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Journal of Integer Sequences, Vol. 17 (2014),
Article 14.1.7
23 11
On the Lcm-Sum Function
Soichi Ikeda and Kaneaki Matsuoka
Graduate School of Mathematics
Nagoya University
Furocho Chikusaku Nagoya 464-8602
Japan
m10004u@math.nagoya-u.ac.jp
m10041v@math.nagoya-u.ac.jp
Abstract
We consider a generalization of the lcm-sum function, and we give two kinds of
asymptotic formulas for the sum of that function. Our results include a generalization
of Bordellès’s results and a refinement of the error estimate of Alladi’s result. We prove
these results by the method similar to those of Bordellès.
1
Introduction
Pillai [6] first defined the gcd-sum function
g(n) =
n
X
gcd(j, n)
j=1
and studied this function. The function g(n) was defined again by Broughan [3]. Broughan
considered
X
Gα (x) =
n−α g(n)
n≤x
for α ∈ R and x ≥ 1, and obtained some asymptotic formulas for Gα (x). The function Gα (x)
was studied by some authors (see, for example, [4, 8]). Some generalizations of the function
g(n) was considered (see, for example, [2, 10]).
1
On the other hand, the lcm-sum function
l(n) :=
n
X
lcm(j, n)
j=1
was considered by some authors. Alladi [1] studied the sum
n
X
(lcm(j, n))r
(r ∈ R, r ≥ 1)
j=1
and obtained
n
XX
(lcm(j, n))r =
n≤x j=1
ζ(r + 2)
x2r+2 + O(x2r+1+ǫ ).
2(r + 1)2 ζ(2)
(1)
We define the functions
La (n) :=
n
X
(lcm(j, n))a
j=1
Ta (x) :=
X
La (n)
n≤x
for a ∈ Z and x ≥ 1.
Bordellès studied the sums T1 (x) and T−1 (x) and obtained
1
l(n) = ((Id2 · (ϕ + τ0 )) ∗ Id)(n),
2
n
XX
ζ(3) 4
x + O(x3 (log x)2/3 (log log x)4/3 ) (x > e),
lcm(j, n) =
8ζ(2)
n≤x j=1
12 n
XX
(log x)3 (log x)2
A
1
=
+
γ + log
+ O(log x),
lcm(j, n)
6ζ(2)
2ζ(2)
2π
n≤x j=1
where Ida (n) = na (a ∈ Z),
τ0 (n) =
(
1, if n = 1;
0, otherwise,
F ∗ G is the usual Dirichlet convolution product, and A is the Glaisher-Kinkelin constant [7,
p. 25]). Gould and Shonhiwa [5] stated that the log-factors in the error term in the second
formula can be removed.
In this paper we study Ta (x) for a ≥ 2 and a ≤ −2. The following theorems are our main
results. These results are proved by the methods similar to those of Bordellès [2, Section 6].
