Mean Values of Generalized gcd-sum and lcm-sum Functions

Olivier Bordellès 2, Allée de la Combe La Boriette 43000 Aiguilhe France

                      borde43@wanadoo.fr

Abstract We consider a generalization of the gcd-sum function, and obtain its average order with a quasi-optimal error term. We also study the reciprocals of the gcd-sum and lcm-sum functions.

1 Introduction and notation

The so-called gcd-sum function, defined by

g (n) = ∑_{j=1}^{n} (n, j)

where (a, b) denotes the greatest common divisor of a and b, was first introduced by Broughan ([3, 4]) who studied its main properties, and showed among other things that g satisfies the convolution identity (see also the beginning of the proof of Lemma 3.1)

g = φ ∗ Id

where F ∗ G is the usual Dirichlet convolution product. By using the following alternative convolution identity

g = μ ∗ (Id · τ) ,

where μ is the Möbius function and τ is the divisor function, we were able in [1] to get the average order of g. Our result can be stated as follow. If θ is the exponent in the Dirichlet divisor problem, then the following asymptotic formula

∑_{n⩽x} g (n) = \frac{x^{2} log x}{2ζ (2)} + \frac{x^{2}}{2ζ (2)} (γ − \frac{1}{2} + log (\frac{A^{12}}{2π} )) + O^{ε} (x^{1+θ+ε})  (1)

holds for any real number ε > 0, where A ≈ 1.282 427 129 … is the Glaisher-Kinkelin constant. The inequality θ ⩾ 1/4 is well-known, and, from the work of Huxley [5] we know that θ ⩽ 131/416 ≈ 0.3149.

The aim of this paper is first to work with a function generalizing the function g and prove an asymptotic formula for its average order similarly as in (1) . In sections 5, 6 and 7 we will establish estimates for the lcm-sum function, and for reciprocals of the gcd-sum and lcm-sum functions. We begin with classical notation.

1. Multiplicative functions. The following arithmetic functions are well-known.

\begin{aligned}Id^{a} (n) & = n^{a} (a ∈ ℤ^{∗}) \\ 1 (n) & = 1 \end{aligned}

and μ, φ, σ_{k} and τ_{k} are respectively the Möbius function, the Euler totient function, the sum of kth powers of divisors function and the kth Piltz divisor function. Recall that τ_{k} can be defined by τ_{k} = \underbrace{1 ∗ ⋯ ∗ 1}_{k times} for any integer k ⩾ 1 and that τ_{2} = τ. We also have σ_{k} = ∑_{d|n} d^{k}

and σ_{0} = τ.

2. Exponent in the Dirichlet-Piltz divisor problem. For any integer k ⩾ 2, θ_{k} is defined to be the smallest positive real number such that the asymptotic formula

∑_{n⩽x} τ^{k} (n) = xP^{k−1} (log x) + O^{ε,k} (x^{θ_{k}+ε})  (2)

holds for any real number ε > 0. Here P_{k−1} is a polynomial of degree k − 1 with real coefficients, the leading coefficient being \frac{1}{(k−1)!}. It is now well-known that \frac{1}{3} ⩽ θ_{3} ⩽ \frac{43}{96} and that \frac{k−1}{2k} ⩽ θ_{k} ⩽ \frac{k−1}{k+2} for k ⩾ 4 (see [6], for example).

By convention, we set

τ_{0} (n) = \begin{cases}1, & if n = 1; \\ 0, & otherwise; \end{cases}

and θ_{1} = 0.

2 A generalization of the gcd-sum function

Definition 2.1. We define the sequence of arithmetic functions f_{k,j} (n) in the following way.

(i) For any integers j, n ⩾ 1, we set

f_{1,j} (n) = \begin{cases}1, & if (n, j) = 1; \\ 0, & otherwise; \end{cases}

f_{2,j} (n) = \begin{cases}(n, j), & if j ⩽ n; \\ 0, & otherwise. \end{cases}

(ii) For any integers j ⩾ 1 and k ⩾ 3, we set

f_{k,j} = f_{2,j} ∗ (Id · τ_{k−2}) .

Definition 2.2. For any integers n, k ⩾ 1, we define the sequence of arithmetic functions g_{k} (n) by

g_{k} (n) = ∑_{j=1}^{n} f_{k,j} (n) .

