On the Gcd-Sum Function

Yoshio Tanigawa Graduate School of Mathematics

Nagoya University Chikusa-ku, Nagoya 464-8602

Japan

tanigawa@math.nagoya-u.ac.jp

Wenguang Zhai School of Mathematical Sciences Shandong Normal University

Jinan 250014, Shandong

P. R. China

zhaiwg@hotmail.com

Abstract

In this paper we give a unified asymptotic formula for the partial gcd-sum function. We also study the mean-square of the error in the asymptotic formula.

1 Introduction

Pillai [6] first defined the gcd-sum (Pillai’s function) by the relation

g(n) := ∑_{j=1}^{n} gcd(j, n),  (1)

The second-named author is supported by National Natural Science Foundation of China (Grant No.

10771127) and National Natural Science Foundation of Shandong Province(Grant No. Y2006A31).

where gcd(a, b) denotes the greatest common divisor of a and b. This is Sequence \underline{A018804} in Sloane’s Online Encyclopedia of Integer Sequences. Pillai [6] proved that

g(n) = n ∑_{d|n} \frac{φ(n)}{d},

where φ(n) is Euler’s function. This fact was proved again by Broughan [2]. He also obtained the asymptotic formula of the partial sum function

G_{α}(x) := ∑_{n≤x} g(n)n^{−α}

for any α ∈ ℝ, which was further improved in Bordellès [1] (the case α = 0) and Broughan [3] (the general case) respectively.

The estimate of G_{α}(x) is closely related to the well-known Dirichlet divisor problem. For any x > 0, define

Δ(x) := ∑_{n≤x} d(n) − x(log x + 2γ − 1),

where d(n) denotes the Dirichlet divisor function and γ is Euler’s constant. Dirichlet first proved that Δ(x) = O(x^{1/2}). The exponent 1/2 was improved by many authors. The latest result reads

Δ(x) ≪ x^{131/416}(log x)^{26947/8320},  (2)

due to Huxley [4]. It is conjectured that

Δ(x) = O(x^{1/4+ε}),  (3)

which is supported by the classical mean square result

∫_{1}^{T} Δ^{2}(x)dx = \frac{ζ^{4}(3/2)}{6π^{2}ζ(3)}T^{3/2} + O(T log^{5} T)  (4)

proved by Tong [7].

In the sequel of this paper, θ denotes the number defined by

θ := inf{a | Δ(x) ≪ x^{a}}.  (5)

Broughan [3] proved that

(1) If α ≤ 1 + θ, then

G_{α}(x) =\frac{x^{2−α} log x}{(2 − α)ζ(2)} + \frac{x^{2−α}}{(2 − α)ζ(2)} (2γ − \frac{1}{2 − α} − \frac{ζ′(2)}{ζ(2)})
+ O(x^{θ+1−α+ε});  (6)

(2) If 1 + θ < α < 2, then

G_{α}(x) =\frac{x^{2−α} log x}{(2 − α)ζ(2)} + \frac{x^{2−α}}{(2 − α)ζ(2)} (2γ − \frac{1}{2 − α} − \frac{ζ′(2)}{ζ(2)})
+ O(1);  (7)

(3) If α = 2, then

G_{α}(x) =\frac{log^{2} x}{2ζ(2)} + \frac{log x}{ζ(2)} (2γ − \frac{ζ′(2)}{ζ(2)}) + O(1);  (8)

(4) If α > 2, then

G_{α}(x) =\frac{x^{2−α} log x}{(2 − α)ζ(2)} + \frac{x^{2−α}}{(2 − α)ζ(2)} (2γ − \frac{1}{2 − α} − \frac{ζ′(2)}{ζ(2)})

+ \frac{ζ^{2}(α − 1)}{ζ(α)} + O(x^{θ+1−α+ε}).  (9)

In this paper we first give a unified asymptotic formula of G_{α}(x). Before stating our result, we introduce the following definitions:

