Mathematical Notes, vol. 69, no. 3, 2001, pp. 347–363. Translated from Matematicheskie Zametki, vol. 69, no. 3, 2001, pp. 383–401. c Original Russian Text Copyright 2001 by Ikonnikova
The Ingham Divisor Problem on the Set of Numbers without kth Powers T. K. Ikonnikova Received June 21, 2000
Abstract—Suppose that k and l are integers such that k ≥ 2 and l ≥ 2 , Mk is a set of numbers without kth powers, and τ (n) = d|n 1 . In this paper, we obtain asymptotic estimates of the sums τ (n)τ (n + 1) over n ≤ x , n ∈ Mk . Key words: Ingham divisor problem, multiplicative arithmetical function, asymptotic relation.
INTRODUCTION
Suppose that τ (n) = d|n 1 , where n ≥ 1 is an integer and S(x) = n≤x τ (n) · τ (n + 1) . In 1927 Ingham [1] stated the following problem: find an optimal remainder in the asymptotic relation 6 τ (n) · τ (n + 1) = A1 x · ln2 x + R(x), where A1 = 2 . S(x) = π
n≤x
In the cited paper Ingham proved that R(x) = O(x ln x) as x → ∞ . In 1931 Esterman [2] strengthened Ingham’s result by proving that S(x) = A1 x · ln2 x + A2 x · ln x + A3 x + O(x11/12 · ln x),
x → ∞,
where A2 and A3 are some constants. In 1979 D. Ismoilov [3] and D. Heath-Brown [4], strengthened Esterman’s result independently of each other: in Esterman’s relation they obtained the remainder O(x5/6+ε ) , where x → ∞ , and ε is an arbitrary fixed positive number. In 2000 A. I. Pavlov [5] proved the following theorem. Theorem. Suppose that F is an arithmetical function satisfying the conditions: 1) F is multiplicative; 2) if F (n) = d|n f (d) , then there exists an α > 0 such that the relation f (n) = O(n−α ) holds as n → ∞ . Then there exist constants A1 , A2 , and A3 such that for any ε > 0 the following relation holds as x → ∞ : F (n) · τ (n) · τ (n + 1) = A1 x ln2 x + A2 x ln x + A3 x + O(x5/6+ε + x1−α/6+ε ). n≤x
Before stating the main results of the present paper, let us recall the following definition. Definition. If k ≥ 2 is an integer, then Mk is the set of all positive integers m such that for any simple p the condition pk m is satisfied. Further, we assume that the integers k ≥ 2 and l ≥ 2 are fixed. The main results of the paper are the following assertions. 0001-4346/2001/6934-0347$25.00
c 2001 Plenum Publishing Corporation
347
348
T. K. IKONNIKOVA
Theorem 1. Let T (k) (x) =
τ (n) · τ (n + 1).
n≤x, n∈Mk
Then there exist constants A(k) , B (k) , and C (k) such that for any fixed ε > 0 the following asymptotic relation holds as x → ∞ : T (k) (x) = A(k) x · ln2 x + B (k) x · ln x + C (k) x + O(x5/6+1/(6k)+ε ), where (k)
A
=
p
k + 1 2k + 1 k 1 1 − 2 − k + k+1 − k+2 p p p p
Theorem 2. Let S (k ,l) (x) =
> 0.
τ (n) · τ (n + 1).
n≤x n∈Mk , n+1∈Ml
Then there exist constants D(k ,l) , E (k ,l) , and F (k ,l) such that for any fixed ε > 0 the following asymptotic relation holds as x → ∞ : S (k ,l) (x) = D(k ,l) x · ln2 x + E (k ,l) x · ln x + F (k ,l) x + O(x5/6+1/(6h)+ε ), where h = min(k, l) and k + 1 2k + 1 k l + 1 2l + 1 l 1 (k ,l) = 1 − 2 − k + k+1 − k+2 − l + l+1 − l+2 > 0. D p p p p p p p p Corollary. If k = l , then τ (n) · τ (n + 1) = D(k) x · ln2 x + E (k) x · ln x + F (k) x + O(x5/6+1/(6k)+ε ), n≤x n∈Mk , n+1∈Mk
where D(k) = D(k ,k) , E (k) = E (k ,k) , F (k) = F (k ,k) , and 2k + 2 4k + 2 2k 1 (k) + k+1 − k+2 > 0. 1− 2 − D = p pk p p p In Theorems 1 and 2 the constants in the arguments of O depend only on ε and k , l . In the proof of both theorems we use the following result due to Heath-Brown [4]. Theorem. Let D(x, q , a) =
τ (n).
n≤x n≡a (mod q)
Then for an arbitrary fixed positive ε we have the following asymptotic relation as x → ∞ : x q x ln 2 + 2γ − 1 t·s·µ D(x, q , a) = 2 q ts q −1 t|(a, q) s|qt q ln s + R(x), t·s·µ +2 ts −1 t|(a, q) s|qt
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THE INGHAM DIVISOR PROBLEM
349
where R(x) = O(x1/3+ε ) uniformly in 1 ≤ q = O(x2/3 ) , γ is the Euler constant, and (a, q) is the greatest common divisor of the numbers a and q . Further, it is easy to prove that if χk (n) =
1 0
if n ∈ Mk , if n ∈ / Mk ,
is the characteristic function of the set Mk , then it is of the form χk (n) =
dk |n µ(d) .
1. PROOF OF THEOREM 2 Using the definition of the function τ (n) and dividing into domains in the same way as in the classical Dirichlet divisor problem, we obtain
S (k ,l) (x) =
(k ,l)
τ (mn + 1) = 2S1
(k ,l)
(x) − S2
(x),
mn≤x mn∈Mk , mn+1∈Ml
where (k ,l)
S1
(x) =
√ m≤ x
(k ,l)
τ (mn + 1),
S2
(x) =
−1
n≤xm mn∈Mk , mn+1∈Ml
√ m≤ x
√ n≤ x mn∈Mk , mn+1∈Ml
τ (mn + 1).
