Numbers with a large prime fa tor
by
R. C. Baker (Provo, Ut.) and G. Harman (Cardi )
1. Introdu tion. Givenalargepositivenumberx,letP(x)denote the
greatest primefa tor of
Y
x<nx+x 1=2
n:
Lower boundsfor P(x) have beengiven by Rama handra [15,16℄, Graham
[9℄,Baker[1℄, Jia[12℄ and Liu[13℄. Thelastpaper ontainsthebound
P(x)>x 0:723
:
In thepresent paperwe give asharperbound.
Theorem 1. For suÆ iently large x, we have
P(x)>x 0:732
:
As in previous papers on the subje t, we ombine sieve methods with
estimates forexponentialsums
(1:1)
X
hH X
m a
m X
v<mnev b
n e
hx
mn
:
Here e() =e 2i
. The paper of Fouvry and Iwanie [7℄ wasan important
stepforwardinthestudyofsums(1.1), andone oftheresultsof [7℄isused
in[13℄withalittleadaptation;seeLemma2,below. Some resultsin[1℄are
stilluseful;see Lemmas 2,3.
Forthespe ialsums(1.1) in whi h
(1:2) a
m
=1;
thebestresultsare duetoLiu [13℄. He uses di erentmethods for
(1:3) v<x
0:6 "
Resear hof the rstauthorsupportedinpartbyagrantfromtheNational S ien e
Foundation.
and
(1:4) vx
0:6 "
:
Here and below,"is a positivenumber,supposedsuÆ ientlysmall.
In the present paper, we give a theorem on bilinear exponential sums
from whi h the results of [13℄ followin both ases (1.3), (1.4) (Theorem 2
and its orollaries;see Se tion3). Theorem2 hasotherappli ations, whi h
willbetaken upelsewhere.
Let y=x 1=2
;L=logx, and
N(d)= X
x<nx+y
djn 1:
Letv be apositive numberin(x 1=2
;x 3=4
℄. Wewritev=x
,
A=fn:v<nev;N(n)=1g; S(A;z)=jfn2A:pjn)pzgj
wherep denotesa primevariable,and
S()=jfp:p2Agj:
Theorem1 willfollowifwe establishthat
(1:5)
X
x 0:732
<pP(x)
N(p)logp>0:
Just asin[1℄,x1, itsuÆ es to prove that
(1:6)
X
d<x 3=5 "
(d)N(d) =
3
5
"
yL+O(y)
and
(1:7)
X
x 3=5 "
<px 0:732
N(p)logp<2yL=5
in order to establish (1.5). The formula (1.6) is given in [13℄, but for the
sakeof ompletenesswe shalldedu eit from Theorem2(see Corollary2).
As onp. 229 of [1℄,(1.7) willfollowfrom thebound
0:732
R
0:6 "
S()d<
2
5
"
yL 1
:
Thuswe seek thesharpest attainable upperbounds forS(). As in[1, 12,
13℄, we use theRosser{Iwanie sieve, at leastfor >0:661. For 0:661,
weusethealternativesievepro eduredevelopedbyHarman[11℄andBaker,
Harman and Rivat [2℄ to give sharperbounds for S(). It is here that we
gaina sizeableadvantageover Liu[13℄.
Throughout the paper, we suppose that x > C(") and write =
exp( 3="), J =[vy 1
x 4
℄. Constants implied by , and O () depend
atmoston". ConstantsimpliedbyO() areabsolute. ThenotationsA.B
and Y Z areabbreviationsforAB(1+O(")) and Y Z Y:
We write
( )= [ ℄ 1=2;
V( )= Y
p<
1 1
p
;
G( )=exp
1
1
log
1
( >0):
We write m M as an abbreviation for M < m 2M. The smallest
primefa tor of nis writtenQ(n),with Q(1)=1.
The rst-named author would like to thank the Institute of Advan ed
StudyatPrin eton,wherethepenultimatedraftofthepaperwasprepared.
2. Exponential sums of form (1.1)
Lemma 1. Let ;
1
;
2
be given real numbers su h that 6= 1 and
1
2
6=0. Let 1M;M
1
;M
2
x. Let A>0,and
S = X
mM
X
m
1
M
1 X
m
2
M
2 b
m
1
;m
2 e(Am
m 1
1 m
2
2 )
where jb
m
1
;m
2 j1:
Writing F =AM
M
1
1 M
2
2 ,L
1
=log(MM
1 M
2
(1+A)), we have
S L 2
1 ((M
1 M
2 )
13=14
M 9=14
F 1=14
+M
1 M
2 M
2=3
+M
1 M
2 M
13=12
F 1=4
+(M
1 M
2 )
3=4
M +(M
1 M
2 )
3=4
M 1=2
F 1=4
):
Proof. Thisisavariantof[7℄,Theorem3;see[2℄foraproof. In[2℄,the
authorsin ludedterms(M
1 M
2 )
23=24
MF 1=6
and(M
1 M
2 )
23=24
M 7=12
F 1=24
.
Theseare super uous,sin e
(M
1 M
2 )
23=24
MF 1=6
=(M
1 M
2 M
13=12
F 1=4
) 4=6
(M
1 M
2 M
2=3
) 1=6
(M 3=4
1 M
3=4
2 M)
1=6
;
(M
1 M
2 )
23=24
M 7=12
F 1=24
=(M 3=4
1 M
3=4
2 M
1=2
F 1=4
) 1=6
(M
1 M
2 M
2=3
) 5=6
M 1=18
:
Lemma 2.Leta
m
(mM),b
n
(nN) be omplexnumbersof modulus
1. We have
(2:1)
X
mM X
nN
v<mnev a
m b
n
x+y
mn
x
mn
yx 5
1=2+" 2 " 3 1=7 1=14 "
Proof. If the sum (2.1) is nonempty, then MN v. (This will fre-
quentlybe usedimpli itly intherest ofthe paper.)
In view of [1℄,Lemma 7,and theproofof [1℄,inequality(50), itsuÆ es
to establish
(2:2) T :=
X
hH X
mM X
nN a
m b
n e
h
mn
vx 6
for H J, MN v, x. (We have dispensed with the summation
onditionv<mnev;thisis permissiblejustasin[1℄, Lemma15.)
WenowgetthedesiredresultbyanappealtoLemma9of[1℄andLemma
2 of[13℄.