M_{α}(x) := \{ \begin{matrix}\frac{x^{2−α} log x}{(2 − α)ζ(2)} + \frac{x^{2−α}}{(2 − α)ζ(2)} (2γ − \frac{1}{2 − α} − \frac{ζ′(2)}{ζ(2)}) , & if α ≠ 2; \\ \frac{log^{2} x}{2ζ(2)} − (\frac{ζ′(2)}{ζ^{2}(2)} + \frac{2γ}{ζ(2)}) log x, & if α = 2, \end{matrix}  (10)

and

c(β) := \{ \begin{matrix}0, & if β ≤ 0; \\ β ∫_{0}^{∞} Δ(x)x^{−β−1}dx, & if 0 < β < 1; \\ 2γ − 1 + ∫_{1}^{∞} Δ(x)x^{−2}dx, & if β = 1; \\ ζ^{2}(β), & if β > 1. \end{matrix}  (11)

We then have

Theorem 1. Suppose α ∈ ℝ is fixed. Then

G_{α}(x) = M_{α}(x) + C(α) + O(x^{θ+1−α+ε}),  (12)

where

C(α) := \{ \begin{matrix}0, & if α ≤ 1; \\ c(α − 1)/ζ(α), & if 1 < α < 2; \\ \frac{c(1)}{ζ(2)} + \frac{2ζ^{′2}(2) − ζ(2)ζ′′(2)}{2ζ^{3}(2)} − \frac{2γζ′(2)}{ζ^{2}(2)}, & if α = 2; \\ \frac{ζ^{2}(α − 1)}{ζ(α)}, & if α > 2. \end{matrix}

Remark 1. Theorem 1 slightly improves Broughan’s result in the case 1 + θ < α ≤ 2.

Define the error term E_{α}(x) by

E_{α}(x) := G_{α}(x) − M_{α}(x) − C(α).

For this error term, we have the following mean square results.

Theorem 2. For any fixed α ∈ ℝ, we have the asymptotic formula

∫_{1}^{T} x^{2α−2}E_{α}^{2}(x)dx = C_{2}T^{3/2} + O(T^{5/4+ε}),  (13)

where

C_{2} = \frac{1}{6π^{2}} ∑_{n=1}^{∞} h_{0}^{2}(n)n^{−3/2}, h_{0}(n) = ∑_{n=ml} μ(m)d(l)m^{−1/2}.

Corollary 1. If α < 7/4, then

∫_{1}^{T} E_{α}^{2}(x)dx = \frac{3C_{2}}{7 − 4α}T^{7/2−2α} + O(T^{13/4−2α+ε}).  (14)

For the upper bound of E_{α}(x), we propose the following conjecture. Conjecture. The estimate

E_{α}(x) ≪ x^{5/4−α+ε}

holds for any α ∈ ℝ.

Throughout this paper, ε denotes an arbitrary small positive number which does not need to be the same at each occurrence. When the summation conditions of a sum are complicated, we write the conditions separately like SC(Σ).

2 Proof of Theorem 1

In this section we prove Theorem 1. First we prove the following

Lemma 2.1. Suppose β ∈ ℝ \ {0} and define D_{β}(x) := ∑_{n≤x} d(n)n^{−β}. Then we have

D_{β}(x) = \frac{x^{1−β} log x}{1 − β} + \frac{x^{1−β}}{1 − β} (2γ − \frac{1}{1 − β} )
+ c(β) + x^{−β}Δ(x) + O(x^{−β})  for β ≠ 1  (15)

and

D_{1}(x) = \frac{log^{2} x}{2} + 2γ log x + c(1) + Δ(x)x^{−1} + O(x^{−1}),  (16)

where c(β) is defined in (1.10).

Proof. First consider the case β < 1. By integration by parts we have

D_{β}(x) = ∑_{0<n≤x} d(n)n^{−β} = ∫_{0}^{x} t^{−β}dD(t)

= ∫_{0}^{x} t^{−β}dH(t) + ∫_{0}^{x} t^{−β}dΔ(t)

= ∫_{0}^{x} t^{−β}(log t + 2γ)dt + t^{−β}Δ(t)|_{0}^{x} + β ∫_{0}^{x} Δ(t)t^{−β−1}dt

= \frac{x^{1−β} log x}{1 − β} + \frac{x^{1−β}}{1 − β} (2γ − \frac{1}{1 − β} ) + x^{−β}Δ(x)

+ β ∫_{0}^{x} Δ(t)t^{−β−1}dt,  (17)

where

D(t) = ∑_{n≤t} d(n)

and

H(t) = t log t + (2γ − 1)t.

