Some Arithmetic Functions Involving Exponential Divisors

Xiaodong Cao Department of Mathematics and Physics Beijing Institute of Petro-Chemical Technology Beijing, 102617 P. R. China

                    caoxiaodong@bipt.edu.cn

Wenguang Zhai School of Mathematical Sciences Shandong Normal University Jinan, 250014 Shandong P. R. China

                      zhaiwg@hotmail.com

Abstract In this paper we study several arithmetic functions connected with the exponential divisors of integers. We establish some asymptotic formulas under the Riemann hypothesis, which improve previous results. We also prove some asymptotic lower bounds.

1 Introduction and results

Suppose n > 1 is an integer with prime factorization n = p_{1}^{a_{1}} ⋯ p_{r}^{a_{r}}. An integer d is called an exponential divisor (e-divisor) of n if d = p_{1}^{b_{1}} ⋯ p_{r}^{b_{r}} with b_{j}|a_{j}(1 ≤ j ≤ r), which is denoted by the notation d|_{e}n. For convenience 1|_{e}1. The properties of the exponential divisors attract the interests of many authors(see, for example, [4, 6, 7, 9, 11, 12, 14, 15, 16, 19, 20, 21, 22]).

An integer n = p_{1}^{a_{1}} ⋯ p_{r}^{a_{r}} is called exponentially squarefree (e-squarefree) if all the exponents a_{1}, ⋯, a_{r} are squarefree. The integer 1 is also considered to be an e-squarefree number.

Suppose f and g are two arithmetic functions and n = p_{1}^{a_{1}} ⋯ p_{r}^{a_{r}}. Subbarao [17] first introduced the exponential convolution (e-convolution) by

(f ⊙ g)(n) = ∑_{b_{1}c_{1}=a_{1}} ⋯ ∑_{b_{r}c_{r}=a_{r}} f(p_{1}^{b_{1}} ⋯ p_{r}^{b_{r}})g(p_{1}^{c_{1}} ⋯ p_{r}^{c_{r}}),

which is an analogue of the classical Dirichlet convolution. The e-convolution ⊙ is commutative, associative and has the identity element μ^{2}, where μ is the Möbius function. Furthermore, a function f has an inverse with respect to ⊙ iff f(1) ≠ 0 and f(p_{1} ⋯ p_{r}) ≠ 0 for any distinct primes p_{1}, ⋯, p_{r}.

The inverse of the constant function f(n) ≡ 1 with respect to ⊙ is called the exponential analogue of the Möbius function and it is denoted by μ^{(e)}. Hence ∑_{d|_{e}n} μ^{(e)}(d) = μ^{2}(n) for n ≥ 1, μ^{(e)}(1) = 1, and μ^{(e)}(n) = μ(a_{1}) ⋯ μ(a_{r}) for n = p_{1}^{a_{1}} ⋯ p_{r}^{a_{r}} > 1. Note that |μ^{(e)}(n)| = 1 or 0, according as n is e-squarefree or not.

An integer d is called an exponential squarefree exponential divisor (e-squarefree edivisor) of n if d = p_{1}^{b_{1}} ⋯ p_{r}^{b_{r}} with b_{j}|a_{j}(1 ≤ j ≤ r) and b_{1}, ⋯, b_{r} are squarefree. Observe that the integer 1 is an e-squarefree and it is not an e-divisor of n > 1. Let t^{(e)} denote the number of e-sqaurefree e-divisors of n.

Now we introduce some notation for later use. As usual, let μ(n) and ω(n) denote the Möbius function , and the number of distinct prime factors of n respectively. If t is real, then {t} denotes the fractional part of t, ψ(t) = {t} − 1/2. Throughout this paper, ε is a small fixed positive constant and m ∼ M means that cM ≤ m ≤ CM for some constants 0 < c < C. For any fixed integers a and b, define the function d(a, b; n) := ∑_{m_{1}^{a}m_{2}^{b}=n} 1 and let Δ(a, b; t) denote the error term in the asymptotic formula of the sum ∑_{n≤x} d(a, b; n).

Many authors have studied the properties of the above three functions; see, for example, [4, 7, 15, 17, 19, 20, 22]. Tóth [20] proved that

∑_{n≤x} μ^{(e)}(n) = A_{1}x + (x^{1/2}exp(−c_{1}(log x)^{Δ})),  (1)

where 0 < Δ < 9/25 and c_{1} > 0 are constants and

A_{1} := m(μ^{(e)}) = ∏_{p} (1 + ∑_{k=2}^{∞} \frac{μ(k) − μ(k − 1)}{p^{k}}) .

