bordelles61. 132KB Jun 04 2011 12:08:26 AM

1

2
3

47

6

Journal of Integer Sequences, Vol. 10 (2007),
Article 07.9.2

23 11

Mean Values of Generalized gcd-sum
and lcm-sum Functions
Olivier Bordell`es
2, All´ee 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) =

n
X

(n, j)

j=1

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 ) ,

1

where µ is the M¨obius 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
x2 log x
x2
g (n) =
+
2ζ (2)
2ζ (2)
n6x

X



1
γ − + log
2



A12
2π



+ Oε x1+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 6 131/416 ≈ 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.
Ida (n) = na
1 (n) = 1

(a ∈ Z∗ )

and µ, ϕ, σk and τk are respectively the M¨obius function, the Euler totient function, the sum
of kth powers of divisors function and the kth Piltz divisor function. Recall that τkP
can be
defined by τk = 1
· · ∗ 1} for any integer k > 1 and that τ2 = τ. We also have σk = d|n dk

| ∗ ·{z
and σ0 = τ.

k times

2. Exponent in the Dirichlet-Piltz divisor problem. For any integer k > 2, θk is defined to be
the smallest positive real number such that the asymptotic formula
X


τk (n) = xPk−1 (log x) + Oε,k xθk +ε
(2)
n6x

holds for any real number ε > 0. Here Pk−1 is a polynomial of degree k − 1 with real coef43
1
. It is now well-known that 31 6 θ3 6 96
and that
ficients, the leading coefficient being (k−1)!
k−1
k−1

6 θk 6 k+2 for k > 4 (see [6], for example).
2k
By convention, we set
τ0 (n) =

(

1, if n = 1;
0, otherwise;

and θ1 = 0.

2

2

A generalization of the gcd-sum function

Definition 2.1. We define the sequence of arithmetic functions fk,j (n) in the following way.
(i) For any integers j, n > 1, we set

(
1, if (n, j) = 1;
f1,j (n) =
0, otherwise;
(
(n, j), if j 6 n;
f2,j (n) =
0,
otherwise.
(ii) For any integers j > 1 and k > 3, we set
fk,j = f2,j ∗ (Id · τk−2 ) .
Definition 2.2. For any integers n, k > 1, we define the sequence of arithmetic functions
gk (n) by
n
X
gk (n) =
fk,j (n) .
j=1

Examples.

n
X

g1 (n) =

g2 (n) =
g3 (n) =

1 = ϕ (n) ,

j=1
(n,j)=1
n
X

(j, n) = g (n) ,

j=1
n X
X

j=1 d|n
d>j

n
(j, d) ,
d

..
.
gk (n) =

n
X
X

X

j=1 dk−2 |n dk−3 |dk−2

···

Xn
(j, d1 ) .
d1

d1 |d2
d1 >j

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
X
n6x

gk (n) =



x2
Rk−1 (log x) + Oε,k x1+θk +ε
2ζ (2)

3

where Rk−1 is a polynomial of degree k − 1 and leading coefficient
gives Rk−1 for k ∈ {1, 2, 3}
1

k
Rk−1

2

1
.
(k−1)!

The following table

3

1
1 X + γ − + log
2



A12
2π



X2
+ αX + β
2

where
 12 
1
A
α = 2γ − + log
2
2π

 12 

 12 2 
′′
A
1
ζ (2)
A
γ − log
− 3γ −
+ γ − log
β=−
2ζ (2)
2π
2
2π


1
12γ1 − 12γ 2 + 6γ − 1
−
4

and

Constant
γ ≈ 0.577 215 664 . . .
γ1 ≈ −0.072 815 845 . . .
A ≈ 1.282 427 129 . . .

3

Name
Euler − Mascheroni
Stieltjes
Glaisher − Kinkelin

Main properties of the function gk

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
gk+1 = gk ∗ Id
and then
gk = ϕ ∗ (Id · τk−1 ) .
Moreover, we have
gk = µ ∗ (Id · τk ) .

