Reciprocals of the Gcd-Sum Functions

Shiqin Chen Experimental Center Linyi University Linyi, 276000, Shandong China

                    shiqinchen2010@163.com

Wenguang Zhai ^{1} Department of Mathematics China University of Mining and Technology Beijing, 100083 China

                      zhaiwg@hotmail.com

Abstract In this paper we study the reciprocals of the gcd-sum function and some related functions and improve some results of Tóth. The harmonic mean of the gcd function is also studied.

1 Introduction

Recently, L. Tóth published a paper [3] about the gcd-sum function P (n), called also Pillai’s function [2] defined by

P (n) = ∑_{k=1}^{n} gcd(k, n).

In this well-written paper, Tóth not only listed many classical results about this function, its analogues and generalizations, but also proved several new results. The aim of this short note is to improve some results in Tóth’s paper.

1This work is supported by the Natural Science Foundation of Beijing (Grant No. 1112010).

Let H(n) denote the harmonic mean of gcd(1, n), gcd(2, n), …, gcd(n, n), namely,

H(n) := n (∑_{j=1}^{n} \frac{1}{gcd(j, n)})^{−1} = \frac{n^{2}}{∑_{d|n} dφ(d)},  (1)

where φ(n) denotes Euler’s function. Theorem 4 of Tóth [3] states that

∑_{n≤x} \frac{H(n)}{n} = C_{1} log x + C_{2} + O(x^{−1+ε}),  (2)

where C_{1} and C_{2} are computable constants, ε > 0 is a small positive constant. As a corollary, he deduced that

∑_{n≤x} H(n) = C_{1}x + O(x^{ε}).  (3)

In this note we shall prove first the following

Theorem 1. We have the asymptotic formula

∑_{n≤x} H(n) = C_{1}x + C_{3} log x + O((log x)^{2/3}),  (4)

where C_{1} and C_{3} are constants.

As a corollary of Theorem 1, we have

Corollary 2.

∑_{n≤x} \frac{H(n)}{n} = C_{1} log x + C_{2} + O(x^{−1}(log x)^{2/3}).  (5)

Remark 3. Tóth [3] studied the average order of ∑_{n≤x} H(n)/n first and then deduced an asymptotic formula of ∑_{n≤x} H(n) through it. However in this note we study ∑_{n≤x} H(n) first, then deduce an asymptotic formula for ∑_{n≤x} H(n)/n.

Now we study the reciprocals of P (n) and some other related functions. We first recall the definitions of some functions.

An integer d is called a unitary divisor of n if d|n and (d, n/d) = 1, notation d∥n. The unitary analogue of the function P is defined by

P^{∗}(n) := ∑_{j=1}^{n}(j, n)_{∗},

where (j, n)_{∗} = max{d ∈ ℕ : d|j, d∥n}, which was first introduced in Tóth [4]. This function P^{∗} (\underline{A145388}) is multiplicative and for every prime power p^{α}(α ≥ 1) we have P^{∗}(p^{α}) = 2p^{α} − 1.

Let n > 1 be an integer of canonical form n = ∏_{j=1}^{r} p_{j}^{a_{j}}. An integer d is called an exponential divisor of n if d = ∏_{j=1}^{r} p_{j}^{b_{j}} such that b_{j}|a_{j}(1 ≤ j ≤ r), notation d|_{e}n. By convention 1|_{e}1. The kernel of n is denoted by κ(n) = ∏_{j=1}^{r} p_{j}.

Two positive integers m > 1, n > 1 have common exponential divisors iff they have the canonical forms n = ∏_{j=1}^{r} p_{j}^{a_{j}} and m = ∏_{j=1}^{r} p_{j}^{b_{j}} with a_{j} ≥ 1, b_{j} ≥ 1(1 ≤ j ≤ r). The greatest common exponential divisor of n and m is (n, m)_{e} = ∏_{j=1}^{r} p_{j}^{(a_{j},b_{j})}. By convention define (1, 1)_{e} = 1 and note that (1, m)_{e} does not exist for m > 1. Now define the function P^{(e)}(n) by the relation

P^{(e)}(n) = ∑_{κ(j)=κ(n)^{1≤j≤n}}(j, n)_{e}.