Note that, from the definition of Δ(x),

Δ(x) = −x log x − (2γ − 1)x

for 0 < x < 1, which implies that the integral ∫_{0}^{1} Δ(t)t^{−β−1}dt is convergent. To treat the last integral in (2.3) we recall the well-known formula (see Voronoï [8])

∫_{0}^{T} Δ(x)dx = T/4 + O(T^{3/4}),  (18)

which combined with integration by parts gives

β ∫_{0}^{x} Δ(t)t^{−β−1}dt ≪ x^{−β}  (β < 0)  (19)

and

β ∫_{x}^{∞} Δ(t)t^{−β−1}dt ≪ x^{−β}  (β > 0).  (20)

Especially, (2.6) shows that the infinite integral ∫_{0}^{∞} Δ(t)t^{−β−1}dt converges in the case β > 0. The assertion of Lemma 2.1 for the case β < 1 follows from (2.3), (2.5) and (2.6).

Next consider the case β > 1. Since the infinite series ∑_{n=1}^{∞} d(n)n^{−β} converges to ζ^{2}(β), we may write

D_{β}(x) = ζ^{2}(β) − ∑_{n>x} d(n)n^{−β}.  (21)

By integration by parts and (2.6) we get

∑_{n>x} d(n)n^{−β} = ∫_{x}^{∞} t^{−β}dH(t) + ∫_{x}^{∞} t^{−β}dΔ(t)
= ∫_{x}^{∞} t^{−β}(log t + 2γ)dt + t^{−β}Δ(t)|_{x}^{∞} + β ∫_{x}^{∞} Δ(t)t^{−β−1}dt
= −\frac{x^{1−β} log x}{1 − β} − \frac{x^{1−β}}{1 − β} (2γ − \frac{1}{1 − β} ) − x^{−β}Δ(x) + O(x^{−β}).  (22)

The assertion of Lemma 2.1 for the case β > 1 follows from (2.7) and (2.8).

Finally consider the case β = 1. We have

D_{1}(x) = 1 + ∑_{1<n≤x} d(n)n^{−1}

= 1 + ∫_{1}^{x} t^{−1}dH(t) + ∫_{1}^{x} t^{−1}dΔ(t)
= 1 + ∫_{1}^{x} t^{−1}(log t + 2γ)dt + Δ(t)t^{−1}|_{1}^{x} + ∫_{1}^{x} Δ(t)t^{−2}dt
= \frac{log^{2} x}{2} + 2γ log x + 1 + Δ(x)x^{−1} − Δ(1) + ∫_{1}^{∞} Δ(t)t^{−2}dt + O(x^{−1})
= \frac{log^{2} x}{2} + 2γ log x + 2γ − 1 + ∫_{1}^{∞} Δ(t)t^{−2}dt + Δ(x)x^{−1} + O(x^{−1}),  (23)

where we used (2.6) and the fact Δ(1) = 2 − 2γ which follows from the definition of Δ(x). This completes the proof of Lemma 2.1.

Proof of Theorem 1. Broughan [3] proved that

G_{α}(x) = ∑_{m≤x} μ(m)m^{−α} ∑_{n≤x/m} d(n)n^{1−α}.  (24)

From (2.10), Lemma 2.1 and some easy calculations, we get

G_{α}(x) = M_{α}(x) + C(α) + E_{α}(x),  (25)

where

E_{α}(x) = x^{1−α} ∑_{m≤x} \frac{μ(m)}{m}Δ(\frac{x}{m}) + O(x^{1−α} log x),  (26)

which is O(x^{θ+1−α+ε}), completing the proof of Theorem 1.

Remark 2. Voronoï [8] actually proved

∫_{0}^{T} Δ(x)dx = \frac{T}{4} + \frac{T^{3/4}}{2\sqrt{2}π^{2}} ∑_{n=1}^{∞} \frac{d(n)}{n^{5/4}} sin(4π\sqrt{nT} − π/4) + O(T^{1/4}).  (27)

Replacing the formula (2.4) in the proof of Lemma 2.1 it is easy to check that log x in the error term of (2.12) can be removed.

3 Proof of Theorem 2

In order to prove Theorem 2 we need the following well-known Voronoï formula (see, e.g., Ivić [5]).

