{ } ≥ − ≥ − 8 and cδ h cn1−γ = 1+h cn1−γ 1 0. { · } ·{ } ≥ − 8 SMALL GAPS BETWEEN THE PIATETSKI-SHAPIRO PRIMES 7 It follows that cδ 0 < h cn1−γ +c(1 nγ )n1−γ 23−γcδ < 1 0 9 ·{ } −{ } ≤ and cδ h cn1−γ +c(1 nγ )n1−γ 1+ 0. { · } −{ } ≥ 8 Noting that (nγ +h)c = n+h cn1−γ +O(n1−2γ), · we get [(nγ +h)c], if 0 h k , [([nγ]+h+1)c] = n+h [cn1−γ] = ≤ ≤ 0 (2.3) · [(nγ +h)c]+1, if k h < 0. ( 0 − ≤ That is, Lemma 2.4. If n X, ∼ δ 1 0Xγ−1 > nγ > 1 4δ Xγ−1 0 − 4 { } − and η < cn1−γ < 2η , 0 0 { } then for h [ k ,k ], we have 0 0 ∈ − s (n) = n+h [cn1−γ]. h · Assume that n X and n Nc. Suppose that 0 h k and let m = s (n), 0 h ∼ ∈ ≤ ≤ i.e, m = [([nγ]+h+1)c]. Clearly mγ h 1 < [nγ] < (m+1)γ h 1 = mγ h 1+O(mγ−1), − − − − − − when χ(mγ) > 0. It follows that [nγ] = [mγ] h. (2.4) − Assume that χ(mγ), ψ(cm1−γ) > 0. Then by Lemma 2.4, [([nγ]+h∗ +1)c] = [([mγ] h+h∗ +1)c] = m+(h∗ h) [cm1−γ] − − · for each 0 h∗ k . 0 ≤ ≤ Furthermore, we must have nγ > 0. In fact, if nγ = 0, then { } { } n =([mγ] h)c = m (h+1) cm1−γ +c(1 mγ )m1−γ +O(m1−2γ) − − · −{ } =m (h+1) [cm1−γ]+c(1 mγ )m1−γ (h+1) cm1−γ +O(m1−2γ). − · −{ } − ·{ } (2.5) 8 HONGZELI ANDHAOPAN Note that 2δ Xγ−1 1 mγ δ Xγ−1 and cm1−γ 2η , (2.5) is impossible 0 0 0 ≥ −{ } ≥ { } ≤ since 2(k +1)η < cδ /2 < 1/4. 0 0 0 Since n Nc, in view of (2.3) and (2.4), we must have ∈ n = [([mγ] h+1)c] = [(mγ h)c]+1 − − provided h 1. Then ≥ mγ h < nγ ((mγ h)c +1)γ. (2.6) − ≤ − But ((mγ h)c +1)γ = (mγ h)+γmγ−1 +O(m−1), − − so in view of (2.4), we must have nγ > mγ 1 2δ Xγ−1. 0 { } { } ≥ − Moreover, we also have n1−γ = (m h [cm1−γ])1−γ = m1−γ +O(m1−2γ). − · Thus Lemma 2.5. Suppose that n X, n Nc and 1 h,h∗ k . If 0 ∼ ∈ ≤ ≤ χ(s (n)γ), ψ(c s (n)1−γ) > 0, h h · then sh∗(n) = sh∗−h(sh(n)). We also have nγ 1 2δ Xγ−1 0 { } ≥ − and cn1−γ = c s (n)1−γ +O(X1−2γ). h { } { · } Conversely, assume that 0 < h k and n = s (m) with χ(mγ),ψ(cm1−γ) > 0. 0 −h ≤ Then nγ [mγ] h+1 < (n+1)γ. ≤ − It is also impossible that nγ = 0, since if so, then { } n = ([mγ] h+1)c = m h [cm1−γ]+c(1 mγ )m1−γ h cm1−γ +O(m1−2γ), − − · −{ } − ·{ } which will lead to a similar contradiction as (2.5). So we also have [mγ] = [nγ]+h, and m = [([mγ]+1)c] = [([nγ]+h+1)c] = s (n). (2.7) h SMALL GAPS BETWEEN THE PIATETSKI-SHAPIRO PRIMES 9 According to Lemma 2.5 and (2.7), for each 0 h k , we get 0 ≤ ≤ 2 ̟(s (n))χ(s (n)γ)ψ(cs (n)1−γ) λ h h h d0,...,dk0 n∼XX, n∈Nc (cid:18)diX|si(n) (cid:19) sj(n)≡1 (mod W) 0≤i≤k0 forany0≤j≤k0 2 = ̟(m)χ(mγ)ψ(cm1−γ) λ +O(logX). (2.8) d0,...,dk0 sj−h(m)m≡X∼1X(mod W) (cid:18)di0|s≤Xi−i≤hk(m0 ) (cid:19) forany0≤j≤k0 Below we just consider the case h = 0, since all other cases are similar. Clearly 2 ̟(n)χ(nγ)ψ(cn1−γ) λ (2.9) d0,...,dk0 sj(n)≡n1X∼(Xmod W) (cid:18)0d≤iX|sii≤(nk0) (cid:19) forany0≤j≤k0 = λ λ ̟(n)χ(nγ)ψ(cn1−γ). 