Examples.

g_{1} (n) = _{∑_{j=1}^{n} ; (n,j)=1}1 = φ (n) ,

g_{2} (n) = ∑_{j=1}^{n} (j, n) = g (n) ,

g_{3} (n) = ∑_{j=1}^{n}_{∑_{d|n} ; d⩾j} \frac{n}{d} (j, d) ,

⋮
g_{k} (n) = ∑_{j=1}^{n} ∑_{d_{k−2}|n} ∑_{d_{k−3}|d_{k−2}} ⋯_{∑_{d_{1}|d_{2}} ; d_{1}⩾j} \frac{n}{d_{1}} (j, d_{1}) .

Now we are able to state the following result.

Theorem 2.3. Let ε > 0 be any real number and k ⩾ 1 any integer. Then, for any real number x ⩾ 1 sufficiently large, we have

∑_{n⩽x} g_{k} (n) = \frac{x^{2}}{2ζ (2)}R_{k−1} (log x) + O_{ε,k} (x^{1+θ_{k}+ε})

where R_{k−1} is a polynomial of degree k − 1 and leading coefficient \frac{1}{(k−1)!}. The following table gives R_{k−1} for k ∈ {1, 2, 3}

\begin{matrix}k & 1 & 2 & 3 \\ R_{k−1} & 1 & X + γ − \frac{1}{2} + log (\frac{A^{12}}{2π} ) & \frac{X^{2}}{2} + αX + β \end{matrix}

where

α = 2γ − \frac{1}{2} + log (\frac{A^{12}}{2π} )
β = −\frac{ζ′′ (2)}{2ζ (2)} + (γ − log (\frac{A^{12}}{2π} ))^{2} − (3γ − \frac{1}{2}) (γ − log (\frac{A^{12}}{2π} ))

− \frac{1}{4} (12γ_{1} − 12γ^{2} + 6γ − 1)

and

Constant  Name

γ ≈ 0.577 215 664 … Euler − Mascheroni

γ_{1} ≈ −0.072 815 845 … Stieltjes

A ≈ 1.282 427 129 … Glaisher − Kinkelin

3 Main properties of the function g_{k}

The following lemma lists the main tools used in the proof of Theorem 2.3.

Lemma 3.1. For any integer k ⩾ 1, we have

g_{k+1} = g_{k} ∗ Id

and then

g_{k} = φ ∗ (Id · τ_{k−1}) .

Moreover, we have

g_{k} = μ ∗ (Id · τ_{k}) .  (3)

Thus, the Dirichlet series G_{k} (s) of g_{k} is absolutely convergent in the half-plane ℜs > 2, and has an analytic continuation to a meromorphic function defined on the whole complex plane with value

G_{k} (s) = \frac{ζ (s − 1)^{k}}{ζ (s)} .

Proof. Broughan already proved the first relation for k = 1 (see [3, Thm. 4.7]), but, for the sake of completeness, we give here another proof.

(g_{1} ∗ Id) (n) = (φ ∗ Id) (n) = ∑_{d|n} dφ (\frac{n}{d})

= ∑_{d|n} d_{∑_{k⩽n/d} ; (k,n/d)=1}1 = ∑_{d|n} d_{n ; ∑_{j=1} ; (j,n)=d}1

= ∑_{j=1}^{n} (j, n) = ∑_{j=1}^{n} f_{2,j} (n) = g_{2} (n) .

For k = 2, we get

(g_{2} ∗ Id) (n) = ∑_{d|n} \frac{n}{d} ∑_{j=1}^{d} (j, d) = ∑_{j=1}^{n}_{∑_{d|n} ; d⩾j} \frac{n}{d} (j, d) = ∑_{j=1}^{n} f_{3,j} (n) = g_{3} (n) .

Now let us suppose k ⩾ 3. We have

g_{k+1} (n) = ∑_{j=1}^{n} f_{k+1,j} (n) = ∑_{j=1}^{n} (f_{2,j} ∗ (Id · τ_{k−1})) (n)

= ∑_{j=1}^{n} (f_{2,j} ∗ Id · τ_{k−2} ∗ Id) (n)

= ∑_{d|n} \frac{n}{d} ∑_{j=1}^{d} (f_{2,j} ∗ (Id · τ_{k−2})) (d) = (g_{k} ∗ Id) (n) .

The second relation is easily shown by induction. For the third, we have using φ = μ ∗ Id

g_{k} = φ ∗ (Id · τ_{k−1}) = μ ∗ (Id ∗ (Id · τ_{k−1}))
= μ ∗ (Id · (1 ∗ τ_{k−1})) = μ ∗ (Id · τ_{k}) .