We write f (x) = O(g(x)), or equivalently f (x) ≪ g(x), where there is a constant C > 0
such that |f (x)| ≤ Cg(x) for all values of x under consideration.
2
Theorem 1. Let Bn be Bernoulli numbers defined by
∞
X
zn
z
=
.
B
n
ez − 1 n=0
n!
If we define
ϕk (n) :=
X
µ(d)
d|n
and
Ma (n) :=
then for a ∈ Z we have
2a
Id ·
d k
n
a−1 1
1 X a+1
1
(n),
Bk+1 ϕk
ϕ + τ0 +
a+1
2
a + 1 k=1 k + 1
La (n) = (Ma ∗ Ida )(n).
Theorem 2. Let x > e and a ∈ N. Then we have
X
ζ(a + 2)
La (n) =
x2a+2 + O(x2a+1 (log x)2/3 (log log x)4/3 )
2 ζ(2)
2(a
+
1)
n≤x
(as x → ∞),
where the implied constant depends on a.
Theorem 3. Let x ≥ 1 and k ∈ N with k ≥ 2. Then we have
∞
X
ζ(k)2
ζ(k)
1+
L−k (n) =
2
ζ(2k)
n=1
(2)
and
ζ(k)2
ζ(k)x−k+1 log x
ζ(k)
+ O(x−k+1 )
1+
−
L−k (n) =
2
ζ(2k)
(k
−
1)ζ(k
+
1)
n≤x
X
(as x → ∞),
where the implied constant depends on k.
We note that the function La (n) is not multiplicative for all a ∈ Z \ {0}, but we can
write La (n) (a ≥ 1) explicitly by Dirichlet convolution. In the proof of Theorem 2 we use
this fact. The error estimates in Theorem 2 are better than (1). Since we have
n
X
gr (n) :=
(gcd(j, n))r > ϕ(n)
j=1
for all r ∈ R, the sum
∞
X
gr (n)
n=1
is divergent for all r. Therefore the behavior of the sum Ta (x) (a ∈ Z and a ≤ −2) is
completely different from that of the sum
X
ga (n).
n≤x
3
2
Lemmas for the proof of theorems
In this section, we collect some auxiliary results and definitions.
Let Bn (x) be Bernoulli polynomials defined by
∞
X
zn
zexz
=
.
B
(x)
n
ez − 1 n=0
n!
The following relations are well-known [7, p. 59].
Bn (x + 1) − Bn (x) = nxn−1 ,
n X
n
Bk xn−k ,
Bn (x) =
k
k=0
Bn (0) = Bn (1) = Bn
Lemma 4. Let m, n ∈ N and
Sn (m) :=
m
X
(n > 1).
ln .
l=1
Then we have
n−1 1 X n+1
mn+1 1 n
Bk+1 mn−k .
+ m +
Sn (m) =
n+1 2
n + 1 k=1 k + 1
Proof. We have
1
(Bn+1 (m + 1) − Bn+1 (1))
n+1
1
=
(Bn+1 (m) + (n + 1)mn − Bn+1 )
n+1
X
n+1 n+1
1
n+1−k
n
Bk m
+ (n + 1)m − Bn+1
=
k
n + 1 k=0
n−1 mn+1 1 n
1 X n+1
=
Bk+1 mn−k .
+ m +
n+1 2
n + 1 k=1 k + 1
Sn (m) =
We use the following lemmas in the proof of Theorem 2.
Lemma 5. Let r, k ∈ N with r > k and x ≥ 1. We have
X
nr ϕk (n) ≤ xr+1 .
n≤x
4
Proof. We have
X
nr ϕk (n) =
n≤x
