ACTA ARITHMETICA
XCV.1 (2000)
On the largest prime factor of integers
by
Chaohua Jia (Beijing) and Ming-Chit Liu (Hong Kong)
1. Introduction. Let Q(x) denote the largest prime factor of
Y
n.
1
x<n≤x+x 2 +ε
We are interested in a lower bound of Q(x). On the Riemann Hypothesis,
one can show that Q(x) > x holds for sufficiently large x.
In 1973, Jutila [11] showed that, for sufficiently large x, Q(x) > xϕ ,
where ϕ = 23 − ε. Balog [1], [2] improved it to ϕ = 0.772. Balog, Harman
11
− ε. Recently,
and Pintz [3] obtained ϕ = 0.82. Heath-Brown [7] got ϕ = 12
17
Heath-Brown and C. Jia [8] showed ϕ = 18 − ε.
In this paper, we use some ideas coming from [5], [6], and [8]–[10] on the
sieve method and a delicate application of the estimate of Deshouillers and
Iwaniec [4] on the mean value of Dirichlet polynomials and ζ function. Then
we can prove the following:
Theorem. Let ε be a sufficiently small positive constant. Then, for sufficiently large x, we have
25
Q(x) > x 26 −ε .
Throughout this paper, we suppose that ε is a sufficiently small positive
constant and that B = B(ε) is a sufficiently large positive constant. We
1
+ 2ε is an integer. Suppose that x (> x0 (ε))
choose ε such that K = 8ε 26
is sufficiently large,
25
ε
1
ε
ε
v = x 26 − 2 , P = x 8 , T0 = x 2 − 6 .
Let c, c1 and c2 denote positive constants which have different values at
different places. m ∼ M means that there are positive constants c1 and c2
such that c1 M < m ≤ c2 M.
2000 Mathematics Subject Classification: 11N36.
Project supported partially by the Natural Science Foundation of China.
[17]
18
C. Jia and M. C. Liu
We often use M (s) (M standing here for any capital letter except P and
L) to denote a Dirichlet polynomial in the form
X a(m)
M (s) =
,
ms
m∼M
where a(m) is a sequence of complex numbers with a(m) = O(1). We also
use P (s) to denote
X 1
P (s) =
,
ps
P <p≤2P
where p denotes a prime number.
All calculations in this paper can be verified on the PC computer. The
paper containing full details is obtainable from the authors.
2. Some preliminary lemmas
Lemma 1. Suppose that M N = v and that M (s), N (s) are Dirichlet
1
polynomials. Let b = 1 + 1/log x, T1 = log2B x. Assume that v/x 2 M 1
x 2 . Then for T1 ≤ T ≤ T0 , we have
\
2T
|M (b + it)N (b + it)P K (b + it)| dt log−B x.
T
This is Lemma 1 of [8].
Lemma 2. Suppose that M N L = v and that M (s), N (s) are Dirichlet
polynomials,
X 1
L(s) =
.
ls
l∼L
√
13
6.5
Let b = 1 + 1/log x, T2 = L. Assume that M v 25 , N v 25 . Then for
T2 ≤ T ≤ T0 , we have
\
2T
|M (b + it)N (b + it)L(b + it)P K (b + it)| dt log−B x.
T
This can be proved in the same way as in Lemma 2 of [8].
Lemma 3. Suppose that M N DL = v and that M (s), N (s), D(s) are
Dirichlet polynomials,
X 1
.
L(s) =
ls
l∼L
√
13
Let b = 1+1/log x, T2 = L. Assume further that M v 25 and that N, D
lie in one of the following regions:
Largest prime factor
(i)
(ii)
(iii)
(iv)
2.6
1
19
6.5
N v 25 , D N − 2 v 25 ;
2.6
26
4
26
v 25 N v 175 , D N − 3 v 75 ;
26
5.2
3
6.5
v 175 N v 25 , D N − 4 v 25 ;
5.2
6.5
13
v 25 N v 25 , D N −2 v 25 .
Then for T2 ≤ T ≤ T0 , we have
\
2T
|M (b + it)N (b + it)D(b + it)L(b + it)P K (b + it)| dt log−B x.
T
1
13
P r o o f. If v/x 2 M v 25 , it can be dealt with by Lemma 1. We
1
1
assume that M v/x 2 . Let M (s)P K (s) = H(s). Then H = M P K x 2 .
It suffices to show
2T
\ 1
1
1
1
1
K
I = M
+ it N
+ it D
+ it L
+ it P
+ it dt
2
2
2
2
2
T
2T
\ 1
1
1
1
+ it N
+ it D
+ it L
+ it dt
= H
2
2
2
2
T
1
x 2 log−B x.
Applying Cauchy’s inequality and the mean value estimate for Dirichlet
polynomials, we obtain
2T
\ 1 2 12 2T\ 1 1 2 1 2 12
H
N
I
+it dt
+it D
+it L
+it dt
2
2
2
2
T
T
2 2 12
2T
\
1
1
1
1
N
x4
+ it D
+ it L
+ it dt .
2
2
2
T
If D N, an application of Theorem 2 of Deshouillers and Iwaniec [4]
yields
2 2
2T
\ 1
1
N
L 1 + it dt
+
it
D
+
it
2
2
2
T
ε
1
3
1
1
7
3
T 4 (T + T 2 N 4 D + T 2 N D 2 + N 4 D 2 ).
Thus, when N, D lie in one of the following regions:
26
(a) N v 175 , D N ;
26
5.2
3
6.5
(b) v 175 N v 25 , D N − 4 v 25 ;
6.5
13
5.2
(c) v 25 N v 25 , D N −2 v 25 ,
20
C. Jia and M. C. Liu
we have
\
2T
T
2 2
ε
N 1 + it D 1 + it L 1 + it dt T 1+ 4 .
0
2
2
2
If D N, changing the roles of N and D, we get
2 2
2T
\ 1
1
1
D
+ it N
+ it L
+ it dt
2
2
2
T
ε
1
3
1
1
7
3
T 4 (T + T 2 D 4 N + T 2 DN 2 + D 4 N 2 ).
Thus, when N, D lie in one of the following regions:
2.6
1
6.5
(d) N v 25 , N D N − 2 v 25 ;
2.6
4
26
26
(e) v 25 N v 175 , N D N − 3 v 75 ,
we have
\
2T
T
2 2
ε
N 1 + it D 1 + it L 1 + it dt T 1+ 4 .
0
2
2
2
Combining the regions (a)–(e), we get the regions in Lemma 3. If N, D
lie in one of these regions, then
1
1
I x 4 T02
+ 8ε
1
x 2 log−B x.
This completes the proof of Lemma 3.
We define w(u) as the continuous solution of the equations
w(u) = 1/u,
1 ≤ u ≤ 2;
0
(uw(u)) = w(u − 1), 2 < u.
In particular, when 2 ≤ u ≤ 3,
1 + log(u − 1)
.
u
Lemma 4. For the function w(u), we have the following bounds:
w(u) =
(i) w(u) ≤ 1/1.763 if u ≥ 2;
(ii) w(u) ≤ 0.5644 if u ≥ 3;
(iii) w(u) ≤ 0.5617 if u ≥ 4.
See Lemma 5 of [8].
Define
(1)

1/u


(1
+ log(u − 1))/u
g(u) =

 0.5644
0.5617
if
if
if
if
1 ≤ u ≤ 2;
2 < u ≤ 3;
3 < u ≤ 4;
4 < u.
Largest prime factor
21
We see that for u ≥ 1,
(2)
w(u) ≤ g(u).
Lemma 5. Let E = {n : t < n ≤ 2t} and z ≤ t. Set
Y
X
P (z) =
p,
S(E, z) =
1.
p<z
t<n≤2t
(n, P (z))=1
Then for sufficiently large t and z, we have
t
t
log t
+O
.
S(E, z) = w
log z log z
log2 z
See Lemma 6 of [8].
3. Sieve method. Let
X
N (d) =
1,
1
x<dp1 ...pK ≤x+x 2 +ε
P <pi ≤2P
A = {n : 2
−K
v < n ≤ 2v, n repeats N (n) times},
Ad = {a : a ∈ A, d | a},
Y
X
1.
P (z) =
p, S(A, z) =
p<z
a∈A
(a, P (z))=1
If we prove that
(3)
X
Φ=
1>0
1
x<pp1 ...pK ≤x+x 2 +ε
P <pi ≤2P
2−K v<p≤2v
then we obtain the assertion of the Theorem.
In the following, we set
(4)
B = {n : v < n ≤ 2v},
X
K
1
1
+ε
2
.
X=x
p
(5)
P <p≤2P
Buchstab’s identity yields
(6)
1
1
Φ = S(A, (2v) 2 ) = S(A, v 25 )−
1
v 25
X
12
<p≤v 25
S(Ap , p)−
12
v 25
X
S(Ap , p).