(k ,l)
First, let us find the asymptotics S1 (x) as x → ∞ . (k ,l) By the definition of S1 (x) and the well-known property of the numbers from Mk , we have (k ,l)
S1
(x) =
√ m≤ x n≤xm−1 mn+1∈Ml
τ (mn + 1)
µ(d) =
dk |mn
µ(d)
√
d≤x1/k
m≤ x
τ (mn + 1).
−1
n≤xm mn+1∈Ml mn≡0 (mod dk )
Suppose that in the last sum mn + 1 = q . Then
τ (mn + 1) =
−1
τ (q).
2≤q≤x+1, q∈Ml q≡1 (mod m), q≡1 (mod dk )
n≤xm mn+1∈Ml , mn≡0 (mod dk )
Therefore, if (m, dk ) = a , then the system of congruences
q ≡ 1 (mod m), q ≡ 1 (mod dk )
is equivalent to the congruence q ≡ 1 (mod dk ma−1 ) . Therefore, (k ,l)
S1
(x) =
µ(d)
a|dk
d≤x1/k
=
d≤x1/k
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µ(d)
a|dk
No. 3
√ 2≤q≤x+1, q∈Ml m≤ x k −1 k (m,d )=a q≡1 (mod d ma )
τ (q)
√ m≤ xa−1 2≤q≤x+1, q∈Ml k k −1 (m,d a )=1 q≡1 (mod d m)
2001
τ (q).
(1)
350
T. K. IKONNIKOVA
In relation (1), let us divide the sum over d into two sums so that in the first sum 1 ≤ d ≤ x1/(6k) , and in the second sum x1/(6k) < d ≤ x1/k . Then (k ,l)
S1
(k ,l)
(x) = S3
(k ,l)
(x) + R1
(x),
where (k ,l)
S3
(x) =
µ(d)
=O
√ m≤ xa−1 2≤q≤x+1, q∈Ml k k −1 (m,d a )=1 q≡1 (mod d m)
a|dk
d≤x1/(6k)
(k ,l) R1 (x)
τ (q), τ (q) .
√ x1/(6k)
2≤q≤x+1 q≡1 (mod dk m)
(k ,l)
The internal sum in R1 (x) is empty for dk m > x ; otherwise, it follows from q ≡ 1 (mod dk m) that q = 1 + dk m · n , where 1 ≤ n ≤ x/(dk m) , since 2 ≤ q ≤ x + 1 . Thus (k ,l) ε 1 R1 (x) = O x √ x1/(6k)
=O x
ε
x1/(6k)
x k d m
√ x1/(6k)
= O x1+ε
For S3
1 dk
= O(x1+ε · (x1/(6k) )1−k ) = O(x5/6+1/(6k)+ε ).
(x) , using the characteristic function of the set Ml , we obtain (k ,l)
S3
(x) =
µ(d)
a|dk
d≤x1/(6k)
=
µ(d)
d≤x1/(6k)
√ 2≤q≤x+1 m≤ xa−1 k (m,dk a−1 )=1 q≡1 (mod d m)
µ(δ)
δ≤(x+1)1/l
a|dk
τ (q)
µ(δ)
δ l |q
√ 2≤q≤x+1 m≤ xa−1 k k −1 (m,d a )=1 q≡1 (mod d m) q≡0 (mod δ l )
τ (q).
√ Let us show that if m ≤ x and δ ≤ x1/(6l) d−k/l , then dk mδl ≤ (x + 1)2/3 . Indeed, it follows from δ ≤ x1/(6l) d−k/l that δl dk ≤ x1/6 ; therefore, √ δl dk m ≤ x1/6 m ≤ x1/6 x = x2/3 ≤ (x + 1)2/3 . Let us divide the sum over δ into three sums so that in the first sum 1 ≤ δ ≤ x1/(6l) d−k/l , in the second sum x1/(6l) d−k/l < δ ≤ x1/(6l) , and in the third sum x1/(6l) < δ ≤ (x + 1)1/l . Then (k ,l)
S3
(k ,l)
(x) = S4
(k ,l)
(x) + R2
(k ,l)
(x) + R3
(x),
where (k ,l)
S4
(x) =
d≤x1/(6k)
µ(d)
δ≤x1/(6l) d−k/l
µ(δ)
a|dk
√ q≤x+1 m≤ xa−1 k k −1 (m,d a )=1 q≡1 (mod d m) q≡0 (mod δ l )
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THE INGHAM DIVISOR PROBLEM
(k ,l) R2 (x)
d≤x1/(6k)
x1/(6l) d−k/l <δ≤x1/(6l)
a|dk
√ m≤ xa−1
=O
(k ,l) R3 (x)
351
d≤x1/(6k)
x1/(6l) <δ≤(x+1)1/l
a|dk
√ m≤ xa−1
=O
τ (q) ,
2≤q≤x+1 q≡1 (mod dk m) q≡0 (mod δ l )
τ (q) .
2≤q≤x+1 q≡1 (mod dk m) q≡0 (mod δ l )
Note that if dk | (q − 1) and δl | q , then (d, δ) = 1 . Similarly, if m | (q − 1) and δl | q , then (m, δ) = 1 . Therefore, the system of congruences q ≡ 1 (mod dk m), q ≡ 0 (mod δl ) has the unique solution q ≡ r (mod dk mδl ) . Moreover, (r, dk mδl ) = δl , since δl | r and (r, dk m) = 1 . Therefore, (k ,l) τ (q) R2 (x) = O √ d≤x1/(6k) x1/(6l) d−k/l <δ≤x1/(6l) a|dk m≤ xa−1
2≤q≤x+1 q≡r (mod dk mδ l )
x+1 +1 =O x dk mδl √ d≤x1/(6k) x1/(6l) d−k/l <δ≤x1/(6l) a|dk m≤ xa−1 1 1 1 1/(6l) −k/l 1−l 1+ε 1+ε (x d ) =O x =O x dk 1/(6l) −k/l δl dk 1/(6k) 1/(6l) 1/(6k) d≤x x d <δ≤x d≤x d−k/l . = O x5/6+1/(6l)+ε
ε
d≤x1/(6k)
O x5/6+1/(6l)+ε
Let us show that
d−k/l
= O(x5/6+1/(6h)+ε ),
d≤x1/(6k)
where h = min(k, l) . Indeed, if k = l , then 5/6+1/(6l)+ε −k/l d = O x5/6+1/(6l)+ε O x d≤x1/(6k)
if k > l , then
−1
d
= O(x5/6+1/(6l)+ε ) ;
d≤x1/(6l)
O x
5/6+1/(6l)+ε
−k/l
d
= O(x5/6+1/(6l)+ε ) ;
d≤x1/(6k)
if k < l , then O x5/6+1/(6l)+ε
−k/l
d
= O(x5/6+1/(6l)+ε (x1/(6k) )1−k/l ) = O(x5/6+1/(6k)+ε ).