(3)

Thus, the Dirichlet series Gk (s) of gk 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
ζ (s − 1)k
.
Gk (s) =
ζ (s)

4

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.
X n
(g1 ∗ Id) (n) = (ϕ ∗ Id) (n) =
dϕ
d
d|n

=

X
d|n

=

n
X

X

d

1=

k6n/d
(k,n/d)=1

(j, n) =

X
d|n

n
X

d

n
X

1

j=1
(j,n)=d

f2,j (n) = g2 (n) .

j=1

j=1

For k = 2, we get
(g2 ∗ Id) (n) =

d
XnX
d|n

d

(j, d) =

j=1

n X
X
n
j=1 d|n
d>j

d

(j, d) =

n
X

f3,j (n) = g3 (n) .

j=1

Now let us suppose k > 3. We have
gk+1 (n) =
=

n
X

j=1
n
X

fk+1,j (n) =

n
X

(f2,j ∗ (Id · τk−1 )) (n)

j=1

(f2,j ∗ Id · τk−2 ∗ Id) (n)

j=1

=

d
XnX
d|n

d

(f2,j ∗ (Id · τk−2 )) (d) = (gk ∗ Id) (n) .

j=1

The second relation is easily shown by induction. For the third, we have using ϕ = µ ∗ Id
gk = ϕ ∗ (Id · τk−1 ) = µ ∗ (Id ∗ (Id · τk−1 ))
= µ ∗ (Id · (1 ∗ τk−1 )) = µ ∗ (Id · τk ) .
The last proposition comes from the equality (3)
gk = µ ∗ (Id · τk ) = µ ∗ Id
· · ∗ Id}
| ∗ ·{z
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
X


nτk (n) = x2 Qk−1 (log x) + Oε,k x1+θk +ε
n6x

where Qk−1 is a polynomial of degree k − 1 and leading coefficient
5

1
.
2(k−1)!

Proof. Using summation by parts and (2), we get
!
Z x X
X
X
τk (n) dt
nτk (n) = x
τk (n) −
n6x

1

n6x

n6t

2

= x Pk−1 (log x) + Oε,k x

1+θk +ε

Writing
Pk−1 (X) =




−

Z

x

tPk−1 (log t) + Oε,k tθk +ε

1

k−1
X





dt.

aj X j

j=0

with ak−1 =

1
,
(k−1)!

X

we obtain

nτk (n) = x

2

k−1
X

j

aj (log x) −

j=0

n6x

and the formula
Z x

k−1
X

aj

j=0

t (log t)j dt = x2

1

j
X

(−1)j−i

i=0

j
X

t (log t)j dt + Oε,k x1+θk +ε

1

2j+1−i × i!

i=0

j=0

x

j!

(easily proved by induction) gives
(
j
k−1
X
X
X
j
2
(−1)j−i
nτk (n) = x
aj (log x) −
n6x

Z

(log x)i − (−1)j

j!
2j+1−i × i!

i

(log x)




j!
2j+1

)



j! aj
+ Oε,k x1+θk +ε
j+1
2
i=0
)
(
j−1
k−1
j
X
X


(log
x)
j!
(−1)j−i j+1−i
(log x)i + Oε,k x1+θk +ε
−
= x2
aj
2
2
× i!
i=0
j=0

+

(−1)j

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)
(
)
X


(log x)2
+ (3γ − 1) log x + 3γ 2 − 3γ − 3γ1 + 1 + Oε xθ3 +ε
τ3 (n) = x
2
n6x
and gives
X
n6x

nτ3 (n) = x2

(

(log x)2
+
4



6γ − 1
4



12γ1 − 12γ 2 + 6γ − 1
log x −
8

)



+ Oε x1+θ3 +ε .

(4)

6

Now we are able to prove Theorem 2.3.
Using (3) we get

X

gk (n) =

n6x

and lemma 3.1 gives

X

µ (d)

d6x

X

mτk (m) ,

m6x/d


 
x 1+θk +ε
x
+ Oε,k
Qk−1 log
gk (n) =
µ (d)
d
d
d
n6x
d6x

X µ (d)


x
2
1+θk +ε
=x
Q
log
+
O
x
.
k−1
ε,k
d2
d
d6x

X

 
x 2

X



Writing
Qk−1 (X) =

k−1
X

bj X j

j=0

with bk−1 =

1
,
2(k−1)!

we get
k−1
X µ (d) X




x j
1+θk +ε
gk (n) = x
b
log
+
O
x
j
ε,k
d2 j=0
d
n6x
d6x
j  
k−1 X
X
X


µ (d)
j
2
(−1)h 2 (log d)h + Oε,k x1+θk +ε
bj (log x)j−h
=x
d
h
j=0 h=0
d6x

X

2

and the equality
∞
X
X
µ (d)
µ (d)
h
h
h µ (d)
(−1)
(log
d)
=
(log
d)
−
(−1)h 2 (log d)h
(−1)
2
2
d
d
d
d6x
d>x
d=1
!