Lemma 3.1. Suppose A > 0 is any fixed constant. If 1 ≪ N ≪ x^{A}, then

Δ(x) = \frac{x^{1/4}}{π\sqrt{2}} ∑_{n≤N} \frac{d(n)}{n^{3/4}} cos (4π\sqrt{nx} − \frac{π}{4} ) + O(x^{ε} + x^{1/2+ε}N^{−1/2}).

It suffices to evaluate the integral ∫_{T}^{2T} x^{2α−2}E_{α}^{2}(x)dx. Let

y = T^{1−ε}.

By (2.12) we have

x^{α−1}E_{α}(x) = ∑_{m≤y} \frac{μ(m)}{m}Δ (\frac{x}{m}) + ∑_{y≤m≤x} \frac{μ(m)}{m}Δ (\frac{x}{m}) + O(log x)
=: F_{1}(x) + F_{2}(x) + O(log x),

say. For F_{2}(x), we have Δ(x/m) ≪ (x/y)^{1/3} ≪ T^{ε}, therefore

F_{2}(x) ≪ ∑_{y<m≤x} \frac{1}{m} |Δ (\frac{x}{m})| ≪ T^{ε}.

For F_{1}(x), we can take N = y in Lemma 3.1 with a suitable A (e.g., A = (1 + ε)/ε), hence we get

F_{1}(x) = E^{∗}(x) + O(T^{ε}),

where

E^{∗}(x) = \frac{x^{1/4}}{\sqrt{2}π} ∑_{m≤y} \frac{μ(m)}{m^{5/4}} ∑_{n≤y} \frac{d(n)}{n^{3/4}} cos (4π\sqrt{\frac{nx}{m}} − \frac{π}{4} ) .

As a result we get an expression of x^{α−1}E_{α}(x):

x^{α−1}E_{α}(x) = E^{∗}(x) + O(T^{ε}).  (28)

Now we consider the mean square of E^{∗}(x). By the elementary formula

cos u cos v = \frac{1}{2}(cos (u − v) + cos (u + v))

we may write

|E^{∗}(x)|^{2} = \frac{x^{1/2}}{2π^{2}} ∑_{m_{1},m_{2}≤y} \frac{μ(m_{1})μ(m_{2})}{(m_{1}m_{2})^{5/4}} ∑_{n_{1},n_{2}≤y} \frac{d(n_{1})d(n_{2})}{(n_{1}n_{2})^{3/4}}

× cos (4π\sqrt{\frac{n_{1}x}{m_{1}}} − \frac{π}{4} ) cos (4π\sqrt{\frac{n_{2}x}{m_{2}}} − \frac{π}{4} )

= S_{1}(x) + S_{2}(x) + S_{3}(x),  (29)

where

S_{1}(x) = \frac{x^{1/2}}{4π^{2}} ∑_{1} \frac{μ(m_{1})μ(m_{2})}{(m_{1}m_{2})^{5/4}} \frac{d(n_{1})d(n_{2})}{(n_{1}n_{2})^{3/4}},
S_{2}(x) = \frac{x^{1/2}}{4π^{2}} ∑_{2} \frac{μ(m_{1})μ(m_{2})}{(m_{1}m_{2})^{5/4}} \frac{d(n_{1})d(n_{2})}{(n_{1}n_{2})^{3/4}} cos (4π\sqrt{x} (\sqrt{\frac{n_{1}}{m_{1}}} − \sqrt{\frac{n_{2}}{m_{2}}} )),
S_{3}(x) = \frac{x^{1/2}}{4π^{2}} ∑_{3} \frac{μ(m_{1})μ(m_{2})}{(m_{1}m_{2})^{5/4}} \frac{d(n_{1})d(n_{2})}{(n_{1}n_{2})^{3/4}} sin (4π\sqrt{x} (\sqrt{\frac{n_{1}}{m_{1}}} + \sqrt{\frac{n_{2}}{m_{2}}} )),

with summation conditions

SC(Σ_{1}) : m_{1}, m_{2}, n_{1}, n_{2} ≤ y, n_{1}m_{2} = n_{2}m_{1},
SC(Σ_{2}) : m_{1}, m_{2}, n_{1}, n_{2} ≤ y, n_{1}m_{2} ≠ n_{2}m_{1},
SC(Σ_{3}) : m_{1}, m_{2}, n_{1}, n_{2} ≤ y,

respectively.