The last proposition comes from the equality (3)

g_{k} = μ ∗ (Id · τ_{k}) = μ ∗ \underbrace{Id ∗ ⋯ ∗ Id}_{k times}

and the Dirichlet series of μ and Id.

4 Proof of Theorem 1

Lemma 4.1. For any integer k ⩾ 1 and any real numbers x > 1 and ε > 0, we have

∑_{n⩽x} nτ_{k} (n) = x^{2}Q_{k−1} (log x) + O_{ε,k} (x^{1+θ_{k}+ε})

where Q_{k−1} is a polynomial of degree k − 1 and leading coefficient \frac{1}{2(k−1)!}.

Proof. Using summation by parts and (2), we get

\begin{aligned}∑_{n⩽x} nτ_{k} (n) & = x ∑_{n⩽x} τ_{k} (n) − ∫_{1}^{x} (∑_{n⩽t} τ_{k} (n)) dt \\ & = x^{2}P_{k−1} (log x) + O_{ε,k} (x^{1+θ_{k}+ε}) − ∫_{1}^{x} (tP_{k−1} (log t) + O_{ε,k} (t^{θ_{k}+ε})) dt. \end{aligned}

Writing

P_{k−1} (X) = ∑_{j=0}^{k−1} a_{j}X^{j}

with a_{k−1} = \frac{1}{(k−1)!}, we obtain

∑_{n⩽x} nτ_{k} (n) = x^{2} ∑_{j=0}^{k−1} a_{j} (log x)^{j} − ∑_{j=0}^{k−1} a_{j} ∫_{1}^{x} t (log t)^{j} dt + O_{ε,k} (x^{1+θ_{k}+ε})

and the formula

∫_{1}^{x} t (log t)^{j} dt = x^{2} ∑_{i=0}^{j} (−1)^{j−i} \frac{j!}{2^{j+1−i} × i!} (log x)^{i} − (−1)^{j} \frac{j!}{2^{j+1}}

(easily proved by induction) gives

∑_{n⩽x} nτ_{k} (n) = x^{2} ∑_{j=0}^{k−1} a_{j} \{(log x)^{j} − ∑_{i=0}^{j} (−1)^{j−i} \frac{j!}{2^{j+1−i} × i!} (log x)^{i}\}

+ ∑_{i=0}^{j} (−1)^{j} \frac{j! a_{j}}{2^{j+1}} + O_{ε,k} (x^{1+θ_{k}+ε})

= x^{2} ∑_{j=0}^{k−1} a_{j} \{\frac{(log x)}{2}^{j} − ∑_{i=0}^{j−1} (−1)^{j−i} \frac{j!}{2^{j+1−i} × i!} (log x)^{i}\} + O_{ε,k} (x^{1+θ_{k}+ε})

which completes the proof of the lemma. Remark. For k = 3, the following result is well-known (see [7, Exer. II.3.4], for example)

∑_{n⩽x} τ_{3} (n) = x \{\frac{(log x)^{2}}{2} + (3γ − 1) log x + 3γ^{2} − 3γ − 3γ_{1} + 1\} + O_{ε} (x^{θ_{3}+ε})

and gives

\begin{matrix}∑_{n⩽x} nτ_{3} (n) = x^{2} \{\frac{(log x)^{2}}{4} + (\frac{6γ − 1}{4}) log x − \frac{12γ_{1} − 12γ^{2} + 6γ − 1}{8}\} + O_{ε} (x^{1+θ_{3}+ε}) . \\ (4) \end{matrix}

Now we are able to prove Theorem 2.3.

Using (3) we get

∑_{n⩽x} g_{k} (n) = ∑_{d⩽x} μ (d) ∑_{m⩽x/d} mτ_{k} (m) ,

and lemma 3.1 gives

\begin{aligned}∑_{n⩽x} g_{k} (n) & = ∑_{d⩽x} μ (d) \{(\frac{x}{d})^{2} Q_{k−1} (log \frac{x}{d}) + O_{ε,k} ((\frac{x}{d})^{1+θ_{k}+ε})\} \\ & = x^{2} ∑_{d⩽x} \frac{μ (d)}{d^{2}}Q_{k−1} (log \frac{x}{d}) + O_{ε,k} (x^{1+θ_{k}+ε}) . \end{aligned}