1
<p≤(2v) 2
By the discussion in Lemma 8 of [8] with the application of Lemma 1, we
22
C. Jia and M. C. Liu
can get the asymptotic formula
X
X
(7)
S(Ap , p) =
v
12
1
v 25 <p≤(2v) 2
X
S(Bp , p) + O
.
log2 v
X
12
1
v 25 <p≤(2v) 2
The discussion in Lemma 9 of [8] yields the asymptotic formula
1
1
X
X
25
25
(8)
S(A, v ) =
· S(B, v ) + O
.
v
log2 v
Applying Buchstab’s identity again, we get
X
(9)
S(Ap , p)
1
12
v 25 <p≤v 25
X
=
1
v 25
X
1
S(Ap , v 25 ) −
1
v 25
12
<p≤v 25
12
<p≤v 25
X
S(Apq , q).
1
v 25
<q<p
1
2v
q< p 2
By the discussion in Lemma 9 of [8], the first sum on the right side in (9)
has an asymptotic formula.
We therefore need to deal with the sum
X
X
S(Apq , q).
12
1
1
v 25 <p≤v 25 v 25 <q<p
v
q< p
12
21
13
Removing the sum with v 25 < pq ≤ v 25 which has an asymptotic formula,
we have to consider the sums
X
X
(10)
Ω1 =
S(Apq , q),
11
1
1
v 25 <p≤v 25 v 25 <q<p
12
(11)
X
Ω2 =
6.5
v 25
12
<p≤v 25
v 25
q< p
X
S(Apq , q).
13
v 25
p <q<p
v
q< p
12
Now we define the deficiency of a sum as defined in Section 3 of [9]. If
X
X
X
X
Σ=
S(Apq , q) ≥ (1 + O(ε))
S(Bpq , q) − C
,
v
log
v
p, q
p, q
then we call the constant C the deficiency of Σ. Of course any constant
greater than C can be used as the deficiency of Σ. If a sum has an asymptotic
formula, then its deficiency is 0.
Largest prime factor
23
When we write
X
XX
XX
S(Apq , q) ≥ 0 =
S(Bpq , q) −
S(Bpq , q),
v p, q
v p, q
p, q
by Lemma 5 and the prime number theorem, we have
v log pq
XX
XX
v
S(Bpq , q) = (1 + O(ε))
w
v p, q
v p, q
log q pq log q
\ dx \ log xyv dy
= (1 + O(ε))X
w
x log x
log y y log2 y
X \ dt \
1 − t − u du
= (1 + O(ε))
.
w
log v t
u
u2
Hence, the deficiency is
\ dt \
t
1 − t − u du
w
.
u
u2
Similarly, we can define the deficiency of the sum in 2n (n ≥ 2) variables.
4. The deficiency of Ω1 . Applying Buchstab’s identity twice, we get
X
X
(12)
Ω1 =
S(Apq , q)
1
11
1
v 25 <p≤v 25 v 25 <q<p
12
v 25
q< p
X
=
1
X
11
1
S(Apq , v 25 )
1
v 25 <p≤v 25 v 25 <q<p
12
X
−
v 25
q< p
X
11
1
1
X
1
v 25 <p≤v 25 v 25 <q<p v 25 <r<q
1
12
2v
v 25
r< pq 2
q< p
X
+
1
v 25
11
<p≤v 25
X
1
v 25
1
S(Apqr , v 25 )
<q<p
12
v 25
q< p
X
1
v 25
X
S(Apqrs , s),
1
v 25
<r<q
<s<r
1
1
2v 2
2v 2
s< pqr
r< pq
where p, q, r, s denote prime numbers.
13
6.5
Note that pq < v 25 , r < q < v 25 in the second term in the above formula. By the discussion in Lemma 9 of [8] with the application of Lemma 2,
we see that the first two terms in the above formula have asymptotic for-
24
C. Jia and M. C. Liu
mulas. Therefore we have to deal with the sum
X
X
X
X
(13)
Λ=
1
v 25
11
<p≤v 25
1
v 25
<q<p
12
1
v 25
<r<q
v
r< pq
v 25
q< p
12
S(Apqrs , s).
1
v 25
<s<r
12
v
s< pqr
We call the above procedure process I. In the following, we shall discuss
13
the deficiency of Λ in some cases. Firstly qrs > v 25 is assumed.
13
I. qrs > v 25 . Now we discuss the deficiency of
X
X
X
X
Γ =
13 1
13
23
13
13
v 25 2
v 75 <p≤v 75 v 75 <q<p
v 25
<r<q
q
12
v 25
q< p
v
r< pq
12
S(Apqrs , s).
qr <s<r
1
v
s< pqr 2
Let Γ ≥ 0. Then the deficiency of Γ is
\
23
75
13
75
dt
t
\
min(t,
12
25 −t)
13
75
du
u
\
u
1 13
2 ( 25 −u)
dr
r
min(r,
\
1
2 (1−t−u−r))
×
13
25 −u−r
\
23
75
≤
13
75
dt
t
\
min(t,
min(r,
12
25 −t)
13
75
du
u
\
1
2 (1−t−u−r))
×
13
25 −u−r
1 − t − u − r − s ds
w
s
s2
\
u
1 13
2 ( 25 −u)
dr
r
1 − t − u − r − s ds
g
s
s2
≤ 0.021170.
13
12
13
Next we assume qrs < v 25 . The sum with v 25 ≤ qrs < v 25 can be
13
12
removed, so that we can assume qrs < v 25 . As in [9], we use qrs < v 25 →
12
13
12
qrs < v 25 to mean that the sum with v 25 ≤ qrs < v 25 is removed.
12
13
II. qrs < v 25 , prs > v 25 . The corresponding deficiency is
\
11
25
5
25
dt
t
\
1
min(t− 25
,
12
25 −t)
1 13
2 ( 25 −t)
du
u
\
u
1 13
2 ( 25 −t)
dr
r
min(r,
\
12
25 −u−r)
13
25 −t−r
1 − t − u − r − s ds
w
,
s
s2
Largest prime factor
25
where we used the fact that
12
r 1 12
1
min r,
−u−r ≤ +
− u − r < (1 − t − u − r).
25
2 2 25
2
We discuss several cases.
2.6
1. r < v 25 . The corresponding deficiency is
11
min(u, 2.6
25
25\−t
\ dt 12
\ 25 ) dr r\
du
1 − t − u − r − s ds
w
.
t 1 13
u 1 13
r 13
s
s2
7.8
2 ( 25 −t)
25
2 ( 25 −t)
25 −t−r
Applying Buchstab’s identity twice, we get
X
X
X
X
7.8
v 25
11
<p≤v 25
13
v 25
p
21
<q< p
X
=
7.8
25
v
11
12
13
v 25
p
7.8
25
11
<p≤v 25
13
v 25
p
X
+
v
7.8
25
12
13
12
12
v 25
p
11
13
v 25
p
12
12
r<v
X
12
v 25
pr <s<r
2.6
25
12
13
v 25
<q< p
v 25
p
r<v
X
13
12
v 25
p
2.6
25
12
r<v
X
13
<r<q
X
v 25
<q< p
1
S(Apqrs , v 25 )
13
<r<q
X
X
<p≤v 25
pr <s<r
<r<q
X
v 25
<q< p
S(Apqrs , s)
13
v 25
2.6
r<v 25
X
X
v
p
X
<p≤v 25
−
13
v 25
12
v 25
1
v 25
v 25 <t<s
pr <s<r
X
X
13
<r<q
2.6
25
1
S(Apqrst , v 25 )
1
v 25
v 25 <t<s
pr <s<r
X
1
v 25
S(Apqrstw , w),
<w<t
where t and w denote prime numbers.
In the second sum above, let m = pq, n = r, d = st. Note that st < s2 <
1
6.5
2
r < r− 2 v 25 . By the discussion in Lemma 9 of [8] with the application of
Lemma 3, we can get an asymptotic formula. We deal with the first sum in
the same way.
The corresponding deficiency of the third sum is
\
11
25
7.8
25
dt
t
\
12
25 −t
1 13
2 ( 25 −t)
du
u
\
min(u,
2.6
25 )
1 13
2 ( 25 −t)
dr
r
\
s
×
1
25
\
r
ds
s
13
25 −t−r
y
dy \
1 − t − u − r − s − y − z dz
w
.
y 1
z
z2
25
26
C. Jia and M. C. Liu
13
1) ptw > v 25 . The deficiency is
\
11
25
7.8
25
dt
t
\
12
25 −t
1 13
2 ( 25 −t)
du
u
\
min(u,
2.6
25 )
1 13
2 ( 25 −t)
\
s
×
1 13
2 ( 25 −t)
\
r
dr
r
1 13
2 ( 25 −t)
\
y
dy
y
ds
s
1−t−u−r−s−y−z
g
z
13
25 −t−y
dz
z2
≤ 0.000108.