Lemma 3. Let (;) be an exponent pair. Let 0:64 < < 0:72. Then
(2.1) holds for all N satisfying
(2:3) vx
1=2+"
N (x
=2 1=4
v 1+
)
1=(1+ )
x
"
:
Proof. Asinthepreviouslemma,itsuÆ esto establish(2.2). A ord-
ingto Lemma 14of [1℄,for N satisfying(2.3) we have
(2:4) T
2
M 2
(NH) 1+
+H 1=2+
N
3=2+ 2
M 1=2
x
1=2++2
:
It is now easy to verify that (2.3) and (2.4) together imply (2.2). This
ompletes theproof ofLemma 3.
We shallapplyLemma3 withthree exponent pairs:
(i) Let = 89 =560+. Then (;1=2+) is an exponent pair (Watt
[19℄). Let
(;)=BA
;
1
2 +
=
1+2
4+4
;
1+2
2+2
:
(ii) (;)=BA 3
B(0;1)=(11=30;8=15).
(iii)(;)=BA 4
B(0;1)=(13=31;16=31):
The orrespondingexpressions in(2.3) arerespe tively
v
(3+2)=(5+6)
x
(1+)=(5+6) "
; v 5=7
x
3=10 "
; v 14=17
x 3=8 "
:
Lemma 4.Wehave
(2:5)
X
hH X
nN X
Mm<M
1
v<mnev b
n e
h
mn
vx 6
for
(2:6) x; 1=2<<3=4 "; M
1
2M; jb
n
j1; H J
if either
1=3 "
or
(2:8) x
1=3 "
N x 3=8 "
and H <vx
" 3=8
N 1=2
:
Proof. ThisisavariantofanargumentofWu[20℄. WeapplyPoisson's
summationformulatotheinnersummation. Asintheproofof[1℄,Lemma4,
theleft-hand sideof (2.5) is
Mv 1=2
H 1=2
x 1=2
X
hH X
nN X
r b
0
n
r e
2
hr
n
1=2
+O
"
HN
Mv 1=2
H 1=2
x 1=2
+L
;
whereb 0
n
;
r
have modulus 1 andthe summationrange forr is
hn 1
max
M 2
1
;
ev
n
2
<r <hn 1
min
M 2
;
v
n
2
:
We mayreadilyverify that
(2:9) HN
Mv 1=2
H 1=2
x 1=2
+L
vx 6
:
Asintheproofof Lemma2, itnowsuÆ es to showthat
(2:10)
X
hH X
nN X
rR b
00
n
0
r e
2
hr
n
1=2
M 1
v 1=2
H 1=2
x 1=2 7
;
where
(2:11) RHxM
2
N 1
and jb 00
n
j1;j 0
r
j1. By astandard divisorargument,the sum(2.10) may
berewrittenas
(2:12)
X
nN X
kRH b
00
n a
k e
2
k
n
1=2
;
wherea
k
x
:
WeapplyLemma1withM
1
R H;M
2
=1andM repla edbyN. Itis
easilyveri edthat (2.10) isa onsequen eof(2.7) or(2.8). This ompletes
the proof of Lemma 4. The reader will note that (2.8) is unne essarily
strong. The reason for the form of the ondition (2.8) will be ome lear
when we prove Corollary1 of Theorem2.
3. Bilinear exponential sums
Lemma 5. Let M N <N
1
M
1
. Let a
n
(M nM
1
) be omplex
numbers. Then
X
N<nN a
n
1
R
1 K( )
X
M<mM a
m e( m)
d
withK( )=min(M
1
M+1;(j j) 1
;( ) 2
) and
1
R
1
K( )d 3log(2+M
1 M):
Proof. This isLemma 2.2of Bombieri and Iwanie [3℄.
Theorem 2. Let ; be given nonzero real numbers, 6=1. Let X;Y
1. Let a
x
(xX) and b
y
(y Y) be omplex numbers of modulus 1.
Let
S = X
xX X
yY a
x b
y e(Ax
y
);
wherethe positivenumber F =AX
Y
satis es
(3:1) F <min(Y
2
;XY 1 3
):
Then for any natural number Q, 1QY 1
,we have
S (XY) 3
fXYQ 1=2
+XY 3=2
F 1=2
Q 1=2
+F 1=6
Q 1=3
XY 13=12
+F 3=8
Q 5=16
XY 11=8
+F 1=4
Q 1=8
X 1=2
Y 3=4
+F 1=3
Q 1=6
X 1=2
Y 7=12
+F 1=8
Q 3=16
X 1=2
Y 7=8
g:
Proof. Wemaysupposethat Y >
0
(). Webegininthe same way as
Liu[13℄,proofofLemma4. ByCau hy'sinequalityanda\Weylshift"(see
e.g. Graham andKolesnik[10℄, Lemma2.5),
jSj 2
X X
xX
X
yY b
y e(Ax
y
)
2
(XY)
2
Q +
XY
Q X
0<jqjQ X
y;y+qY b
y b
y+q X
xX e(Ax
t(y;q)):
Here t(y;q) = (y +q)
y
. After splitting the range of q into dyadi
intervals,weobtain
Y
jSj 2
(XY)
2
Q +
XY
Q
X
qQ1 X
y;y+qY b
y b
y+q X
xX e(Ax
t(y;q))
forsome Q
1
,1Q
1
Q.
There arenowtwo ases to onsider.
Case 1: Q
1
Y 2
. Now
d
dx (Ax
t(y;q)) FQ
1 Y
1
X 1
Y
by(3.1). Lemmata 4.2 and4.8 of[17℄ yield
X
e(Ax
t(y;q)) (FQ
1 )
1
XY
and
Y
jSj 2
(XY)
2
Q +
XY 2
Q Q
1 (FQ
1 )
1
XY
(XY) 2
(Q 1
+YF 1
Q 1
);
sothat thetheoremis true inthis ase.
Case 2: Q
1
>Y 2
. We apply asharp form of thePoisson summation
formulato theinnermost sum(Min [14℄,Theorem2.2):
X
xX e(Ax
t)= X
u2I C
1 j(At)
u 1=2
je(C
2 (At)
2
u 1 2
) (3:2)
+O
min
X 2
Y
Q
1 F
1=2
; 1
kg
1 (y;q)k
+ 1
kg
2 (y;q)k
+ XY
Q
1 F
+(XY)
:
Here I = [C
3 AX
1
jtj, C
4 AX
1
jtj℄, = 1=(2(1 ) ), 2 (2 1) 6= 0,
g
1
= AX 1
t and g
2
= A(2X) 1
t. (The onstants C
1
;C
2
;::: depend
onlyon .)