Writing

Q_{k−1} (X) = ∑_{j=0}^{k−1} b_{j}X^{j}

with b_{k−1} = \frac{1}{2(k−1)!}, we get

∑_{n⩽x} g_{k} (n) = x^{2} ∑_{d⩽x} \frac{μ (d)}{d^{2}} ∑_{j=0}^{k−1} b_{j} (log \frac{x}{d})^{j} + O_{ε,k} (x^{1+θ_{k}+ε})

= x^{2} ∑_{j=0}^{k−1} ∑_{h=0}^{j} \binom{j}{h}b_{j} (log x)^{j−h} ∑_{d⩽x} (−1)^{h} \frac{μ (d)}{d^{2}} (log d)^{h} + O_{ε,k} (x^{1+θ_{k}+ε})

and the equality

∑_{d⩽x} (−1)^{h} \frac{μ (d)}{d^{2}} (log d)^{h} = ∑_{d=1}^{∞} (−1)^{h} \frac{μ (d)}{d^{2}} (log d)^{h} − ∑_{d>x} (−1)^{h} \frac{μ (d)}{d^{2}} (log d)^{h}

= [\frac{d^{h}}{ds^{h}} (\frac{1}{ζ (s)})]_{[s=2]} + O (\frac{(log x)^{h}}{x} ) ,

implies

∑_{n⩽x} g_{k} (n) = x^{2} ∑_{j=0}^{k−1} ∑_{h=0}^{j} \binom{j}{h} ([\frac{d^{h}}{ds^{h}} (\frac{1}{ζ (s)})]_{[s=2]}) b_{j} (log x)^{j−h}

+ O (x (log x)^{k−1}) + O_{ε,k} (x^{1+θ_{k}+ε})

= x^{2} ∑_{j=0}^{k−1} ∑_{h=0}^{j} \binom{j}{h} ([\frac{d^{h}}{ds^{h}} (\frac{1}{ζ (s)})]_{[s=2]}) b_{j} (log x)^{j−h} + O_{ε,k} (x^{1+θ_{k}+ε}) ,

and writing

[\frac{d^{h}}{ds^{h}} (\frac{1}{ζ (s)})]_{[s=2]} = \frac{A_{h}}{2ζ (2)^{h+1}}

with A_{h} ∈ ℝ (and A_{0} = 2), we obtain

∑_{n⩽x} g_{k} (n) = \frac{x^{2}}{2ζ (2)} ∑_{j=0}^{k−1} ∑_{h=0}^{j} \binom{j}{h}\frac{A_{h}b_{j} (log x)^{j−h}}{ζ (2)^{h}} + O_{ε,k} (x^{1+θ_{k}+ε})

which is the desired result. The leading coefficient is \binom{k−1}{0}A_{0}b_{k−1} = \frac{1}{(k−1)!}. The particular cases are easy to check.

(i) For k = 1, the result is well-known (see [2, Exer. 4.14])

∑_{n⩽x} g_{1} (n) = ∑_{n⩽x} φ (n) = \frac{x^{2}}{2ζ (2)} + O (x log x) .

(ii) For k = 2, see [1].

(iii) For k = 3, we use (4) and the computations made above. The proof of Theorem 2.3 is now complete.

5 Sums of reciprocals of the gcd

The purpose of this section is to prove the following estimate.

Theorem 5.1. For any real number x > e sufficiently large, we have

∑_{n⩽x} (∑_{j=1}^{n} \frac{1}{(j, n)}) = \frac{ζ (3)}{2ζ (2)} x^{2} + O (x (log x)^{2/3} (log log x)^{4/3}) .

Proof. For any integer n ⩾ 1, we set

G (n) = ∑_{j=1}^{n} \frac{1}{(j, n)}.

With a similar argument used in the proof of the identity g = φ ∗ Id (see lemma 3.1), it is easy to check that

G = φ ∗ Id^{−1},

and thus

∑_{n⩽x} G (n) = ∑_{d⩽x} \frac{1}{d} ∑_{m⩽x/d} φ (m) .

The well-known result (see [8], for example)

∑_{n⩽x} φ (n) = \frac{x^{2}}{2ζ (2)} + O (x (log x)^{2/3} (log log x)^{4/3}) ,

combined with some classical computations, allows us to conclude the proof of Theorem 5.1.

6 The lcm-sum function

Definition 6.1. For any integer n ⩾ 1, we define

l (n) = ∑_{j=1}^{n} [n, j]

where [a, b] is the least common multiple of a and b. Lemma 6.2. We have the following convolution identity

l = \frac{1}{2} ((Id^{2} · (φ + τ_{0})) ∗ Id) .