Tóth [20] proved that if the Riemann hypothesis (RH) is true, then

∑_{n≤x} μ^{(e)}(n) = A_{1}x + O(x^{91/202+ε}).  (2)

Subbarao [17] and Wu [22] studied the asymptotic properties of the sum ∑_{n≤x} |μ^{e}(n)|. Tóth [20] proved that if RH is true, then

∑_{n≤x} |μ^{(e)}(n)| = B_{1}x + O(x^{1/5+ε}),  (3)

where

B_{1} := ∏_{p}(1 + ∑_{α=4}^{∞} \frac{μ^{2}(a) − μ^{2}(a − 1)}{p^{a}}).

For the function t^{(e)}(n), Tóth [20] proved the asymptotic formula

∑_{n≤x} t^{(e)}(n) = C_{1}x + C_{2}x^{1/2} + O(x^{1/4+ε}),  (4)

where

C_{1} := ∏_{p} (1 + ∑_{k=2}^{∞} \frac{2^{ω(k)} − 2^{ω(k−1)}}{p^{k}} ) ,

C_{2} := ζ(\frac{1}{2}) ∏_{p} (1 + ∑_{k=4}^{∞} \frac{2^{ω(k)} − 2^{ω(k−1)} − 2^{ω(k−2)} + 2^{ω(k−3)}}{p^{k}} ) .

Pétermann [13] proved the formula (4) with a better error term O(x^{1/4}), which is the best unconditional result up to date. In order to further reduce the exponent 1/4, we have to know more information about the distribution of the non-trivial zeros of the Riemann zeta-function.

In this short paper we shall prove the following theorem.

Theorem 1. If RH is true, then

∑_{n≤x} μ^{(e)}(n) = A_{1}x + O (x^{\frac{37}{94} +ε}) ,  (5)
∑_{n≤x} |μ^{(e)}(n)| = B_{1}x + B_{2}x^{\frac{1}{5}} + O(x^{\frac{38}{193} +ε}),  (6)
∑_{n≤x} t^{(e)}(n) = C_{1}x + C_{2}x^{1/2} + O (x^{\frac{3728}{15469} +ε}) ,  (7)

where B_{2} is a computable constant.

Remark 1. Numerically, we have

\frac{37}{94} = 0.39361 ⋯, \frac{91}{202} = 0.45049 ⋯,

\frac{38}{193} = 0.1968 ⋯ < 1/5, \frac{3728}{15469} = 0.240 ⋯ < 1/4.

Let Δ_{μ^{(e)}}(x), Δ_{|μ^{(e)}|}(x), Δ_{t^{(e)}}(x) denote the error terms in (5), (6) and (7) respectively. We have the following result.

Theorem 2. We have

Δ_{μ^{(e)}}(x) = Ω(x^{1/4}),  (8)
Δ_{|μ^{(e)}|}(x) = Ω(x^{1/8}),  (9)
Δ_{t^{(e)}}(x) = Ω(x^{1/6}).  (10)

2 Some generating functions

In this section we shall study the generating Dirichlet series of the functions μ^{(e)}(n), |μ^{(e)}(n)| and t^{(e)}(n), respectively. We consider only |μ^{(e)}(n)| and t^{(e)}(n), since Tóth [20] already proved the following formula, which is enough for our purpose,

∑_{n=1}^{∞} \frac{μ^{(e)}(n)}{n^{s}} = \frac{ζ(s)}{ζ^{2}(2s)}U(s) (ℜs > 1),  (11)

where U(s) := ∑_{n=1}^{∞} \frac{u(n)}{n^{s}} is absolutely convergent for ℜs > 1/5.

