Chen primes in arithmetic progressions

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by  Paweł Karasek

Chen primes in arithmetic progressions
Pawe l Karasek
Abstract
6 Wefinda lower bound for thenumberof Chen primes in thearithmetic progression amodq,
1 where (a,q)=(a+2,q)=1. Ourestimate is uniform for q≤logMx,where M >0 is fixed.
0
2

n
a
1 Introduction
J

1
3
The famoustwinprimeconjectureassertsthatthereexistinfinitely manyprimespsuchthatp+2

]
is also a prime. It is a common perception that our current methods are insufficient to prove this
T
supposition, although in some way one is able to get very close to it. For example in [1] and [2], J.
N
Chen proved the following famous result.
.
h
t Theorem 1.1. There are infinitely many primes p such that p+2 has at most two prime factors,
a
each greater or equal than p1/10.
m
[ Infact,Chenshowedthatthenumberofsuchprimesinaninterval[1,x]isgreaterthancx/log2x,

where c is some positive constant. In [3], B. Green and T. Tao showed that there are infinitely many
2
v 3-term arithmetic progressions among them. This result was extended by B. Zhou in [4], where it is
3 proven that in the same circumstances one can find arithmetic progressions of any given length.
7 ItisnaturaltoaskwhetherChenprimes(ortwinprimes)areequallydistributedamongarithmetic
8
progressions as primes do. For example, using Cram´er’s random model we are led to the following
2
conjecture about twin primes (cf. section 2 for the notation).
0
.
1 Conjecture 1.2. Let a,q be positive integers such that (a(a+2),q)=1. Then
0
6 x
#{twin primes p : p≡a (mod q)}=(2Π +o (1)) .
1 2 q ϕ (q)log2x
2
:
v
i The goal of this paper is to prove a uniform (in a Siegel−Walfisz manner) lower bound for an
X
analogous count of Chen primes. We will accomplish this using techniques developed by Chen with
r slight modifications (specifically, we shall follow the discussion in [7]). The main result can be stated
a
as follows:
Theorem 1.3. Let M > 0 and let a,q be positive integers such that (a,q) = (a +2,q) = 1 and
q ≤logMx. Then
x
Λ (n)1 (n+2)1 ≫ .
a,q P2 (n+2,P(x1/8))=1 M ϕ2(q)logx
x/2≤Xn≤x−2
After removing the small contribution of the prime powers, we can also convert this into a more
elegant form.
1
1 INTRODUCTION
Corollary 1.4. Let M > 0 and let a,q be positive integers such that (a,q) = (a+2,q) = 1 and
q ≤logMx. Then
x
1 ≫ .
n∈PCh M ϕ (q)log2x
n6x 2
X
n a (q)
≡
The proof of Theorem 1.3 is based on the following observation.
Lemma 1.5. For x2/3 <n6x we have
1 1
1 (n)≥1− 1 − 1 − 1 .
P2 2 p|n 2 n=p1p2p3 p2|n
p≤Xx1/3 p1≤x1X/3<p2≤p3 p≤Xx1/3
Proof. This is an easy case analysis.
By Lemma 1.5 one can reduce our task into the following estimation.
1 1 x
A − A − A − A ≫ , (1.1)
1 2,p 3 4,p
2 2 ϕ (q)logx
2
x1/8≤Xp≤x1/3 x1/8≤Xp≤x1/3
where
A := Λ (n)1 , (1.2)
1 a,q (n+2,P(x1/8))=1
x/2≤Xn≤x−2
A := Λ (n)1 , (1.3)
2,p a,q (n+2,P(x1/8))=1
x/2≤Xn≤x−2
pn+2
|
A := Λ (n) 1 . (1.4)
3 a,q n+2=p1p2p3
x/2≤Xn≤x−2 x1/8≤p1≤Xx1/3<p2≤p3
A := Λ (n)1 , (1.5)
4,p a,q (n+2,P(x1/8))=1
x/2p≤2Xnn≤+x2−2
|
To this end, we prove the following estimates.