We now discuss the remaining cases which are similar to case 1), and
write the deficiencies in brackets.
12
13
12
13
2) ptw < v 25 , psw > v 25 (0.000006). 3) psw < v 25 , prw > v 25
12
13
12
13
(0.000010). 4) prw < v 25 , prt > v 25 (0.000033). 5) prt < v 25 , pqt > v 25
12
(0.000013). 6) pqt < v 25 (0.000012).
Therefore the total deficiency in 1. is 0.000182.
2.6
26
2. v 25 < r ≤ v 175 . The corresponding deficiency is
\
9.4
25
39
175
dt
t
\
1
min(t− 25
,
12
25 −t)
max( 21 ( 13
25 −t),
2.6
25 )
du
u
\
min(u,
26
175 )
max( 12 ( 13
25 −t),
min(r,
dr
r
2.6
25 )
\
12
25 −u−r)
×
13
25 −t−r
2
1−t−u−r−s
w
s
ds
.
s2
13
1) s < r− 3 v 75 . The corresponding deficiency is
\
9.4
25
52
175
dt
t
\
12
25 −t
max( 2.6
25 ,
26
25 −3t)
du
u
\
min(u,
max( 2.6
25 ,
26
175 )
dr
r
26
25 −3t)
\
13
2
75 − 3 r
×
w
13
25 −t−r
1−t−u−r−s
s
ds
.
s2
We use the discussion in 1. Let m = pq, n = r, d = st. Note that st < s2 <
4
26
r− 3 v 75 . We have to deal with a sum whose deficiency is
\
9.4
25
52
175
dt
t
\
12
25 −t
max( 2.6
25 ,
26
25 −3t)
du
u
\
min(u,
max( 2.6
25 ,
\
s
×
1
25
26
175 )
26
25 −3t)
dy
y
\
y
1
25
dr
r
\
13
2
75 − 3 r
13
25 −t−r
ds
s
1−t−u−r−s−y−z
w
z
dz
.
z2
Largest prime factor
27
We discuss the following cases.
13
12
13
i) ptw > v 25 (0.000029). ii) ptw < v 25 , psw > v 25 (0.000006).
12
13
12
13
iii) psw < v 25 , prw > v 25 (0.000175). iv) prw < v 25 , prt > v 25 (0.000041).
12
13
12
v) prt < v 25 , pqt > v 25 (0.000015). vi) pqt < v 25 (0.000013).
Hence, the total deficiency in 1) is 0.000279.
2
13
2) s > r− 3 v 75 . The corresponding deficiency is
\
9.4
25
39
175
dt
t
\
1
,
min(t− 25
12
25 −t)
max( 21 ( 13
25 −t),
2.6
25 )
\
min(u,
du
u
26
175 )
max( 21 ( 13
25 −t),
min(r,
2.6
25 )
dr
r
\
12
25 −u−r)
×
max( 13
25 −t−r,
13
2
75 − 3 r)
1 − t − u − r − s ds
w
.
s
s2
We have to estimate the deficiency of the sum
X
X
X
(14) Σ =
39
13 1
13 1
9.4
p
v 175 <p≤v 25
v 25 2
v 25 2
<q<
p
v
2.6
25
1
2.6
v 25
<q< p
S(Apqrs , s).
13
v 25
<r<q
p
v 25
12
v 25
X
pr <s<r
26
<r<v 175
13
v 75
2
r3
12
v 25
<s< qr
We shall employ Buchstab’s identity in the following way:
1 2v 2
(15)
S(Apqrs , s) = S Apqrs ,
pqrs
X
+
S(Apqrsk , k),
1
2v
2
s≤k<max s,
pqrs
2
providing pqrs < v.
The first term on the right side of (15) counts prime numbers and the
second term counts almost-prime numbers. If pqrs3 > 2v, then the second
term is 0.
An application of the decomposition in (15) yields
X
X
X
(16) Σ =
39
9.4
13 1
13 1
p
v 175 <p≤v 25
v 25 2
v 25 2
p
2.6
<q<
v 25 <q<
1
v 25
12
v 25
<r<q
p
2.6
v 25
p
26
<r<v 175
1 2v 2
S Apqrs ,
pqrs
X
13
v 25
pr <s<r
13
v 75
2
r3
12
v 25
<s< qr
28
C. Jia and M. C. Liu
X
+
X
39
v 175 <p≤v
9.4
25
13
v 25
p
v
12
2.6
25
X
13
p
<q<
v 25
p
1
v 25
12
v 25
<q< p
2.6
v 25
12
<r<q
26
<r<v 175
X
X
13
v 25
2v
s≤k< pqrs
pr <s<r
13
v 75
2
r3
21
S(Apqrsk , k)
12
v 25
<s< qr
2v
s< pqr
13
= Σ1 + Σ2 .
We call the above procedure process II. We call Σ1 the prime term and
Σ2 the almost-prime term.
i) The deficiency of the prime term is
\
9.4
25
39
175
dt
t
\
1
min(t− 25
,
12
25 −t)
max( 21 ( 13
25 −t),
2.6
25 )
du
u
\
min(u,
26
175 )
max( 21 ( 13
25 −t),
min(r,
2.6
25 )
dr
r
\
12
25 −u−r)
×
max( 13
25 −t−r,
13
2
75 − 3 r)
ds
s(1 − t − u − r − s)
≤ 0.019316.
ii) The deficiency of the almost-prime term is
\
9.4
25
39
175
dt
t
\
1
min(t− 25
,
12
25 −t)
max( 21 ( 13
25 −t),
2.6
25 )
du
u
\
min(u,
26
175 )
13
max( 12 ( 25
−t),
2.6
25 )
\
1
2 (1−t−u−r−s)
×
s
min(r,
dr
r
max( 13
25 −t−r,
13
\
39
175
dt
t
\
1
min(t− 25
,
12
25 −t)
max( 21 ( 13
25 −t),
2.6
25 )
du
u
min(u,
\
26
5.5
t
175 , 25 + 2 −u)
13
max( 12 ( 25
−t),
13
2
75 − 3 r)
1−t−u−r−s−k
w
k
a) kps > v 25 . The corresponding deficiency is
9.4
25
\
12
1
25 −u−r, 3 (1−t−u−r))
2.6
25 )
dr
r
ds
s
dk
.
k2
Largest prime factor
\
1
3 (1−t−u−r))
min(r,
×
max( 13
25 −t−r,
13
2
1
75 − 3 r, 25 −t+u+r)
\
1
2 (1−t−u−r−s)
×
max(s,
29
w
13
25 −t−s)
ds
s
1−t−u−r−s−k
k
dk
,
k2
12
where we used the fact that 13 (1 − t − u − r) ≤ 25
− u − r.
13
12
α) kqrs > v 25 (0.003549). β) kqrs < v 25 (0.001012).
12
b) kps < v 25 (0.000156).
Thus the total deficiency in 2) is 0.024033 and that in 2. is 0.024312.
26
3. r > v 175 . The corresponding deficiency is
\
58
175
5
25
dt
t
\
1
min(t− 25
,
12
25 −t)
max( 12 ( 13
25 −t),
26
175 )
du
u
\
u
max( 12 ( 13
25 −t),
26
175 )
dr
r
\
12
25 −u−r)
min(r,
×
w
13
25 −t−r
1 − t − u − r − s ds
.
s
s2
1) The deficiency of the prime term is 0.017964.
2) The deficiency of the almost-prime term is
\
58
175
5
25
dt
t
\
1
min(t− 25
,
12
25 −t)
max( 12 ( 13
25 −t),
26
175 )
du
u
\
u
max( 12 ( 13
25 −t),
min(r,
26
175 )
dr
r
\
12
1
25 −u−r, 3 (1−t−u−r))
×
13
25 −t−r
\
1
2 (1−t−u−r−s)
×
s
ds
s
1−t−u−r−s−k
w
k
dk
.
k2
12
13
25
(0.000960). b) kps < v 25 . There are two subcases:
a) kps > v
13
12
25
α) kqrs > v (0.003065); β) kqrs < v 25 (0.000757).
Therefore the total deficiency in 3. is 0.022746 and that in II is 0.047240.
12
13
III. prs < v 25 , pqs > v 25 . The corresponding deficiency is
\
10
25
14
75
dt
t
\
min(t,
12
25 −t)
1 14
2 ( 25 −t)
du
u
\
1
u− 25
13
25 −t−u
dr
r
min(r,
\
12
25 −t−r)
13
25 −t−u
1 − t − u − r − s ds
w
.
s
s2
30
C. Jia and M. C. Liu
26
1. q < v 175 . Applying process I again and writing m = prs, n = q and
d = t, we have to deal with a sum whose deficiency is
\
10
25
46
175
dt
t
\
min( 12
25 −t,
26
175 )
1 14
2 ( 25 −t)
du
u
\
1
u− 25
13
25 −t−u
\
s
×
1
25
dr
r
min(r,
\
y
dy
y
1
25
\
12
25 −t−r)
13
25 −t−u
ds
s
1−t−u−r−s−y−z
w
z
12
13
dz
.
z2
13
1) pqw > v 25 (0.000334). 2) pqw < v 25 , pqt > v 25 (0.000006). 3) pqt <
12
25
v (0.000013).