The function P^{(e)}(see Tóth [5]) is multiplicative and for any prime power p^{α} (α ≥ 1) we have

P^{(e)}(p^{α}) = ∑_{1≤j≤α} p^{(j,α)} = ∑_{d|α} p^{d}φ(α/d).  (6)

Now we recall another analogue of P. Let n > 1 be an integer and we say an integer j is regular (mod n) if there exists an integer x such that j^{2}x ≡ j (mod n). Tóth [6] introduced the function (\underline{A176345})

P̃(n) := ∑_{j∈Reg_{n}} gcd(j, n),

where Reg_{n} = {1 ≤ j ≤ n : j is regular(mod n)}. Tóth[6] showed that P̃ is multiplicative and

P̃(n) = n ∏_{p|n}(2 − 1/p).  (7)

The above mentioned functions have been extensively studied and many papers about them have been published. See Tóth [3] and references therein.

Theorem 6 of Tóth [3] states that

∑_{n≤x} \frac{1}{P (n)} = K(log x)^{1/2} + O((log x)^{−1/2}),  (8)

∑_{n≤x} \frac{1}{P^{∗}(n)} = K^{∗}(log x)^{1/2} + O((log x)^{−1/2}),  (9)

∑_{n≤x} \frac{1}{P̃(n)} = K̃(log x)^{1/2} + O((log x)^{−1/2}),  (10)

∑_{n≤x} \frac{1}{P^{(e)}(n)} = K^{(e)} log x + O(1),  (11)

where K, K^{∗}, K̃, K^{(e)} are computable constants.

In this note we shall prove the following result.

Theorem 4. Suppose N ≥ 1 is a fixed integer, then

∑_{n≤x} \frac{1}{P (n)} = ∑_{j=0}^{N} K_{j}(log x)^{1/2−j} + O((log x)^{−1/2−N}),  (12)

∑_{n≤x} \frac{1}{P^{∗}(n)} = ∑_{j=0}^{N} K_{j}^{∗}(log x)^{1/2−j} + O((log x)^{−1/2−N}),  (13)

∑_{n≤x} \frac{1}{P̃(n)} = ∑_{j=0}^{N} K̃_{j}(log x)^{1/2−j} + O((log x)^{−1/2−N}),  (14)

∑_{n≤x} \frac{1}{P^{(e)}(n)} = K_{0}^{(e)} log x + K_{1}^{(e)} + O(x^{−1} log^{2} x),  (15)

where K_{j}, K_{j}^{∗}, K̃_{j}, K_{j}^{(e)}(j ≥ 0) are computable constants.

Notation. Throughout this paper, ε denotes a sufficiently small positive constant. ζ(s) denotes the Riemann zeta-function. For any real number t, [t] denotes the greatest integer not exceeding t, {t} = t − [t], and ψ(t) = {t} − 1/2. For any complex number z, σ_{z}(n) = ∑_{d|n} d^{z} and d_{z}(n) denotes the generalized divisor function, μ(n) denotes the Möbius function.

2 Proof of Theorem 1

Define for ℜs > 1 that

D_{H}(s) := ∑_{n=1}^{∞} H(n)n^{−s}.  (16)

Since H(n) is multiplicative, by Euler’s product we get

D_{H}(s) = ∏_{p} (1 + ∑_{α=1}^{∞} H(p^{α})p^{−αs}) .  (17)

We evaluate H(p^{α}) first for any prime p and α ≥ 1. By (1) we have

H(p^{α}) = p^{2α} × (1 + ∑_{j=1}^{α} p^{2j−1}(p − 1))^{−1}  (18)

= p^{2α} × (1 + \frac{p − 1}{p} ∑_{j=1}^{α} p^{2j})^{−1}

= p^{2α} × (1 + \frac{p^{2α+2} − p^{2}}{p(p + 1)} )^{−1}

= p^{2α} × (\frac{p^{2α+1} + 1}{p + 1})^{−1}

= \frac{p^{2α}(p + 1)}{p^{2α+1} + 1} = \frac{1 + p^{−1}}{1 + p^{−2α−1}} .