x
A ≥(4.394−o(1))Π (section 4), (1.6)
1 2
2ϕ (q)logx
2
x
A ≤(7.168+o(1))Π (section 5), (1.7)
2,p 2
2ϕ (q)logx
2
x1/8≤Xp≤x1/3
x
A ≤(1.456+o(1))Π (section 6). (1.8)
3 2
2ϕ (q)logx
2
Notice that the right hand side sum in (1.1) can be handled easily. Indeed, we have
1
A ≪xlogx ≪x7/8+o(1). (1.9)
4,p p2
x1/8≤Xp≤x1/3 x1/8≤Xp≤x1/3
Acknowledgement. This work is heavily influenced by the article [7] from T. Tao’s blog for which
I am very grateful. I would also like to thank M. Radziejewski, J. Kaczorowski and P. Achinger for
valuable discussions and many corrections.
2
3 PRELIMINARIES
2 Notation
• p always denotes a prime number and x will be a real number greater than 3;
• P(y):= p for y ≥2,
p<y
• Λ (n):Q=1 Λ(n),
a,q n a(q)
≡
• P, P are the sets of primes and almost primes respectively,
2
• P :={p:∃ p ,p >p1/8, p+2=p p } ∪ {twin primes},
Ch p1,p2∈P 1 2 1 2
• Π := (1− 1 )≈0.66,
2 p>2 (p 1)2
−
Q
• ϕ (n):=n 1− 2 ×(1−1 /2),
2 pn:p=2 p 2n
| 6 |
(cid:16) (cid:17)
• ∆(f;a (d))Q:= f(n)− 1 f(n),
n a(d) φ(d) n:(n,d)=1
≡
• we will treat MPas an absolute conPstant, so in the further sections the dependence on M will
not be emphasized in any way.
3 Preliminaries
Theorem 3.1 (Siegel−Walfisz). For any A>0 there exists a constant C >0 such that
A
x
Λ(n)= +O(xexp(−C logx))
A
φ(d)
n x
n=Xa≤(d) p
for every residue class a (d) satisfying (a,d)=1 and for all x≥2 such that d≤logAx.
Theorem 3.2 (Bombieri−Vinogradov). Let x≥2. Then
max ∆(Λ1[1,x];a (d)) ≪A xlog−Ax
d Da∈(Z/dZ)×
X≤ (cid:12) (cid:12)
(cid:12) (cid:12)
for all A>0, where D ≤x1/2log−Bx for some sufficiently big B =B(A).
Let us define the real functions f(s) and F(s) in the following way:
2eγ 2eγ
F(s)= , f(s)= log(s−1).
s s
Theorem 3.3 (Jurkat−Richert). Consider s > 1 and z,D ≥ 2 which satisfy z = D1/s. Let Q be a
finite subset of P and let Q be the product of the primes in Q. Furthermore, let h be a multiplicative
function that for some ε with 0<ε<1/200 satisfies the inequality
logz
(1−h(p)) 1 <(1+ε) (3.1)
−
logu
pu∈YPp(cid:31)<Qz
≤
and
0≤h(p)<1.
3
3 PRELIMINARIES
for every prime p. For a fixed integer α, let {Ed}∞d=1 be a family of subsets of Z+ of the form
{n:n≡αmodd}. Let us consider a finitely supported non-negative real sequence (a ) and let r
n ∞n=1 d
be defined by the equality
a =h(d)X +r .
n d
nX∈Ed
for every square-free d≤D, some X >0 and some error terms r . Then, for any 1<s≤3, there is
d
an upper bound
a ≤(F(s)+εe14 s)XV(z)+ |r | + |α|, (3.2)
n − d
n6∈SXp<zEp d≤QDX:µ2(d)=1
and for any 2≤s≤4 there is a lower bound
a ≥(f(s)−εe14 s)XV(z)+ |r | − |α|, (3.3)
n − d
n6∈SXp<zEp d≤QDX:µ2(d)=1
where V(z)= (1−h(p)).
p<z
Proof. This isQa special case of the Jurkat-Richert theorem from [6].
