A Vinogradov-type problem in almost primes

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by  Paweł Karasek

A VINOGRADOV-TYPE PROBLEM IN ALMOST PRIMES
PAWEL KARASEK
6 Abstract. We prove a generalisation of Vinogradov’s theorem by finding for m>3 and
1 fixed positive integers c1,...,cm,r1,...,rm the asymptotics of the number of sequences
0 (n1,...,nm)∈Nmsuchthatc1n1+···+cmnm=N andΩ(ni)=riforeveryi=1,...,m
2 undertheassumptionthatatleastthreeoftheri areequalto1.

n
a
J

1
1 1. Introduction

One ofthe mostfamous problems ofadditive combinatoricswhich are alreadysolvedis the
]
T so called weak (or ternary) Goldbach conjecture, which can be stated in the following form:
N
Theorem 1.1. Every odd number N greater than 1 is a sum of at most three primes.
.
h
t The assertion of Theorem 1.1 was proven to be correct for all sufficiently large N in
a
m 1937 by Vinogradov [1]. Later Chen and Wang [2] gave an effective proof i.e. for all N >
exp(exp(11.503)). This threshold was lowered to N > exp(3100) by Liu and Wang [3], but it
[
was still too weak to prove ternary Goldbach conjecture for all lower positive integers using

1 computer calculations1. In 2012 and 2013 Helfgott in [4], [5] gave new bounds which were
v strong enough to verify all remaining cases directly.
1
The proof of the ineffective version of Theorem 1.1 gave us also the precise asymptotic of
9
the numberofsolutionsoftheequationp +p +p =N forp ,p ,p beingprimeswhereN is
5 1 2 3 1 2 3
2 a variable. For technical reasons it is easier to attach a weight logp to each power of a prime
0 p and deal with the sum
.
1
0
R (N)= Λ(n )Λ(n )Λ(n ),
6 3 1 2 3
1 n1+nn1X,2n+2n,n33=N
:
v
where Λ(n) denotes the von Mangoldt function.
i
X In this paper we calculate the asymptotics of the number of solutions in almost primes
r of the more general equation c n + +c n = N, where the c are some fixed positive
a 1 1 ··· m m i
integers. Formally, we assume that the number Ω(n ) of prime divisors of n is equal to r ,
i i i
where (r ,...,r ) is a sequence of positive integers independent of N. There are obstacles
1 m
which force us to assume that at least three of the r are equal to 1. By using the Hardy-
i
Littlewoodcirclemethodandsomecombinatorialargumentsweareabletoprovethefollowing
result
Theorem 1.2. Fixm>3. Letc ,...,c besomepositive integerssatisfying (c ,...,c )=1,
1 m 1 m
and let r ,...,r bea sequenceof positive integers which contains at least threeelements equal
1 m
to 1. Then for N >20
1However,acompleteproofwaspresentedin[6]ontheassumptionthatthegeneralisedRiemannhypothesis
istrue.
1
2 PAWEL KARASEK
1 1 1
1 =
(m 1)!(r 1)!...(r 1)!c ...c ×
Ω(cn11n)1=+rn·11·,·X,.+.....c,,Ωmn(mnnmm=)=Nrm − 1− m− 1 m
Nm 1
logm−N(loglogN)r1+···+rm−m(Sc1,...,cm(N)+o(1)),
where
S (N)=
c1,...,cm
µ p ...µ p µ p ...µ p
1+ (c1,p) (cm,p) (p 1) 1 (c1,p) (cm,p) .
 ϕ(cid:16) p (cid:17)...ϕ(cid:16) p (cid:17) −   − ϕ(cid:16) p (cid:17)...ϕ(cid:16) p (cid:17)
pY|N (c1,p) (cm,p) pY∤N (c1,p) (cm,p)
 (cid:16) (cid:17) (cid:16) (cid:17)   (cid:16) (cid:17) (cid:16) (cid:17)
The first part of the proof is based on standard arguments developed by Vinogradov to
calculate the number of solutions of b n + +b n = N in weighted primes, where we
1 1 m m
···
let the b depend somehow on N (precisely, we assume that b N1/12m). After that, we
i i
≪
transmute the von Mangoldt weights into standard indicators and show how this affects the
asymptotic. Thefinalstep(the mosttechnicallyinvolved)istousethe definitionofthe b and
i
some combinatorialarguments to get the desired asymptotics.