Hence we have

1 + ∑_{α=1}^{∞} H(p^{α})p^{−αs}  (19)

= 1 + ∑_{α=1}^{∞} \frac{1 + p^{−1}}{1 + p^{−2α−1}} p^{−αs}

= 1 + (1 + p^{−1}) ∑_{α=1}^{∞} p^{−αs} ∑_{m=0}^{∞} (−p^{−2α−1})^{m}

= 1 + (1 + p^{−1}) ∑_{m=0}^{∞}(−1)^{m}p^{−m} ∑_{α=1}^{∞} p^{−αs−2αm}

= 1 + (1 + p^{−1}) ∑_{m=0}^{∞}(−1)^{m}p^{−m} \frac{p^{−s−2m}}{1 − p^{−s−2m}}

= 1 + (1 + p^{−1})\frac{p^{−s}}{1 − p^{−s}} + (1 + p^{−1}) ∑_{m=1}^{∞}(−1)^{m}p^{−m} \frac{p^{−s−2m}}{1 − p^{−s−2m}}

= \frac{1 + p^{−1−s}}{1 − p^{−s}} + (1 + p^{−1}) ∑_{m=1}^{∞}(−1)^{m}p^{−m} \frac{p^{−s−2m}}{1 − p^{−s−2m}} ,

which implies that

(1 − p^{−s})(1 − p^{−s−1})(1 + ∑_{α=1}^{∞} H(p^{α})p^{−αs})  (20)

= 1 − p^{−2s−2} + (1 − p^{−s})(1 − p^{−s−1})(1 + p^{−1}) ∑_{m=1}^{∞}(−1)^{m}p^{−m} \frac{p^{−s−2m}}{1 − p^{−s−2m}}

= 1 + O(p^{−2σ−2} + (1 + p^{−σ})(1 + p^{−σ−1})p^{−σ−3}) = 1 + O(p^{−2σ−2} + p^{−σ−3} + p^{−2σ−3} + p^{−3σ−4}),

where σ = ℜs.

From (17), (20) and noting

ζ(s) = ∏_{p} (1 − p^{−s})^{−1} (ℜs > 1)  (21)

we get

D_{H}(s) = ζ(s)ζ(s + 1)G(s) (ℜs > 1),  (22)

where

G(s) = ∏_{p}(1 − p^{−s})(1 − p^{−s−1}) (1 + ∑_{α=1}^{∞} H(p^{α})p^{−αs})  (23)

such that if we expand G(s) into a Dirichlet series

G(s) = ∑_{n=1}^{∞} g(n)n^{−s},  (24)

then this Dirichlet series is absolutely convergent for ℜs > −1/2.

From (22) and (24) we have

H(n) = ∑_{d|n} σ_{−1}(d)g(n/d).  (25)

For σ_{−1}(n) we have

∑_{n≤x} σ_{−1}(n) = ∑_{nm≤x} \frac{1}{n} = ∑_{n≤x} \frac{1}{n} ∑_{m≤x/n} 1  (26)

= ∑_{n≤x} \frac{1}{n} [\frac{x}{n}] = ∑_{n≤x} \frac{1}{n} (\frac{x}{n} − \frac{1}{2} − ψ(\frac{x}{n}))

= x ∑_{n≤x} n^{−2} − \frac{1}{2} ∑_{n≤x} n^{−1} − ∑_{n≤x} \frac{1}{n}ψ(\frac{x}{n})

= \frac{π^{2}x}{6} − \frac{log x}{2} + O((log x)^{2/3}),

where in the last step we used the well-known bound(see Walfisz [7])

