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Journal of Integer Sequences, Vol. 13 (2010),
Article 10.3.8

23 11

Mean Values of a Gcd-Sum Function
Over Regular Integers Modulo n
Deyu Zhang and Wenguang Zhai1
School of Mathematical Sciences
Shandong Normal University
Jinan 250014
Shandong

P. R. China
zdy 78@yahoo.com.cn
zhaiwg@hotmail.com
Abstract
In this paper we study the mean value of a gcd-sum function over regular integers
modulo n. In particular, we improve the previous result under the Riemann hypothesis
(RH). We also study the short interval problem for it without assuming RH.

1

Introduction

In general, an element k of a ring R is said to be (von Neumann) regular if there is an x ∈ R
such that k = kxk. Let n > 1 be an integer with prime factorization n = pν11 · · · pνrr . An
integer k is called regular (mod n) if there exists an integer x such that k 2 x ≡ k (mod n),
i.e., the residue class of k is a regular element (in the sense of J. von Neumann) of the ring
Zn of residue classes (mod n).
Let Regn = {k : 1 ≤ k ≤ n and k is regular (mod n)}. T´oth [11] first defined the
gcd-sum function over regular integers modulo n by the relation
X

gcd(k, n),
(1)
P˜ (n) =
k∈Regn

1
This work is supported by National Natural Science Foundation of China(Grant Nos. 10771127,
10826028) and Research Award Foundation for Young Scientists of Shandong Province (No. BS2009SF018).

1

where gcd(a, b) denotes the greatest common divisor of a and b. It is sequence A176345 in
Sloane’s Encyclopedia. This is analogous to the gcd-function, called also Pillai’s arithmetical
function,
n
X
P (n) =
gcd(k, n),
k=1

which has been studied recently by several authors, see [2, 3, 4, 5, 6, 9, 12]; it is Sloane’s
sequence A018804. T´oth [11] proved that P˜ (n) is multiplicative and for every n ≥ 1,
P˜ (n) = n

Y
p|n

1
(2 − ).
p

(2)

He also obtained the following asymptotic formula
X
n≤x

P˜ (n) =

x2

(K1 log x + K2 ) + O(x3/2 δ(x)),
2ζ(2)

(3)

where the function δ(x) and constants K1 and K2 are given by
δ(x) = exp(−A(log x)3/5 (log log x)−1/5 ),
!
∞
Y
X
1
µ(n)
=
,
K1 =
1−
nψ(n)
p(p + 1)
p

n=1
K2 = K1
where ψ(n) = n

Q



1 2ζ ′ (2)
2γ − −
2
ζ(2)

p|n (1

!

−

∞

X
µ(n)(log n − α(n) + 2β(n))

nψ(n)

n=1

(4)

,

(5)

+ p1 ) denotes the Dedekind function, and
α(n) =

X log p
X log p
, β(n) =
.

p−1
p2 − 1
p|n

p|n

It is very difficult to improve the exponent 32 in the error term of (3) unless we have
substantial progress in the study of the zero free region of ζ(s). Therefore it is reasonable to
get better improvements by assuming the truth of the Riemann hypothesis (RH). Let d(n)
denote the Dirichlet divisor function and
X
∆(x) :=
d(n) − x(log x + 2γ − 1).
(6)
n≤x

Dirichlet first proved that ∆(x) = O(x1/2 ). The exponent 1/2 was improved by many
authors. The latest result reads
∆(x) ≪ xθ+ǫ ,

θ = 131/416,
2

(7)

due to Huxley [7]. T´oth [11] proved that if RH is true, then the error term of (3) can be
replaced by O(x(7−5θ)/(5−4θ) exp(B log x(log log x)−1 )). For θ = 131/416 one has (7 − 5θ)/(5 −
4θ) ≈ 1.4505.
In this paper, we will use the DirichletPconvolution method to study the mean value
˜
of P (n), and we find that the estimate of n≤x P˜ (n) is closely related to the square-free
divisor problem. Let d(2) (n) denote the number of square-free divisors of n. Note that
d(2) (n) = 2ω(n) , where ω(n) is the number of distinct prime factors of n. Let
X
D(2) (x) =
d(2) (n).
n≤x

It was shown by Mertens [8] that

1
D(2) (x) =
x log x +
ζ(2)

!
2γ − 1 2ζ ′ (2)
− 2
x + ∆(2) (x),
ζ(2)
ζ (2)

(8)

where ∆(2) (x) = O(x1/2 log x). The exponent 21 is also difficult to be improved, because
it is related to the zero distribution of ζ(s). One way of making progress is to assume the
Riemann hypothesis (RH). Many authors investigated this problem, and the best result
under the Riemann hypothesis is
∆(2) (x) ≪ xλ+ǫ ,

(9)

where λ = 4/11, due to Baker [1].
In this paper, we shall prove the following results.
Theorem 1. For any real numbers x ≥ 1 and ǫ > 0, if
∆(2) (x) ≪ xλ+ǫ ,
then we have
X

P˜ (n) =

n≤x

x2
(K1 log x + K2 ) + O(x1+λ+ǫ ),
2ζ(2)

(10)

where K1 , K2 are defined by (4) and (5).