For xedq Q
1
,wehave
(3:3) 2
X
j=1 X
yY min
(X 2
YQ 1
1 F
1
) 1=2
; 1
kg
j (y;q)k
(1+FQ
1 X
1
Y 1
)((X 2
YQ 1
1 F
1
) 1=2
+F 1
Q 1
1 XY
2+
)
by a variant of Lemma 9 of Vinogradov [18℄, Chapter I. Thus if we sum
(3.2) over y, inter hange summations, and apply a partial summation, we
ndthat
(3:4) (XY) 2
jSj 2
(XY) 2
Q 1
+X 2
Y 3=2
Q 1=2
F 1=2
+F 1
Q 1
X 2
Y 3
+F 1=2
Q 1=2
XY 1=2
+XY 2
+XYQ 1
(1+FQ
1 X
1
Y 1
)(AQ
1 Y
1
)
(XYF 1
Q 1
1 )
+1=2
jS
1 j:
Here,forsome xed uFQ
1 X
1
Y 1
,
S
1
= X
y X
1 b
y b
y+q e(C
2 (At)
2
u 1 2
);
wheretheouter summation istaken overa subintervalof (Y;2Y℄.
By Lemma 5,we have
(XY)
jS
1 j
X
yY
X
qQ b
y+q
e( q)e(C
2 (At)
2
u 1 2
)
forsomerealnumber independentofyandq. ApplyingCau hy'sinequal-
ityand a Weylshift, we have
(3:5) (XY)
2
jS
1 j
2
(YQ
1 )
Q
2 2
+ YQ
1
Q
2
X
1q1Q2 jS
2 (q
1 )j
with
S
2 (q
1 )=
X
yY X
q;q+q1Q1 b
y+q b
y+q+q1 e(t
1 ):
Here
t
1
=t
1 (y;q;q
1 )=C
2 u
1 2
A 2
(t(y;q+q
1 )
2
t(y;q) 2
):
The hoi eofQ
2
(1Q
2
Q
1
)isatourdisposal. Forreasonswhi hshould
shortlybe ome lear,we take
Q
2
=min(Q (1 )=2
1
;(Q
1 Y
2
F 1
) 1=3
):
We now estimate S
2
for a given value of q
1
. (We suppress dependen e
on q
1
.) Writing y+q =z,weget
S
2
= X
q;q+q
1
Q
1 X
z qY b
z b
z+q1 e(T)
with
T =T(z;q)=C
2 u
1 2
A 2
(t(z q;q+q
1 )
2
t(z q;q) 2
):
ApplyingLemma5 on emore gives
(XY)
jS
2 j
X
Y=2<z<2Y
X
1
e(T(z;q))
:
A nal appli ationof Cau hy'sinequalityand a Weylshiftyields
(3:6) (XY) 2
jS
2 j
2
(YQ
1 )
Q
3 2
+ YQ
1
Q
3
X
1q
2
Q
3 X
1 jS
3 (q;q
2 )j
withQ
3
=Q 2
2
Q 1
1
and
S
3 (q;q
2 )=
X
Y=2<z<2Y
e(T(z;q) T(z;q+q
2 )):
Nowit iseasy to verify that, inthelastsum,
d
dz
(T(z;q) T(z;q+q
2
))Fq
1 q
2 Q
1
1 Y
2
FQ 3
2 Q
1
1 Y
2
Y
:
By Lemmata 4.2and 4.8of [17℄,wehave
S
3 (q;q
2
) (Fq
1 q
2 Q
1
Y 2
) 1
:
With(3.6), (3.5), thisyields
(XY) 3
jS
2 j
2
Y
2
Q 2
1
Q 2
2 +
Y 3
Q 3
1
Q 2
2 q
1 F
;
(XY) 4
jS
1 j
2
Y
2
Q 2
1
Q
2 +
Y 5=2
Q 5=2
1
Q 3=2
2 F
1=2 :
Usingthede nitionof Q
2
,we get
(XY) 5
jS
1 j
2
Y 2
Q 3=2
1 +Y
4=3
Q 5=3
1 F
1=3
+Y 5=2
Q 7=4
1 F
1=2
+Y 3=2
Q 2
1 :
Sin e Q
1
<Y,the lastterm may be omitted. Re alling (3.4), and noting
that
(AQ
1 Y
1
)
(XYF 1
Q 1
1 )
=X 1=2
;
we obtain
(XY) 5
jSj 2
(XY) 2
Q 1
+X 2
Y 3=2
Q 1=2
F 1=2
+F 1
Q 1
X 2
Y 3
+F 1=2
Q 1=2
XY 1=2
+XY 2
+X 2
Y 3=2
Q 1
F 1=2
Q 1=2
1
(1+FQ
1 X
1
Y 1
)
(YQ 3=4
1 +Y
2=3
Q 5=6
1 F
1=6
+Y 5=4
Q 7=8
1 F
1=4
)
X 2
Y 2
Q 1
+X 2
Y 3=2
Q 1=2
F 1=2
+F 1
Q 1
X 2
Y 3
+F 1=2
Q 1=2
XY 1=2
+XY 2
+X 2
Y 5=2
Q 3=4
F 1=2
+F 1=3
Q 2=3
X 2
Y 13=6
+F 3=4
Q 5=8
X 2
Y 11=4
+XY 3=2
Q 1=4
F 1=2
+XY 7=6
F 2=3
Q 1=3
+XY 7=4
F 1=4
Q 3=8
=T
1
+:::+T
11
; say :
ClearlyT
4
T
9 and T
2
T
6
=T 1=3
1 T
2=3
8
max(T
1
;T
8
). We maysuppose
that T
3
<X 2
Y 2
; onsequently, Y <FQ and T
5
<T
11
. The result follows
inCase2. This ompletestheproofof Theorem2.
Corollary 1.Wehave, for 1=2<3=4 ",
(3:7)
X
Mm<M
1 X
nN
v<mnev b
n
x+y
mn
x
mn
yx 5
whenever jb
n
j1, M
1
2M and
(3:8) N x
3=8 "
:
Proof. Asintheproof ofLemma2,itsuÆ es to showthat(2.5) holds
when x,H J. Inview ofLemma 4,we maysupposethat
1=3 " 3=8 "
and that
(3:10) Hvx
" 3=8
N 1=2
:
As in the proof of Lemma 4, we only have to prove that the left-hand
sideof(2.12) is
(3:11) M
1
v 1=2
H 1=2
x 1=2 6
:
We applyTheorem2 with
X=N; Y H 2
xv 2
N; F Hxv 1
; Q=[x 1=4
℄:
It is easy to dedu e from (3.10) that Q <Y 1
and that (3.1) holds. The
bound(3.11) mayreadilybe veri ed,and Corollary 1is proved.