Hence, the total deficiency in 1. is 0.000353.
5.2
26
2. v 175 < q < v 25 .
3
6.5
1) s < q − 4 v 25 . Applying process I again and writing m = prs, n = q
and d = t, we have to deal with a sum whose deficiency is
\
58
175
5.2
25
dt
t
\
min( 12
25 −t,
max( 12 ( 14
25 −t),
min(r,
×
5.2
25 )
26
26
175 , 25 −4t)
du
u
\
×
13
25 −t−u
12
6.5
3
25 −t−r, 25 − 4 u)
13
25 −t−u
min(y,
\
1
u− 25
\
ds
s
1
2 (1−t−u−r−s−y))
1
25
\
s
1
25
dr
r
dy
y
1−t−u−r−s−y−z
w
z
dz
,
z2
6.5
3
where we used the fact that s < 12
25 −t−r and s < 25 − 4 u ⇒ t+u+r +3s =
1
(t + r + s) + (u + 2s) < 1 ⇒ s < 2 (1 − t − u − r − s).
13
13
12
13
i) pqw > v 25 : a) pstw > v 25 (0.000324); b) pstw < v 25 , prtw > v 25
12
13
12
(0.000108); c) prtw < v 25 , prsw > v 25 (0.000010); d) prsw < v 25 , prst >
12
12
13
13
v 25 (0.000006); e) prst < v 25 (0.000363). ii) pqw < v 25 , pqt > v 25 (0.000300).
12
13
12
iii) pqt < v 25 , pqtw > v 25 (0.000322). iv) pqtw < v 25 (0.000006).
Hence, the total deficiency in 1) is 0.001439.
3
6.5
2) s > q − 4 v 25 (0.002038).
Thus the total deficiency in 2. is 0.003477.
5.2
6.5
3. v 25 < q (< v 25 ).
13
1) s < q −2 v 25 . Applying process I again and writing m = prs, n = q
and d = t, we have to deal with a sum whose deficiency is
Largest prime factor
\
6.8
25
5.2
25
dt
t
\
min(t,
12
25 −t)
5.2
25
du
u
min(r,
\
1
u− 25
13
25 −t−u
dr
r
\
12
13
25 −t−r, 25 −2u)
×
13
25 −t−u
min(y,
\
s
ds
s
\
1
2 (1−t−u−r−s−y))
×
31
1
25
1
25
dy
y
1−t−u−r−s−y−z
w
z
13
12
dz
.
z2
13
i) pqw > v 25 (0.001662). ii) pqw < v 25 , pqt > v 25 (0.000020).
12
13
12
iii) pqt < v 25 , pqtw > v 25 (0.000019). iv) pqtw < v 25 (0).
Therefore the total deficiency in 1) is 0.001701.
13
2) s > q −2 v 25 .
i) The deficiency of the prime term is 0.009678.
ii) The deficiency of the almost-prime term is
\
6.8
25
5.2
25
dt
t
\
min(t,
12
25 −t)
5.2
25
du
u
\
1
2u−t− 25
13
25 −2u
dr
r
min(r,
\
12
25 −t−r)
13
25 −2u
\
1
2 (1−t−u−r−s)
×
s
ds
s
1−t−u−r−s−k
w
k
dk
,
k2
r
2 12
where we used the fact that min r, 12
25 − t − r ≤ 3 + 3 25 − t − r <
1
3 (1 − t − u − r).
13
12
13
a) kps > v 25 (0.000006). b) kps < v 25 , kpr > v 25 (0.000653). c) kpr <
12
13
12
v 25 : α) kqrs > v 25 (0.002882); β) kqrs < v 25 (0.001823).
Hence, the total deficiency in 2) is 0.015042 and that in 3. is 0.016743.
Therefore the total deficiency in III is 0.020573.
12
13
IV. pqs < v 25 , pqr > v 25 . The corresponding deficiency is
\
9
25
13
75
dt
t
\
min(t,
11
25 −t)
1 13
2 ( 25 −t)
du
u
\
u
13
25 −t−u
dr
r
\
12
25 −t−u
×
1
25
26
w
1−t−u−r−s
s
ds
.
s2
1. r < v 175 . Applying process I again and writing m = pqs, n = r and
d = t, we have to deal with a sum whose deficiency is
32
C. Jia and M. C. Liu
\
9
25
dt
t
32.5
175
\
min(t,
11
25 −t)
max( 21 ( 13
25 −t),
\
s
×
65
175 −t)
dy
y
1
25
du
u
\
min(u,
26
175 )
13
25 −t−u
\
12
25 −t−u
dr
r
1
25
\
1
2 (1−t−u−r−s−y))
min(y,
1
25
ds
s
1−t−u−r−s−y−z
w
z
12
13
dz
.
z2
13
1) prtw > v 25 (0.000164). 2) prtw < v 25 , prsw > v 25 (0.000008).
12
13
12
13
3) prsw < v 25 , prst > v 25 (0.000009). 4) prst < v 25 , prstw > v 25
12
(0.000519). 5) prstw < v 25 (0.000269).
Hence, the total deficiency in 1. is 0.000969.
26
5.2
2. v 175 < r < v 25 .
3
6.5
1) s < r− 4 v 25 . Applying process I again and writing m = pqs, n = r
and d = t, we have to deal with a sum whose deficiency is
\
51
175
13
75
dt
t
\
min(t,
11
25 −t)
max( 12 ( 13
25 −t),
26
175 )
du
u
\
min( 12
25 −t−u,
×
\
min(u,
max( 13
25 −t−u,
6.5
3
25 − 4 r)
1
25
min(y,
×
5.2
25 )
ds
s
\
1
2 (1−t−u−r−s−y))
1
25
\
s
1
25
26
175 )
dr
r
dy
y
1−t−u−r−s−y−z
w
z
13
12
dz
.
z2
13
i) prtw > v 25 (0.000387). ii) prtw < v 25 , prsw > v 25 (0.000052).
13
12
13
12
iii) prsw < v 25 , prst > v 25 (0.000009). iv) prst < v 25 , prstw > v 25
12
(0.000293). v) prstw < v 25 (0.000015).
Hence, the total deficiency in 1) is 0.000756.
3
6.5
2) s > r− 4 v 25 (0.000006).
Thus the total deficiency in 2. is 0.000762.
5.2
3. r > v 25 (0.000108).
Therefore the total deficiency in IV is 0.001839.
12
13
V. pqr < v 25 , pqrs > v 25 . The corresponding deficiency is
\
10
25
13
100
dt
t
\
min(t,
11
25 −t)
1 13
3 ( 25 −t)
du
u
min(u,
\
12
25 −t−u)
1 13
2 ( 25 −t−u)
dr
r
\
r
13
25 −t−u−r
1 − t − u − r − s ds
w
.
s
s2
Largest prime factor
33
26
1. s < v 175 . Applying process I again and writing m = pqr, n = s and
d = t, we have to deal with a sum whose deficiency is
\
10
25
13
100
dt
t
\
min(t,
1 13
3 ( 25 −t)
\
min(r,
×
11
25 −t)
26
175 )
\
s
ds
s
13
25 −t−u−r
1
25
du
u
dy
y
min(u,
\
12
25 −t−u)
1 13
2 ( 25 −t−u)
min(y,
dr
r
\
1
2 (1−t−u−r−s−y))
1
25
1 − t − u − r − s − y − z dz
w
.
z
z2
13
12
13
1) prtw > v 25 (0.000700). 2) prtw < v 25 , pqtw > v 25 (0.000849).
12
13
12
13
3) pqtw < v 25 , pqsw > v 25 (0.000515). 4) pqsw < v 25 , pqst > v 25 :
12
12
13
i) pqrw > v 25 (0.000017); ii) pqrw < v 25 (0.000374). 5) pqst < v 25 , pqrt >
13
12
13
13
v 25 : i) qrstw > v 25 (0.000065); ii) qrstw < v 25 , prstw > v 25 (0.000148);
12
12
13
iii) prstw < v 25 (0.000910). 6) pqrt < v 25 : i) pqstw > v 25 (0.000193);
12
13
12
ii) pqstw < v 25 , pqrtw > v 25 (0.000008); iii) pqrtw < v 25 (0.000006).
Thus the total deficiency in 1. is 0.003785.
26
2. s > v 175 (0.000027).
Therefore the total deficiency in V is 0.003812.