Remark 3.4. In[6, §10.3]onecanalsofindaproofthatourparticularchoiceofhinthenextsection
satisfies condition (3.1) with Q being the set of first y primes where y depends on ε. In this situation
we can also get ε→0 by letting Q→∞.
Lemma 3.5. Let
b(n)=1 1 .
x/2+2 n x n=p1p2p3
≤ ≤
x1/8≤p1≤Xx1/3<p2≤p3
Then
1
max |Ra,d|:= max b(n)− b(n) ≪A xlog−Ax
dX≤Da∈(Z/dZ)× dX≤Da∈(Z/dZ)×(cid:12)(cid:12)(cid:12)n≡aXmodd ϕ(d)Xn (cid:12)(cid:12)(cid:12)
for any A>0 provided that D ≤x1/2log−Bx f(cid:12)(cid:12)or some sufficiently large B =(cid:12)(cid:12)B(A).
Proof. Follows from Theorem 22.3 from [5]. We also note that for each integer d ≥ x1/8 we have
b(n)≪x7/8,sothe totalcontributionofingredientsofthisformisatmostx7/8+o(1) which
n:(n,d)>1
is neglible.
P
Lemma 3.6. Let d ≥ 1 be a fixed integer and let f : (0,∞)d → C be a fixed compactly supported,
Riemann integrable function. Then for x>1 we have
1 logp logp dt ...dt
1 d 1 d
f ,..., = f(t ,...,t ) +o (1).
1 d x
p1X,...,pd p1...pd (cid:18)logx logx (cid:19) Z(0,∞)d t1...td →∞
Proof. Follows from Mertens’ second theorem combined with elementary properties of the Riemann
integral.
4
4 ESTIMATING A
1
4 Estimating A
1
According to Theorem 3.3 let us define E as the set of positive integers from the residue class
d
−2 (d). Let us also define a multiplicative function g(d) such that g(2) = 0 and g(p) = 1/(p−1).
The precise value in non-square-free numbers is not relevant. Put z = x1/8 and D = x1/2−ǫx, where
ǫ :=(logx) 1/2. WetakeQtobethelargestsetoftheformasinRemark3.4satisfyingQ≤loglogx.
x −
We have
A = Λ (n).
1 a,q
x/2≤Xn≤x−2
n6∈Sp<zEp
We define
x
Λ (n)= Λ(n)=g (d) +r(a,q),
a,q a,q 2 d
x/2n≤X∈nE≤dx−2 x/n2≡≤X−n2≤(xd−)2
n a (q)
≡
where g (d) := 1 g(d)/ϕ(q) and r(a,q) is a remainder term. Here we notice that if p|d,q for
a,q (q,d)=1 d
somep, then p|n+2 whichcontradictsn+2≡a+2 (q) joinedwith (a+2,q)=1. In sucha case one
(a,q)
has r =0.
d
According to Theorem 3.3, we put X = x and h(d) = g(d)1 . If (q,d) = 1, then by the
2ϕ(q) (q,d)
prime number theorem we get
x
r(a,q) =∆(Λ1 ;c (qd))+O .
d [x/2,x−2] q,d ϕ(qd)exp(Clog1/10x)!
for some residue class c ∈(Z/qdZ) and some positive constant C.
q,d ×
From the Bombieri−Vinogradovtheorem one gets
|∆(Λ1[x/2,x 2];cq,d (qd))|≤ max |∆(Λ1[x/2,x 2];c (d))|≪xlog−M−10x. (4.1)
− c (Z/dZ) −
d QD d qQD ∈ ×
µ2≤X(d)=1 ≤X
Notice that the estimate above is uniform because
qQD ≪x1/2x−1/(logx)1/2logMxloglogx=o x1/2log−B(M+10)x , (4.2)
(cid:16) (cid:17)
where B(M +10) is a real number large enough such that Theorem 3.2 can be used with exponent
M +10.