d≤x1/(6k) (k ,l)
Now let us estimate R3 (k ,l) R3 (x)
=O
(x) . Recall that
√ d≤x1/(6k) x1/(6l) <δ≤(x+1)1/l a|dk m≤ xa−1
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2≤q≤x+1 q≡1 (mod dk m) q≡0 (mod δ l )
τ (q) .
352
T. K. IKONNIKOVA
It follows from the congruence q ≡ 1 (mod dk m) that q = 1 + dk m · u , where 1 ≤ u ≤ x/(dk m) . l l l Similarly, it follows from the congruence q ≡ 0 (mod δ ) that q = δ · ν , where 1 ≤ ν ≤ (x + 1)/δ . Thus, if τ3 (N ) = x1 ,x2 ,x3 ≥1 1 , then x1 x2 x3 =N (k ,l) ε 1 R3 (x) = O x √ x1/(6l) <δ≤(x+1)1/l ν≤(x+1)/δ l d≤x1/(6k) m≤ x
=O x
ε
τ3 (δ ν − 1)
u≥1 dk ·m·u=δ l ν−1
l
x1/(6l) <δ≤(x+1)1/l ν≤(x+1)/δ l
=O x
ε
x1/(6l) <δ≤(x+1)1/l
x+1 δl
= O(x1+ε (x1/(6l) )1−l )
= O(x5/6+1/(6l)+ε ). Here we have used the relation τ3 (N ) = O(N ε ) , where N → ∞ and ε > 0 is fixed. Thus (k ,l) (k ,l) S1 (x) = S4 (x) + O(x5/6+1/(6h)+ε ). (k ,l) S4 (x)
Now let us find the asymptotics of (k ,l) µ(d) S4 (x) = d≤x1/(6k)
in relation (2). We have µ(δ) a|dk
1/(6l) −k/l
δ≤x d (δ ,d)=1
δ≤x1/(6l) d−k/l (δ ,d)=1
√ m≤ xa−1 k (m,d δa−1 )=1
a|dk
√ m≤ xa−1 q≤x+1, q≡r (mod dk mδ l ) k (m,d δa−1 )=1 (r ,dk mδ l )=δ l
Therefore, using the Heath-Brown theorem, we obtain (k ,l) µ(d) µ(δ) S4 (x) = d≤x1/(6k)
(2)
τ (q).
x+1 (dk mδl )2
x+1 dk mδl · ln k l 2 + 2γ − 1 × t·s·µ ts (d mδ ) t|δ l s|dk mδ l t−1 k l d mδ ln s t·s·µ +2 ts t|δ l s|dk mδ l t−1 1/3+ε x +O
√ d≤x1/(6k) δ≤x1/(6l) d−k/l a|dk m≤ xa−1 (k ,l)
= x(ln x + 2γ − 1) · S5
(k ,l)
(x) + x · S6
√ (x) + O x1/3+ε x · x1/(6l)
d−k/l
d≤x1/(6k) (k ,l)
= x(ln x + 2γ − 1) · S5 where (k ,l) S5 (x)
=
d≤x1/(6k)
(k ,l)
S6
µ(d) d2k
(x) = 2
d≤x1/(6k)
×
(k ,l)
(x) + x · S6
δ≤x1/(6l) d−k/l (δ , d)=1
µ(d) d2k
µ(δ) δ2l k a|d
δ≤x1/(6l) d−k/l (δ ,d)=1
t|δ l s|dk mδ l t−1
t·s·µ
√ m≤ xa−1 (m,dk δa−1 )=1
µ(d) δ2l k
dk mδl ts
(x) + O(x5/6+1/(6h)+ε ),
a|d
√ m≤ xa−1 (m,dk δa−1 )=1
ln
1 m2 l
dk mδl , t·s·µ ts −1
t|δ s|dk mδ l t
1 m2
s . dk mδl
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THE INGHAM DIVISOR PROBLEM
353 (k ,l)
Replacing the variable s by dk mδl /(ts) in the internal sum for S5 (k ,l)
S5
(x) =
d≤x1/(6k)
=
d≤x1/(6k)
µ(d) d2k
δ≤x1/(6l) d−k/l (δ ,d)=1
µ(d) dk
δ≤x1/(6l) d−k/l (δ ,d)=1
µ(δ) δ2l k a|d
µ(δ) δl k a|d
(x) , we obtain
1 m2 l
√ m≤ xa−1 (m,dk δa−1 )=1
t·
t|δ s|dk mδ l t−1
1 m l
√ m≤ xa−1 (m,dk δa−1 )=1
t|δ s|dk mδ l t−1
dk mδl · µ(s) ts
µ(s) . s
Switching over to external summation over s and replacing t by δl /t , we obtain (k ,l)
S5
µ(s) s 2/3
(x) =
s≤x
=
d≤x1/(6k)
µ(s) s 2/3
µ(s) b s2 2/3 b|s
s≤x
µ(d) dk
b|s d≤x1/(6k)
s≤x
=
µ(δ) δl k l
δ≤x1/(6l) d−k/l (δ ,d)=1
µ(d) dk
d≤x1/(6k)
a|d
δ≤x1/(6l) d−k/l (δ ,d)=1
µ(d) dk
t|δ
√ m≤ xa−1 k (m,d δa−1 )=1 mdk t≡0 (mod s)
µ(δ) δl k
δ≤x1/(6l) d−k/l (δ ,d)=1
a|d
1 m
√ m≤ xa−1 t|δ l k (m,d δa−1 )=1 (td ,s)=b m≡0 (mod sb−1 ) k
µ(δ) δl
m≤T (m,a)=1
−1
µ(d) ln d ϕ(a) τ (a) 1 = (ln T + γ) − +O , m a d T
T → ∞,
d|a
which is uniform in a ≥ 1 We obtain (k ,l)
S5
(x) =
µ(s) b s2 2/3 b|s
s≤x
d≤x1/(6k)
×
µ(d) dk
a|dk t|δ l (dk δa−1 ,sb−1 )=1 (tdk ,s)=b
−
q|dk δa−1