In big O or notation the dependence on absolute constants will not be emphasized in
≪
any way.
Acknowledgement. The author would like to thank Jacek Pomykal a, Maciej Radziejewski
and Piotr Achinger for valuable comments and many corrections.
2. Major arcs
In the next two sections, we are going to find the asymptotics of the sum
Λ(n )...Λ(n )
1 m
n1,..X.,nm6N
b1n1+···+bmnm=N
with varying coefficients b via the circle method. Let Q = logBN for some B > 0, which is
i
going to be fixed later. Then for q 6 Q and a, such that (a,q) = 1, we define a major arc in
the usual manner:
a Q
Ma,q := α∈R:kα− qkR/Z 6 N .
(cid:26) (cid:27)
Let us also denote the sum of all major arcs by
q
:= .
a,q
M M
q6Q a=1
[ (a,[q)=1
Put S(x,α)= Λ(n)e(nα), S (x,α)= Λ(n)e(nb α) and m=R/Z(cid:31) .
n6x i n6bxi i M
Theorem 2.1.PFix c ,...,c N. Let b ,.P..,b 6Nδ for any δ (0, 1 ) and let b =c η
1 m ∈ 1 m ∈ 12m i i i
for i = 1,...,m, where η is a positive integer with prime divisors greater than Q. Let us
i
A VINOGRADOV-TYPE PROBLEM IN ALMOST PRIMES 3
further assume, that (b ,...,b )=1.Then for every ε>0, we have
1 m
1 Nm 1 1 Nm 1
Λ(n )...Λ(n )= − S (N)+ O − +
1 m (m 1)!b ...b c1,...,cm b ...b Qm 2 ε
n1,..X.,nm6N − 1 m 1 m (cid:18) − − (cid:19)
b1n1+···+bmnm=N
m
S (N,α)e( Nα)dα.
i
m −
Z i=1
Y
Proof. Let us define u(y) = e(ny), u (y) = e(nb y) and put α = a +y. Recall
n6N i n6N i q
bi
the well known identity (for proof see [9]) for a,q N coprime and q <Q:
P P
∈
µ(q)
(2.1) S(N,α)= u(y)+O (1+N y )N√qexp c logN ,
ϕ(q) | | −
(cid:16) (cid:16) p (cid:17)(cid:17)
where c is some positive constant. From S (N,α) =S(N/b ,b α) and b 6N1/12m we simply
i i i i
get
µ q
(2.2) S (N,α)= (bi,q) u (y)+O (1+N y )N√qexp c logN .
i ϕ(cid:16) q (cid:17) i | | bi − r bi!!
(bi,q)
(cid:16) (cid:17)
Applying 1 Q, y 6 Q, q 6Q and b <Nδ we can estimate
≪ | | N i
N N NQ3/2
(2.3) (1+N y ) √qexp c log exp c logN ,
1
| | bi − r bi!≪ bi −
(cid:16) p (cid:17)
wherec issomepositiveconstant. UsingQ exp(ε√logN)foranyε>0wefinallyconclude
1
≪
µ q
(2.4) S (N,α)= (bi,q) u (y)+O N exp C logN
i ϕ(cid:16)(biq,q)(cid:17) i (cid:18)bi (cid:16)− p (cid:17)(cid:19)
(cid:16) (cid:17)
for some positive constantC. Right now we are ready to estimate the contributionof a single
major arc to the integral.
m m µ q
(2.5) S (N,α)e( Nα)dα= (bi,q) e aN
ZMa,qiY=1 i − iY=1ϕ(cid:16)(biq,q)(cid:17) (cid:18)− q (cid:19)×
(cid:16) (cid:17)
Q m Q
N N
Q ui(y)e(−Ny)dy+O Q fω1(y)...fωm(y)e(−Ny)dy,
Z−N iY=1 ω∈{X0,1}m Z−N 
ω=(1,...,1) 
6
 
where
u (y) µ q /ϕ q if ω =1
(2.6) fωj(y):=(Nbiiexp×−C(cid:16)√(bil,oqg)(cid:17)N (cid:16)(bi,q)(cid:17) if ωjj =0 .