By Theorem 3.3 one has
x
A1 ≥(f(4−8ǫx)−O(ε)) V(z)−O(xlog−M−10x). (4.3)
2ϕ(q)
From Mertens’ third theorem we conclude
2Π2 1 −1
V(z)=(1+o(1)) 1− . (4.4)
eγlogz p−1
p6=Y2:p|q(cid:18) (cid:19)
This can be simplified further to
5
5 ESTIMATING A
2,P
x
A ≥(log3−o(1))Π . (4.5)
1 2
2ϕ (q)logz
2
5 Estimating A
2,p
For z ≤p≤x1/3 we have
A = Λ (n)1 .
2,p a,q pn+2
|
x/2≤Xn≤x−2
n S E
6∈ p′<z p′
We apply the Jurkat−Richerttheorem, but this time taking D/p instead of D. For any square-free d
we have
g(p)x
Λ (n)1 = Λ(n)=g(d)1 +r(a,q), (5.1)
a,q p|n+2 (d,q)=12ϕ(q) pd
x/2n≤X∈nE≤dx−2 xn/≡2≤−Xn2≤(xp−d)2
n a (q)
≡
and hence
logD/p g(p)x
A ≤ F +O(ε) V(z)+O |∆(Λ1 ;c (pqd))| . (5.2)
2,p logz 2ϕ(q)  [x/2,x−2] q,pd 
(cid:18) (cid:18) (cid:19) (cid:19) d≤XQD/p
 
Since every d≤QD has at most O(logx) prime factors, one can conclude
|∆(Λ1[x/2,x 2];cq,pd (pqd))|≪logx |∆(Λ1[x/2,x 2];cq,d (qd))|≪xlog−M−9x.
− −
z≤pX≤x1/3d≤XQD/p d≤XQD
(5.3)
From (4.4), (5.2) and (5.3)
F logD/p +O(ε)
A2,p ≤Π2eγϕ (xq)logz (cid:16) (cid:16) logzp(cid:17) (cid:17) +O(xlog−M−9x). (5.4)
2
z≤pX≤x1/3 z≤pX≤x1/3
By Lemma 2.5 we have
F(logD/p) 8/3 dt 8/3 1 dt
logz = F(4−8ǫ −t) +o(1)=2eγ +o(1)= (5.5)
x
p t 4−t−8ǫ t
z≤pX<x1/3 Z1 Z1 x
8/3 dt eγlog6
(2eγ +o(1)) = +o(1).