We have

∫_{T}^{2T} S_{1}(x)dx = \frac{B(T)}{4π^{2}} ∫_{T}^{2T} x^{1/2}dx,  (30)

where

B(T) = ∑_{1} \frac{μ(m_{1})μ(m_{2})}{(m_{1}m_{2})^{5/4}} \frac{d(n_{1})d(n_{2})}{(n_{1}n_{2})^{3/4}}.

We evaluate B(T). It is written as

B(T) = ∑_{1} \frac{μ(m_{1})μ(m_{2})d(n_{1})d(n_{2})(m_{1}m_{2})^{−1/2}}{(n_{1}m_{2}n_{2}m_{1})^{3/4}}

= ∑_{n≤y^{2}} h^{2}(n; y)n^{−3/2},

where

h(n; y) = ∑_{m,l≤y^{n=ml}} μ(m)d(l)m^{−1/2}.

Let

h_{0}(n) = ∑_{n=ml} μ(m)d(l)m^{−1/2}, h_{1}(n) = ∑_{n=ml} d(l)m^{−1/2}.

Obviously,

h(n; y) = h_{0}(n), n ≤ y,
|h(n; y)| ≤ h_{1}(n), |h_{0}(n)| ≤ h_{1}(n), n ≥ 1.

Since h_{1}(n) is a multiplicative function, by using Euler’s product, it is easy to show that

∑_{n=1}^{∞} h_{1}^{2}(n)n^{−s} = ζ^{4}(s)M(s), ℜs > 1,

where M(s) is regular for ℜs > 1/2. Thus

∑_{n≤x} h_{1}^{2}(n) ≪ x log^{3} x.

From the above estimates we get

B(T) = ∑_{n≤y} h_{0}^{2}(n)n^{−3/2} + O ( ∑_{y<n≤y^{2}} h_{1}^{2}(n)n^{−3/2})

= ∑_{n=1}^{∞} h_{0}^{2}(n)n^{−3/2} + O (∑_{n>y} h_{1}^{2}(n)n^{−3/2})

= ∑_{n=1}^{∞} h_{0}^{2}(n)n^{−3/2} + O(y^{−1/2} log^{3} y).  (31)

Next we consider the integral of S_{2}(x). By the first derivative test we get

∫_{T}^{2T} S_{2}(x)dx ≪ T ∑_{2} \frac{d(n_{1})d(n_{2})}{(m_{1}m_{2})^{5/4}(n_{1}n_{2})^{3/4}} \frac{1}{|\sqrt{\frac{n_{1}}{m_{1}}} − \sqrt{\frac{n_{2}}{m_{2}}} |}

≪ T ∑_{2} \frac{d(n_{1})d(n_{2})}{(n_{2}m_{1}n_{1}m_{2})^{3/4}} \frac{1}{|\sqrt{n_{1}m_{2}} − \sqrt{n_{2}m_{1}}|}

≪ T ∑_{\begin{matrix}n,m≤y^{2} \\ n≠m \end{matrix}} \frac{d_{3}(m)d_{3}(n)}{n^{3/4}m^{3/4}} \frac{1}{|\sqrt{n} − \sqrt{m}|}

≪ T ∑_{\begin{matrix}n,m≤y^{2} \\ |\sqrt{n}−\sqrt{m}|≥ \frac{1}{2} (mn)^{1/4} \end{matrix}} \frac{d_{3}(m)d_{3}(n)}{n^{3/4}m^{3/4}} \frac{1}{|\sqrt{n} − \sqrt{m}|}

+ T ∑_{\begin{matrix}n,m≤y^{2} \\ 0<|\sqrt{n}−\sqrt{m}|< \frac{1}{2} (mn)^{1/4} \end{matrix}} \frac{d_{3}(m)d_{3}(n)}{n^{3/4}m^{3/4}} \frac{1}{|\sqrt{n} − \sqrt{m}|}

≪ T (∑_{n≤y^{2}} d_{3}(n)n^{−1})^{2} + T ∑_{\begin{matrix}n,m≤y^{2} \\ n≠m \end{matrix}} \frac{d_{3}(m)d_{3}(n)}{n^{1/2}m^{1/2}} \frac{1}{|n − m|}

≪ T log^{9} T + T ∑_{n≤y^{2}} d_{3}^{2}(n)n^{−1} ≪ T log^{9} T,  (32)

where we used the well-known Hilbert’s inequality and the estimates

∑_{n≤x} d_{3}(n) ≪ x log^{2} x, ∑_{n≤x} d_{3}^{2}(n) ≪ x log^{8} x.