Corollary 2. If RH is true, then
X
n≤x

P˜ (n) =

x2
(K1 log x + K2 ) + O(x15/11+ǫ ).
2ζ(2)

(11)

Remark. Note that 15/11 ≈ 1.3636, which improves the previous result.
In order to avoid assuming the truth of the Riemann hypothesis, we study the short
interval problem for it.

3

Theorem 3. For
xθ+3ǫ ≤ y ≤ x,
we have
X

P˜ (n) =

x 0. Write
G(s) =

∞
X
g(n)
n=1

ns

,

(16)

then we can easily get
X

g(n) ≪ nǫ ,

|g(n)| = O(xǫ ).

(17)

n≤x

Notice that
∞
X
d(2) (m)
ζ 2 (s)
=
.
ζ(2s) m=1 ms

(18)

By the Dirichlet convolution, we have
X
X
X
X
P˜ ∗ (n) =
d(2) (m)g(ℓ) =
g(ℓ)
d(2) (m),
n≤x

mℓ≤x

ℓ≤x

m≤x/ℓ

and formula (8) applied to the inner sum gives



 x

′
X
X
x
2ζ
(2)
x
∗
λ+ǫ
log( ) + 2γ − 1 −
+O ( )
P˜ (n) =
g(ℓ)
ζ(2)ℓ
ℓ
ζ(2)
ℓ
n≤x
ℓ≤x
x
=
ζ(2)

(

x
=
ζ(2)

(
)
 ∞
∞


2ζ ′ (2) X g(ℓ) X g(ℓ) log ℓ
−1+ǫ
log x + 2γ − 1 −
−
+ O(x
) + O xλ+ǫ ,
ζ(2) ℓ=1 ℓ
ℓ
ℓ=1

2ζ ′ (2)
log x + 2γ − 1 −
ζ(2)

X
ℓ≤x

g(ℓ) X g(ℓ) log ℓ
−
ℓ
ℓ
ℓ≤x

if we notice by (17) that both of the infinite series

∞
P

ℓ=1

g(ℓ)
ℓ

)

+O x

and

convergent.
From (15), (16) and the definitions of K1 , K2 , we have

∞
X
Y
g(ℓ)
1
= K1 ,
= G(1) =
1− 2
ℓ
p
+
p
p
ℓ=1
∞
X
g(ℓ) log ℓ
ℓ=1

ℓ

=

n=1

5

X |g(ℓ)|
ℓ≤x

∞
P

ℓ=1

∞
X
µ(n)(log n − α(n) + 2β(n))

nψ(n)

λ+ǫ

g(ℓ) log ℓ
ℓ

ℓλ+ǫ

!

.

are absolutely

(19)

(20)

1 2ζ ′ (2)
2γ − −
2
ζ(2)



− K2 .

1
(log x − )K1 + K2
2



+ O(xλ+ǫ ).

= K1
Then
x
P˜ ∗ (n) =
ζ(2)
n≤x

X





(21)

From the definitions of P˜ ∗ (n)and Abel’s summation formula, we can easily get

X

P˜ (n)

=

n≤x

X

P˜ ∗ (n)n =

1

n≤x
2

=

Z

x

td

X
n≤t

!

P˜ ∗ (n)

x
(K1 log x + K2 ) + O(x1+λ+ǫ ).
2ζ(2)

Corollary 2 follows by taking λ = 4/11.

3

Proof of Theorem 3

From the proof of Theorem 1, we have
F (s) =

∞ ˜∗
X
P (n)
n=1

ns

ζ 2 (s)
=
G(s).
ζ(2s)

(22)

Let
ζ 2 (s)G(s) =

∞
X
h(n)
n=1

ns

,

ℜs > 1.

(23)

Then we have
Lemma 5. For any real numbers x ≥ 1 and ǫ > 0, we have



X
1 2ζ ′ (2)
h(n) = x
log x − +
K1 + K2 + O(xθ+ǫ ),
2
ζ(2)
n≤x

(24)

where θ is defined in (7).
Proof. Recall that

∞
X
g(n)
n=1

Then we have

h(n) =

ns

X

= G(s),

g(n) ≪ nǫ .

d(m)g(ℓ),

h(n) ≪ nǫ .

n=mℓ

6

(25)

Thus from (6),(7) we get
X
X
X
X
h(n)
=
d(m)g(ℓ) =
g(ℓ)
d(m)
n≤x

mℓ≤x

m≤ x

ℓ≤x

ℓ

 x
o
nx 
x
log( ) + 2γ − 1 + O ( )θ+ǫ
=
g(ℓ)
ℓ
ℓ
ℓ
ℓ≤x
(
)
!
X g(ℓ) X g(ℓ) log ℓ
X |g(ℓ)|
= x (log x + 2γ − 1)
−
+ O xθ+ǫ
ℓ
ℓ
ℓθ+ǫ
ℓ≤x
ℓ≤x
ℓ≤x
)
(
∞
∞
X


g(ℓ) X g(ℓ) log ℓ
−
+ O(x−1+ǫ ) + O xθ+ǫ
= x (log x + 2γ − 1)
ℓ
ℓ
ℓ=1
ℓ=1

X

Then Lemma 5 follows from the above formula and (19), (20).
Lemma 6. For any real numbers x ≥ 1 and x < u ≤ 2x, we have
X
P˜ ∗ (n) = M (u) − M (x) + E(u, x),

(26)

x