∑_{n≤x} \frac{1}{n}ψ(\frac{x}{n}) ≪ (log x)^{2/3}.  (27)

From (25)-(27) we have

∑_{n≤x} H(n) = ∑_{m≤x} g(m) ∑_{n≤x/m} σ_{−1}(n)  (28)

= \frac{π^{2}x}{6} ∑_{m≤x} \frac{g(m)}{m} − \frac{log x}{2} ∑_{m≤x} g(m) + \frac{1}{2} ∑_{m≤x} g(m) log m

+O (log^{2/3} x ∑_{m≤x} |g(m)|) .

Recall that the Dirichlet series ∑_{n=1}^{∞} g(n)n^{−s} is absolutely convergent for ℜs > −1/2. Hence for any U > 1 we have

∑_{U<n≤2U} |g(n)| ≪ U^{−1/2+ε}.

It follows that the infinite series ∑_{m≥1} g(m)m^{−1}, ∑_{m≥1} |g(m)|, ∑_{m≥1} g(m) log m are all convergent and that

∑_{m≤x} \frac{g(m)}{m} = ∑_{m=1}^{∞} \frac{g(m)}{m} + O(x^{−3/2+ε}),  (29)

∑_{m≤x} g(m) = ∑_{m=1}^{∞} g(m) + O(x^{−1/2+ε})

∑_{m≤x} g(m) log m ≪ 1, ∑_{m≤x} |g(m)| ≪ 1.

Now Theorem 1 follows from (28) and (29).

3 Proof of Theorem 4

In this section we shall prove Theorem 4. We only prove (12) and (15) since the proofs of (12), (13) and (14) are the same.

3.1 The generalized divisor problem

Suppose k ≥ 2 (k ∈ ℕ) is a fixed integer. The divisor function d_{k}(n) denotes the number of ways n can be written as a product of k natural-number factors. It is an important problem in the analytic number theory to study the mean value of the divisor function d_{k}(n). It is well-known that d_{k}(n) are the coefficients of the Dirichlet series

ζ^{k}(s) = ∑_{n=1}^{∞} d_{k}(n)n^{−s} (ℜs > 1).

Now suppose z is a fixed complex number and let d_{z}(n) denote the coefficients of the Dirichlet series

ζ^{z}(s) := ∑_{n=1}^{∞} d_{z}(n)n^{−s} (ℜs > 1),  (30)

where ζ^{z}(s) = e^{z log ζ(s)} such that for log s we take the main value log 1 = 0. The function d_{z}(n) is called the generalized divisor function.

Suppose A > 0 is an arbitrary but fixed real number and N ≥ 1 is an arbitrary but fixed integer. Then uniformly for |z| ≤ A we have

∑_{n≤x} d_{z}(n) = x ∑_{j=1}^{N} c_{j}(z)(log x)^{z−j} + O(x(log x)^{ℜz−N−1}),  (31)

where the functions c_{1}(z), ⋯, c_{N}(z) are regular in the region |z| ≤ A.

The above result is Theorem 14.9 of Ivić [1].

3.2 Proof of (12)

Define P_{1}(n) = n/P (n). Then by Euler’s product we have for ℜs > 1 that

∑_{n=1}^{∞} P_{1}(n)n^{−s} = ∏_{p} (1 + ∑_{α=1}^{∞} P_{1}(p^{α})p^{−αs}) .  (32)

We evaluate P_{1}(p^{α})(α ≥ 1) first. The formula (14) of [3] reads

P (p^{α}) = (α + 1)p^{α} − αp^{α−1},

which implies that

P_{1}(p^{α}) = \frac{p^{α}}{(α + 1)p^{α} − αp^{α−1}} = \frac{1}{α + 1} + O(p^{−1}).  (33)