12
VI. pqrs < v 25 . The corresponding deficiency is
\
9
25
1
25
dt
t
\
min(t,
10
25 −t)
1
25
du
u
min(u,
\
11
25 −t−u)
1
25
min(r,
×
dr
r
\
12
25 −t−u−r)
1
25
1 − t − u − r − s ds
w
.
s
s2
Applying process I again, we have to deal with a sum whose deficiency is
\
9
25
1
25
dt
t
\
min(t,
10
25 −t)
1
25
du
u
min(r,
×
min(u,
\
11
25 −t−u)
1
25
\
12
25 −t−u−r)
1
25
ds
s
dr
r
\
s
1
25
dy
y
\
y
1
25
1−t−u−r−s−y−z
w
z
dz
,
z2
r 1 12
12
where we used
the
fact
that
s
≤
min
r,
−t−u−r
≤
+
−t−u−r
=
25
2
2
25
1
1 12
1
2 25 − t − u < 4 (1 − t − u − r) ⇒ y < 2 (1 − t − u − r − s − y).
13
12
13
1. prstw > v 25 (0.000453). 2. prstw < v 25 , pqstw > v 25 (0.000324).
13
12
13
12
3. pqstw < v 25 , pqrtw > v 25 (0.000303). 4. pqrtw < v 25 , pqrsw > v 25
13
12
(0.000153). 5. pqrsw < v 25 , pqrst > v 25 (0.000046).
34
C. Jia and M. C. Liu
12
13
6. pqrst < v 25 , pqrstw > v 25 . Applying process I once more, we have
the deficiency
\
8
25
13
150
dt
t
\
min(t,
9
25 −t)
1 13
5 ( 25 −t)
du
u
min(r,
\
×
1
25
1 13
4 ( 25 −t−u)
\
11
25 −t−u−r)
×
α
\
10
25 −t−u)
min(u,
1 13
3 ( 25 −t−u−r)
ds
s
dr
r
min(s,
\
12
25 −t−u−r−s)
1 13
2 ( 25 −t−u−r−s)
1−t−u−r−s−y−z−α−β
w
β
1
≤
1.763
\
8
25
dt
t
13
150
min(r,
\
min(t,
9
25 −t)
1 13
5 ( 25 −t)
\
11
25 −t−u−r)
×
1 13
3 ( 25 −t−u−r)
\
y
×
13
25 −t−u−r−s−y
ds
s
du
u
min(u,
\
10
25 −t−u)
1 13
4 ( 25 −t−u)
min(s,
\
12
25 −t−u−r−s)
1 13
2 ( 25 −t−u−r−s)
\
y
dy
y
13
25 −t−u−r−s−y
dz
z
\
z
1
25
dα
α
dβ
β2
dr
r
dy
y
1 dz
25 log(25z) − 25 +
z z
≤ 0.000359,
where we used the fact that β ≤ 13 (1−t−u−r −s−y −z −α) and Lemma 4.
12
7. pqrstw < v 25 . Applying process I twice, we have the deficiency
\
7
25
0.5617
1
25
dt
t
\
min(t,
8
25 −t)
1
25
du
u
min(r,
min(u,
1
25
\
10
25 −t−u−r)
×
1
25
min(y,
\
9
25 −t−u)
\
ds
s
dr
r
min(s,
12
25 −t−u−r−s−y)
×
1
25
\
11
25 −t−u−r−s)
1
25
dy
y
1 dz
25 log(25z) − 25 +
z z
≤ 0.000495,
where we used the fact that β ≤ 15 (1−t−u−r −s−y −z −α) and Lemma 4.
Therefore the total deficiency in VI is 0.002133. Summing up the above
estimates, we conclude that the total deficiency of Ω1 is 0.096767.
Largest prime factor
35
5. The deficiency of Ω2 . Now we consider the sum
X
X
Ω2 =
S(Apq , q).
v
6.5
25
12
<p≤v 25
13
v 25
p <q<p
v
q< p
12
The corresponding deficiency is
\
12
25
6.5
25
min(t,
dt
t
\ (1−t))
1
2
13
25 −t
1 − t − u du
w
.
u
u2
We discuss several cases.
2.6
I. q < v 25 . The corresponding deficiency is
\
12
25
10.4
25
\
2.6
25
dt
t
13
25 −t
1 − t − u du
w
.
u
u2
Let m = p, n = q and d = r. Applying process I with Lemma 3, we have to
deal with a sum whose deficiency is
\
12
25
10.4
25
dt
t
\
2.6
25
13
25 −t
du
u
\
u
dr
r
1
25
\
r
1
25
1−t−u−r−s
w
s
ds
.
s2
13
1. ps > v 25 .
1
6.5
1
1) s < q − 4 r− 2 v 50 . Applying process I again and writing m = p, n = q
and d = rst, we have to deal with a sum whose deficiency is
\
12
25
78
175
dt
t
\
65
min( 2.6
25 , 6t− 25 )
13
25 −t
\
s
×
1
25
13
dy
y
du
u
\
19.5
min(u, 2t− u
2 − 25 )
min(y,
13
25 −t
\
dr
r
1
2 (1−t−u−r−s−y))
1
25
12
min(r,
\
6.5
u
r
50 − 4 − 2 )
13
25 −t
ds
s
1−t−u−r−s−y−z
w
z
dz
.
z2
i) pw > v 25 (0.000508). ii) pw < v 25 (0).
1
1
6.5
2) s > q − 4 r− 2 v 50 (0.007963).
Thus the total deficiency in 1. is 0.008471.
12
13
12
13
2. ps < v 25 , pr > v 25 (0.000996). 3. pr < v 25 , prs > v 25 (0.000831).
12
4. prs < v 25 (0).
Therefore the total deficiency in I is 0.010298.
36
C. Jia and M. C. Liu
2.6
26
II. v 25 < q < v 175 . The corresponding deficiency is
\
12
25
65
175
dt
t
\
1 − t − u du
w
.
u
u2
26
175
max( 13
25 −t,
2.6
25 )
Let m = p, n = q and d = r. Applying process I with Lemma 3, we have to
deal with a sum whose deficiency is
\
12
25
dt
t
65
175
\
26
175
max( 13
25 −t,
2.6
25 )
\
u
du
u
1
25
dr
r
\
1
2 (1−t−u−r))
min(r,
1
25
1 − t − u − r − s ds
w
.
s
s2
13
1. ps > v 25 .
2
1
13
1) s < q − 3 r− 2 v 75 . Let m = p, n = q and d = rst. Applying process I
with Lemma 3, we have to deal with a sum the deficiency of which is
\
12
25
33.8
75
dt
t
\
26
min( 175
,
9
91
4 t− 100 )
2.6
25
min(r,
\
min(u, 2t− 43 u− 52
75 )
du
u
13
25 −t
\
13
2
r
75 − 3 u− 2 )
×
13
25 −t
min(y,
ds
s
\
s
dy
y
1
25
\
1
2 (1−t−u−r−s−y))
×
dr
r
g
1
25
1−t−u−r−s−y−z
z
dz
z2
≤ 0.000486.
2
1
13
2) s > q − 3 r− 2 v 75 (0.071032).
Hence, the total deficiency in 1. is 0.071518.
12
13
2. ps < v 25 , pr > v 25 .
26
4
1) s < q − 3 r−1 v 75 . Let m = ps, n = q and d = rt. Applying process I
with Lemma 3, we have to deal with a sum whose deficiency is
\
11
25
68
175
dt
t
\
26
min( 175
,
3
4
4 t− 25 )
max( 13
25 −t,
2.6
25 )
du
u
min(u,
\
23
4
75 − 3 u)
13
25 −t
min( 12
25 −t,
×
\
dr
r
26
4
75 − 3 u−r)
1
25
ds
s
min(s,
\
1
2 (1−t−u−r−s))
1
25
dy
y
Largest prime factor
min(y,
\
1
2 (1−t−u−r−s−y))
×
1
25
37
1−t−u−r−s−y−z
g
z
dz
z2
≤ 0.000290.
4
26
2) s > q − 3 r−1 v 75 .
i) The deficiency of the prime term is 0.005291.
ii) The deficiency of the almost-prime term is
\
11
25
65
175
dt
t
\
26
175
max( 13
25 −t,
3
10
7 (t− 75 ))
du
u
\
u
dr
r
4
10
max( 13
25 −t, t− 3 u− 75 )
\
12
25 −t
×
1
max( 25
,
26
4
75 − 3 u−r)
\
1
2 (1−t−u−r−s)
×
s
13
25
ds
s
1−t−u−r−s−k
w
k
12
dk
.
k2
13
12
a) kp > v (0.001783). b) kp < v 25 , kps > v 25 (0.001370). c) kps < v 25
(0.000068).
Thus the total deficiency in 2. is 0.008802.
12
13
3. pr < v 25 , prs > v 25 .