δ≤x1/(6l) d−k/l (δ ,d)=1
µ(δ) δl
√ b x ϕ(dk δa−1 ) +γ · ln dk δa−1 sa
τ (dk δa−1 )sa µ(q) ln q √ +O . q b x
Replacing a by dk /a and removing the brackets in the last relation, we obtain µ(d) 1 µ(s) µ(δ) (k ,l) ln x + γ b S5 (x) = 2 k 2 s d δl 2/3 1/(6k) 1/(6l) −k/l b|s
s≤x
×
a|dk t|δ l −1 k (δa, sb )=1 (td ,s)=b
MATHEMATICAL NOTES
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No. 3
d≤x
ϕ(δa) δa
2001
√ f ≤b(sa)−1 x a|dk t|δ l k −1 −1 k (d δa ,sb )=1 (td ,s)=b (f ,d δa )=1 k
Let us use the asymptotics
1 m
δ≤x d (δ ,d)=1
1 . f
354
T. K. IKONNIKOVA
+
µ(s) b s2 2/3 b|s
s≤x
×
d≤x1/(6k)
µ(d) dk
µ(q) q
q|aδ a|dk t|δ l (δa, sb−1 )=1 (tdk ,s)=b
1 1 +O √ s x 2/3 s≤x
d≤x1/(6k)
1 dk
δ≤x1/(6l) d−k/l (δ ,d)=1
ln
µ(δ) δl
ba sdk q
δ≤x1/(6l) d−k/l
τ (δ) τ (a) k d 1 . δl a k l a|d
t|δ
Here we have used the relation τ (δa) = τ (δ) · τ (a) , since (δ , a) = 1 . Let D (k ,l) =
∞ ∞ ∞ µ(s) µ(d) µ(δ) ϕ(δ) b · s2 dk δl δ s=1 d=1
b|s
(k ,l) D1
δ=1 (δ , d)=1
∞ ∞ ∞ µ(s) µ(d) µ(δ) = b s2 dk δl s=1 d=1
b|s
δ=1 (δ , d)=1
k
a|d (aδ , sb−1 )=1
ϕ(a) a
t|δ (tdk ,s)=b
µ(q)
q|δa a|dk t|δ l −1 k (aδ , sb )=1 (td ,s)=b
1,
l
q
ln
ba . sdk q
(k ,l)
Then, passing to infinite sums in the expression for S5 (x) , we obtain 1 (k ,l) (k ,l) (k ,l) ln x + γ D (k ,l) + D1 + R4 (x), S5 (x) = 2 (k ,l)
where R4 (x) contains the “remainders” of the infinite series. (k ,l) are convergent. Note that t | δl and (d, δ) = 1 ; Let us show that the series for D(k ,l) and D1 k k k therefore, (t, d ) = 1 . Since b | td and (t, d ) = 1 , we have b = b1 b2 , where b1 | t and b2 | dk . Since s is square-free, b | s and b = b1 b2 , it follows that b1 and b2 are also square-free. Therefore, b1 | δ and b2 | d . In view of this, we see that for any ε > 0 there exists a C = C(ε) > 0 such that |D (k ,l) | < C
∞ ∞ ∞ 1 1 1 1 (b1 b2 )ε dε1 δ1ε l−1 k−1 k l 2 s d δ b b2 d1 =1 1 δ1 =1 1 s=1 b b |s 1 1 2
≤C
∞ s=1
∞ ∞ 1 1 1 1 , k−ε l−ε s2 bh−1−ε d δ 1 1 d =1 δ =1 b|s 1
1
which proves the convergence of the series defining D(k ,l) , since h = min(k, l) ≥ 2 . Using the fact that ba k ε ε ln sdk q ≤ ln b + ln s + ln a + ln q + ln d = O(ln s + ln d) = O(s · d ), we obtain (k ,l) |D1 |
∞ ∞ ∞ 1 1 1 1 ε ε ε ε ≤C d δ s b . s2 bh−1 dk δl 1 1 s=1 d =1 1 δ =1 1
b|s
(k ,l)
Therefore, the series for D1
1
1
is also convergent. By a similar argument, we obtain (k ,l)
R4
(x) = O(x−1/6+1/(6h)+ε ). MATHEMATICAL NOTES
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THE INGHAM DIVISOR PROBLEM
Thus (k ,l) S5 (x)
=
1 (k ,l) ln x + γ D (k ,l) + D1 + O(x−1/6+1/(6h)+ε ). 2
(k ,l)
Similarly, by transforming S6 (k ,l)
S6 (k ,l)
where D2
(k ,l)
(k ,l)
(x) , we obtain
1 (k ,l) (k ,l) ln x + γ D2 (x) = −2 − 2D3 + O(x−1/6+1/(6h)+ε ), 2
and D3
S1
355
are constants. Thus
1 (k ,l) D x ln2 x + G(k ,l) · x ln x + H (k ,l) · x + O(x5/6+1/(6h)+ε ), 2
(x) =
where (k ,l)
G
=
1 (k ,l) (k ,l) D(k ,l) + D1 2γ − − D2 , 2 (k ,l)
H (k ,l) = (2γ 2 − γ)D(k ,l) + (2γ − 1)D1 (k ,l)
Now let us show that the asymptotics of S2 (k ,l)
S2
(k ,l)
− 2γD2
(k ,l)
− 2D3
.