Inserting (33) into (32) and recalling (21) we get that

∑_{n=1}^{∞} P_{1}(n)n^{−s} = ζ^{1/2}(s)G_{1}(s), ℜs > 1,  (34)

where

G_{1}(s) = ∏_{p}(1 − p^{−s})^{1/2} (1 + ∑_{α=1}^{∞} P_{1}(p^{α})p^{−αs})  (35)

such that if we expand G_{1}(s) into a Dirichlet series

G_{1}(s) = ∑_{n=1}^{∞} g_{1}(n)n^{−s},  (36)

then this Dirichlet series is absolutely convergent for ℜs > 1/2. For the function g_{1} we have the trivial estimate

∑_{m≤y} |g_{1}(m)| ≤ y^{1/2+ε}.  (37)

From (37) we get by partial summation that

∑_{m≤y} |g_{1}(m)|m^{−1} ≪ 1, ∑_{m>y} |g_{1}(m)|m^{−1} ≪ y^{−1/2+ε}  (38)

and for any fixed constant C that

∑_{m≤y} \frac{g_{1}(m) log^{C} m}{m} = ∑_{m=1}^{∞} \frac{g_{1}(m) log^{C} m}{m} + O(y^{−1/2+ε}).  (39)

Let e_{C} denote the value of the infinite series in (39).

Suppose β is a real number which is not a non-negative integer. By Taylor’s expansion we have

(1 − u)^{β} = ∑_{ℓ=0}^{N} d_{ℓ}^{(β)}u^{ℓ} + O(|u|^{N+1}), |u| ≤ 1/2,  (40)

where d_{ℓ}^{(β)} = (−1)^{l}β(β − 1) ⋯ (β − ℓ + 1)/ℓ!.

By the hyperbolic approach, (31) with A = z = 1/2, (38) and (39) with y = \sqrt{x} and (40) we get for any fixed N ≥ 1 that

∑_{n≤x} P_{1}(n) = ∑_{nm≤x} d_{1/2}(n)g_{1}(m)  (41)

= ∑_{m≤\sqrt{x}} g_{1}(m) ∑_{n≤x/m} d_{1/2}(n) + ∑_{n≤\sqrt{x}} d_{1/2}(n) ∑_{\sqrt{x}<m≤x/n} g_{1}(m)

= ∑_{m≤\sqrt{x}} g_{1}(m) ∑_{n≤x/m} d_{1/2}(n) + O(x^{3/4+ε})

= ∑_{m≤\sqrt{x}} g_{1}(m) (\frac{x}{m} ∑_{j=1}^{N} c_{j}(1/2)(log \frac{x}{m})^{1/2−j} + O(\frac{x}{m}(log x)^{−1/2−N}))

+O(x^{3/4+ε})

= x ∑_{j=1}^{N} c_{j}(1/2)(log x)^{1/2−j} ∑_{m≤\sqrt{x}} \frac{g_{1}(m)}{m} (1 − \frac{log m}{log x})^{1/2−j}

+O(x(log x)^{−1/2−N})

= x ∑_{j=1}^{N} c_{j}(1/2)(log x)^{1/2−j} ∑_{ℓ=0}^{N} d_{ℓ}^{(1/2−j)} ∑_{m≤\sqrt{x}} \frac{g_{1}(m)}{m} \frac{log^{ℓ} m}{log^{ℓ} x}

+O (x ∑_{j=1}^{N}(log x)^{1/2−j} ∑_{m≤\sqrt{x}} \frac{|g_{1}(m)|}{m} \frac{log^{N+1} m}{log^{N+1} x})

+O(x(log x)^{−1/2−N})

= x ∑_{j=1}^{N} c_{j}(1/2)(log x)^{1/2−j} ∑_{ℓ=0}^{N} d_{ℓ}^{(1/2−j)}e_{ℓ} log^{−ℓ} x

+O(x(log x)^{−1/2−N})

= x ∑_{j=1}^{N} K_{j}(log x)^{1/2−j} + O(x(log x)^{−1/2−N}),

where

K_{j} = ∑_{j_{1}≥1,ℓ≥0^{j=j_{1}+ℓ}} c_{j_{1}}(1/2)d_{ℓ}^{(1/2−j_{1})}e_{ℓ} = ∑_{ℓ=0}^{j−1} c_{j−ℓ}(1/2)d_{ℓ}^{(1/2−j+ℓ)}e_{ℓ} (1 ≤ j ≤ N).