1,d1,...,dk0 1,e1,...,ek0 d1,...,dkX0,e1,...,ek0 n,s1(n),...,sk0nX(∼nX)≡1 (mod W) [di,ei]|si(n)for1≤i≤k0 (2.10) Fix d ,...,d ,e ,...,e with d d ,e e R. We need to consider 1 k0 1 k0 1··· k0 1··· k0 ≤ ̟(n)χ(nγ)ψ(cn1−γ). (2.11) n∼X n,s1(n),...,sk0X(n)≡1 (mod W) [di,ei]|si(n)for1≤i≤k0 First, we claim that the sum (2.11) is 0 unless those [d ,e ] are pairwise coprime. i i In fact, assume that [d ,e ] and [d ,e ] have a common prime divisor p. Clearly i1 i1 i2 i2 p ∤ W, i.e., p > k . Recall that for each 1 i k , 0 0 ≤ ≤ s (n) = n+i [cn1−γ] i · provided χ(nγ),ψ(cn1−γ) > 0. Thus we must have p [cn1−γ] and p n. It is | | impossible since n is prime now. Moreover, clearly [d ,e ] is coprime to W for each i i 1 i k . 0 ≤ ≤ 10 HONGZELI ANDHAOPAN Below we assume that W,[d ,e ],...,[d ,e ] are pairwise coprime. Clearly 1 1 k0 k0 ̟(n)χ(nγ)ψ(cn1−γ) = ̟(n)χ(nγ)ψ(cn1−γ) n∼X n∼X n,si(n)≡X1 (mod W) n≡1 X(mod W) [di,ei]|si(n)for1≤i≤k0 [cn1−γ]≡0 (mod W) n+i·[cn1−γ]≡0 (mod [di,ei]) for1≤i≤k0 1 W−1 r [cn1−γ] = ̟(n)χ(nγ)ψ(cn1−γ) e 0 W W nX∼X (cid:18) rX0=0 (cid:18) (cid:19)(cid:19) n≡1 (mod W) 1 [di,ei]−1 r (n+i[cn1−γ]) i e · [d ,e ] [d ,e ] 1≤Yi≤k0(cid:18) i i rXi=0 (cid:18) i i (cid:19)(cid:19) 1 = ̟(n)χ(nγ)ψ(cn1−γ) W[d ,e ] [d ,e ] 1 1 ··· k0 k0 0≤rX0≤W−1 nX∼X 0≤ri≤[di,ei]−1n≡1 (mod W) k0 r r k0 r i e n i +[cn1−γ] 0 + i . (2.12) · [d ,e ] W [d ,e ] (cid:18) i=1 i i (cid:18) i=1 i i (cid:19)(cid:19) X X Fix r ,r ,...,r and let 0 1 k0 k0 r r k0 r i i 0 i θ = , θ = + . 1 2 [d ,e ] W [d ,e ] i i i i i=1 i=1 X X Lemma 2.6. For any H 2, ≥ e(hx) e( θ x ) = c(θ) +O(Φ(x;H)logH), (2.13) − { } h+θ |h|≤H X where c(θ) θ and Φ(x;H) = (1+H x )−1. | | ≤ k k k k Proof. (2.13) is an easy exercise for the Fourier series. We leave its proof to the (cid:3) reader. Let H = X2σ0. Note that Φ(cn1−γ;H) H−1 if ψ(cn1−γ) > 0. Then ≪ ̟(n)χ(nγ)ψ(cn1−γ)e(nθ +[cn1−γ]θ ) 1 2 n∼X X n≡1 (mod W) e nθ +(h+θ ) cn1−γ = ̟(n)χ(nγ)ψ(cn1−γ) c(θ ) 1 2 · +O(H−1logH) . 2 · h+θ n∼X (cid:18) |h|≤H (cid:0) 2 (cid:1) (cid:19) X X n≡1 (mod W) (2.14)

Small gaps between the Piatetski-Shapiro primes

Similar Small gaps between the Piatetski-Shapiro primes