(cid:0) (cid:1)
4 PAWEL KARASEK
By the obviousinequality u (y) 6 N, if atleastone coordinateof ω is equalto 0,then for
| i | bi
such ω we get
NQ 1 Nm−1
(2.7) f (y)...f (y)e( Ny)dy .
Z−NQ ω1 ωm − ≪ b1...bme(C−ε)√logN
Summing over every admissible a,q one gets
(2.8)
m m µ q Q m
S (N,α)e( Nα)dα= (bi,q) e aN N u (y)e( Ny)dy +
ZMYi=1 i − qX6QiY=1ϕ(cid:16)(biq,q)(cid:17) aX6q (cid:18)− q (cid:19)×Z−NQ iY=1 i −
(a,q)=1
(cid:16) (cid:17)
1 Nm 1
−
O
b1...bm (cid:18)e(C−3ε)√logN(cid:19)
Put c (N) = e aN . We can show that the sum over q on the right hand side
q a6q − q
(a,q)=1
(cid:16) (cid:17)
of the equation (P2.8) equals
m µ q m µ p m µ p
(2.9) (bi,q) c (N)= 1+ (ci,p) (p 1) 1 (ci,p) +
ϕ(cid:16) q (cid:17) q  ϕ(cid:16) p (cid:17) −   − ϕ(cid:16) p (cid:17)
qX6QiY=1 (bi,q) pY|N iY=1 (ci,p) pY∤N iY=1 (ci,p)
(cid:16) (cid:17)  (cid:16) (cid:17)   (cid:16) (cid:17)
m µ q
O (ci,q) c (N) =S (N)+O 1 .
qX>QYi=1ϕ(cid:16)(ciq,q)(cid:17) q  c1,...,cm (cid:18)Qm−2−ε(cid:19)
Moreover,  (cid:16) (cid:17) 
1
S (N) < 1+ 1.
| c1,...,cm | ϕ(p)m 1 ≪
p (cid:18) − (cid:19)
Y
Right now we only need to estimate the integral on the right hand side of (2.8). Let us
consider the following subsets of R/Z:
k 1 k 1
J(j) = , + ,
k b − b N1/3 b b N1/3
(cid:20) j j j j (cid:21)
k 1 k 1
I(j) = , + ,
k b − b N1/2 b b N1/2
(cid:20) j j j j (cid:21)
for j = 1,...,m and k = 0,...,b 1. The distance between two fractions of the form k/b
i j
−
satisfies
k k 1 1 1 1 1
(2.10) 1 2 > >max , > +
b − b b b b Nδ b Nδ b N1/3 b N1/3
(cid:12) j1 j2(cid:12) j1 j2 (cid:26) j1 j2 (cid:27) j1 j2
(cid:12) (cid:12)
for N >21−33δ(cid:12)(cid:12) and arbit(cid:12)(cid:12)rary k1,k2 Z. Consequently, we can assume that N is so large that
∈
every two intervals J(j) centered in different points k/b have empty intersections.
