Small gaps between the Piatetski-Shapiro primes

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by  Hongze Li

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)