(x) as x → ∞ is of the form
(x) = D(k ,l) x ln x + M (k ,l) · x + O(x5/6+1/(6h)+ε ),
where M (k ,l) is some constant independent of x . Recall that (k ,l)
S2
(k ,l)
Similarly for S1 (k ,l)
S2
(x) =
=
√ m≤ x
√ m≤ x
√ n≤ x mn∈Mk , mn+1∈Ml
µ(d)
µ(d)
µ(d) =
dk |mn
n≤ x mn+1∈Ml
τ (mn + 1).
a|dk
√ m≤ x k (m,d )=a
√ 2≤q≤m x+1 q∈Ml q≡1 (mod dk ma−1 )
a|dk
µ(d)
a|dk
µ(d)
Vol. 69
a|dk
2001
√ m≤ x n≤ x mn+1∈Ml mn≡0 (mod dk )
τ (q)
√ √ m≤ xa−1 2≤q≤ma x+1 k −1 q∈M l (m,d a )=1 q≡1 (mod dk m)
No. 3
√
τ (q)
τ (q)
√ √ m≤ xa−1 2≤q≤ma x+1 q∈Ml (m,dk a−1 )=1 q≡1 (mod dk m)
x1/(6k)
MATHEMATICAL NOTES
√ √ m≤ xa−1 2≤q≤ma x+1 k −1 q∈M l (m,d a )=1 q≡1 (mod dk m)
µ(d)
d≤x1/k
d≤x1/(6k)
+
τ (mn + 1)
√
d≤x1/k
=
(x) , we obtain
d≤x1/k
=
(x) =
τ (q).
τ (mn + 1)
356
T. K. IKONNIKOVA
√ √ Since m ≤ xa−1 , we have ma x + 1 ≤ x + 1 ; therefore, the second summand can be estimated (k ,l) in the same way as the similar summand in the sum S1 (x) . Namely, τ (q) = O(x5/6+1/(6k)+ε ). O x1/(6k)
Thus (k ,l)
S2
(x) =
√ √ 2≤q≤ma x+1 m≤ xa−1 k (m,dk a−1 )=1 q∈Ml , q≡1 (mod d m)
µ(d)
d≤x1/(6k)
√ √ a|dk m≤ xa−1 2≤q≤ma x+1 k k −1 (m,d a )=1 q∈Ml , q≡1 (mod d m)
Using the sieve for the second time, we obtain (k ,l) µ(d) µ(δ) S2 (x) = d≤x1/(6k)
τ (q) + O(x5/6+1/(6k)+ε ).
√ √ a|dk m≤ xa−1 2≤q≤ma x+1 l (m,dk δa−1 )=1 q≡0 (mod δ ) k q≡1 (mod d m)
δ≤(x+1)1/l
τ (q) + O(x5/6+1/(6k)+ε ). (3)
In relation (3), dividing the sum over δ into two sums so that in the first sum 1 ≤ δ ≤ x1/(6l) d−k/l , and in the second sum x1/(6l) d−k/l < δ ≤ (x + 1)1/l , we obtain (k ,l)
S2 where (k ,l)
S7
(x) =
(k ,l) R5 (x)
=O
µ(d)
d≤x1/(6k)
(k ,l)
(x) = S7
µ(δ)
a|dk
δ≤x1/(6l) d−k/l (δ ,d)=1
(k ,l)
(x) + R5
(x),
√ √ q≤ma x+1 m≤ xa−1 k l k −1 (m,d δa )=1 q≡r (mod d mδ ) (r ,dk mδ l )=δ l
√ √ d≤x1/(6k) x1/(6l) d−k/l <δ≤(x+1)1/l a|dk m≤ xa−1 2≤q≤ma x+1 q≡0 (mod δ l ) q≡1 (mod dk m)
τ (q),
τ (q)
+ O(x5/6+1/(6k)+ε ) = O(x5/6+1/(6h)+ε ). (k ,l)
On applying the Heath-Brown theorem to S7 (x) , we obtain √ 1 √ (k ,l) (k ,l) (k ,l) ln x + 2γ − 1 · S8 (x) + x · S9 (x) + O(x5/6+1/(6h)+ε ), S7 (x) = x 2 where (k ,l) S8 (x)
=
d≤x1/(6k)
(k ,l)
S9
(x) =
d≤x1/(6k)
×
µ(d) d2k µ(d) d2k
δ≤x1/(6l) d−k/l (δ , d)=1
δ≤x1/(6l) d−k/l (δ , d)=1
√ m≤ xa−1 k (m,d δa−1 )=1
1 m l
µ(δ) a δ2l k a|d
µ(δ) a δ2l k
√ m≤ xa−1 k (m,d δa−1 )=1
1 m l
dk mδl , t·s·µ ts −1
t|δ s|dk mδ l t
a|d
dk mδl t·s·µ ts −1
t|δ s|dk mδ l t
s2 ma ln . (dk mδl )2
MATHEMATICAL NOTES
Vol. 69
No. 3
2001
THE INGHAM DIVISOR PROBLEM
Just as above, we obtain (k ,l)
S8
(x) =
√
357
xD(k ,l) + O(x1/3+1/(6h)+ε ) ;
(k ,l)
The expression for S9 (x) differs from all previous sums by the presence of ln m . Therefore, to (k ,l) find the asymptotics of S9 (x) , it is necessary to use the asymptotics ϕ(a) (T ln T − T ) + O(τ (a) ln(aT )), T → ∞, ln m = a m≤T (m,a)=1
which is uniform in a ≥ 1 . As a result, we obtain
√ 1 (k ,l) (k ,l) √ S9 (x) = D(k ,l) x ln x − D (k ,l) + 2D2 x + O(x1/3+1/(6h)+ε ). 2 Thus (k ,l) S2 (x) = D(k ,l) x ln x + M (k ,l) · x + O(x5/6+1/(6h)+ε ), where
(k ,l)
M (k ,l) = (2γ − 2)D(k ,l) − 2D2 Returning to S
(k ,l)
.