We first consider the function |μ^{(e)}(n)|. The function μ^{(e)} is multiplicative and μ^{(e)}(p^{a}) = μ(a) for every prime power p^{a}, namely for every prime p, μ^{(e)}(p) = 1, μ^{(e)}(p^{2}) = −1, μ^{(e)}(p^{3}) = −1, μ^{(e)}(p^{4}) = 0, μ^{(e)}(p^{5}) = −1, μ^{(e)}(p^{6}) = 1, μ^{(e)}(p^{7}) = 1, μ^{(e)}(p^{8}) = 0, μ^{(e)}(p^{9}) = 0, μ^{(e)}(p^{10}) = 1, μ^{(e)}(p^{11}) = −1, · · · . Hence by the Euler product we have for ℜs > 1 that

∑_{n=1}^{∞} \frac{|μ^{(e)}(n)|}{n^{s}} = ∏_{p} (1 + ∑_{m=1}^{∞} \frac{|μ(m)|}{p^{ms}}) .  (12)

Applying the product representation of Riemann zeta-function

ζ(s) = ∏_{p}(1 + p^{−s} + p^{−2s} + ⋯) = ∏_{p}(1 − p^{−s})^{−1} (ℜs > 1),  (13)

we have for ℜs > 1

ζ(s)ζ(5s) = ∏_{p} ((1 − p^{−s})(1 − p^{−5s}))^{−1} .  (14)

Let

f_{|μ^{(e)}|}(z) : = 1 + ∑_{m=1}^{∞} |μ(m)|z^{m}  (15)

= 1 + z + z^{2} + z^{3} + z^{5} + z^{6} + z^{7} + z^{10} + z^{11} + ∑_{m=12}^{∞} |μ(m)|z^{m}.

For |z| < 1, it is easy to verify that

f_{|μ^{(e)}|}(z)(1 − z)(1 − z^{5})  (16)

= (1 + z + z^{2} + z^{3} + z^{5} + z^{6} + z^{7} + z^{10} + z^{11} + ∑_{m=12}^{∞} |μ(m)|z^{m}) (1 − z − z^{5} + z^{6})

= 1 − z^{4} − z^{8} + z^{9} + ∑_{m=12}^{∞} c_{m}^{(1)}z^{m},

where

c_{m}^{(1)} := |μ(m)| − |μ(m − 1)| − |μ(m − 5)| + |μ(m − 6)|. (m ≥ 12)

Furthermore, we have

f_{|μ^{(e)}|}(z)(1 − z)(1 − z^{5})(1 − z^{4})^{−1}(1 − z^{8})^{−1}  (17)

= (1 − z^{4} − z^{8} + z^{9} + ∑_{m=12}^{∞} c_{m}^{(1)}z^{m}) (1 + z^{4} + z^{8} + ⋯)(1 + z^{8} + z^{16} + ⋯)

= 1 + z^{9} + ∑_{m=12}^{∞} C_{m}^{(1)}z^{m}.

We get from (12), (13), (15) and (17) by taking z = p^{−s} that

∑_{n=1}^{∞} \frac{|μ^{(e)}(n)|}{n^{s}} = \frac{ζ(s)ζ(5s)}{ζ(4s)ζ(8s)}V (s) (ℜs > 1),  (18)

where V (s) := ∑_{n=1}^{∞} \frac{v(n)}{n^{s}} is absolutely convergent for ℜs > \frac{1}{9}.

We now consider the function t^{(e)}(n), which is also multiplicative and t^{(e)}(p^{a}) = 2^{ω(a)} for every prime power p^{a}. Therefore for every prime p, t^{(e)}(p) = 1, t^{(e)}(p^{2}) = t^{(e)}(p^{3}) = t^{(e)}(p^{4}) = t^{(e)}(p^{5}) = 2, t^{(e)}(p^{6}) = 4, t^{(e)}(p^{7}) = t^{(e)}(p^{8}) = t^{(e)}(p^{9}) = 2, t^{(e)}(p^{10}) = 4, · · · .

By the Euler product we have for ℜs > 1 that

∑_{n=1}^{∞} \frac{t^{(e)}(n)}{n^{s}} = ∏_{p} (1 + ∑_{m=1}^{∞} \frac{2^{ω(m)}}{p^{ms}} ) .  (19)

Let

f_{t^{(e)}}(z) := 1 + ∑_{m=1}^{∞} 2^{ω(m)}z^{m}  (20)

= 1 + z + 2z^{2} + 2z^{3} + 2z^{4} + 2z^{5} + 4z^{6} + 2z^{7} + 2z^{8} + 2z^{9} + 4z^{10} + 2z^{11} + ∑_{m=12}^{∞} 2^{ω(m)}z^{m}.