Corollary 2.Theformula (1.6) holds.
Proof. An examinationof the proof of [1℄, Proposition1, reveals that
itsuÆ es to prove that (3.7) holdswheneverjb
n
j1,M
1
2M and
(3:12) M >x
1=4 6
; v<x 3=5 "
:
Now(3.12) impliesN vM 1
x 7=20+2"
:Hen e Corollary2 followsfrom
Corollary1.
4. The Rosser{Iwanie sieve. Let () be de ned to be 2 3 in
(3=5 ";27=44℄, (2+1)=14in(27=44;0:642℄,(3+2)=(5+6) (1+)=(5+
6) in (0:642;0:671℄, 5=7 3=10 in (0:671;357=520℄ and 14=17 3=8 in
(357=520;0:7℄. Let
a=x
1=2+"
; b=x
() "
; ba 1
=x g
; I =[a;b℄:
Then
(4:1) Thebound (2.1) holdsfor N 2I:
This is a onsequen e of Lemma 2 for 0:642. We use the remarks
following Lemma 3 forthe remaining intervals. When 0:661, we shall
applytheRosser{Iwanie sieveasin[1℄toboundS(A;z)fromabove. Here
z=D 1=3
,the\levelofdistribution"Dbeingde nedasfollows. For0:661<
0:7, let
D
0
=x 5"
min(b 3
a 1
;x 3=4
a 2
)
and
D=max(D
0
;x
3=8+g 5"
)=x
%()
; say :
For0:7<0:732, let D=x
%()
, where%()=3=8 4".
The interval of in whi h D = D
0
is rather short, 0:6825 < <
0:6854:::;butthe devi eseems worth in luding,partlybe ause analogous
Lemma 6.Let A
1
;:::;A
t
bepositive numbers withA
1
::: A
t
1,
(4:2) A
1 :::A
2j A
3
2j+1
D 1+
(0j (t 1)=2):
Then for 0:661<0:7, either
(4:3) A
1 :::A
t
<x 3=8 2"
or some setS f1;:::;tg satis es
(4:4)
Y
i2S A
i 2I:
For 0:7<0:73, the inequality (4.3) holds.
Proof. The lemma is obvious for > 0:7, sin e (4.2) implies that
A
1 :::A
t
D
1+
x 3=8 3"
. Now let 2 (0:661;0:7℄. Suppose that
neither(4.3) nor(4.4) holds;weshall obtaina ontradi tion.
Suppose rst that D = x
3=8+g 5"
. By Lemma 5 of Fouvry [6℄, with
Y
i
=A
i
,W =D 1+2
,U =a, V =b,we have
A
1 :::A
t
D 1+2
ab 1
<x 3=8 2"
:
Thisis absurd.
Next, suppose that D = D
0
. As in [1℄, p. 215, we partition A
1 :::A
t
into two produ ts P and Q, with
P D 1=2+
0
; QD 1=2+
0 :
SupposeifpossiblethatP b. ThenP <a. Now
A
1 :::A
t
=PQ<aD 1=2+
0
<x 3=8 2"
:
Thisis absurd,soP >b. Similarly,Q>b.
Let P 0
bethe subprodu tof P formedfrom those A
i
that ex eed ba 1
;
de ne Q 0
similarly. Sin e (4.4) never holds, it is lear that P 0
b and
Q 0
b. If P 0
Q 0
=A
j
1 :::A
j
r
with j
1
>:::>j
r
then, from(4.2),
D 1+
P 0
Q 0
A
jr
b 2
ba 1
;
whi h isabsurd. This ompletestheproofof Lemma6.
Lemma 7.Let z=D 1=3
. Then
S()S(A;z). 2
%()
y
L :
Proof. Inview of Lemma6 and (4.1), thisfollows inexa tlythe same
wayas[1℄,Lemma 16. (The ondition(63) of[1℄ isobviously satis ed.)
5. An alternativesieve. Inthisse tionandthenextwesupposethat
Let
B=fn:v<nevg; !(n)=y=n (n2B):
ForE =A orB,let E
m
=fn:mn2Eg:We write
S(B
m
;u)= X
mn2B
Q(n)u
!(nm)
wheneverQ(m)u. ThequantityS(B;u)willa tasa model forS(A;u).
We letw(u) beBu hstab'sfun tion, sothat
w(u)=1=u (1u2); (uw(u)) 0
=w(u 1) (u>2):
Asu!1,wehave
(5:1) w(u)=e
+O(G(1=u)):
SeeCheerand Goldston[4℄.
Lemma 8.For mv 1=2
and x
"
v=m, Q(m), we have
S(B
m
;)=w
log (v=m)
log
y
mlog (1+O
"
(L 1
)):
Proof. Let (u) = jfn u : Q(n) gj. For u 2 [ =m;ev=m℄, we
have
(u)=w
log (v=m)
log
u
log (1+O
"
(L 1
))
(see Friedlander [8℄). Consequently,
S(B
m
;)= y
m X
mn2B
Q(n)
1
n
= y
m ev=m
R
v=m
d (u)
u
= y
m
(u)
u
ev=m
v=m +
ev=m
R
v=m (u)
u 2
du
=w
log (v=m)
log
y
mlog (1+O
"
(L 1
)):
This ompletes theproof ofLemma 8.
Lemma 9. For M x 3=8 "
,we have
(5:2)
X
mM a
m jA
m j=y
X
mM a
m
m +O
"
(yx 4
)
whenever ja j1.
Proof. The left-hand sideof (5.2) is
X
mM a
m X
v<mnev
X
k
x<mnkx+y 1
= X
mM a
m X
n
v<mnev
y
mn
x+y
mn
+
x
mn
=y X
mM a
m
m
X
n
v<mnev 1
n +O
"
(yx 4
)
byCorollary 1ofTheorem2. Wemaynoweasily ompletetheproof( om-
pare[1℄, p.210).
Lemma 10. Let M x 3=8 2"
, 0 a
m
1, a
m
=0 unless Q(m) x
(m=1;:::;M). Then
X
mM a
m S(A
m
;x
)= X
mM a
m S(B
m
;x
)
1+O
G
"
+O
"
(yx 3
):
Proof. We applyLemma 8 of Baker, Harmanand Rivat [2℄, with z;y
repla edbyx
,x
"
,takingE =A
m
andY =y=m. Inthenotationof[2℄,we
have =", hen e 50
log (x
)>1. We dedu e that, whenever a
m 6=0,
S(A
m
;x
)= y
m V(x
)
1+O
G
"
+O
X
d<x
"
pjd)p<x
jA
md j
y
md
:
(Note thatE
d
=A
md
be ause (m;d)=1 here.)