k j
A VINOGRADOV-TYPE PROBLEM IN ALMOST PRIMES 5
AccordingtoAppendixAwedefineJ (N)asanumberoftuplesoftheform(n ,...,n )
b1,...,bm 1 m ∈
Nm which fulfils b n + +b n =N. From I(j) J(j) one can decompose
1 1 ··· m m k ⊂ k
m
(2.11) J (N)= u (y)e( Ny)dy =: =
b1,...,bm ZR/Zi=1 i − ZR/Z
Y
NQ m bj−1 bN11/3 −NQ
+ O + + + + ,
Z−NQ Xj=1 Xk=1 (cid:12)(cid:12)ZIk(j) ZJk(j)(cid:31)Ik(j)(cid:12)(cid:12) ZNQ Z−bN11/3 ZS
(cid:12) (cid:12)
whereb:=min{b1,...,bm}andS(cid:12)(cid:12)denotesthesetR(cid:12)(cid:12)/Z(cid:31) mj=1 bkj=−01Jk(j). FrombasicDirichlet
kernel estimations one gets
S S
1 if y J(i)
N1/3 6∈ k
 1 if y I(i)
(2.12) u (y) N1/2 6∈ k ,
| i |≪|b1iy| if y ∈ −bN11/3,bN11/3 (cid:31){0}
N always
(cid:2) (cid:3)
(the second inequality from the bottom is true because for N sufficiently large we have
|biy|6 Nb1i/3 < 12, which gives kbiykR/Z =|biy|) and thus
Nm−1+31 if y ∈ mj=1 kbi=−11Ik(j)
(2.13) m u (y) Nm2−1+13 if y ∈Smj=1Skbi=−11Jk(j)(cid:31)Ik(j) .
(cid:12)(cid:12)(cid:12)(cid:12)iY=1 i (cid:12)(cid:12)(cid:12)(cid:12)≪Nb1.m.13.bm|y1|m iiff yy ∈∈(cid:2)SS−bN11S/3,bN11/3(cid:3)(cid:31){0}
The two inequaliti(cid:12)es from th(cid:12)e top follow from the fact that if (b ,...,b )=1, then for every
1 m
fractionofthe formk/b =0there exists atleastoneinterval J(j′) whichis notcenteredinit.
j 6 l
From (2.13) we can see that
1 Q 1 m 1
m bN1/3 −N 1 bN1/3 dy 1 N −
(2.14) N 3 , + ,
ZS ≪ ZNQ Z−bN11/3!≪ b1...bm ZNQ ym ≪ b1...bm (cid:18)Q(cid:19)
m bj−1 + m bj−1 Nm−1+13 1 +Nm2−1+13 1 N(1+δ)m−76.
Xj=1 kX=1 (cid:12)(cid:12)ZIk(j) ZJk(j)(cid:31)Ik(j)(cid:12)(cid:12)≪Xj=1 Xk=1 (cid:18) bjN1/2 bjN1/3(cid:19)≪
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Combining (2.11), (2.12), (2.13), (2.14) andrecalling the assumption δ < 1 , we conclude
12m
that
m m 1
1 N −
(2.15) u (y)e( Ny)dy =J (N)+ O
ZR/Zi=1 i − b1,...,bm b1...bm (cid:18)Q(cid:19) !
Y
From (2.8), (2.9), (2.15) and Theorem A.2 we get
(2.16)
m 1 Nm 1 1 Nm 1
S (N,α)e( Nα)dα= − S (N)+ O − .
i − (m 1)!b ...b c1,...,cm b ...b Qm 2 ε
ZMiY=1 − 1 m 1 m (cid:18) − − (cid:19)
(cid:3)
6 PAWEL KARASEK
We end the discussion in this section by proving a useful lemma on the function S :
c1,...,cm
Lemma 2.2. If(c ,...,c )=1, then it is truethat S (N)=0 iff c + +c +N 0
1 m c1,...,cm 6 1 ··· m ≡
mod 2 and
(N,c ,...,c )= =(c ,...,c ,N,c ,...c )= =(c ,...,c ,N)=1.