From (41) we get (12) immediately by partial summation and some easy calculations.

3.3 Proof of (15)

Now we prove (15). Define P_{1}^{(e)}(n) = n/P^{(e)}(n). By Euler’s product we have for ℜs > 1 that

∑_{n=1}^{∞} P_{1}^{(e)}(n)n^{−s} = ∏_{p} (1 + ∑_{α=1}^{∞} P_{1}^{(e)}(p^{α})p^{−αs}) .  (42)

Suppose p is a prime. From (6) it is easy to see that

P^{(e)}(p) = p, P^{(e)}(p^{2}) = p^{2} + p, P^{(e)}(p^{3}) = p^{3} + 2p, P^{(e)}(p^{α}) = p^{α} + O(p^{α/2}) (α ≥ 4).

Hence

P_{1}^{(e)}(p) = 1, P_{1}^{(e)}(p^{2}) = \frac{1}{1 + p^{−1}} , P_{1}^{(e)}(p^{3}) = \frac{1}{1 + 2p^{−2}} ,  (43)

P_{1}^{(e)}(p^{α}) = \frac{p^{α}}{p^{α} + O(p^{α/2})} = \frac{1}{1 + O(p^{−α/2})} = 1 + O(p^{−α/2}) (α ≥ 4).

From (43) we get

1 + ∑_{α=1}^{∞} P_{1}^{(e)}(p^{α})p^{−αs}  (44)

= 1 + ∑_{α=1}^{∞} p^{−αs} + ∑_{α=1}^{∞}(P_{1}^{(e)}(p^{α}) − 1)p^{−αs}

= 1 + ∑_{α=1}^{∞} p^{−αs} + ∑_{α=2}^{∞}(P_{1}^{(e)}(p^{α}) − 1)p^{−αs}

= \frac{1}{1 − p^{−s}} + (\frac{1}{1 + p^{−1}} − 1)p^{−2s} + (\frac{1}{1 + 2p^{−2}} − 1)p^{−3s} + ∑_{α=4}^{∞}(P_{1}^{(e)}(p^{α}) − 1)p^{−αs}

= \frac{1}{1 − p^{−s}} + (\frac{1}{1 + p^{−1}} − 1)p^{−2s} + O (p^{−3σ−2} + ∑_{α=4}^{∞} p^{−α/2−ασ})

= \frac{1}{1 − p^{−s}} + (\frac{1}{1 + p^{−1}} − 1)p^{−2s} + O (p^{−3σ−2} + p^{−4σ−2})

= \frac{1}{1 − p^{−s}} − p^{−1−2s} + O (p^{−2σ−2} + p^{−3σ−2} + p^{−4σ−2})

= \frac{1}{1 − p^{−s}} − p^{−1−2s} + O (p^{−2σ−2} + p^{−4σ−2}) .

Hence we get

(1 − p^{−s})(1 + ∑_{α=1}^{∞} P_{1}^{(e)}(p^{α})p^{−αs})  (45)
= 1 − p^{−1−2s}(1 − p^{−s}) + O (p^{−2σ−2}(1 + p^{−2σ})(1 + p^{−σ}))
= 1 − p^{−1−2s} + p^{−1−3s} + O (p^{−2σ−2}(1 + p^{−3σ}))

and

(1 − p^{−s})(1 + p^{−1−2s})(1 + ∑_{α=1}^{∞} P_{1}^{(e)}(p^{α})p^{−αs})  (46)
= 1 − p^{−2−4s} + p^{−1−3s} + p^{−2−5s} + O (p^{−2σ−2}(1 + p^{−3σ})(1 + p^{−1−2σ}))
= 1 + p^{−1−3s} + O (p^{−2σ−2}(1 + p^{−3σ})(1 + p^{−1−2σ}))