By a divisor argument ( ompare [2℄) there are numbers
j
x
for
whi h
X
mM a
m S(A
m
;x
)
=yV(x
) X
mM a
m
m
1+O
G
"
+O
X
j<x 3=8 "
j
jA
j j
y
j
=yV(x
) X
mM a
m
m
1+O
G
"
+O
"
(yx 3
)
byLemma9.
CombiningLemma8 with(5.1), we have
S(B
m
;x
)= y
e
+O
G
"
:
In viewof theapproximation
V(x
)= e
L
f1+O
"
(L 1
)g;
we obtainthedesired result.
Let I betheintervalde ned at thebeginningofSe tion 4.
Lemma 11.For N 2I,ja
m
j1, jb
n
j1,we have
X
mn2A
mM;nN a
m b
n
=y
X
mn2B
mM;nN a
m b
n
mn +O
"
(yx 5
):
Proof. As inLemma9,theleft sideis
X
mn2B
mM;nN a
m b
n
y
mn
x+y
mn
+
x
mn
=y
X
mn2B
mM;nN a
m b
n
mn +O
"
(yx 5
)
from (4.1). Thisestablishes Lemma11.
Lemma 12. Let h 1 be given and suppose that D f1;:::;hg and
M 2I, M
1
<2M. Then
(5:3)
X
p
1 :::
X
p
h
S(A
p
1 :::p
h
;p
1 )
= X
p1 :::
X
ph
S(B
p
1 :::p
h
;p
1 )+O
"
(yx 4
):
Here
indi ates that p
1
;:::;p
h
satisfy
x
p
1
<:::<p
h
; (5:4)
M Y
j2D p
j
<M
1 (5:5)
together with no morethan "
1
further onditions of the form
(5:6) R
Y
j2F p
j
S:
Proof. The left-hand sideof (5.3) is
X
p :::
X
p
t
X
j=1 X
q :::
X
q 1;
where t < 1
, and the inner summation extends over primes q
1
;:::;q
j
satisfying
p
1
q
1
:::q
j
; p
1 :::p
h q
1 :::q
j 2A:
This would be the type of sum estimated in Lemma 11, if we ould dis-
entangle the intera tions between the variables p
1
;:::;p
h
;q
1
;:::;q
j . The
pro edure for doing this via the trun ated Perron formula in [1℄ (proof of
Lemma11) maybeapplied here. A ordinglythe left-handsideof (5.3) is
(5:7)
X
p1 :::
X
ph
t
X
j=1 X
q1 :::
X
qj
y
p
1 :::p
h q
1 :::q
j +O
"
(yx 4
):
Naturally we mayalso obtaintheformula(5.7) forthesum
X
p
1 :::
X
p
h
S(B
p1:::p
h
;p
1 )
and Lemma12 follows.
Lemma 13. Let M a=2 and N x 3=8 2"
=(2a). Let M M
1
2M
and N N
1
2N. Let x
<zba 1
. Suppose that f1;:::;hg partitions
into two sets C and D. Then
(5:8)
X
p
1 :::
X
p
h
S(A
p1:::p
h
;z) = X
p
1 :::
X
p
h
S(B
p1:::p
h
;z)(1+O(")):
Here
indi ates that p
1
;:::;p
h
satisfy
(5:9) zp
1
<:::<p
h
;
(5:10) M
Y
j2C p
j
<M
1
; N
Y
j2D p
j
<N
1
together with no morethan "
1
further onditions of the form (5.6).
We remarkthatthe ase h=0,C and D emptyispermitted.
Proof. Let uswritep =(p
1
;:::;p
h
) and m = Q
j2C p
j , n=
Q
j2D p
j .
Bu hstab's identity
S(E;z
1
)=S(E;z
2 )
X
z
2
p<z
1 S(E
p
;p)
appliesto bothA
mn and B
mn (2z
2
<z
1
). In parti ular,
X
p
S(A
mn
;z)= X
p
S(A
mn
;x
) X
p
X
x
q1<z S(A
mnq1
;q
1 ):
The rstterm ontheright hasan asymptoti formulabyLemma10.
The subsum of the other term on the right for whi h mq
1
a has an
asymptoti formulabyLemma12, sin emq 2Mzb.
To the residual sum in whi h mq
1
<a, we applyBu hstab on e more.
Ifwe ontinue inthisfashion,the jth step isthe identity
X
j :=
X
p
X
(5:11) S(A
mnq
1 :::q
j
;q
j )
= X
p
X
(5:11) S(A
mnq1:::qj
;x
) X
p
X
(5:12) S(A
mnq1:::qj+1
;q
j+1 )
withsummation onditions
x
q
j
<:::<q
1
<z; mq
1 :::q
j
<a;
(5:11)
x
q
j+1
<q
j
<:::<q
1
<z; mq
1 :::q
j
<a:
(5:12)
The rst sum on theright has an asymptoti formulaby Lemma 10, sin e
thevariablessatisfy
mnq
1 :::q
j
<a(x 3=8 2"
=a)=x 3=8 2"
:
The subsum of the se ond sum on the right,given by mq
1 :::q
j+1
a,
hasan asymptoti formulabyLemma12, sin e
mq
1 :::q
j+1
<aq
j+1
<az b:
The residual sum is P
j+1
. After O(
1
) steps the residual sum is learly
empty, giving a de omposition of P
p S(A
mn
;z) into a main term and an
errorterm, sayE.
A orresponding de omposition with the error term applies to
P
p S(B
mn
;z), and to omplete the proof we must show that E is of a -
eptable size. Just asin[2℄,proof ofLemma 12,wehave
E=O
1
2 1=
G
"
X
p
S(A
mn
;z)
=O
"
X
p
S(A
mn
;z)
sin e
1
2 1=
G
"
<exp
1
"
log
"
<exp
1
:
This ompletes theproof ofLemma 13.