13
2
1) s < q − 3 v 75 . Let m = pr, n = q and d = st. Applying process I with
Lemma 3, we have to deal with a sum whose deficiency is
\
11
25
65
175
dt
t
\
26
175
max( 13
25 −t,
2.6
25 )
du
u
\
12
25 −t
1 13
2 ( 25 −t)
\
s
×
1
25
dr
r
dy
y
min(r,
\
13
2
75 − 3 u)
13
25 −t−r
\
ds
s
y
1
25
1−t−u−r−s−y−z
w
z
13
12
13
dz
.
z2
i) ptw > v 25 (0.000088). ii) ptw < v 25 , psw > v 25 (0.000006).
12
13
12
iii) psw < v 25 , prw > v 25 (0.000011). iv) prw < v 25 (0).
2
13
2) s > q − 3 v 75 (0.000571).
Therefore the total deficiency in 3. is 0.000676 and that in II is 0.080996.
26
5.2
III. v 175 < q < v 25 . Now we consider the deficiency of the sum
X
X
S(Apq , q).
Σ=
v
7.8
25
12
13
5.2
<p≤v 25 v 25
25
p <q<v
26
v 175 <q
38
C. Jia and M. C. Liu
We write
(17)
where
Σ = Ψ1 + Ψ2 ,
X
Ψ1 =
7.8
v 25
X
11.5
<p≤v 25
26
X
X
8.7
v 25
11.5
<p≤v 25
5.2
25
p <q<v
v 175 <q<
Ψ2 =
S(Apq , q),
13
v 25
9.55
v 25
1
p2
v
9.55
25
1
p2
X
S(Apq , q) +
<q<v
11.5
v 25
5.2
25
X
12
<p≤v 25
26
v 175
S(Apq , q).
5.2
<q<v 25
The deficiency of Ψ2 is
\
11.5
25
(18)
8.7
25
dt
t
\
12
\ dt
1 − t − u du 25
g
+
u
u2 11.5 t
5.2
25
t
9.55
25 − 2
25
\
5.2
25
g
26
175
1 − t − u du
≤ 0.157504.
u
u2
Next we shall discuss the deficiency of Ψ1 . Applying Buchstab’s identity
twice, we have
X
X
1
(19) Ψ1 =
S(Apq , v 25 )
v
7.8
25
<p≤v
11.5
25
13
5.2
v 25
25
p <q<v
26
v 175 <q<
v
7.8
25
<p≤v
11.5
25
9.55
25
1
p2
X
X
5.2
v 25
25
p <q<v
v 25 <r<q
X
−
v
13
26
v 175 <q<
X
X
+
7.8
v 25
v
11.5
<p≤v 25
13
v 25
1
9.55
25
1
p2
p <q<v
1
S(Apqr , v 25 )
X
1
v 25
5.2
25
<r<q
9.55
26
v 25
v 175 <q<
1
p2
X
S(Apqrs , s).
1
v 25
<s<r
1
2v
s< pqr 2
The first sum has an asymptotic formula. For the second sum
X
X
X
1
S(Apqr , v 25 ),
v
7.8
25
<p≤v
11.5
25
13
5.2
v 25
25
p <q<v
26
v 175 <q<
v
9.55
25
1
p2
1
v 25 <r<q
Largest prime factor
39
13
3
6.5
if pr ≤ v 25 , then we have an asymptotic formula. If r ≤ q − 4 v 25 , then
writing m = p, n = q and d = r, we also have an asymptotic formula.
3
6.5
13
Hence, we assume that pr > v 25 , r > q − 4 v 25 and have to deal with the
sum
X
X
X
1
S(Apqr , v 25 ).
Λ=
v
7.8
25
<p≤v
11.5
25
13
13
5.2
v 25
25
p <q<v
26
v 175 <q<
v
v 25
p <r<q
9.55
25
1
p2
6.5
25
3
q4
v
<r
Now we use the discussion in [6]. Applying Buchstab’s identity to the
p-summation in Λ, we have
X
X
X
1
Λ=
S(Anqr , v 25 )
v
7.8
25
<n≤v
1
(n, P (v 25
11.5
25
13
5.2
v 25
25
n <q<v
))=1
9.55
26
v 25
v 175 <q<
1
n2
X
−
v
7.8
25
X
<zν≤v
11.5
25
1
v 25
≤z<ν
(ν, P (z))=1
13
v 25
n <r<q
6.5
v 25
3 <r
q4
X
1
S(Azνqr , v 25 ),
13
13
v 25
zν <r<q
6.5
v 25
3 <r
q4
5.2
v 25
25
zν <q<v
9.55
26
v 25
v 175 <q<
1
(zν) 2
where n, ν denote integers and z denotes a prime number.
We write
X
X
X
1
S(Anqr , v 25 )
v
7.8
25
<n≤v
1
(n, P (v 25
11.5
25
13
13
v 25
n <r<q
6.5
v 25
3 <r
q4
5.2
v 25
25
n <q<v
))=1
9.55
26
25
v
v 175 <q<
1
n2
7.8
= #{nqrl ∈ A : v 25 < n ≤ v
26
v 175 < q < v
9.55
25
1
11.5
25
1
13
5.2
, (n, P (v 25 )) = 1, v 25 /n < q < v 25 ,
13
6.5
3
1
/n 2 , v 25 /n < r < q, v 25 /q 4 < r, (l, P (v 25 )) = 1}.
13
12
6.5
Note that v 25 < nr ⇒ ql v 25 and that r < q < v 25 . By the discussion
in Lemma 9 of [8] with the application of Lemma 2, we can get an asymptotic
formula. The idea of transferring the application of the sieve method from
the l-summation to the n-summation comes from [6].
Now we shall deal with the deficiencies of the following two sums:
X
X
X
1
(20) Γ1 =
S(Azνqr , v 25 ),
v
7.8
25
<zν≤v
1
v 25
11.5
25
≤z<ν
(ν, P (z))=1
13
5.2
v 25
25
zν <q<v
9.55
26
v 25
v 175 <q<
1
(zν) 2
13
v 25
zν <r<q
6.5
v 25
3 <r
q4
40
C. Jia and M. C. Liu
X
(21) Γ2 =
7.8
v 25
X
11.5
<p≤v 25
13
v 25
X
p <q<v
1
v 25
5.2
25
X
S(Apqrs , s).
1
v 25
<r<q
<s<r
1
v
s< pqr 2
9.55
26
v 25
v 175 <q<
1
p2
By Lemma 5, the deficiency of Γ1 is
\
11.5
25
\
min( 5.2
25 ,
dt
7.8
25
9.55
t
25 − 2 )
26
175 )
max( 13
25 −t,
\
t
2
du
u
t−z
w
z
1
25
dz
z2
\
u
×
25w(25(1 − t − u − r))
max( 13
25 −t,
6.5
3
25 − 4 u)
dr
r
and that of Γ2 is
\
11.5
25
7.8
25
dt
t
\
min( 5.2
25 ,
9.55
t
25 − 2 )
max( 13
25 −t,
26
175 )
du
u
\
u
dr
r
1
25
min(r,
\
1
2 (1−t−u−r))
×
1
25
1−t−u−r−s
w
s
ds
.
s2
We refer to [6] and [10]. Firstly we consider the deficiency of Γ1 in some
cases.
13
1. rν > v 25 . The corresponding deficiency is
\
11.5
25
8.9
25
\
9.55
t
25 − 2
dt
max( 14
25 −t,
26
175 )
du
u
\
t+u− 13
25
1
25
t−z
w
z
dz
z2
\
u
×
max( 13
25 −t+z,
25w(25(1 − t − u − r))
6.5
3
25 − 4 u)
We use Buchstab’s identity
X
1
S(Azνqr , v 25 ) = S(Azνqr , r) +
S(Azνqrs , s)
1
v 25
≤s<r
1
2v
s< zνqr 2
to produce two sums
Λ1 =
XXXX
z
ν
q
r
S(Azνqr , r),
dr
.
r
Largest prime factor
Λ2 =
XXXX
z
ν
q
X
r
41
S(Azνqrs , s).
1
v 25
≤s<r
1
2v
s< zνqr 2
1) The deficiency of Λ1 is
\
\
11.5
25
t
9.55
25 − 2
dt
8.9
25
max( 14
25 −t,
26
175 )
du
u
\
t+u− 13
25
1
25
t−z
w
z
dz
z2
\
u
×
max( 13
25 −t+z,
w
6.5
3
25 − 4 u)
1−t−u−r
r
dr
.
r2
Applying process II to the ν-summation, we have to consider the prime
term and the almost-prime term. For convenience, we call them the z-prime
term and the z-almost-prime term respectively.
i) The deficiency of the z-prime term is
\
\
11.5
25
9.55
t
25 − 2
dt
8.9
25
max( 14
25 −t,
26
175 )
du
u
\
t+u− 13
25
1
25
dz
z(t − z)
\
u
×
13
−t+z,
max( 25
6.5
3
25 − 4 u)
1−t−u−r
g
r
dr
≤ 0.002064.
r2
ii) The deficiency of the z-almost-prime term. Now ν is replaced by eβ,
where z < e < β, (β, P (e)) = 1, e is a prime number and β is an integer.