Proof. We have

∑_{j=1}^{n} \frac{j}{(n, j)} = ∑_{d|n} \frac{1}{d}_{∑_{j=1}^{n} ; (n,j)=d}j = ∑_{d|n} \frac{1}{d}_{∑_{k⩽n/d} ; (k,n/d)=1}kd = ∑_{d|n}_{∑_{k⩽n/d} ; (k,n/d)=1}k,

with

∑_{k⩽N} ; (k,N)=1k = ∑_{d|N} dμ (d) ∑_{m⩽N/d} m

= \frac{1}{2} ∑_{d|N} dμ (d) \{\frac{N}{d} (\frac{N}{d} + 1)\}

= \frac{N}{2} ∑_{d|N} μ (d) (\frac{N}{d} + 1) = \frac{N}{2} (φ + τ_{0}) (N) ,

and hence

∑_{j=1}^{n} \frac{j}{(n, j)} = \frac{1}{2} ∑_{d|n} \frac{n}{d} (φ + τ_{0}) (\frac{n}{d}) = \frac{1}{2} ((Id · (φ + τ_{0})) ∗ 1) (n) ,

and we conclude by noting that

l (n) = n ∑_{j=1}^{n} \frac{j}{(n, j)}

which completes the proof, since Id is completely multiplicative. Theorem 6.3. For any real number x > e sufficiently large, we have the following estimate

∑_{n⩽x} (∑_{j=1}^{n} [n, j]) = \frac{ζ (3)}{8ζ (2)} x^{4} + O (x^{3} (log x)^{2/3} (log log x)^{4/3}) .

Proof. Using lemma 6.2, we get

\begin{aligned}∑_{n⩽x} l (n) & = \frac{1}{2} ∑_{d⩽x} d ∑_{m⩽x/d} m^{2} (φ + τ_{0}) (m) \\ & = \frac{1}{2} ∑_{d⩽x} d ∑_{m⩽x/d} m^{2}φ (m) + O (x^{2}) \end{aligned}

and the estimation (see [8])

∑_{n⩽x} n^{2}φ (n) = \frac{x^{4}}{4ζ (2)} + O (x^{3} (log x)^{2/3} (log log x)^{4/3})

implies

∑_{n⩽x} l (n) = \frac{1}{2} ∑_{d⩽x} d \{\frac{1}{4ζ (2)} (\frac{x}{d})^{4} + O ((\frac{x}{d})^{3} (log x)^{2/3} (log log x)^{4/3})\} + O (x^{2})

= \frac{x^{4}}{8ζ (2)} ∑_{d=1}^{∞} \frac{1}{d^{3}} + O (x^{3} (log x)^{2/3} (log log x)^{4/3}) + O (x^{2}) ,

which is the desired result.

7 Sum of reciprocals of the lcm

We will prove the following result. Theorem 7.1. For any real number x > 1 sufficiently large, we have

∑_{n⩽x} (∑_{j=1}^{n} \frac{1}{[n, j]}) = \frac{(log x)^{3}}{6ζ (2)} + \frac{(log x)^{2}}{2ζ (2)} (γ + log (\frac{A^{12}}{2π} )) + O (log x) .

Some useful estimates are needed. Lemma 7.2. Set C_{φ} = log (\frac{A^{12}}{2π} ) ≈ 1.147 176 … For any real number x ⩾ 1, we have

(i) : ∑_{n⩽x} \frac{φ (n)}{n^{2}} = \frac{log x}{ζ (2)} + \frac{C_{φ}}{ζ (2)} + O (\frac{log ex}{x}) .
(ii) : ∑_{n⩽x} \frac{φ (n)}{n^{2}} log (\frac{x}{n}) = \frac{(log x)^{2}}{2ζ (2)} + \frac{C_{φ} log x}{ζ (2)} + O (1) .
(iii) : \frac{1}{2} ∑_{n⩽x} \frac{φ (n)}{n^{2}} (log (\frac{x}{n}))^{2} = \frac{(log x)^{3}}{6ζ (2)} + \frac{C_{φ} (log x)^{2}}{2ζ (2)} + O (log x) .