Lemma 14. Let x 3"
P <max(b 2
=a;x 7=8
). Then
(5:13)
X
pP S(A
p
;ba 1
)= X
pP S(B
p
;ba 1
)(1+O(")):
Remark. b 2
=a>x 7=8
for<29=48=0:60416:::
Proof of Lemma 14. In view of Lemma 13 we need only on ern
ourselveswith the ase
7=8 3" 2
We an beginthe proof as inLemma 13, butwe annot ontinue when we
rea h a point where
(5:14) Pq
1 :::q
n
>x 3=8 2"
; q
1 :::q
n
<a:
At su ha stage we note that
(5:15)
X
q
j
Q
j
pP S(A
pq1:::qn
;q
n )=
X
q
j
Q
j
r
S(A
rq1:::qn
;(2P) 1=2
);
wherer has noprime fa torsbelowq
n
,and v=(2P)rq
1 :::q
n
<ev=P. In
view of(5.14) we thusobtain
(5:16) r vx 2" 3=8
=x
+2" 3=8
<x
7=8 3"
=x 3=8 2"
=a;
sin e29=48<5=8. We therefore an applyBu hstab to theright-handside
of (5.15) to obtain
(5:17)
X
q
j
;r S(A
rq1:::qn
;ba 1
)
X
q
j
;r
ba 1
s<(2P) 1=2
S(A
rq1:::qns
;s);
wherethe rstsum anbeestimatedbyLemma13. These ondsum ounts
numbersrq
1 :::q
n
st2A,whereP=(se)t<2eP=s. Wenotethat(2P) 1=2
<b,sowe an applyLemma12 whensa. If s<athen
P=(se)>P=(ae)>a and 2Pe=s2Pea=b<b;
sot2[a;b℄andthusLemma12is appli ableinthis asealso. Wetherefore
obtaina suitableformula forboth sumsin (5.17), whi h establishes (5.13)
asrequired.
Lemma 15. Let x 3"
P < max(b 2
=a;x 7=8
);P > ev=b 2
and ba 1
< Q
b. Then
(5:18)
X
pP
qQ S(A
pq
;q)= X
pP
qQ S(B
pq
;q)(1+O(")):
Proof. If Q a we an apply Lemma 12 so we hen eforth suppose
Q<a. We rst onsiderthe ase
x 7=8
<x 3"
P <b 2
=a
(so<29=48). Hereweworkina similarmannerto Lemma14 to obtain
X
pP
qQ S(A
pq
;q)= X
r S(A
qr
;(2P) 1=2
) (5:19)
= X
r S(A
qr
;ba 1
)
X
1 1=2
S(A
qrs
;s):
Now
r v
PQ
<
vx
7=8+3"
a
b
=x
6+2" 27=8
<x
7=8 3"
= x
3=8 2"
a
sin e 29=48 < 17=28 (= 0:607:::). As Q < a we may therefore apply
Lemma 13 to the rst sum in (5.19). The se ond sum in (5.19) ounts
numbersqrst2A. AsintheproofofLemma14 wedis overthatoneofs;t
must belong to theinterval[a;b℄.
Wenowassumethatx 3"
P x 7=8
. WeapplyBu hstabdire tlyto the
left-handsideof (5.18) to obtain
(5:20)
X
pP
qQ S(A
pq
;ba 1
)
X
pP
ba 1
r<q S(A
pqr
;r):
Lemma13 an beappliedtothe rstsumin(5.20). Nowforthehypothesis
of Lemma 15 to hold we must have x 7=8
> v=b 2
so < 39=64. It then
follows that(ba 1
) 2
>a. We an thereforeapplyLemma 12 to those parts
of the se ond sum with qr b. For the remaining portion of the sum we
notethat it ounts numbers pqrs2A where
s<
ev
Pqr
ev
(ev=b 2
)b
=b;
and
s>
v
8PQ 2
v
8x
7=8 3"
a 2
= x
1=8+5"
8
>a:
Hen e Lemma12 isagain appli ableand this ompletes theproof.
6. Implementing the alternative sieve. We would like to give an
upperboundforS(A;v 1=2
)oftheformu()y=(logx)whereu()isassmall
aspossible. For<0:6wehaveobtainedthe\ orre t"value1. Themethod
we now present gives a ontinuous fun tion u() starting with u(0:6) = 1.
Sadly,bythetime rea hes 39=64 (=0:609:::),thevalueof u() isnearly
2. From this point onwards we are giving an upper bound for S(A;b).
Although this upper boundis very lose to the expe ted value of S(A;b),
thesubstitutionofbforv 1=2
ausesthemarked deteriorationinthequality
ofourbound. Nevertheless,ourmethodherestillprodu esasuperiorresult
to that obtained from the Rosser{Iwanie sieve (see Lemma 17 below) up
to about=0:661. Theoreti ally thealternative sieve willbe better up to
=0:7, but the al ulations ne essary be ome impra ti al. The values of
We begin by usingthe Bu hstab identityto write
(6:1) S(A;v 1=2
)=S(A;ba 1
)
X
ba 1
pb S(A
p
;p)
X
b<p<v 1=2
S(A
p
;p):
The ontribution from the rst two terms on the right-hand side of (6.1)
equals S(A;b). We willonly be able to give a non-trivial estimate for the
naltermwhen<39=64. We startbygivingalowerboundforthisterm.
We write
y=x 3"
max(b 2
=a;x 7=8
):
Then,for<39=64;
(6:2)
X
b<p<v 1=2
S(A
p
;p) X
ev=b 2
<p<y S(A
p
;p)
= X
ev=b 2
<p<y S(A
p
;ba 1
)
X
ev=b 2
<p<y
ba 1
q<min(p;(ev=p) 1=2
) S(A
pq
;q):
Weobtainanasymptoti formulaforthe rsttermin(6.2)fromLemma14,
andforthese ondtermfromLemma15,sin ep>ev=b 2
gives(ev=p) 1=2
<b.
Thus
(6:3)
X
b<p<v 1=2
S(A
p
;p) X
ev=b 2
<p<y S(B
p
;p)(1+O(")):
Now we an applyLemma13 to S(A;ba 1
) in(6.1), andLemma 12 to
X
apb S(A
p
;p):
Thisleavesa term
X
ba 1
p<a S(A
p
;p)=
X
ba 1
p<a S(A
p
;ba 1
)+ X
ba 1
q<p<a S(A
pq
;q) (6:4)
=
X
ba 1
p<a S(A
p
;ba 1
)+ X
ba 1
q<p<a S(A
pq
;ba 1
)
X
ba 1
r<q<p<a
pqr 2
ev
S(A
pqr
;r)
= X
1 +
X
2 X
3
; say.