For the integral of S_{3}(x), by the first derivative test again, we get

∫_{T}^{2T} S_{3}(x)dx ≪ T ∑_{3} \frac{d(n_{1})d(n_{2})}{(m_{1}m_{2})^{5/4}(n_{1}n_{2})^{3/4}} \frac{1}{|\sqrt{\frac{n_{1}}{m_{1}}} + \sqrt{\frac{n_{2}}{m_{2}}} |}

≪ T ∑_{3} \frac{d(n_{1})d(n_{2})}{(m_{1}m_{2})^{5/4}(n_{1}n_{2})^{3/4}} \frac{1}{(\frac{n_{1}}{m_{1}} )^{1/4}(\frac{n_{2}}{m_{2}} )^{1/4}}

≪ T ∑_{3} \frac{d(n_{1})d(n_{2})}{m_{1}m_{2}n_{1}n_{2}} ≪ T log^{6} T.  (33)

From (3.2)-(3.6) we get

∫_{T}^{2T} |E^{∗}(x)|^{2}dx = \frac{1}{4π^{2}} ∑_{n=1}^{∞} h_{0}^{2}(n)n^{−3/2} ∫_{T}^{2T} x^{1/2}dx
+ O(T^{3/2}y^{−1/2} log^{3} T + T log^{9} T)
= \frac{1}{4π^{2}} ∑_{n=1}^{∞} h_{0}^{2}(n)n^{−3/2} ∫_{T}^{2T} x^{1/2}dx + O(T^{1+ε}).  (34)

From (3.1), (3.7) and Cauchy’s inequality we get

∫_{T}^{2T} x^{2α−2}|E_{α}(x)|^{2}dx = \frac{1}{4π^{2}} ∑_{n=1}^{∞} h_{0}^{2}(n)n^{−3/2} ∫_{T}^{2T} x^{1/2}dx + O(T^{5/4+ε}),  (35)

which implies Theorem 2 by a splitting argument.

Remark 3. The referee kindly indicated that the average order of h_{1}^{2}(n) is also derived by Shiu’s theorem. Indeed, since h_{1}(n) ≪ d_{3}(n) ≪ n^{ε} we have

∑_{n≤x} h_{1}^{2}(n) ≪ \frac{x}{log x} exp (∑_{p≤x} \frac{h_{1}^{2}(p)}{p})

≪ \frac{x}{log x} exp (∑_{p≤x} (\frac{4}{p} + \frac{4}{p^{3/2}} + \frac{1}{p^{2}} ))

≪ x(log x)^{3}.

4 Acknowledgments

The authors express their gratitude to the referee for careful reading of the manuscript and many valuable suggestions, especially pointing out Pillai’s paper and Remark 3 to the authors.

References

[1] O. Bordellès, A note on the average order of the gcd-sum function, J. Integer Sequences 10 (2007), Article 07.3.3.

[2] K. Broughan, The gcd-sum function, J. Integer Sequences 4 (2001), Article 01.2.2.

[3] K. Broughan, The average order of the Dirichlet series of the gcd-sum function, J. Integer Sequences 10 (2007), Article 07.4.2.

[4] M. N. Huxley, Exponential sums and Lattice points III, Proc. London Math. Soc. 87 (2003), 591–609.

[5] A. Ivić, The Riemann Zeta-Function, John Wiley and Sons, 1985.

[6] S. S. Pillai, On an arithmetic functions, J. Annamalai Univ. II (1937), 243–248.

[7] K. C. Tong, On divisor problem III, Acta Math. Sinica 6 (1956), 515–541.

[8] G. Voronoï, Sur une fonction transcendante et ses applications a la sommation de quelques séries, Ann. Sci. École Norm Sup. 21 (1904), 207–267, 459–533.

2000 Mathematics Subject Classification: Primary 11N37; Secondary 11M06. Keywords: gcd-sum function, Dirichlet divisor problem, mean square estimate.

(Concerned with sequence \underline{A018804}.)

Received March 30 2008; revised version received May 9 2008. Published in Journal of Integer Sequences, June 3 2008.

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