The corresponding deficiency is
\
\
11.5
25
8.9
25
9.55
t
25 − 2
dt
max( 14
25 −t,
\
1
2 (t−z)
×
z
26
175 )
du
u
t−z−e
w
e
13
\
t+u− 13
25
1
25
de
e2
dz
z
\
u
max( 13
25 −t+z,
6.5
3
25 − 4 u)
12
1−t−u−r
w
r
13
dr
.
r2
a) qre > v 25 (0.000029). b) qre < v 25 , qrze > v 25 (0.000229). c)
12
13
12
13
qrze < v 25 : α) rβ > v 25 (0.000006); β) rβ < v 25 , rzβ > v 25 (0.000644);
12
γ) rzβ < v 25 (0.000463).
Therefore the total deficiency of Λ1 is 0.003435.
42
C. Jia and M. C. Liu
2) The deficiency of Λ2 is
\
\
11.5
25
9.55
t
25 − 2
dt
8.9
25
14
max( 25
−t,
26
175 )
du
u
\
t+u− 13
25
1
25
\
u
×
max( 13
25 −t+z,
\
8.9
25
max( 25 −t,
175 )
max( 13
25 −t+z,
8.9
25
\
1
2 (1−t−u−r))
1
25
1−t−u−r−s
w
s
ds
.
s2
13
25
\
×
\
dz
z2
i) sν > v 25 (0.000102). ii) sν < v 25 , szν > v 25 (0.001319).
12
iii) szν < v 25 . The corresponding deficiency is
t
9.55
t+u− 13
25 \− 2
\ 25 t − z dz
du
dt
w
u
z
z2
14
26
1
u
11
25
min(r,
12
13
11
25
t−z
w
z
dr
r
6.5
3
25 − 4 u)
6.5
3
25 − 4 u)
dr
r
\
min( 12 (1−t−u−r),
12
25 −t)
1
25
1−t−u−r−s
w
s
ds
.
s2
a) The deficiency of the z-prime term is 0.001807.
b) The deficiency of the z-almost-prime term is
9.55
t
t+u− 13
25 \− 2
\
\ 25 dz 12 (t−z)
t − z − e de
du
w
dt
u
z
e
e2
14
26
1
z
max( 25 −t,
175 )
25
\
u
×
max( 13
25 −t+z,
6.5
3
25 − 4 u)
dr
r
\
min( 12 (1−t−u−r),
12
25 −t)
1
25
13
12
1 − t − u − r − s ds
w
.
s
s2
13
α) rβ > v 25 (0.000006). β) rβ < v 25 , rzβ > v 25 (0.000471). γ) rzβ <
13
12
13
v , rszβ > v 25 (0.000396). δ) rszβ < v 25 , qszβ > v 25 (0.000006).
12
λ) qszβ < v 25 (0.000611).
Therefore the total deficiency of Λ2 is 0.004718 and that in 1. is 0.008153.
13
12
2. rν < v 25 , qν > v 25 . The corresponding deficiency is
73.7
9.55
t
t+u− 13
175
25 \− 2
\
\ 25
du
t − z dz
dt
w
u
z
z2
30
3
5.5
8.9
14
1
12
25
25
max( 25 −t,
175 )
max( 25 , t− 4 u−
25
)
\
12
25 −t+z
×
max( 13
25 −t,
3
6.5
25w(25(1 − t − u − r))
6.5
3
25 − 4 u)
dr
.
r
Note that z < q − 4 v 25 . Let m = rν, n = q and d = z. Then we have an
asymptotic formula. Hence, the deficiency is 0.
Largest prime factor
43
12
3. qν < v 25 . The corresponding deficiency is
11.5
9.55
t
25
25 ,\ 25 − 2 )
2
\ min( 5.2
\t
du
t − z dz
dt
w
u
z
z2
7.8
13
26
12
max( 25 −t,
25
175 )
t+u− 25
\
u
×
25w(25(1 − t − u − r))
max( 13
25 −t,
6.5
3
25 − 4 u)
dr
.
r
13
1) qrz > v 25 .
i) The deficiency of Λ1 is
\
11.5
25
7.8
25
\
min( 5.2
25 ,
dt
13
max( 25
−t,
9.55
t
25 − 2 )
26
6.5
t
175 , 25 − 4 )
\
t
2
du
u
13
25 −2u
t−z
g
z
dz
z2
\
u
×
max( 13
25 −t,
6.5
3
13
25 − 4 u, 25 −u−z)
1−t−u−r
g
r
dr
r2
≤ 0.004218.
ii) The deficiency of Λ2 is
\
11.5
25
\
min( 5.2
25 ,
dt
7.8
25
max( 13
25 −t,
9.55
t
25 − 2 )
26
6.5
t
175 , 25 − 4 )
du
u
\
u
×
13
max( 25
−t,
6.5
3
13
25 − 4 u, 25 −u−z)
\
t
2
w
13
25 −2u
dr
r
min(r,
t−z
z
\
dz
z2
1
2 (1−t−u−r))
1
25
1−t−u−r−s
w
s
ds
.
s2
13
a) szν > v 25 (0.000236).
12
b) szν < v 25 .
13
12
13
α) qrs > v 25 (0.000146). β) qrs < v 25 : β1 ) qsν > v 25 (0.000006);
12
β2 ) qsν < v 25 (0.007355).
Therefore the total deficiency in 1) is 0.011961.
12
6.5
2) qrz < v 25 . If ν ≤ v 25 , writing m = qrz and n = ν, we have an asymp6.5
totic formula. Hence, we assume ν > v 25 . The corresponding deficiency is
11.5
9.55
t
5.5
u
min(t− 6.5
25
25 ,\ 25 − 2 )
25\ , 25 − 4 )
\ min( 5.2
du
t − z dz
dt
w
u
z
z2
7.8
13
26
12
25
max( 25 −t,
175 )
t+u− 25
min(u,
×
\
12
25 −u−z)
max( 13
25 −t,
25w(25(1 − t − u − r))
6.5
3
25 − 4 u)
dr
.
r
44
C. Jia and M. C. Liu
26
i) r ≤ v 175 . The corresponding deficiency is
9.55
t
5.5
u
11.5
min(t− 6.5
25
25 ,\ 25 − 2 )
25\ , 25 − 4 )
\ min( 5.2
du
t − z dz
dt
w
u
z
z2
65
26
12
175
t+u− 25
175
\
min( 12
25 −u−z,
×
25w(25(1 − t − u − r))
13
max( 25
−t,
2.6
4
26
175 )
6.5
3
25 − 4 u)
26
3
dr
.
r
6.5
Note that v 25 < r. If z < r− 3 v 75 (r < z − 4 v 25 ), writing m = qν, n = r
3
6.5
and d = z, we have an asymptotic formula. Hence, we assume r ≥ z − 4 v 25 .
The deficiency is
11.5
t
5.5
u
min(t− 6.5
25
25 \− 2
25\ , 25 − 4 ) \ 9.55
du
t − z dz
dt
g
u
z
z2
26
71.5
26
175
175
175
\
min( 12
25 −u−z,
×
13
max( 25
−t,
26
175 )
25g(25(1 − t − u − r))
6.5
3
6.5
3
25 − 4 u, 25 − 4 z)
dr
r
≤ 0.000754.
ii) v
26
175
5.2
< r (< v 25 ). The corresponding deficiency is
11.5
9.55
t
58
min(t− 6.5
−u) 25
25 ,\ 25 − 2 )
25
\ min( 5.2
\ , 175
du
t − z dz
dt
w
u
z
z2
7.8
13
26
12
max( 25 −t,
25
175 )
t+u− 25
min(u,
×
\
12
25 −u−z)
max( 13
25 −t,
3
6.5
4
25w(25(1 − t − u − r))
26
175 )
dr
.
r
26
If z < r− 4 v 25 (r < z − 3 v 75 ), writing m = qν, n = r and d = z, we have an
26
4
asymptotic formula. Hence, we assume r ≥ z − 3 v 75 . The deficiency is
11.5
9.55
t
58
min(t− 6.5
−u) 25
25 ,\ 25 − 2 )
25
\ min( 5.2
\ , 175
du
t − z dz
dt
g
u
z
z2
26
7.8
13
12
25
max( 25 −t,
175 )
t+u− 25
\
26
26
4
12
max(max( 13
25 −t, 175 , 75 − 3 z), min(u, 25 −u−z))
×
max( 13
25 −t,
26
26
4
175 , 75 − 3 z)
25g(25(1 − t − u − r))
dr
r
≤ 0.001733.