X
nr−k
n≤x
=
X
X
µ(d)dk
d|n
µ(d)dk dr−k + (2d)r−k + · · · + (d⌊x/d⌋)r−k
d≤x
=
X
µ(d)dr
d≤x
≤x
r+1
X
j r−k
j≤x/d
.
Lemma 6. Let r ∈ N and x > e. We have
X
nr ϕ(n) =
n≤x
xr+2
+ O(xr+1 (log x)2/3 (log log x)4/3 )
(r + 2)ζ(2)
(as x → ∞),
where the implied constant depends on r.
Proof. We can obtain the lemma by the estimate [11]
X
ϕ(n) =
n≤x
x2
+ O(x(log x)2/3 (log log x)4/3 )
2ζ(2)
and the partial summation formula.
3
Proof of Theorem 1 and Theorem 2
Proof of Theorem 1. We have
n X
j=1
j
gcd(n, j)
a
X 1
=
da
d|n
=
X 1
da
d|n
n
X
ja
j=1
gcd(j,n)=d
X
k≤n/d
gcd(k,n/d)=1
5
(kd)a =
X
d|n
X
k≤n/d
gcd(k,n/d)=1
ka.
By Lemma 4 we have
X
X
ka =
ka
k≤N
k≤N
gcd(k,N )=1
=
X
d|N
X
µ(d) =
d|gcd(k,N )
X
da µ(d)
d|N
X
ma
m≤N/d
a+1
a
N
1 N
1
+
+
d µ(d)
a+1 d
2 d
a
a−k a−1 N
1 X a+1
Bk+1
+
a + 1 k=1 k + 1
d
a X
a X
N
N
N
=
µ(d) +
µ(d)+
a+1
d
2
d|N
Na
+
a+1
=N
a
d|N
a−1 X
k=1
k
X
a+1
d
Bk+1
µ(d)
k+1
N
1
1
1
ϕ + τ0 +
a+1
2
a+1
d|N
a−1 X
k=1
a+1
Bk+1 ϕk (N ).
k+1
Hence we obtain
n X
a
j
La (n) = n
gcd(n, j)
j=1
a−1 X n 2a 1
1
1 X a+1
=
Bk+1 ϕk (n/d) · da
ϕ + τ0 +
·
d
a+1
2
a + 1 k=1 k + 1
a
d|n
= (Ma ∗ Ida )(n).
6
Proof of Theorem 2. By Lemma 5, Lemma 6 and Theorem 1, we have
X
X
X
X
La (n) =
(Ma ∗ Ida )(n) =
da
Ma (m)
n≤x
n≤x
=
X
d≤x
d
X
a
d≤x
=
m
2a
m≤x/d
1
a+1
X
da
d≤x
X
m≤x/d
a−1 1
1 X a+1
1
Bk+1 ϕk (m)
ϕ + τ0 +
a+1
2
a + 1 k=1 k + 1
m2a ϕ(m) + O(xa+1 )+
m≤x/d
a−1 X
X
1 X a+1
Bk+1
da
m2a ϕk (m)
+
a + 1 k=1 k + 1
d≤x
m≤x/d
2a+2
1 X a
x
1
=
+
d
a + 1 d≤x
(2a + 2)ζ(2) d
2a+1
x
2/3
4/3
+
(log(x/d)) (log log(x/d))
+O
d
X
a+1
a
2a+1
+ O(x ) + O
d (x/d)
d≤x
2a+2
=
X 1
x
+ O(x2a+1 (log x)2/3 (log log x)4/3 )+
a+2
(a + 1)(2a + 2)ζ(2) d≤x d
+ O(x2a+1 ).
This implies the theorem.
4
Proof of Theorem 3
Proof of Theorem 3. Since we have
L−k (n) =
n
X
j=1
=
n
1
1 X (gcd(n, j))k
1 X k
=
=
d
(lcm(n, j))k
nk j=1
jk
nk
d|n
1 X k
d
nk
d|n
=
1 X
nk
d|n
X
i≤ n
d
)=1
gcd(i, n
d
X
i≤ n
d
gcd(i, n
)=1
d
1
ik d k
1
,
ik
7
n
X
j=1
gcd(j,n)=d
1
jk
we obtain
∞
X
∞
∞
X
1 X
L−k (n) =
nk
n=1
n=1
∞
X
1
= ζ(k)
jk
j=1
= ζ(k)
XX 1
1
=
ik
j k dk
d=1 j=1
X
1
ik
i≤ n
d
)=1
gcd(i, n
d
d|n
∞
X
i≤j
gcd(i,j)=1
∞
X
X
1
(
nk
n=1
i≤j
X
i≤j
gcd(i,j)=1
1
ik
1).
gcd(i,j)=1
ij=n
Also we have