and

(1 − p^{−s})(1 + p^{−1−2s})(1 − p^{−1−3s})(1 + ∑_{α=1}^{∞} P_{1}^{(e)}(p^{α})p^{−αs})  (47)
= 1 − p^{−2−6s} + O (p^{−2σ−2}(1 + p^{−3σ})(1 + p^{−1−2σ})(1 + p^{−1−3σ}))
= 1 + O (p^{−2−6σ} + p^{−2σ−2}(1 + p^{−3σ})(1 + p^{−1−2σ})(1 + p^{−1−3σ})) .

We write for σ > 1

1 + ∑_{α=1}^{∞} P_{1}^{(e)}(p^{α})p^{−αs}  (48)

= \frac{1}{1 − p^{−s}} × \frac{1}{1 + p^{−1−2s}} × \frac{1}{1 − p^{−1−3s}}

× ((1 − p^{−s}) × (1 + p^{−1−2s}) × (1 − p^{−1−3s}) × (1 + ∑_{α=1}^{∞} P_{1}^{(e)}(p^{α})p^{−αs}))

= \frac{1}{1 − p^{−s}} × \frac{1 − p^{−1−2s}}{1 − p^{−2−4s}} × \frac{1}{1 − p^{−1−3s}}

× ((1 − p^{−s}) × (1 + p^{−1−2s}) × (1 − p^{−1−3s}) × (1 + ∑_{α=1}^{∞} P_{1}^{(e)}(p^{α})p^{−αs})) .

From (48) we may write for σ > 1 that

∑_{n=1}^{∞} P_{1}^{(e)}(n)n^{−s} = \frac{ζ(s)ζ(3s + 1)}{ζ(2s + 1)}G_{P_{1}^{(e)}}(s),  (49)

where

G_{P_{1}^{(e)}}(s) = ζ(4s + 2) ∏_{p} ((1 − p^{−s})(1 + p^{−1−2s})(1 − p^{−1−3s})(1 + ∑_{α=1}^{∞} P_{1}^{(e)}(p^{α})p^{−αs})) . (50)

From (47) we see that if we write the function G_{P_{1}^{(e)}}(s) into a Dirichlet series

G_{P_{1}^{(e)}}(s) = ∑_{n=1}^{∞} g_{P_{1}^{(e)}}(n)n^{−s},  (51)

then this Derichlet series is absolutely for σ > −1/6. This fact implies that

∑_{n≤x} |g_{P_{1}^{(e)}}(n)| ≪ 1, ∑_{n≤x} \frac{g_{P_{1}^{(e)}}(n)}{n} = ∑_{n=1}^{∞} \frac{g_{P_{1}^{(e)}}(n)}{n} + O(x^{−7/6+ε}).  (52)

From (49), (51) and (52) we get

∑_{n≤x} P_{1}^{(e)}(n) = ∑_{n_{1}n_{2}^{2}n_{3}^{3}n_{4}≤x} \frac{μ(n_{2})}{n_{2}n_{3}}g_{P_{1}^{(e)}}(n_{4})  (53)

= ∑_{n_{4}≤x} g_{P_{1}^{(e)}}(n_{4}) ∑_{n_{2}^{2}n_{3}^{3}≤ \frac{x}{n_{4}}} \frac{μ(n_{2})}{n_{2}n_{3}} ∑_{n_{1}≤_{\frac{x}{n^{2}_{2}n^{3}_{3}n_{4}}}} 1

= ∑_{n_{4}≤x} g_{P_{1}^{(e)}}(n_{4}) ∑_{n_{2}^{2}n_{3}^{3}≤ \frac{x}{n_{4}}} \frac{μ(n_{2})}{n_{2}n_{3}} (\frac{x}{n_{2}^{2}n_{3}^{3}n_{4}} + O(1))