Thus the deficiency in 2) is 0.002487 and that in 3. is 0.014448. Therefore
the deficiency of Γ1 is 0.022601.
Largest prime factor
45
Next we consider the deficiency of Γ2 .
13
1. ps > v 25 . The deficiency is
\
11.5
25
9
25
dt
t
\
t
14
min( 9.55
25 − 2 , 2t− 25 )
max( 13
25 −t,
du
u
26
175 )
\
1
min(u, t−u− 25
)
13
25 −t
dr
r
\
1
2 (1−t−u−r))
min(r,
×
13
25 −t
1−t−u−r−s
g
s
ds
s2
≤ 0.013694.
12
13
2. ps < v 25 , pr > v 25 . The corresponding deficiency is
\
11
25
7.8
25
dt
t
\
min( 5.2
25 ,
9.55
t
25 − 2 )
max( 13
25 −t,
26
175 )
du
u
\
u
13
25 −t
dr
r
\
min( 12 (1−t−u−r),
×
12
25 −t)
1
25
1−t−u−r−s
w
s
ds
.
s2
13
1) qrs > v 25 (0.000236).
12
2) qrs < v 25 .
i) The deficiency of the prime term is 0.036198.
ii) The deficiency of the almost-prime term is
\
11
25
7.8
25
dt
t
\
min( 5.2
25 ,
9.55
t
25 − 2 )
max( 13
25 −t,
26
175 )
du
u
\
u
13
25 −t
dr
r
min( 12
25 −t,
\
1
25
1
2 (1−t−u−r−s)
×
s
\
12
1
25 −u−r, 3 (1−t−u−r))
1−t−u−r−s−k
w
k
12
13
ds
s
13
dk
.
k2
13
a) kp > v 25 (0.002569). b) kp < v 25 , kps > v 25 : α) kqrs > v 25
12
12
13
(0.000774); β) kqrs < v 25 (0.004275). c) kps < v 25 : α) kqrs > v 25 (0.000402);
12
β) kqrs < v 25 (0.003268).
Therefore the total deficiency in 2. is 0.047722.
13
12
3. pr < v 25 , prs > v 25 . The corresponding deficiency is
\
11
25
7.8
25
dt
t
\
min( 5.2
25 ,
9.55
t
25 − 2 )
13
max( 25
−t,
26
175 )
du
u
\
12
25 −t
1 13
2 ( 25 −t)
dr
r
min(r,
\
1
2 (1−t−u−r))
13
25 −t−r
1 − t − u − r − s ds
w
.
s
s2
46
C. Jia and M. C. Liu
3
6.5
1) s < q − 8 v 50 . Let m = pr, n = q and d = st. Applying process I with
Lemma 3, we have to deal with a sum whose deficiency is
\
11
25
7.8
25
dt
t
\
min( 5.2
25 ,
9.55
t
25 − 2 )
max( 13
25 −t,
min(r,
26
175 )
du
u
\
6.5
3
50 − 8 u)
×
\
12
25 −t
max( 12 ( 13
25 −t),
ds
s
13
25 −t−r
\
s
1
25
dy
y
\
y
1
25
dr
r
19.5
3
50 −t+ 8 u)
1−t−u−r−s−y−z
w
z
13
dz
.
z2
12
i) prw > v 25 (0.000643). ii) prw < v 25 (0).
Hence, the total deficiency in 1) is 0.000643.
3
6.5
2) s > q − 8 v 50 .
i) The deficiency of the prime term is 0.021656.
ii) The deficiency of the almost-prime term is
\
11
25
7.8
25
dt
t
\
min( 5.2
25 ,
9.55
t
25 − 2 )
max( 13
25 −t,
26
175 )
du
u
\
12
25 −t
1 13
2 ( 25 −t)
dr
r
max(max( 13
25 −t−r,
\
6.5
3
1
50 − 8 u), min(r, 3 (1−t−u−r)))
×
13
−t−r,
max( 25
\
1
2 (1−t−u−r−s)
×
s
6.5
3
50 − 8 u)
1−t−u−r−s−k
w
k
13
12
ds
s
dk
.
k2
13
13
a) kp > v 25 (0.000349). b) kp < v 25 , kps > v 25 : α) kqrs > v 25
12
12
(0.001240); β) kqrs < v 25 (0.004339). c) kps < v 25 (0.000786).
Therefore the total deficiency in 2) is 0.028370 and that in 3. is 0.029013.
12
4. prs < v 25 . The corresponding deficiency is
\
10
25
7.8
25
dt
t
\
min( 5.2
25 ,
9.55
t
25 − 2 )
max( 13
25 −t,
26
175 )
du
u
\
11
25 −t
1
25
dr
r
min(r,
\
12
25 −t−r)
1
25
1 − t − u − r − s ds
w
.
s
s2
12
Let m = prs, n = q and d = t. Note that t < s < min(r, v 25 /pr) ≤
12
1
6.5
3
(v 25 /p) 2 < v 25 q − 4 . Applying process I with Lemma 3, we have to deal
with a sum whose deficiency is
Largest prime factor
\
10
25
7.8
25
dt
t
\
min( 5.2
25 ,
9.55
t
25 − 2 )
max( 13
25 −t,
26
175 )
min(y,
×
\
11
25 −t
du
u
1
25
dr
r
min(r,
\
1
2 (1−t−u−r−s−y))
1
25
13
\
12
25 −t−r)
1
25
47
ds
s
\
s
1
25
dy
y
1−t−u−r−s−y−z
w
z
dz
.
z2
12
1) prsw > v 25 (0.000373). 2) prsw < v 25 (0.000063).
Thus the total deficiency in 4. is 0.000436. Hence, the deficiency of Γ2
is 0.090865. The deficiency of Ψ1 is 0.113466 and that of Ψ2 is 0.157504.
Therefore the total deficiency in III is 0.270970.
5.2
IV. q > v 25 . The deficiency is
12
25
\ dt min(t, \21 (1−t)) 1 − t − u du
g
≤ 0.529493.
t
u
u2
6.5
13
5.2
25
max( 25 −t,
25
)
Therefore the deficiency of Ω2 is 0.891757.
6. Proof of the Theorem. The above discussion yields
X
X
1
X Φ≥
· S(B, v 25 ) −
S(Bp , p) −
S(Bp , p)
v
1
12
12
1
v 25 <p≤v 25
v 25 <p≤(2v) 2
X
X
− 0.891757 ·
log v
log v
1
X
X
=
· S(B, (2v) 2 ) − 0.988524 ·
v
log v
X
X
= (1 + O(ε)) ·
− 0.988524 ·
log v
log v
X
≥ 0.011475 ·
.
log v
Thus (3) holds and so the proof of the Theorem is complete.
− 0.096767 ·
Acknowledgements. We would like to thank the referee for his helpful
comments and news on the recent progress in this problem. In an unpub19
lished manuscript, Harman got the exponent ϕ = 20
. In his doctoral thesis
24
(Oxford, 1998), J. K. Haugland got ϕ = 25 .
References
[1]
A. B a l o g, Numbers with a large prime factor , Studia Sci. Math. Hungar. 15 (1980),
139–146.
48
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
C. Jia and M. C. Liu
A. B a l o g, Numbers with a large prime factor II , in: Topics in Classical Number
Theory, Colloq. Math. Soc. János Bolyai 34, Elsevier, North-Holland, Amsterdam,
1984, 49–67.
A. B a l o g, G. H a r m a n and J. P i n t z, Numbers with a large prime factor IV ,
J. London Math. Soc. (2) 28 (1983), 218–226.
J.-M. D e s h o u i l l e r s and H. I w a n i e c, Power mean-values for Dirichlet’s polynomials and the Riemann zeta-function, II , Acta Arith. 43 (1984), 305–312.
G. H a r m a n, On the distribution of αp modulo one, J. London Math. Soc. (2) 27
(1983), 9–18.
—, On the distribution of αp modulo one II , Proc. London Math. Soc. (3) 72 (1996),
241–260.
D. R. H e a t h - B r o w n, The largest prime factor of the integers in an interval, Sci.
China Ser. A 26 (1996), 385–411 (in Chinese); Sci. China Ser. A 39 (1996), 449–476.
D. R. H e a t h - B r o w n and C. J i a, The largest prime factor of the integers in an
interval , II , J. Reine Angew. Math. 498 (1998), 35–59.
C. J i a, On the distribution of αp modulo one, J. Number Theory 45 (1993), 241–253.
—, On the distribution of αp modulo one, II , preprint.
M. J u t i l a, On numbers with a large prime factor , J. Indian Math. Soc. (N.S.) 37
(1973), 43–53.
Institute of Mathematics
Academia Sinica
Beijing, 100080, China
E-mail: [email protected]
Department of Mathematics
The University of Hong Kong
Pokfulam Road, Hong Kong
E-mail: [email protected]
Received on 26.4.1999
and in revised form on 10.10.1999
(3590)