2 m 1 i 1 i+1 m 1 m 1
··· − ··· −
Moreover, S (N) 1 for every N such that S (N)=0.
c1,...,cm ≫ c1,...,cm 6
Proof. The first part of the lemma follows almost directly from the definition of S .
c1,...,cm
The second part follows from the fact that there exist at most finitely many primes which
divide m c . Let z be a real number which is greater than all of them. From the first part
i=1 i
of this lemma one can see that for N such that S (N)=0 we have
Q c1,...,cm 6
1 1 1
S (N)>2 1 1 1 1 (cid:3)
c1,...,cm − p 1 − (p 1)2 − (p 1)m 1 ≫
2<pY|pN6z(cid:18) − (cid:19)2<pY∤pN6z(cid:18) − (cid:19)pY>z(cid:18) − − (cid:19)
3. Minor arcs
In this section we will estimate the integral
m
(3.1) S (N,α)e( Nα)dα.
i
m −
Z i=1
Y
Usually,somevariationsoftheVinogradovLemma(lemma3.4inourcase)areusedtoestablish
results of this type. Recall the following result which is going to be helpful further:
Lemma 3.1 (R. C. Vaughan [3]). Let X,Y,α R where X,Y > 1. Let us assume that
∈
α a 6 1 for some a,q N such that (a,q)=1. Then
| − q| q2 ∈
XY 1 XY
min , +X +q log(2Xq).
n6X (cid:26) n knαkR/Z(cid:27)≪(cid:18) q (cid:19)
X
We will also need the following version of the Vaughan’s identity:
Lemma 3.2. For every real x>0, U,V >2 we have the identity
S(x,α)=S S S +S ,
I,1 I,2 II 0
− −
where
S = µ(d) logne(ndα),
I,1
d6U n6x
X Xd
S = Λ(d) µ(δ) e(ndδα),
I,2
d6V δ6U n6x
X X Xdδ
SII =  µ(δ) Λ(n)e(ndα),
d>U δ6U n>V
XXδ|d nXd6x
 
S = Λ(n)e(nα).
0
n6V
X
A VINOGRADOV-TYPE PROBLEM IN ALMOST PRIMES 7
Let us use Lemma 3.2 by putting x = N and b α instead of α. The method used here is
bi i
well described in [3], however we will present it for the sake of completeness.
The inner sum in S is equal to
I,1
e(ndb α) n dy = dNbi e(ndb α)1 dy = dNbi e(ndb α)dy
i i y<n i
y y y
nX6dNbi Z1 Z1 nX6dNbi Z1 y<Xn6dNbi
which gives us
(3.2)
N
dbi dy N 1
S = µ(d) e( y dα) e(ndb α) logN min , =
I,1 i
dX6U Z1 ⌈ ⌉ n6XdNbi−y y ≪ dX6U (cid:26)dbi kdbiαkR/Z(cid:27)
N 1
logN min , .
d6XUbi (cid:26) d kdαkR/Z(cid:27)
bi|d
Moreover,one can see that
(3.3) S = µ(δ )Λ(δ )e(δ δ db α)=
I,2 1 2 1 2 i
δ16XU,δ26V d6Xδ1δN2bi
µ(δ )Λ(δ )e(ndb α)= µ(δ )Λ(δ )e(ndb α)=
1 2 i 1 2 i
δ16XU,δ26V d6Xδ1δN2bi δn1X6δ2U=Vn δ16δ1Uδ,2δ=26Xn,V,ddn,n66bNiUV
N 1
=  µ(δ1)Λ(δ2) e(ndbiα) logUV min , =
nX6UV δ16δ1XUδ2,δ=2n6V d6XnNbi ≪ nX6UV (cid:26)nbi knbiαkR/Z(cid:27)
 
N 1
logUV min , .
d6XUVbi (cid:26) d kdαkR/Z(cid:27)
bi|d
To estimate the value of S in a similar manner let us define the set
II
:= U,2U,4U,...,2kU :2kUV <N/b 62k+1UV .
i
Y { }
Thus
SII = S(Z), where S(Z)=  µ(δ) Λ(n)e(ndbiα).
ZX∈Y Z<Xd62ZδXδ6|dU V<Xn6dNbi
 
By the Cauchy-Schwarzinequality
2
S(Z)2 6 τ(d)2 Λ(n)e(ndb α) .