For |z| < 1, it is easy to check that

f_{t^{(e)}}(z)(1 − z)(1 − z^{2})(1 − z^{6})^{2}

= 1 − z^{4} − 2z^{7} − 2z^{8} + 2z^{9} + ∑_{m=10}^{∞} c_{m}^{(2)}z^{m}

and

f_{t^{(e)}}(z)(1 − z)(1 − z^{2})(1 − z^{6})^{2}(1 − z^{4})^{−1}  (21)

= (1 − z^{4} − 2z^{7} − 2z^{8} + 2z^{9} + ∑_{m=10}^{∞} c_{m}^{(2)}z^{m}) (1 + z^{4} + z^{8} + ⋯)

= 1 − 2z^{7} + ∑_{m=8}^{∞} C_{m}^{(2)}z^{m}.

Combining (13) and (19)–(21) we get

∑_{n=1}^{∞} \frac{t^{(e)}(n)}{n^{s}} = \frac{ζ(s)ζ(2s)ζ^{2}(6s)}{ζ(4s)}W (s) (ℜs > 1),  (22)

where W (s) := ∑_{n=1}^{∞} \frac{w(n)}{n^{s}} is absolutely convergent for ℜs > \frac{1}{7}.

3 Proof of Theorem 2

We see from (11) that the generating Dirichlet series of the function μ^{(e)}(n) is \frac{ζ(s)}{ζ^{2}(2s)}U(s), which has infinitely many poles on the line ℜs = \frac{1}{4}, whence the estimate (8) follows. From the expression (18) we see that the generating Dirichlet series of the function |μ^{(e)}(n)| is \frac{ζ(s)ζ(5s)}{ζ(4s)}V (s), which has infinitely many poles on the line ℜs = \frac{1}{8}, whence the estimate (9) follows. Via (22), we get the estimate (10) with the help of Theorem 2 of Küleitner and Nowak [8] (or by Balasubramanian, Ramachandra and Subbarao’s method in [1]).

4 The Proofs of (5) and (7)

We follow the approach of Theorem 1 in Montgomery and Vaughan [10]. Throughout this section, we assume RH.

We first prove (5). It is well-known that the characteristic function of the set of squarefree integers is

μ^{2}(n) = |μ(n)| = ∑_{d^{2}|n} μ(d),  (23)

We write

Δ(x) := ∑_{n≤x} μ^{2}(d) − \frac{x}{ζ(2)} := D(x) − \frac{x}{ζ(2)}.  (24)

Define the function a_{1}(n) by

a_{1}(n) = ∑_{md^{2}=n} μ^{2}(m)μ(d),  (25)

it is easy to see that

\frac{ζ(s)}{ζ^{2}(2s)} = \frac{∑{∞ \atop n=1} \frac{μ^{2}(n)}{n^{s}}}{ζ(2s)} = ∑_{n=1}^{∞} \frac{a_{1}(n)}{n^{s}}, ℜs > 1,  (26)

Thus

T_{1}(x) := ∑_{n≤x} a_{1}(n) = ∑_{md^{2}≤x} μ^{2}(m)μ(d) = ∑_{d≤x^{\frac{1}{2}}} μ(d)D(\frac{x}{d^{2}} ).  (27)

Suppose 1 < y < x^{1/2} is a parameter to be determined. We now write T_{1}(x) in the form

T_{1}(x) = S_{1}(x) + S_{2}(x),  (28)

where

S_{1}(x) = ∑_{d≤y} μ(d)D (\frac{x}{d^{2}} ) ,  (29)

and

S_{2}(x) = ∑_{y<d≤x^{\frac{1}{2}}} μ(d)D (\frac{x}{d^{2}} ) = ∑_{d>y^{md^{2}≤x}} μ^{2}(m)μ(d).  (30)

We first evaluate S_{1}(x). From (24) we get

S_{1}(x) = ∑_{d≤y} μ(d) (\frac{x}{d^{2}ζ(2)} + Δ (\frac{x}{d^{2}} ))  (31)

= \frac{x}{ζ(2)} ∑_{d≤y} \frac{μ(d)}{d^{2}} + ∑_{d≤y} μ(d)Δ (\frac{x}{d^{2}} ) .

