de Bordeaux 19(2007), 1–25
Sign changes of error terms related to
arithmetical functions
parPaulo J. ALMEIDA
R´esum´e. Soit H(x) = P
n≤x φ(n)
n −
6
π2x. Motiv´e par une
con-jecture de Erd¨os, Lau a d´evelopp´e une nouvelle m´ethode et il a d´emontr´e que #{n≤T : H(n)H(n+ 1)<0} ≫ T.Nous consi-d´erons des fonctions arithm´etiquesf(n) =P
d|n bd
d dont l’addition
peut ˆetre exprim´ee commeP
n≤xf(n) =αx+P(log(x)) +E(x).
IciP(x) est un polynˆome,E(x) =−P
n≤y(x)
bn nψ
x n
+o(1) avec ψ(x) = x− ⌊x⌋ −1/2. Nous g´en´eralisons la m´ethode de Lau et d´emontrons des r´esultats sur le nombre de changements de signe pour ces termes d’erreur.
Abstract. LetH(x) =P
n≤x φ(n)
n −
6
π2x. Motivated by a
con-jecture of Erd¨os, Lau developed a new method and proved that #{n ≤T : H(n)H(n+ 1) <0} ≫ T. We consider arithmetical functions f(n) =P
d|n bd
d whose summation can be expressed as
P
n≤xf(n) =αx+P(log(x)) +E(x), whereP(x) is a polynomial,
E(x) =−P
n≤y(x)
bn nψ
x n
+o(1) andψ(x) =x− ⌊x⌋ −1/2. We generalize Lau’s method and prove results about the number of sign changes for these error terms.
1. Introduction
We say that an arithmetical functionf(x) has asign change on integers
atx =n, if f(n)f(n+ 1) <0. The number of sign changes on integers of
f(x) on the interval [1, T] is defined as
Nf(T) = #{n≤T, n integer :f(n)f(n+ 1)<0}.
We also define zf(T) = #{n≤T, n integer :f(n) = 0}. Throughout this work,ψ(x) =x− ⌊x⌋ −1/2 andf(n) will be an arithmetical function such
Manuscrit re¸cu le 8 janvier 2006.
that
f(n) =X d|n
bd
d for some sequence of real numbersbn.
The motivation for our work was a paper by Y.-K. Lau [5], where he proves that the error term,H(x), given by
X
n≤x φ(n)
n =
6
π2x+H(x)
has a positive proportion of sign changes on integers solving a conjecture stated by P. Erd¨os in 1967.
An important tool that Lau used to prove his theorem, was that the error termH(x) can be expressed as
(1) H(x) =− X n≤ x log5x
µ(n)
n ψ
x
n
+O
1 log20x
, (S. Chowla [3])
We generalize Lau’s result in the following way
Theorem 1.1. Suppose H(x) is a function that can be expressed as
(2) H(x) =− X
n≤y(x)
bn nψ
x
n
+O
1
k(x)
,
where each bn is a real number and
(i) y(x) increasing, x14 ≪y(x)≪ x
(logx)5+D2
, for some D >0, and
(3) X
n≤x
b4n≪xlogDx;
(ii) k(x) is an increasing function, satisfying limx→∞k(x) =∞. (iii) H(x) =H(⌊x⌋)−α{x}+θ(x), where α6= 0 andθ(x) =o(1).
Let ≺ ∈ {<,=,≤}. If #{1 ≤ n≤ T : αH(n) ≺0} ≫ T then there exists a positive constant c0 and c0T disjoint subintervals of [1, T], with each of
them having at least two integers, m and n, such that αH(m) > 0 and αH(n)≺0. In particular,
(1) #{n≤T :αH(n)>0} ≫T;
Theorem 1.2. Let f(n) be an arithmetical function and suppose the se-quence bn satisfies condition (3) and
(4) X
n≤x
bn=Bx+O
x
logAx
for some B real, D >0 and A >6 +D
2, respectively. Let α= P∞
n=1nbn2,
γb = lim x→∞
X
n≤x bn
n −Blogx
and H(x) =X n≤x
f(n)−αx+Blog 2πx
2 +
γb 2 .
If α 6= 0, then Theorem 1.1 is valid for the error term H(x). Moreover, if f(n) is a rational function, then, except when α = 0, or B = 0 and α is rational, we have
NH(T)≫T if and only if #{n≤T :αH(n)<0} ≫T.