(cid:12) i (cid:12)
| | (cid:12) (cid:12)
Z<Xd62Z Z<Xd62Z(cid:12)(cid:12)V<Xn6dNbi (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
8 PAWEL KARASEK
From τ(n)2 nlog3n for n>2 and the following identity
n6x ≪
P 2
Λ(n)e(ndb α) = Λ(n )Λ(n )e((n n )db α),
(cid:12) i (cid:12) 1 2 1 2 i
(cid:12) (cid:12) −
(cid:12)(cid:12)V<Xn6dNbi (cid:12)(cid:12) V<n1X,n26dNbi
(cid:12) (cid:12)
one gets (cid:12) (cid:12)
(cid:12) (cid:12)
(3.4) S(Z)2 Zlog3N Λ(n )Λ(n ) e((n n )db α)
1 2 1 2 i
| | ≪ − ≪
V<n1X,n26ZNbi Z<Xd62Z
1
Zlog5N min Z, =
n1,nX26ZNbi (cid:26) k(n1−n2)biαkR/Z(cid:27)
1 N N 1
Zlog5N min Z, log5NZ+ min , ,
nXb6i|nNZ −n6Xbdi6|dNZ−n (cid:26) kdαkR/Z(cid:27)≪ bi  dXb6i|NdZ (cid:26) d kdαkR/Z(cid:27)
 
2
Let us combine (3.2), (3.3), (3.4) and put U,V := N 5. Let us say that α a 6 1 for
bi | − q| q2
some a,q N such that (a,q)=1, q 6N. Using Lem(cid:16)ma(cid:17)3.1 one gets2
∈
(3.5) d6XUbimin(cid:26)Nd ,kdαk1R/Z(cid:27), n6XUVbimin(cid:26)Nn,knα1kR/Z(cid:27)≪(cid:18)Nq +N45bi51 +q(cid:19)logN,
N 1 N N
Z+ min , + +q+Z logN.
dX6NZ (cid:26) d kdαkR/Z(cid:27)≪(cid:18) q Z (cid:19)
Since
N N N
S(Z)2 log6N + +q+Z ,
| | ≪ b q Z
i (cid:18) (cid:19)
one gets
1 N N 1 N
SII log3N + + Nq+√NZ log4N +N45 + Nq .
≪ √bi Z (cid:18)√q √Z (cid:19)≪ √bi (cid:18)√q (cid:19)
X∈Y p p
The best result we are able to obtain is the following estimate.
2Notethattheestimationsareratherweak,especiallyifwewanttousethemto(3.2),(3.3)and(3.4)sums.
Therestricionsonbi|dundereverysummandweresimplycancelled. Thereasonbehindsuchamanouveristhe
lackofany visiblepossibilityto usethem. Ontheother hand, one has somechance to improvethese results,
someother ideasarerequiredthough.
A VINOGRADOV-TYPE PROBLEM IN ALMOST PRIMES 9
Lemma 3.3. Let α a 6 1 for some a,q N such that (a,q)=1, q 6N. Then
| − q| q2 ∈
N
Si(N,α) log4N +N54 + Nq .
≪ √q
(cid:18) (cid:19)
p
Proof. We use the bounds on SI,1, SI,2 i SII and the obvious fact that S0 N25. (cid:3)
≪
Using the fact that for any α R there exists a positive integer q 6 N such that
∈ Q
α a 6 Q 6 1 for some a N which fulfils the condition (a,q)=1 we get
| − q| qN q2 ∈
Lemma 3.4. Let B >0 and a positive integer bi 6N316. Hence for every α m we have
∈
N
S (N,α) .
i ≪ logB2−4N
We shall prove the following
Theorem 3.5. Under assumptions of the Theorem (2.1) and the extra assumption η =η =
1 2
η =1 we have
3
m 1 Nm 1
−
S (N,α)e( Nα)dα .