To treat S_{2}(x), we let

g_{y}(s) = ζ^{−1}(s) − ∑_{d≤y} \frac{μ(d)}{d^{s}}, (s = σ + it),  (32)

so that for ℜs = σ > 1

g_{y}(s) = ∑_{d>y} \frac{μ(d)}{d^{s}}.  (33)

Hence

\frac{g_{y}(2s)ζ(s)}{ζ(2s)} = ∑_{n=1}^{∞} \frac{b_{1}(n)}{n^{s}},  (34)

for σ > 1, where

b_{1}(n) = ∑_{d>y^{md^{2}=n}} μ^{2}(m)μ(d)  (35)

Let \frac{1}{2} < σ < 2, δ = \frac{ε}{10}. Assume RH, from 14.25 in Titchmarsh [18] we have

∑_{d≤y} \frac{μ(d)}{d^{s}} = ζ^{−1}(s) + O (y^{\frac{1}{2} −σ+δ}(|t|^{δ} + 1)) .  (36)

In addition, RH implies that ζ(s) ≪ |t|^{δ} + 1 and ζ^{−1}(s) ≪ (|t|^{δ} + 1) uniformly for σ > \frac{1}{2} + δ and |s − 1| > ε, and ∑_{n≤y} μ(n) ≪ y^{\frac{1}{2} +δ}. From (32), (33) and (36), we have

g_{y}(2s) ≪ y^{− \frac{1}{2}} (|t|^{δ} + 1), (σ ≥ \frac{1}{2} + δ).

Thus

g_{y}(2s)\frac{ζ(s)}{ζ(2s)} ≪ y^{− \frac{1}{2}} (|t|^{3δ} + 1), (σ ≥ \frac{1}{2} + δ, |s − 1| > ε).  (37)

From (30), (34), (35) and Perron’s formula([18], Lemma 3.19), we obtain

S_{2}(x) = ∑_{n≤x} b_{1}(n) = \frac{1}{2πi} ∫_{1+ε−ix^{2}}^{1+ε+ix^{2}} g_{y}(2s)\frac{ζ(s)}{ζ(2s)}x^{s}s^{−1}ds + O(x^{δ}),  (38)

since b(n) ≪ n^{δ} by a divisor argument. If we move the line of integration to σ = \frac{1}{2} + δ, then by the residue theorem

\frac{1}{2πi} ∫_{1+ε−ix^{2}}^{1+ε+ix^{2}} g_{y}(2s)\frac{ζ(s)}{ζ(2s)}x^{s}s^{−1}ds  (39)

= I_{1} + I_{2} − I_{3} + Res_{s=1} g_{y}(2s)\frac{ζ(s)}{ζ(2s)}x^{s}s^{−1},

where

I_{1} = \frac{1}{2πi} ∫_{\frac{1}{2} +δ+ix^{2}}^{1+ε+ix^{2}} g_{y}(2s)\frac{ζ(s)}{ζ(2s)}x^{s}s^{−1}ds, I_{2} = \frac{1}{2πi} ∫_{\frac{1}{2} +δ−ix^{2}}^{\frac{1}{2} +δ+ix^{2}} g_{y}(2s)\frac{ζ(s)}{ζ(2s)}x^{s}s^{−1}ds,

I_{3} = \frac{1}{2πi} ∫_{\frac{1}{2} +δ−ix^{2}}^{1+ε−ix^{2}} g_{y}(2s)\frac{ζ(s)}{ζ(2s)}x^{s}s^{−1}ds.

From (37) it is not difficult to see that

I_{j} ≪ y^{− \frac{1}{2}} x^{\frac{1}{2} +8δ}, (j = 1, 2, 3).  (40)

Combining (38)–(40), we get

S_{2}(x) = \frac{x}{ζ(2)} ∑_{d>y} \frac{μ(d)}{d^{2}} + O (y^{− \frac{1}{2}} x^{\frac{1}{2} +ε} + x^{ε}) .  (41)

Finally, combining (27), (28), (31) and (41) we obtain

T_{1}(x) = \frac{x}{ζ^{2}(2)} + ∑_{d≤y} μ(d)Δ (\frac{x}{d^{2}} ) + O (x^{\frac{1}{2} +ε}y^{− \frac{1}{2}} + x^{ε}) .  (42)

In [5], Jia proved the estimate Δ(u) ≪ u^{\frac{17}{54} +ε}. Inserting this estimate into (42) and on taking y = x^{\frac{10}{57}} , we get

T_{1}(x) = \frac{x}{ζ^{2}(2)} + O (x^{\frac{37}{94} +ε}) .  (43)

The asymptotic formula (5) follows form (25)-(27) and (43) by the well-known convolution method.