(4−t)t 2
Z1
Combining (5.4) and (5.5) we get
x
A ≤(log6+o(1))Π . (5.6)
2,p 2
2ϕ (q)logz
2
z≤pX≤x1/3
6
6 ESTIMATING A
3
6 Estimating A
3
We have
A = Λ (n) 1 = (6.1)
3 a,q n+2=p1p2p3
x/2≤Xn≤x−2 z≤p1≤xX1/3<p2≤p3
Λ(n−2) 1 ≤logx b(n)1 +O(x0.51),
n=p1p2p3 (n 2,P(√x))=1
xn/2+aX+2≤2n(≤q)x x1/8≤p1≤Xx1/3<p2≤p3 n anX+∈2Z(q) −
≡ ≡
where b(n) is defined as in Lemma 3.5. The error term comes from the numbers of the form pk for
k ≥2. One can rewrite the above sum into
b(n)1 ,
n a+2 (q)
≡
n Z
n6∈SXp<∈√xEp′
where E denotes the residue class 2 (d). We use the Jurkat−Richert theorem again. By Lemma
d′
3.5 for some c ∈ (Z/dqZ) we get (remember that conditions n ≡ 2 (d) and n ≡ a+2 (q) force
′d,q ×
(d,q)=1)
g(d)
b(n)1 = b(n)= 1 b(n)+R = (6.2)
n≡a+2 (q) ϕ(q) (d,q)=1 c′d,q,dq
nX∈Ed′ n≡X2 (d) nX∈Z
n a+2 (q)
≡
g(d)1 b(n)−g(d)1 R +R ,
(d,q)=1 (d,q)=1 a,q c′d,q,dq
n≡Xa (q)
where
|R −g(d)1 R |≤ max |R | +
c′d,q,dq (d,q)=1 a,q c (Z/dZ) c′,d
d QD d qQD ′∈ ×
µ2≤X(d)=1 µ2≤X(d)=1
1
|Ra,q| ≪xlog−M−10x;
ϕ(d)
d QD
µ2≤X(d)=1
we used trivial inequality |Ra,q| ≤ d D|Ra,d| ≪ xlog−M−11x above. Similarily to the previous
situation of this kind, let us emphasize≤that the upper bound here is uniform with respect to q. By
using the inequality from Theorem 3P.3 with the level of distribution QD1/(1+ǫx) we get
b(n)1(n 2,P(√x))=1 ≤(F(1+ǫx)+O(ε))V(D1/(1+εx)) b(n)+O(xlog−M−10x).
n Z − n Z
n aX+∈2 (q) n aX+∈2 (q)
≡ ≡
(6.3)
Again, by using the Mertens’ third theorem one gets
1
1 1 1 −
V(D1/(1+ǫx))= Π (1+o(1)) 1− . (6.4)
2 2 eγlogz p−1
p6=Y2:p|q(cid:18) (cid:19)
By F(s)−s−→−1−→+ 2eγ and
7
REFERENCES REFERENCES
1
b(n)= b(n)+R ,
a,q
ϕ(q)
n Z n Z
n aX+∈2 (q) X∈
≡
where |Ra,q|≪xlog−M−11x, we have
1
n Z b(n)1(n−2,P(√x))=1 ≤(1+o(1))Π2ϕ2(q)logz n Zb(n)+O(xlog−M−10x). (6.5)
n aX+∈2 (q) X∈
≡
By (6.1)−(6.5) and Lemma 3.6 we get
x dt dt x
1 2
b(n)≤(1+o(1)) ≤(0.364+o(1)) . (6.6)
2logx t t (1−t −t ) 2logx
nX∈Z Z1/8≤t1≤1/3<t2<1−t1−t2 1 2 1 2
References
[1] J. Chen, On the representation of a large even integer as the sum of a prime and the product of
at most two primes, Kexue Tongbao. 17 (1966).
[2] J. Chen, On the representation of a large even integer as the sum of a prime and the product of
at most two primes, Sci. Sin. 16 (1973).
[3] B.GreenandT.Tao,Restriction theoryoftheSelbergsieve, withapplications, J.Th´eor.Nombres
Bordeaux 18 (2006).
[4] B. Zhou, The Chen primes contain arbitrarily long arithmetic progressions, Acta Arith. 138
(2009).
[5] J. Friedlander and H. Iwaniec, Opera de cribro. American Mathematical Society Colloquium
Publications, 57. American Mathematical Society, Providence, RI, 2010.
[6] M. Nathanson, Additive Number Theory: The Classical Bases, Graduate Texts in Mathematics,
164. Springer-Verlag,1996.
[7] T. Tao, 254A, Supplement 5: The linear sieve and Chen’s theorem (op-
tional), blog post, available on https://terrytao.wordpress.com/2015/01/29/
254a-supplement-5-the-linear-sieve-and-chens-theorem-optional/
Institute of Mathematics, Polish Academy of Sciences, S´niadeckich 8, 00-656 Warsaw,
Poland
E-mail address: [email protected]
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