Zmi=1 i − ≪ b1...bmlogB2−6N
Y
Proof. We have
m m
(3.6) S (N,α)e( Nα)dα6 max S (N,α) S (N,α)S (N,α)dα6
Zmi=1 i − i=3α∈m| i |ZR/Z| 1 2 |
Y Y
1/2 1/2
Nm 3
− max S (N,α) S (N,α)2dα S (N,α)2dα
3 1 2
b4...bm × α∈m| | ZR/Z| | ! ZR/Z| | ! ≪
Nm 3 N N Nm 1 1
− −
log(N/b )log(N/b ) ,
b4...bmlogB2−4N √b1b2 1 2 ≪ b1...bmlogB2−6
which follows from the Cauchy-Schwarz inequality, the Plancherel identity, the constancy of
b ,b ,b and the basic fact that Λ(n)2 xlogx for x>1.
1 2 3 n6x ≪
(cid:3)
P
Right now we can combine the results from last two sections to establish the following
Theorem 3.6. Let us consider the constants c ,...,c N. Let b ,...,b 6 Nδ for some
1 m 1 m
∈
δ (0, 1 ) and let b =c η for every i=1,...,m where the η are some positive integers all
∈ 12m i i i i
of whose prime divisors are greater than Q. Let us further assume that (b ,...,b ) = 1 and
1 m
η ,η ,η =1. Hence for some A>0 we have
1 2 3
1 Nm 1 1 Nm 1
Λ(n )...Λ(n )= − S (N)+ O −
n1,..X.,nm6N 1 m (m−1)!b1...bm c1,...,cm b1...bm (cid:18)logAN(cid:19)
b1n1+···+bmnm=N
10 PAWEL KARASEK
Proof. Follows easily from 2.1 and 3.3 upon taking B =2A+12 and ε= 1. (cid:3)
2
Corollary 3.7. Under assumptions of Theorem 3.6 for sufficiently large N we have
S (N)=0 iff R (N;b ,...,b )=0.
c1,...,cm m 1 m
Proof. Follows from Lemma 2.2 and Theorem 3.6. (cid:3)
4. Reducing the logarithmic weights
Firstly we are going to show that the contribution of the numbers of the form pk for
some k > 2 which appear in the support of the von Mangoldt function does not change the
asymptotics of R (N;b ,...,b ) given in Theorem 3.6. Let us define
m 1 m
logn, if n P
(4.1) θ(n)= ∈ .
(0, otherwise
Then for some A>0 one has
(4.2) Λ(n )...Λ(n ) θ(n )...θ(n )6
1 m 1 m
−
n1,..X.,nm6N n1,..X.,nm6N
b1n1+···+bmnm=N b1n1+···+bmnm=N
m m
Λ(n )...Λ(n )6logmN 16
1 m
Xi=1b1n1n+1Ω,·.·(.·X.n+,nib)mm>6n2mN=N Xi=1nb11ω,n.(.1n.,+in)·m=··X+61,bNΩm(,nnniim6)>=√2NN
logmN m 1=Nm−32 logmN 6 Nm−23+δmlogmN.
b ...b
1 m
Xi=1k16√N,kX2,...,km−16N
We are going to study the asymptotics of the function
m
r (N;b ,...,b )= 1 1 .
m 1 m b1n1+···+bmnm=N nl∈P
n16bN1,X...,nm6bNm Yl=1
We can also define the function
R (N;b ,...,b )= 1 θ(n )...θ(n ).
m 1 m b1n1+···+bmnm=N 1 m
n16bN1,X...,nm6bNm
g
We showed that R = 1 Nm−1 (S (N)+o(1)). Let us prove the following
m (m−1)!b1...bm c1,...,cm
Theorem 4.1. Under assumptions of Theorem 3.6 we have
g
m
N
r (N;b ,...,b ) log =R (N;b ,...,b )(1+o(1)).
m 1 m m 1 m
b
i
i=1
Y
g
Proof. ObviouslyR =0iffr =0andinsuchacasethetheoremfollowstrivially. Notethat
m m
g m
N
(4.3) R (N;b ,...,b )6r (N;b ,...,b ) log .
m 1 m m 1 m
b
i
i=1
Y
g