 h 
(log x)h
1
d
+O
,
=
dsh ζ (s) [s=2]
x

X

h

implies
!

j    h 
k−1 X
X
d
j
1
gk (n) = x2
bj (log x)j−h
h
h
ds
ζ (s) [s=2]
j=0 h=0
n6x




+ O x (log x)k−1 + Oε,k x1+θk +ε
!

j    h 
k−1 X
X


d
j
1
j−h
1+θk +ε
b
(log
x)
+
O
x
,
= x2
j
ε,k
h
dsh ζ (s) [s=2]
j=0 h=0

X

and writing



dh
dsh



1
ζ (s)



=

[s=2]

7

Ah
2ζ (2)h+1

with Ah ∈ R (and A0 = 2), we obtain
k−1 j  


x2 X X j Ah bj (log x)j−h
1+θk +ε
gk (n) =
+
O
x
ε,k
2ζ (2) j=0 h=0 h
ζ (2)h
n6x

X




k−1
0

which is the desired result. The leading coefficient is
cases are easy to check.

A0 bk−1 =

1
.
(k−1)!

The particular

(i) For k = 1, the result is well-known (see [2, Exer. 4.14])
X
X
x2
+ O (x log x) .
g1 (n) =
ϕ (n) =
2ζ (2)
n6x
n6x
(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 X
1
ζ (3) 2
2/3
4/3
.
x + O x (log x) (log log x)
=
(j, n)
2ζ (2)
j=1
n6x
Proof. For any integer n > 1, we set
G (n) =

n
X
j=1

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

X
n6x

G (n) =

X1 X
d6x

d

ϕ (m) .

m6x/d

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


X
x2
2/3
4/3
+ O x (log x) (log log x)
,
ϕ (n) =
2ζ (2)
n6x
combined with some classical computations, allows us to conclude the proof of Theorem 5.1.

8

6

The lcm-sum function

Definition 6.1. For any integer n > 1, we define
n
X

l (n) =

[n, j]

j=1

where [a, b] is the least common multiple of a and b.
Lemma 6.2. We have the following convolution identity
l=
Proof. We have
n
X
j=1



1
(Id2 · (ϕ + τ0 )) ∗ Id .
2

n
X1 X
X1 X
X
j
j=
=
kd =
(n, j)
d j=1
d
d|n

d|n

k6n/d
(k,n/d)=1

(n,j)=d

d|n

X

k,

k6n/d
(k,n/d)=1

with
X

k6N
(k,N )=1

k=

X

dµ (d)

d|N

X

m

m6N/d


N
+1
d
d|N


NX
N
N
=
+1 =
(ϕ + τ0 ) (N ) ,
µ (d)
2
d
2
1X
dµ (d)
=
2



N
d



d|N

and hence

n
X
j=1

n 1
1Xn
j
=
(ϕ + τ0 )
= ((Id · (ϕ + τ0 )) ∗ 1) (n) ,
(n, j)
2
d
d
2
d|n

and we conclude by noting that
l (n) = n

n
X
j=1

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 X
ζ (3) 4
2/3
4/3
3
x + O x (log x) (log log x)
.
[n, j] =
8ζ (2)
j=1
n6x

9

Proof. Using lemma 6.2, we get
X

l (n) =

n6x

1X X 2
d
m (ϕ + τ0 ) (m)
2 d6x
m6x/d

=

1X
2

d6x

d

X

m2 ϕ (m) + O x2

m6x/d




and the estimation (see [8])


x4
2/3
4/3
3
+ O x (log x) (log log x)
n ϕ (n) =
4ζ (2)
n6x

X

2

implies
1X
d
l (n) =
2 d6x
n6x

X


∞


 


1  x 4
x 3
2/3
4/3
+ O x2
+O
(log x) (log log x)
4ζ (2) d
d





x4 X 1
2/3
4/3
3
+
O
x
(log
x)
(log
log
x)
+ O x2 ,
=
3
8ζ (2) d=1 d

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
!