(x) , we have
S (k ,l) (x) = D(k ,l) x ln2 x + E (k ,l) · x ln x + F (k ,l) · x + O(x5/6+1/(6h)+ε ), where (k ,l)
E (k ,l) = (4γ − 2)D(k ,l) + 2D1
(k ,l)
F (k ,l) = (4γ 2 − 4γ + 2)D(k ,l) + (4γ − 2)D1
(k ,l)
− 2D2
, (k ,l)
− (4γ − 2)D2
(k ,l)
− 4D3
.
Let us show that the constant D(k ,l) is positive. To do this, let us first prove that k + 1 2k + 1 k l + 1 2l + 1 l 1 (k ,l) = 1 − 2 − k + k+1 − k+2 − l + l+1 − l+2 . D p p p p p p p p Recall that D (k ,l) =
∞ ∞ ∞ µ(s) µ(d) µ(δ) ϕ(δ) b · s2 dk δl δ s=1 b|s
d=1
δ=1 (δ ,d)=1
k
a|d (aδ , sb−1 )=1
ϕ(a) a
1.
l
t|δ (tdk ,s)=b
If t | δl and (δ , d) = 1 then (t, dk ) = 1 . Therefore, since (tdk , s) = b and (t, dk ) = 1 , we have b = b1 · b2 , where b1 | t and b2 | dk . Moreover, since b | s and s ∈ M2 , it follows that b1 ∈ M2 and b2 ∈ M2 ; hence b1 | δ , b2 | d . Let b3 = sb−1 . Then s = b1 b2 b3 , where b1 , b2 , b3 are pairwise coprime. Therefore, if d = d1 · b2 and δ = δ1 · b1 , then ∞ ∞ µ(s) µ(d1 b2 ) b b D (k ,l) = 1 2 2 s (d1 b2 )k s=1 b1 b2 b3 =s ∞
×
δ1 =1 (δ1 b1 ,d1 b2 )=1
=
∞ µ(s) s=1
×
s2
d1 =1
µ(δ1 b1 ) ϕ(δ1 b1 ) · (δ1 b1 )l δ1 b1
b1 b2 b3 =s ∞
δ1 =1 (δ1 ,d1 b2 b3 )=1
MATHEMATICAL NOTES
∞
b1 b2
d1 =1 (d1 ,b1 b3 )=1
No. 3
2001
k
a|(d1 b2 ) (aδ1 b1 ,b3 )=1
ϕ(a) a
a|(d1 b2 )k (a, b3 )=1
ϕ(a) a
1 l
t|(δ1 b1 ) k k−1 (tb−1 ,b3 )=1 1 d 1 b2
µ(d1 b2 ) (d1 b2 )k
µ(δ1 b1 ) ϕ(δ1 b1 ) · (δ1 b1 )l δ1 b1
Vol. 69
t|δ1l bl−1 1 (t,b3 )=1
1.
358
T. K. IKONNIKOVA
Therefore, D(k ,l) =
∞ µ(s) µ(b1 ) ϕ(b1 ) µ(b2 ) · k−1 s2 b1 bl−1 −1 b2 s=1 b |s 1 b2 |sb1
1
∞
×
d1 =1 (d1 ,s)=1
µ(d1 ) dk1
∞ δ1 =1 (δ1 ,d1 s)=1
µ(δ1 ) ϕ(δ1 ) · δ1 δ1l
ϕ(a) 1. a l l−1
a|(d1 b2 )k
t|δ1 b1
In the last relation, the conditions (a, b3 ) = 1 and (t, b3 ) = 1 are not are not taken into account, since a | dk1 bk2 , (d1 , b3 ) = 1 and (b2 , b3 ) = 1 ; similarly, t | δ1l bl−1 1 , (δ1 , b3 ) = 1 and (b1 , b3 ) = 1 . Since (b1 , δ1 ) = 1 , by using the relation
l 1 = τ (bl−1 1 ) · τ (δ1 ),
t|δ1l bl−1 1
we obtain D (k ,l) =
∞ µ(b2 ) µ(s) µ(b1 ) ϕ(b1 ) · · τ (bl−1 1 ) l−1 k−1 2 s b1 b −1 b2 s=1 b |s 1 b2 |sb1
1
∞
×
d1 =1 (d1 ,s)=1
µ(d1 ) dk1
∞ δ1 =1 (δ1 ,d1 s)=1
µ(δ1 ) ϕ(δ1 ) · · τ (δ1l ) l δ δ1 1
a|(d1 b2 )k
ϕ(a) . a
Since µ(n) = 0 , we have the relation ϕ(d) 1 = . 1+k 1− d p k p|n
d|n
Indeed, since n ∈ M2 , we have n = p1 · · · ps , where pi = pj for i = j . Then ϕ(d) = d k
d|n
ϕ(pβ1 1 )
0≤β1 ,...,βs ≤k
pβ1 1
···
ϕ(pβs s ) pβs s
=
s ϕ(p0 ) i=1
i p0i
+ ··· +
ϕ(pki ) pki
=
p|n
1 . 1+k 1− p
Moreover, (d1 , b2 ) = 1 ; therefore, ϕ(a1 ) ϕ(a2 ) ϕ(a) 1 1 = · . = 1+k 1− 1+k 1− a a1 a2 p p k k k k
a|d1 b2
a1 |d1
a2 |b2
p|d1
p|b2
Therefore, D (k ,l) =
∞ µ(b2 ) µ(s) µ(b1 ) ϕ(b1 ) 1 l−1 · · τ (b ) 1 + k 1 − 1 s2 b1 p bl−1 bk−1 −1 2 s=1 b |s 1 p|b b2 |sb1
1
×
∞ δ1 =1 (δ1 ,s)=1
µ(δ1 ) ϕ(δ1 ) · τ (δ1l ) l δ δ1 1
∞ d1 =1 (d1 ,δ1 s)=1
2
µ(d1 ) dk1
p|d1
1 1+k 1− p
MATHEMATICAL NOTES
Vol. 69
.