= x ∑_{n_{4}≤x} \frac{g_{P_{1}^{(e)}}(n_{4})}{n_{4}} ∑_{n_{2}^{2}n_{3}^{3}≤ \frac{x}{n_{4}}} \frac{μ(n_{2})}{n_{2}^{3}n_{3}^{4}} + O( ∑_{n_{4}≤x} |g_{P_{1}^{(e)}}(n_{4})| × log^{2} x)

= x ∑_{n_{4}≤x} \frac{g_{P_{1}^{(e)}}(n_{4})}{n_{4}} ∑_{n_{2}≤\sqrt{\frac{x}{n_{4}}}} \frac{μ(n_{2})}{n_{2}^{3}} ∑_{n_{3}≤(_{\frac{x}{n_{4}n^{2}_{2}}} )^{1/3}} \frac{1}{n_{3}^{4}} + O(log^{2} x)

= x ∑_{n_{4}≤x} \frac{g_{P_{1}^{(e)}}(n_{4})}{n_{4}} ∑_{n_{2}≤\sqrt{\frac{x}{n_{4}}}} \frac{μ(n_{2})}{n_{2}^{3}} (ζ(4) + O(\frac{n_{4}n_{2}^{2}}{x})) + O(log^{2} x)

= xζ(4) ∑_{n_{4}≤x} \frac{g_{P_{1}^{(e)}}(n_{4})}{n_{4}} ∑_{n_{2}≤\sqrt{\frac{x}{n_{4}}}} \frac{μ(n_{2})}{n_{2}^{3}} + O(log^{2} x)

= xζ(4) ∑_{n_{4}≤x} \frac{g_{P_{1}^{(e)}}(n_{4})}{n_{4}} (\frac{1}{ζ(3)} + O(\frac{n_{4}}{x})) + O(log^{2} x)

= x\frac{ζ(4)}{ζ(3)} ∑_{n_{4}≤x} \frac{g_{P_{1}^{(e)}}(n_{4})}{n_{4}} + O(log^{2} x)

= x\frac{ζ(4)}{ζ(3)} ∑_{n_{4}=1}^{∞} \frac{g_{P_{1}^{(e)}}(n_{4})}{n_{4}} + O(log^{2} x),

where we used the following easy estimates(y ≥ 2)

∑_{n≤y} n^{−4} = ζ(4) + O(y^{−3}),  (54)

∑_{n≤y} μ(n)n^{−3} = \frac{1}{ζ(3)} + O(y^{−2}),  (55)
∑_{n≤y} n^{−1} ≪ log y.  (56)

Now (15) follows from (53) by partial summation.

4 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.

References

[1] A. Ivić, The Riemann Zeta-Function, John Wiley & Sons, 1985.

[2] S. S. Pillai, On an arithmetic function, J. Annamalai Univ. 2 (1933), 243–248.

[3] L. Tóth, A survey of gcd-sum functions, J. Integer Sequences, 13 (2010), Article 10.8.1.

[4] L. Tóth, The unitary analogue of Pillai’s arithmetical function, Collect. Math. 40 (1989), 19–30.

[5] L. Tóth, On certain arithmetic functions involving exponential divisors, Annales Univ. Sci. Budapest., Sect. Comp. 24 (2004), 285–294.

[6] L. Tóth, A gcd-sum function over regular integers (mod n), J. Integer Sequences 12 (2009), Article 09.2.5.

[7] A. Walfisz, Weylsche Exponentialsummen in der neueren Zahlentheorie, Berlin, 1963.

2010 Mathematics Subject Classification: Primary 11A25; Secondary 11N37. Keywords: gcd-sum function, regular integers, harmonic mean, unitary divisor.

(Concerned with sequences \underline{A018804}, \underline{A145388} and \underline{A176345}.)

Received February 13 2011; revised versions received May 15 2011; June 28 2011; August 12 2011. Published in Journal of Integer Sequences, September 25 2011.

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