 12 
n
X X
(log x)3 (log x)2
A
1
=
+
γ + log
+ O (log x) .
[n, j]
6ζ (2)
2ζ (2)
2π
j=1
n6x
Some useful estimates are needed.
 12 
A
≈ 1.147 176 . . . For any real number x > 1, we have
Lemma 7.2. Set Cϕ = log
2π


X ϕ (n)
Cϕ
log x
log ex
(i) :
.
+
+O
=
n2
ζ (2) ζ (2)
x
n6x
 x  (log x)2 C log x
X ϕ (n)
ϕ
(ii) :
log
+
+ O (1) .
=
2
n
n
2ζ (2)
ζ (2)
n6x
1 X ϕ (n)   x 2 (log x)3 Cϕ (log x)2
log
(iii) :
+
+ O (log x) .
=
2 n6x n2
n
6ζ (2)
2ζ (2)
10

Proof. (i) . Using ϕ = µ∗ Id, we get
X ϕ (n)
n6x

n2

X µ (d) X 1
d2
m
d6x
m6x/d

 
x
X µ (d)
d
=
log
+
γ
+
O
d2
d
x
d6x
=

= (log x + γ)

X µ (d)
d6x

d2

−

X µ (d) log d

log x
γ
ζ ′ (2)
+O
=
+
−
ζ (2) ζ (2) (ζ (2))2
Recall that

d2

d6x



log ex
x

+O



1X1
x d6x d

!

.

ζ ′ (2)
= γ − Cϕ .
ζ (2)

(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) =

n
X
j=1

1
.
[n, j]

Since
n
n
1X X 1
1 X (n, j)
=
d
L (n) =
n j=1 j
n
j
j=1
d|n

(j,n)=d

X 1
1X
1X
=
=
d
n
kd
n
d|n

k6n/d
(k,n/d)=1

11

d|n

X 1
,
k

k6n/d
(k,n/d)=1

we get
X

L (n) =

n6x

X1X
n
n6x
d|n

=

X 1
k

k6n/d
(k,n/d)=1

X1 X 1 X 1
d
h k6h k
d6x
h6x/d

(k,h)=1

X 1 X 1 X µ (δ) X 1
=
d
h
δ
m
d6x
h6x/d

δ|h

X 1 X µ (δ)
=
d
δ2
d6x
δ6x/d

m6h/δ

X

a6x/(dδ)

1X 1
a m6a m

X X 1 µ (δ) X 1 X 1
=
d δ2
a m6a m
d6x δd6x
a6x/(dδ)
X 1 X n X 1 X 1
,
dµ
=
2
n
d
a
m
m6a
n6x
d|n

a6x/n

and the convolution identity ϕ = µ ∗ Id implies that
X

L (n) =

n6x

X ϕ (n) X 1 X 1
.
2
n
a
m
n6x
m6a
a6x/n

Thus
 
X ϕ (n) X 1 
1
L (n) =
log a + γ + O
2
n
a
a
n6x
n6x
a6x/n



X ϕ (n) 1 
x 2
x
log
+ O (1)
+ γ log
=
2
n
2
n
n
n6x

X

=
where Cϕ = log

8



A12
2π



(log x)3 Cϕ (log x)2 γ (log x)2
+
+
+ O (log x)
6ζ (2)
2ζ (2)
2ζ (2)

, which concludes the proof.

Acknowledgment

I am indebted to the referee for his valuable suggestions.

12

References
[1] O. Bordell`es, A note on the average order of the gcd-sum function, J. Integer Sequences
10 (2007), Article 07.3.3.
[2] O. Bordell`es, Th`emes d’Arithm´etique, Editions Ellipses, 2006.

[3] K. A. Broughan, The gcd-sum function, J. Integer Sequences 4 (2001), Article 01.2.2.
Errata, April 16, 2007, http://www.cs.uwaterloo.ca/journals/JIS/VOL4/BROUGHAN/errata1.pdf
[4] K. A. Broughan, The average order of the Dirichlet series of the gcd-sum function, J.
Integer Sequences 10 (2007), Article 07.4.2.
[5] M. N. Huxley, Exponential sums and lattice points III, Proc. London Math. Soc. 87
(2003), 591–609.
[6] A. Ivi´c, E. Kr¨atzel, M. K¨
uhleitner, and W. G. Nowak, Lattice points in large regions and
related arithmetic functions: Recent developments in a very classic topic, in W. Schwarz
and J. Steuding, eds., Conference on Elementary and Analytic Number Theory, Frank
Steiner Verlag, 2006, pp. 89–128.
[7] G. Tenenbaum, Introduction `
a la Th´eorie Analytique et Probabiliste des Nombres, Soci´et´e
Math´ematique 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.
Return to Journal of Integer Sequences home page.

13