 X
∞
X
1 

k 
n
 i≤j
n=1
gcd(i,j)=1
ij=n
and the relation [9, 1.2.8]



∞
X


1X 1 



1 = 1 +
1

k 
2
n

n=2
gcd(i,j)=1
ij=n

∞
X
X
1 

nk 
n=1
Therefore we obtain


∞
X
X
1 

nk 
 i≤j
n=1
gcd(i,j)=1
ij=n
This implies (2).
By the relation

gcd(i,j)=1
ij=n

 ζ(k)2
1
 = ζ(2k) .


2
2

ζ(k)
1
ζ(k)
1
1
 = 1 + 2 ζ(2k) − 1 = 2 1 + ζ(2k) .

X
ζ(k)2
ζ(k)
1+
−
L−k (n),
L−k (n) =
2
ζ(2k)
n>x
n≤x
X
8
the remaining task is to estimate the sum
X 1 X
L−k (n) =
nk
n>x
n>x
X
d|n
=
∞ X
X
d=1 h> x
d
=
∞ X
X
d=1 h> x
d
=
X
P
i≤ n
d
gcd(i, n
)=1
d
n>x
L−k (n). We have
∞
XX 1
1
=
k
ik
x (hd)
d=1
h> d
1 X 1
(hd)k j≤h j k
X
X
j≤h
gcd(j,h)=1
1
jk
µ(δ)
δ| gcd(j,h)
1 X X µ(δ)
(hd)k
mk δ k
h
δ|h m≤
δ
∞
∞
X
1 X X µ(δ) X 1
dk δ=1 x lk δ 2k m≤l mk
d=1
l> dδ
=
∞
X
q=1
∞
X
X1X 1
1 X k
d
µ(δ)
k
k
q 2k dδ=q
x l m≤l m
l> q
X 1
l1−k
1 X k
−k
d µ(δ)
ζ(k) −
=
+ O(l )
q 2k dδ=q
lk
k−1
q=1
l> xq
X 1
X 1 X
l1−k
k
−k
d µ(δ)
ζ(k) −
+ O(l ) +
=
2k
k
q
l
k
−
1
x
q<x
dδ=q
l> q
X 1
X 1 X
l1−k
k
−k
d µ(δ)
ζ(k) −
+ O(l )
+
k
q 2k dδ=q
k−1
x l
q≥x
l> q
=: S1 + S2 ,
9
say. We have
X 1
X 1 X
l1−k
k
−k
d µ(δ)
ζ(k) −
+ O(l )
S1 =
k
q 2k dδ=q
k−1
x l
q<x
l> q
X 1 X
ζ(k)
dk µ(δ)
=
(x/q)−k+1 + O((x/q)−k )−
2k
q
k
−
1
q<x
dδ=q
( xq )−2k+2
−2k+1
+ O((x/q)
)
−
(k − 1)(2k − 2)
X 1 X
ζ(k)
k
−k+1
−k
d µ(δ)
=
(x/q)
+ O((x/q) )
q 2k dδ=q
k−1
q<x
X 1 X µ(d) ζ(k)
−k+1
−k
=
(x/q)
+ O((x/q) )
k
k
q
d
k
−
1
q<x
d|q
−k+1 X
X X |µ(d)| X µ(d)
ζ(k)x
−k
−1
+O x
q
=
k − 1 q<x
dk
dk
q<x
d|q
d|q
−k+1
=
=
=
ζ(k)x
k−1
XX
j −1
d<x j< x
d
µ(d)
+ O(x−k+1 )
dk+1
ζ(k)x−k+1 X x µ(d)
+ O(x−k+1 )
log
k − 1 d<x
d dk+1
ζ(k)x−k+1 log x
+ O(x−k+1 )
(k − 1)ζ(k + 1)
and
X 1 X
X 1
l1−k
k
−k
d µ(δ)
ζ(k) −
+ O(l )
2k
k
q
l
k
−
1
x
q≥x
dδ=q
l> q
∞
X
X 1 X
1
l1−k
k
−k
d µ(δ)
ζ(k) −
≪
+ O(l )
2k
k
q
l
k
−
1
q≥x
dδ=q
l=1
≪
X
q −k
q≥x
X |µ(d)|
d|q
≪ x−k+1 .
dk
Therefore we obtain
X
n>x
L−k (n) =
ζ(k)x−k+1 log x
+ O(x−k+1 ).
(k − 1)ζ(k + 1)
This completes the proof.
10
References
[1] K. Alladi, On generalized Euler functions and related totients, in New Concepts in
Arithmetic Functions, Matscience Report 83, Madras, 1975.
[2] O. Bordellès, Mean values of generalized gcd-sum and lcm-sum functions, J. Integer
Sequences 10 (2007), Article 07.9.2.
[3] K. A. Broughan, The gcd-sum function, J. Integer Sequences 4 (2001), Article 01.2.2.
[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] H. W. Gould and T. Shonhiwa, Functions of GCD’s and LCM’s, Indian J. Math. 39
(1997), 11–35.
[6] S. S. Pillai, On an arithmetic function, J. Annamalai Univ. 2 (1933), 243–248.
[7] H. M. Srivastava and J. Choi, Series Associated with Zeta and Related Functions, Kluwer
Academic Publishers, 2001.
[8] Y. Tanigawa and W. Zhai, On the gcd-sum function, J. Integer Sequences 11 (2008),
Article 08.2.3.
[9] E. C. Titchmarsh, The Theory of the Riemann Zeta-function, Second Edition, Oxford
University Press, 1986.
[10] L. Tóth, A survey of gcd-sum functions, J. Integer Sequences 13 (2010), Article 10.8.1.
[11] A. Walfisz, Weylsche Exponentialsummen in der neueren Zahlentheorie, Leipzig BG
Teubner, 1963.
2010 Mathematics Subject Classification: Primary 11A25; Secondary 11N37.
Keywords: arithmetic function, lcm-sum function, least common multiple.
(Concerned with sequences A018804 and A051193.)
Received August 19 2013; revised versions received November 14 2013; November 23 2013;
December 19 2013. Published in Journal of Integer Sequences, December 27 2013.
Return to Journal of Integer Sequences home page.
11