Now we prove the asymptotic formula (7). We define the functions a_{2}(n) by the following identity

\frac{ζ(s)ζ(2s)}{ζ(4s)} = ∑_{n=1}^{∞} \frac{a_{2}(n)}{n^{s}}, ℜs > 1.  (44)

Hence

a_{2}(n) = ∑_{d^{4}m=n} μ(d)d(1, 2; m).

Thus by the same approach as (42), we easily get that for 1 ≤ y ≤ x^{\frac{1}{4}}

T_{2}(x) := ∑_{n≤x} a_{2}(n)  (45)

= \frac{ζ(2)}{ζ(4)}x + \frac{ζ(\frac{1}{2})}{ζ(2)}x^{\frac{1}{2}} + ∑_{d≤y} μ(d)Δ (1, 2; \frac{x}{d^{4}} ) + O (x^{\frac{1}{2} +ε}y^{− \frac{3}{2}} + x^{ε}) .

Graham and Kolesnik [2] proved that Δ(1, 2; u) ≪ u^{\frac{1057}{4785} +ε}. Inserting this estimate into (45) and on taking y =^{\frac{2671}{15469}} , we get

T_{2}(x) = ∑_{n≤x} a_{2}(n) = \frac{ζ(2)}{ζ(4)}x + \frac{ζ(\frac{1}{2})}{ζ(2)}x^{\frac{1}{2}} + O (x^{\frac{3728}{15469} +ε}) .  (46)

The asymptotic formula (7) follows from (22), (44) and (46) by the convolution method.

5 Proof of (6)

Throughout this section we assume RH.

Define the function a_{3}(n) by

\frac{ζ(s)ζ(5s)}{ζ(4s)} = ∑_{n=1}^{∞} \frac{a_{3}(n)}{n^{s}}, ℜs > 1.  (47)

Similar to (45) we have for some 1 ≤ y ≤ x^{\frac{1}{4}} that

T_{3}(x) := ∑_{n≤x} a_{3}(n)

= \frac{ζ(5)}{ζ(4)}x + \frac{ζ(\frac{1}{5})}{ζ(\frac{4}{5})}x^{\frac{1}{5}} + ∑_{d≤y} μ(d)Δ (1, 5; \frac{x}{d^{4}} ) + O (x^{\frac{1}{2} +ε}y^{− \frac{3}{2}} + x^{\frac{1}{5} +ε}y^{− \frac{3}{10}} + x^{ε}) .

Unfortunately, we can not improve Tóth’s exponent \frac{1}{5} in (3) by the above formula even if we use the conjectural bound Δ(1, 5; t) ≪ t^{1/12+ε}. So we use a different approach to prove (6).

Let q_{4}(n) denote the characteristic function of the set of 4-free numbers, then

∑_{n=1}^{∞} \frac{q_{4}(n)}{n^{s}} = \frac{ζ(s)}{ζ(4s)}, ℜs > 1.

We write

Δ_{4}(x) := ∑_{n≤1} q_{4}(n) − \frac{x}{ζ(4)}.  (48)

From (47) and (48), it follows from the Dirichlet hyperbolic approach that for some 1 ≤ y ≤ x^{\frac{1}{5}} ,

B_{3}(x) = ∑_{n≤x} a_{3}(n) = ∑_{md^{5}≤x} q_{4}(m)  (49)

= ∑_{d≤y} ∑_{m≤ \frac{x}{d^{5}}} q_{4}(m) + ∑_{m≤ \frac{x}{y^{5}}} q_{4}(m) ∑_{y<d≤( \frac{x}{m})^{\frac{1}{5}}} 1

= \frac{x}{ζ(4)} ∑_{d≤y} \frac{1}{d^{5}} + x^{\frac{1}{5}} ∑_{m≤ \frac{x}{y^{5}}} \frac{q_{4}(m)}{m^{\frac{1}{5}}} − y ∑_{m≤ \frac{x}{y^{5}}} q_{4}(m)

+ ∑_{d≤y} Δ_{4} (\frac{x}{d^{5}} ) − ∑_{m≤ \frac{x}{y^{5}}} q_{4}(m)ψ ((\frac{x}{m})^{\frac{1}{5}} ) + ψ(y) ∑_{m≤ \frac{x}{y^{5}}} q_{4}(m).