Proof. (i) . Using φ = μ∗ Id, we get

∑_{n⩽x} \frac{φ (n)}{n^{2}} = ∑_{d⩽x} \frac{μ (d)}{d^{2}} ∑_{m⩽x/d} \frac{1}{m}

= ∑_{d⩽x} \frac{μ (d)}{d^{2}} \{log (\frac{x}{d}) + γ + O (\frac{d}{x})\}

= (log x + γ) ∑_{d⩽x} \frac{μ (d)}{d^{2}} − ∑_{d⩽x} \frac{μ (d) log d}{d^{2}} + O (\frac{1}{x} ∑_{d⩽x} \frac{1}{d})

= \frac{log x}{ζ (2)} + \frac{γ}{ζ (2)} − \frac{ζ′ (2)}{(ζ (2))^{2}} + O (\frac{log ex}{x}) .

Recall that \frac{ζ′ (2)}{ζ (2)} = γ − C_{φ}.

(ii) and (iii) . Abel’s summation and estimate (i) . We leave the details to the reader.

Now we are able to show Theorem 7.1. For any integer n ⩾ 1, we set

L (n) = ∑_{j=1}^{n} \frac{1}{[n, j]}.

Since

\begin{aligned}L (n) & = \frac{1}{n} ∑_{j=1}^{n} \frac{(n, j)}{j} = \frac{1}{n} ∑_{d|n} d_{∑_{j=1}^{n} ; (j,n)=d} \frac{1}{j} \\ & = \frac{1}{n} ∑_{d|n} d_{∑_{k⩽n/d} ; (k,n/d)=1} \frac{1}{kd} = \frac{1}{n} ∑_{d|n}_{∑_{k⩽n/d} ; (k,n/d)=1} \frac{1}{k}, \end{aligned}

we get

∑_{n⩽x} L (n) = ∑_{n⩽x} \frac{1}{n} ∑_{d|n}_{∑_{k⩽n/d} ; (k,n/d)=1} \frac{1}{k}

= ∑_{d⩽x} \frac{1}{d} ∑_{h⩽x/d} \frac{1}{h}_{∑_{k⩽h} ; (k,h)=1} \frac{1}{k}

= ∑_{d⩽x} \frac{1}{d} ∑_{h⩽x/d} \frac{1}{h} ∑_{δ|h} \frac{μ (δ)}{δ} ∑_{m⩽h/δ} \frac{1}{m}

= ∑_{d⩽x} \frac{1}{d} ∑_{δ⩽x/d} \frac{μ (δ)}{δ^{2}} ∑_{a⩽x/(dδ)} \frac{1}{a} ∑_{m⩽a} \frac{1}{m}

= ∑_{d⩽x} ∑_{δd⩽x} \frac{1}{d} \frac{μ (δ)}{δ^{2}} ∑_{a⩽x/(dδ)} \frac{1}{a} ∑_{m⩽a} \frac{1}{m}

= ∑_{n⩽x} \frac{1}{n^{2}} ∑_{d|n} dμ (\frac{n}{d}) ∑_{a⩽x/n} \frac{1}{a} ∑_{m⩽a} \frac{1}{m},

and the convolution identity φ = μ ∗ Id implies that

∑_{n⩽x} L (n) = ∑_{n⩽x} \frac{φ (n)}{n^{2}} ∑_{a⩽x/n} \frac{1}{a} ∑_{m⩽a} \frac{1}{m}.

Thus

∑_{n⩽x} L (n) = ∑_{n⩽x} \frac{φ (n)}{n^{2}} ∑_{a⩽x/n} \frac{1}{a} \{log a + γ + O (\frac{1}{a})\}

= ∑_{n⩽x} \frac{φ (n)}{n^{2}} \{\frac{1}{2} (log \frac{x}{n})^{2} + γ (log \frac{x}{n}) + O (1)\}

= \frac{(log x)^{3}}{6ζ (2)} + \frac{C_{φ} (log x)^{2}}{2ζ (2)} + \frac{γ (log x)^{2}}{2ζ (2)} + O (log x)

where C_{φ} = log (\frac{A^{12}}{2π} ) , which concludes the proof.

8 Acknowledgment

I am indebted to the referee for his valuable suggestions.

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[7] G. Tenenbaum, Introduction à la Théorie Analytique et Probabiliste des Nombres, Société Mathématique de France, 1995.

[8] A. Walfisz, Weylsche Exponentialsummen in der neueren Zahlentheorie, Leipzig BG Teubner, 1963.

2000 Mathematics Subject Classification: Primary 11A25; Secondary 11N37. Keywords: greatest common divisor, average order, Dirichlet convolution.

Received July 2 2007; revised version received August 27 2007. Published in Journal of Integer Sequences, August 27 2007.

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