We an applyLemma13to P
1 and
P
2
to obtainasymptoti formulae. We
thenwant a lowerboundfor P
3
. Clearly,ifanyprodu tof 2or3 of p;q;r
liesintheintervalI we anapplyLemma12. Otherwiseitmaybepossible
points. We write
(p;q;r)=(x 1
;x 2
;x 3
);
and similarlyintrodu e
4
;
5
;::: whenperformingfurtherde ompositions
of P
3
aswe bringinnewprime variables s=x
4
;t=x
5
;:::We put
E
n
=f =(
1
;:::;
n )2R
n
:g
n
<
n 1
<:::<
1
< 1=2+";
1
+:::+
n 1 +2
n
+1=Lg;
where g = g(), as de ned at the start of Se tion 4. The sum over p;q;r
in P
3
thus orrespondsto 2E
3
. Given a set Z R n
we let Z
=f 2
R n
: 62Zg. A point of E
n
issaid to be bad if
nosum X
j2S
j
(S f1;:::;ng) liesin[ 1=2+"; () "℄:
The other pointsof E
n
are alledgood. These orrespond to partsof sums
forwhi h Lemma12 an beapplied. We writeA
n
forthesetof bad points
of E
n ,
W =f(
1
;
2
;
3 )2A
3
:either
2 +
3
>7=8 4" or
(
1
;
2
;
3
;
4 )2A
4
forat leastone
4
2[g;
3
) forwhi h
2 +
3 +
4
7=8 4"g;
U =A
3
\W
;Z =E
3
\A
3
;
X
1
=f(
1
;
2
;
3
;
4 )2E
4
\A
4 :(
1
;
2
;
3
)2Ug;
X
2
=f(
1
;
2
;
3
;
4 )2A
4 :(
1
;
2
;
3 )2Ug
(we notethat
2 +
3 +
4
7=8 4"inX
2
bythe previousde nitions),
V =f(
1
;
2
;
3
;
4
;
5 )2A
5 :(
1
;
2
;
3
;
4 )2X
2 g:
We rstobservethatE
3
partitionsintoW,U andZ. Sin eallthepoints
ofZ aregood,weobtainthedesired asymptoti formulaforthepartof P
3
with 2Z using Lemma 12. In ontrast, the part with 2 W must be
dis arded,that is,thetrivialestimate
X
2W S(A
pqr
;r)0
appears to be theonlyone a essible withtheavailabletools.
To seethis, applyafurtherde ompositionto anysubsetW 0
of W,
X
2W 0
S(A
pqr
;r)= X
2W 0
S(A
pqr
;ba 1
) X
S(A
pqrs
;s)
= X
1 X
2
; say :
In P
2
,(p;q;r;s)=(x 1
;:::;x 4
);
( ; ; )2W 0
; 2[g; ):
IfW 0
overlapsthe region
2 +
3
>7=8 4";
we annot always split pqr into two subprodu ts respe tively less than a
and x 3=8 2"
=(2a), so Lemma 13 will not handle P
1
. Neither an we use
Lemma 12 to handle P
1
, by de nitionof A
3
. If, on the other hand, W 0
doesnotoverlapthishalf-spa e,itoverlapstheset of(
1
;
2
;
3
)forwhi h
(
1
;
2
;
3
;
4 )2A
4
forat leastone
4
2[g;
3
),su hthat
2 +
3 +
4
>7=8 4":
We annotsplitp;q;r;sintotwosubprodu tswhi hLemma13willhandle,
so a further appli ation of Bu hstab to P
2
o ers no way out; nor will
Lemma12 help,byde nitionofA
4 .
Wenowturnourattentiontothesumover 2U. ApplyingBu hstab's
identitywe obtain
(6:5)
X
2U S(A
pqr
;r)
= X
2U S(A
pqr
;ba 1
)+ X
2X
1 S(A
pqrs
;s)+ X
2X
2 S(A
pqrs
;s):
The rst term on the right side of (6.5) may be estimated by Lemma 13
sin e
2 +
3
7=8 4" for all 2 U. The sum over 2 X
1 may
beestimatedbyLemma12 sin eall aregoodinX
1
. We applyBu hstab
on emore to thesum over 2X
2 :
(6:6)
X
2X
2 S(A
pqrs
;s)
= X
2X2 S(A
pqrs
;ba 1
)
X
2E5
( 1; 2; 3; 4)2X2 S(A
pqrst
;t):
We an apply Lemma 13 to the rst sum on the right side of (6.6) sin e
2 +
3 +
4
7=8 4"inX
2
,andwe an applyLemma12 forall the
good inthe nalsum. Thisleavesa nal sumto onsider:
(6:7)
X
2V S(A
pqrst
;t):
We are unableto give a formula for any part of this sum without further
appli ationsof Bu hstab's identity. Dis ardingthissum would give rise to
a lossof
(6:8) y
L R
w
1
2
3
4
5
5
d
1
1 :::
d
4
4 d
5
2
5
;
after usingthe standardpro edure forrepla ing sumsover primesby inte-
grals( f. [1℄,p.227).
Toexplainwhat ismeantbya\loss"of thequantity(6.8), we illustrate
byasimpler example. From (6.1),
S(A;v 1=2
)=S(A;ba 1
)
X
apb S(A
p
;p) X
p2P S(A
p
;p);
where P = [ba 1
;a)[(b;v 1=2
). Suppose we were to dis ard the lastsum.
Sin e
S(B;v 1=2
)=S(B;ba 1
)
X
apb S(B
p
;p) X
p2P S(B
p
;p);
andsin ewehaveasymptoti formulaeforwhatwedonotdis ard,wewould
ndthat
S(A;v 1=2
).S(B;v 1=2
)+ X
p2P S(B
p
;p)
.S(B;v 1=2
)+ y
L R
P 0
w
1
1
d
1
2
1 :
Here P 0
= [5=2 4; 1=2℄[ [2 3;=2℄. Our \loss" by this rude
pro edurewouldbetheintegraloverP 0
. Usinginsteadtheapproa habove,
whi h resultsindis arding
P 00
=[2 3;7 4℄[
max
9 14
2
; 7 8
8
;
2
inR 1
;W in R 3
and V inR 5
,we areled bya similarargument to a lossof
R
P 00
w
1
1
d
1
2
1
+K()+R 0
();
where
K()= R
W w
1
2
3
3
d
1 d
2 d
3
1
2
2
3
; (6:9)
R 0
()= R
V w
1
2
3
4
5
5
d
1 d
2 d
3 d
4 d
5
1
2
3
4
2
5 :
Sin e uw(u)=1 for1u2,and
uw(u)=1+log (u 1) (2u3);
a straightforward al ulation now leadsto
(6:10) S(A;v
1=2
). y
(M()+K()+R 0
()):
Here
(6:11) M()= 1
1+log
7 4
2 3
16 9
9 14
+ ()
R
2
1+log(v 1)
v
dv
for3=5 "<<29=48;
(6:12) M()= 1
1+log
7 4
2 3
16 7
7 8
+ ()
R
2
1+log(v 1)
v
dv
for29=48 <39=64;
(6:13) M()= 1
1+log2+ ()
R
2
1+log(v 1)
v
dv
for39=64.