Applying partial summation formula, we get

∑_{m≤ \frac{x}{y^{5}}} \frac{q_{4}(m)}{m^{\frac{1}{5}}}  (50)

= \frac{∑_{m≤ \frac{x}{y^{5}}} q_{4}(m)}{(\frac{x}{y^{5}} )^{\frac{1}{5}}} + \frac{1}{5} ∫1^{\frac{x}{y^{5}}} \frac{∑_{m≤t} q_{4}(m)}{t^{\frac{6}{5}}}dt

= \frac{y}{x^{\frac{1}{5}}} ∑_{m≤ \frac{x}{y^{5}}} q_{4}(m) + \frac{1}{5} ∫1^{\frac{x}{y^{5}}} \frac{\frac{t}{ζ(4)} + Δ_{4}(t)}{t^{\frac{6}{5}}}dt

= \frac{y}{x^{\frac{1}{5}}} ∑_{m≤ \frac{x}{y^{5}}} q_{4}(m) + \frac{1}{4ζ(4)} (\frac{x^{\frac{4}{5}}}{y^{4}} − 1) + \frac{1}{5} ∫1^{\frac{x}{y^{5}}} \frac{Δ_{4}(t)}{t^{\frac{6}{5}}}dt.

In addition, we have by Euler-Maclaurin formula

∑_{d≤y} \frac{1}{d^{5}} = ζ(5) − \frac{1}{4y^{4}} − ψ(y)y^{−5} + O(y^{−6}).  (51)

If RH is true, Graham and Pintz [3] showed that

Δ_{4}(x) = O(x^{\frac{7}{38} +ε}),  (52)

and so

∫1^{\frac{x}{y^{5}}} \frac{Δ_{4}(t)}{t^{\frac{6}{5}}}dt = ∫_{1}^{∞} \frac{Δ_{4}(t)}{t^{\frac{6}{5}}}dt − ∫ ∞_{\frac{x}{y^{5}}} \frac{Δ_{4}(t)}{t^{\frac{6}{5}}}dt  (53)

= ∫_{1}^{∞} \frac{Δ_{4}(t)}{t^{\frac{6}{5}}}dt + O (x^{− \frac{3}{190} +ε}y^{\frac{3}{38}} ) .

Combining (49)–(53) and taking on y = x^{31/193}, we obtain

T_{3}(x) = ∑_{md^{5}≤x} q_{4}(m)  (54)

= \frac{ζ(5)}{ζ(4)}x + (\frac{1}{5} ∫_{1}^{∞} \frac{Δ_{4}(t)}{t^{\frac{6}{5}}}dt − \frac{1}{4ζ(4)}) x^{\frac{1}{5}}

− ∑_{m≤ \frac{x}{y^{5}}} q_{4}(m)ψ ((\frac{x}{m})^{\frac{1}{5}} ) + O (x^{\frac{7}{38} +ε}y^{\frac{3}{38}} + xy^{−6})

= \frac{ζ(5)}{ζ(4)}x + C_{3}x^{1/5} + O(x^{\frac{7}{38} +ε}y^{\frac{3}{38}} + xy^{−5}),

= \frac{ζ(5)}{ζ(4)}x + C_{3}x^{1/5} + O(x^{\frac{38}{193} +ε}),

where we used the trivial estimate

∑_{m≤ \frac{x}{y^{5}}} q_{4}(m)ψ ((\frac{x}{m})^{\frac{1}{5}} ) ≪ \frac{x}{y^{5}}

and where

C_{3} := \frac{1}{5} ∫_{1}^{∞} \frac{Δ_{4}(t)}{t^{\frac{6}{5}}}dt − \frac{1}{4ζ(4)}.

Now the asymptotic formula (6) follows from (18), (47), (49) and (54) by the convolution approach.

6 Acknowledgments

The authors express their gratitude to the referee for a careful reading of the manuscript and many valuable suggestions, which highly improve the quality of this paper.

This work is supported by National Natural Science Foundation of China (Grant No. 10771127) and Research Award Foundation for Young Scientists of Shandong Province (No. BS2009SF018).

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2000 Mathematics Subject Classification: Primary 11N37. Keywords: exponential divisor, error term, asymptotic lower bound.

Received January 29 2010; revised version received March 10 2010. Published in Journal of Integer Sequences, March 12 2010.

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