No. 3
2001
THE INGHAM DIVISOR PROBLEM
359
Since
−1 µ(b2 ) k 1 1 k+1 1+k 1− 1+k 1− 1 − k−1 + k = = k−1 p pk−1 p p p −1 b2 −1 p|b p|b p|s
b2 |sb1
b2 |sb1
2
2
pb1
and ∞ d1 =1 (d1 ,δ1 s)=1
1 µ(d1 ) 1+k 1− = p dk1 p|d1
=
∞
d1 =1 p|d1 (d1 ,δ1 s)=1
pδ1 s
it follows from τ (bl−1 1 ) = D
(k ,l)
p|b1
l and τ (δ1l ) =
−1 1 1+k 1− pk p
p|δ1 (l
k k+1 1 − k + k+1 , p p
+ 1) that
∞ k µ(s) (1 − p)l k+1 = 1 − k−1 + k s2 pl p p s=1 b1 |s p|b1
∞
×
p|s pb1
(1 − p)(l + 1) k k+1 + 1 − . pk pk+1 p(l+1)
δ1 =1 p|δ1 (δ1 ,s)=1
pδ1 s
Therefore, after a few manipulations, we obtain D (k ,l) =
−1 ∞ k k k µ(s) k+1 k+1 k+1 + + + 1 − 1 − pk pk+1 s=1 s2 pk−1 pk pk pk+1 p p|s −1 (1 − p)l k k+1 × 1 − k−1 + k pl p p b1 |s p|b1 −1 ∞ (1 − p)(l + 1) k k+1 × . 1 − k + k+1 pl+1 p p 1−
δ1 =1 p|δ1 (δ1 ,s)=1
Since −1 −1 (1 − p)l k k k+1 (1 − p)l k+1 = 1 − k−1 + k 1+ 1 − k−1 + k pl p p pl p p b1 |s p|b1
p|s
and −1 (1 − p)(l + 1) k k+1 1 − k + k+1 pl+1 p p
∞
δ1 =1 p|δ1 (δ1 ,s)=1
(1 − p)(l + 1) 1− l+1 p p = (1 − p)(l + 1) 1+ 1− pl+1 1+
p|s
MATHEMATICAL NOTES
Vol. 69
No. 3
2001
k k+1 + k+1 pk p k k+1 + k+1 pk p
−1 −1 ,
360
T. K. IKONNIKOVA
we have D
(k ,l)
−1 k (l + 1)(1 − p) k+1 = · 1+ 1 − k + k+1 pl+1 p p p −1 k+1 (1 − p)l k k+1 k 1 − k−1 + k 1+ 1 − k−1 + k ∞ −1 pl p p p p · · × −1 2 k + 1 k p s=1 p|s 1 − k + k+1 1 + (l + 1)(1 − p) 1 − (k + 1) + k p p pl+1 pk pk+1 k (l + 1)(1 − p) k+1 = 1 − k + k+1 + p p pl+1 p −1 k (1 − p)l k+1 k+1 k 1+ 1 − k−1 + k 1 − k−1 + k −1 pl p p p p · × 1 + 2 · −1 k + 1 k p 1 − k + k+1 1 + (l + 1)(1 − p) 1 − (k + 1) + k p p l+1 k k+1 p p p k l+1 l+1 k+1 1 − k + k+1 − l + l+1 = p p p p p k+1 k (1 − p)l 1 − + + 1 pk−1 pk pl · × 1 − 2 k+1 k (l + 1)(1 − p) p 1 − k + k+1 + p p pl+1 k l+1 l+1 k+1 1 − k + k+1 − l + l+1 = p p p p p k+1 k l l 1 − k−1 + k − l−1 + l 1 p p p p · × 1 − 2 k+1 k l+1 l+1 p 1 − k + k+1 − l + l+1 p p p p k l+1 l+1 1 k l l k+1 k+1 1 − k + k+1 − l + l+1 − 2 1 − k−1 + k − l−1 + l = p p p p p p p p p p k l+1 l+1 1 k+1 k l l k+1 1 − k + k+1 − l + l+1 − 2 + k+1 − k+2 + l+1 − l+2 = p p p p p p p p p p k + 1 2k + 1 k l + 1 2l + 1 l 1 1 − 2 − k + k+1 − k+2 − l + l+1 − l+2 . = p p p p p p p p k k+1 1 − k + k+1 p p
Thus D (k ,l) =
p
1−
k + 1 2k + 1 k l + 1 2l + 1 l 1 − + − − + − . p2 pk pk+1 pk+2 pl pl+1 pl+2
Let us show that if k ≥ 2 and l ≥ 2 , then for any prime p the following inequality is valid: 1−
k + 1 2k + 1 k l + 1 2l + 1 l 1 − k + k+1 − k+2 − l + l+1 − l+2 > 0. 2 p p p p p p p
(4)
Note that 1−
k + 1 2k + 1 k l + 1 2l + 1 l 1 − k + k+1 − k+2 − l + l+1 − l+2 2 p p p p p p p MATHEMATICAL NOTES
Vol. 69
No. 3
2001
THE INGHAM DIVISOR PROBLEM
361
1 k + 1 2k + 1 k 1 l + 1 2l + 1 l 1 1 − k + k+1 − k+2 + − 2 − l + l+1 − l+2 − 2 2p2 p p p 2 2p p p p 1 = k+2 (pk (p − 1)(p + 1) − (2k + 2)p(p − 1) + 2k(p − 1)) 2p 1 + l+2 (pl (p − 1)(p + 1) − (2l + 2)p(p − 1) + 2l(p − 1)) 2p p−1 p−1 = k+2 (p(pk − 2k) + p(pk−1 − 2) + 2k) + l+2 (p(pl − 2l) + p(pl−1 − 2) + 2l). 2p 2p =
Since pk ≥ 2k ≥ 2k for k ≥ 2 (and similarly for l), we obtain inequality (4). Therefore, the inequality D (k ,l) > 0 is proved. Thus the proof of Theorem 2 is complete. The proof of Theorem 1 is much simpler than that of Theorem 2. Since no new ideas are invoked, we will only give a brief outline of this proof. 2. PROOF OF THEOREM 1 Using the definition of the function τ (n) , we obtain (k) (k) τ (mn + 1) = 2T1 (x) − T2 (x), T (k) (x) = mn≤x mn∈Mk
where
(k)
T1 (x) =
√ m≤ x n≤xm−1 mn∈Mk
(k)
τ (mn + 1),
T2 (x) =
√ √ m≤ x n≤ x mn∈Mk
τ (mn + 1).