In (6.11){(6.13), ()== () 1. In(6.13),
M()= 1
() w
()
sin e
()
2[3;4℄:
We an improveon (6.10),althoughtheimprovement onlybe omessig-
ni ant for 2[0:64;0:661℄. Let V
1
be the subsetof V for whi h the sum
1 +
2 +
3 +
4 +
5 +
5
de omposesintotwosubsumslessthan 1=2 2"
and 7=8 4";forinstan e, 2V
1
whenever 2V and
1 +
2 +
3 +
4 +2
5
<7=8 4":
For the subsum of (6.7) with 2 V
1
, two more appli ations of Bu hstab
may be handled in essentially the same way as (6.6). We do not extra t
all the information this yields, owing to the large number of ways seven
variablesmight ombine to give a value in[ 1=2+"; () "℄. However,
5
R
g
6
R
g w
1
:::
7
7
d
6 d
7
6
2
7
<0:57
5
R
g
6
R
g d
6
6 d
7
2
7
=0:57
1
g log
5
g
1
g +
1
5
for 2 V, sin e one may verify that
1
:::
7
> 3
7
. (For the
boundw(u)0:57foru3,see [4℄.)
Let V
2
bethat partof V
1 where
0:57
1
g log
5
g
1
g +
1
5
<
1
5 w
1
:::
5
5
and V the omplement of V in V. Our dis ussion shows that in (6.10),
R 0
()mayberepla ed by
R ()= y
L
R
V
3 w
1
:::
5
5
d
1 :::d
5
1 :::
4
2
5 (6:14)
+0:57 R
V
2
1
g log
5
g 1
g +
1
5
d
1
1 :::
d
5
5
:
We re ordthis on lusionina lemma.
Lemma 16. For 3=5 "<0:661 we have
S(A;v 1=2
). y
L
(M()+K()+R ()):
HereM(), K(),R () arede ned by (6.11){(6.13), (6.9) and (6.14).
We have ta itly assumed so far thatit will always be to ouradvantage
to e e t the rst ve de ompositionswhen these arepossible. For0:65
thisisnotalways the ase. Toallowforthis, we de ne
I
1 (
1
;
2
;
3 )=
R
(
1
;:::;
5 )2V
w
1
:::
5
5
d
4
4 d
5
2
5
where the indi ates that the integral is to be subdivided further as in
(6.14). Now let
I
2 (
1 )
=
R
( 1; 2; 3)2U min
1
3 w
1
2
3
3
;I
1 (
1
;
2
;
3 )
d
2
2 d
3
3
;
I
3 (
1 )=
R
( 1; 2; 3)2W w
1
2
3
3
d
2
2 d
3
2
3
and
I
4 ()=
1=2
R
g
min
1
1 w
1
1
;I
2 (
1 )+I
3 (
1 )
d
1
1 :
AveryslightimprovementinLemma16isattainedbyrepla ingK()+R ()
byI
4 ().
We nishthisse tionbyrevisitingtheboundsobtainedfromtheRosser{
Iwanie sieve. We an ombine Lemma 7 with Lemma 12 to obtain the
followingresult.
Lemma 17. For 2(0:661;0:7℄, we have
Ly 1
S(). 2
%()
()
R
w
d
2
R
B w
1
2
2
d
1 d
2
1
2
2 :
Here
B =f(
1
;
2
):%()=3<
1
< 1=2; 1=2<
2
< ()g:
Proof. We note that S(A;z) ounts many numbers not ounted by
S(). Forsome ofthesewe an applyLemma12andsoobtainanimproved
boundbyremovingthe\dedu tible" terms. To bepre ise,we have
(6:15) S()S(A;z)
X
pm2A
p2I
Q(m)>p 1
X
p
1 p
2 m2A
z<p
1
<a;p
2 2I
Q(m)>p
2 1:
We note that for the values of in the lemma we have z < a and so the
dedu tiblesumsarenon-empty. Sin eLemma 12 anbe appliedto bothof
these sumswe an repla esumsbyintegrals in(6.15) and useLemma7 for
S(A;z) to ompletetheproof.
7. Completion of the proof. The graph of our upper bound for
S()Ly 1
isshown inDiagram1 (asu()).
Diagram1
FromLemma 17 we obtain
(7:1)
0:7
R
0:661
S()d<0:1386yL 1
:
UsingLemma16 wenote that
(7:2)
0:661
R
M()d<0:1256
and
(7:3)
0:661
R
0:6
min(I
4
();K()+R ())d<0:0125:
We remarkthat it is straightforward to obtain very a urate estimates for
theintegrals in(7.1) and (7.2). The estimate forthe integral in(7.3) is an
upper bound, but a more pre ise estimate is more diÆ ult to a hieve. We
thusobtain
0:7
R
0:6
S()d<0:2767yL 1
:
Sin e,byLemma7,
L
y
R
0:7
S()d 16
3
R
0:7
d = 8
3 (
2
0:49);
thisindi atesthat ourtheoremholdsforanyexponent lessthan
0:49+ 3
8
(0:4 0:2767)
1=2
>0:732;
whi h establishesTheorem1.
We nish by remarking that it appears to be very diÆ ult to make
any further progress without new exponential sum estimates. To in rease
the exponent just by 0:001 would require us to make a saving of nearly
0:004 between0:6 and 0:7. The onlyroomfor improvement seems to be in
K()+R ()for0:64. Evenat =0:64,wehave (K()+R ())<0:42.
Assumingthatwewillhavetoswit htothemethodofSe tion4by=0:67,
it appears unlikely that we an make the ne essary average saving of 0:13
between0:64 and 0:67withouta newidea.
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DEPARTMENTOFMATHEMATICS MATHEMATICSINSTITUTE
BRIGHAMYOUNGUNIVERSITY UNIVERSITYOFWALES
PROVO,UTAH84602 SENGHENNYDDROAD
U.S.A. CARDIFFCF24AG,U.K.
Re eivedon1.8.1994
andinrevisedformon21.2.1995 (2649)