(k)
First, let us show that the asymptotics for T1 (x) as x → ∞ is of the form (k)
T1 (x) =
1 (k) A x ln2 x + P (k) · x ln x + Q(k) · x + O(x5/6+1/(6k)+ε ), 2
where P (k) and Q(k) are some constants independent of x . Using the characteristic function of Mk and making the substitution mn + 1 = q , we obtain (k) µ(d) τ (q). (5) T1 (x) = l|dk
d≤x1/k
√ 2≤q≤x+1 m≤ xl−1 k (m,dk l−1 )=1 q≡1 (mod d m)
In relation (5), let us divide the sum over d into two sums so that the first sum satisfies the inequalities 1 ≤ d ≤ x1/(6k) and the second sum, x1/(6k) < d ≤ x1/k . Then (k)
(k)
(k)
T1 (x) = T3 (x) + R1 (x), where
(k)
T3 (x) =
d≤x1/(6k)
µ(d)
√ q≤x+1 l|dk m≤ xl−1 k k −1 (m,d l )=1 q≡1 (mod d m)
τ (q),
(k)
(6)
(k)
R1 (x) = O(x5/6+1/(6k)+ε ).
Now let us find the asymptotics of T3 (x) in relation (6). We apply the particular case of the Heath-Brown theorem for a = 1 in the form ϕ(q) ϕ(q) 2 µ(δ) · ln δ · x + O(x1/3+ε ). τ (n) = 2 x · ln x + (2γ − 1) 2 − · q q q δ n≤x n≡1 (mod q)
MATHEMATICAL NOTES
δ|q
Vol. 69
No. 3
2001
362
T. K. IKONNIKOVA
We obtain (k)
(k)
(k)
T3 (x) = x(ln x + 2γ − 1) · T4 (x) − 2x · T5 (x) + O(x5/6+1/(6k)+ε ), where
(k)
T4 (x) =
µ(d)
d≤x1/(6k)
(k)
T5 (x) =
d≤x1/(6k)
l|dk
√ m≤ xl−1 k −1 (m,d l )=1
µ(d) dk k l|d
ϕ(dk m) , (dk m)2
√ m≤ xl−1 k −1 (m,d l )=1
1 µ(δ) ln δ . m k δ δ|d m
(k)
In the sum for T4 (x) , let us switch to external summation over δ and use the asymptotics µ(d) ln d 1 ϕ(a) τ (a) = (ln T + γ) − +O , T → ∞, m a d T m≤T (m,a)=1
d|a
which is uniform in a ≥ 1 We obtain 1 µ(δ) (k) ln x + γ b T4 (x) = 2 δ2 2/3 δ≤x
+
µ(δ) b δ2 2/3 b|δ
δ≤x
b|δ
d≤x1/(6k) (d,δ)=b
d≤x1/(6k) (d,δ)=b
µ(d) dk
µ(d) dk
l|dk (l,δb−1 )=1
µ(q)
q|l l|dk (l,δb−1 )=1
q
ϕ(l) l · ln
lb dk δq
+ O(x−1/2+1/(6k)+ε ).
(k)
Then, passing to infinite sums in the expression for T4 (x) , we obtain 1 (k) (k) ln x + γ A(k) + A1 + O(x−1/6+1/(6k)+ε ), T4 (x) = 2 (k)
where A(k) and A1 are some constants, which can be written out in a similar way to the corresponding constants in Theorem 2. Similarly, we obtain 1 (k) (k) (k) ln x + γ A2 + A3 + O(x−1/6+1/(6k)+ε ). T5 (x) = 2 Thus
1 (k) A · x ln2 x + P (k) · x ln x + Q(k) · x + O(x5/6+1/(6k)+ε ). 2 Reasoning in the same way as above, we obtain (k)
T1 (x) =
(k)
T2 (x) = A(k) · x ln x + S (k) · x + O(x5/6+1/(6k)+ε ). Returning to T (k) (x) , we can now write T (k) (x) = A(k) · x ln2 x + B (k) · x ln x + C (k) · x + O(x5/6+1/(6k)+ε ). In the same way as in the proof of Theorem 2, we can show that k + 1 2k + 1 k 1 (k) 1 − 2 − k + k+1 − k+2 > 0. A = p p p p p This completes the proof of Theorem 1.
MATHEMATICAL NOTES
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No. 3
2001
THE INGHAM DIVISOR PROBLEM
363
ACKNOWLEDGMENTS The author wishes to thank A. I. Pavlov for setting the problem and valuable advice. REFERENCES 1. A. E. Ingham, “Some asymptotic formulae in the theory of numbers,” J. London Math. Soc., 2 (1927), no. 3, 202–208. ¨ 2. T. Esterman, “Uber die Darstellung einer Zahl als Differenz von zwei Produkten,” J. Reine Angew. Math., 164 (1931), 173–182. 3. D. Ismoilov, “On the asymptotics of the representation of numbers as a difference of two products,” Dokl. Akad. Nauk Tajik SSR, 22 (1979), no. 2, 75–79. 4. D. R. Heath-Brown, “The fourth power moment of the Riemann zeta function,” Proc. London Math. Soc., 38 (1979), no. 3, 385–422. 5. A. I. Pavlov, “On the Ingham Divisor Problem,” Mat. Zametki [Math. Notes], 68 (2000), no. 3, 429– 438. Moscow Pedagogical State University
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No. 3
2001