Notice that this class of arithmetical functions is closed for addition, i.e., if f(n) and g(n) are members of the class then also is (f +g)(n). In the case considered by Lau, it was known that H(x) has a positive proportion of negative values (Y.-F. S. P´etermann [6]), so the second part of Theorem 1.2 generalizes Lau’s result. Another example is f(n) = φ(nn).
Using a result of U. Balakrishnan and Y.-F. S. P´etermann [2] we are able to apply Theorem 1.1 to more general arithmetical functions:
Theorem 1.3. Let f(n) be an arithmetical function and suppose the se-quence bn satisfies condition (3) and
(5)
∞
X
n=1
bn ns =ζ
β(s)g(s)
for some β real, D >0, and a function g(s) with a Dirichlet series expan-sion absolutely convergent forσ >1−λ, for someλ >0. Letα=ζβ(2)g(2) and
H(x) =
( P
n≤xf(n)−αx, if β <0,
P
n≤xf(n)−αx−
P⌊β⌋
j=0Bj(logx)β−j if β >0,
where the constants Bj are well defined. If α 6= 0, then Theorem 1.1 is valid for the error termH(x).
Theorem 1.3 is valid for the following examples, wherer6= 0 is real:
φ(n)
n
r
,
σ(n)
n
r
,
φ(n)
σ(n)
r
2. Main Lemma
The main tool used by Y.-K. Lau was his Main Lemma, where he proved that ifH(x) =P
n≤x φ(n)
n −
6
π2x then
Z 2T
T
Z t+h
t
H(u) du
2
dt≪T h,
for sufficiently largeT and any 1 ≤h ≪ log4T. Lau’s argument depends essentially on the formula (1). In this section, we obtain a generalization of Lau’s Main Lemma.
Main Lemma. Suppose H(x) is a function that can be expressed as (2) and satisfies conditions (i) and (ii) of Theorem 1.1. Then, for all largeT and h≤min logT, k2(T)
, we have
(6)
Z 2T
T
Z t+h
t
H(u) du
2
dt≪T h32. For any positive integerN, define
(7) HN(x) =−
X
d≤N bd
dψ
x
d
.
The Main Lemma will follow from the next result.
Lemma 2.1. Assume the conditions of the Main Lemma and takeD >0
satisfying condition (i). LetE = 4 +D2, then
(a) For any δ >0, largeT, any Y ≪T and N ≤y(T), we have
Z T+Y
T
(H(u)−HN(u))2 du≪ Y N1−δ +
Y
k2(T) +y(T +Y) (logT)
E ; (b) For all large T,N ≤y(T) and 1≤h≤min logT, k2(T)
, we have
Z 2T
T
Z t+h
t
HN(u) du
2
dt≪T h32 +N3(logN)E. Now we prove the Main Lemma:
Proof. Take N =T14 and δ >0 small. Cauchy’s inequality gives us
Z t+h
t
H(u) du
2
≤2
Z t+h
t
HN(u) du
2
+ 2
Z t+h
t
(H(u)−HN(u)) du
2
Since N = T14 then, for sufficiently large T, N3logEN ≪ T. So, using part (b) of Lemma 2.1 we have
Z 2T
T
Z t+h
t
HN(u) du
2
dt≪T h32 +N3logEN ≪T h 3 2.
Using Cauchy’s inequality and interchanging the integrals,
Z 2T
T
Z t+h
t
(H(u)−HN(u)) du
2
dt
≤h
Z 2T
T
Z t+h
t
(H(u)−HN(u))2 du
dt
≤h
Z 2T+h
T
Z min(u,2T)
max(u−h,T)
(H(u)−HN(u))2dt
!
du
≤h2
Z 2T+h
T
(H(u)−HN(u))2 du
≪h2
T+h N1−δ +
T+h
k2(T) +y(2T+h) (logT)
E
≪T+T h≪T h32
sincey(2T+h)≪ (logTT)E+1 and h≤min(logT, k2(T)). Hence
Z 2T
T
Z t+h
t
H(u) du
2
dt≪T h32.
3. Step I
In this section, we will prove part (a) of Lemma 2.1. Using expression (2) and Cauchy’s inequality, we obtain
Z T+Y
T
(H(u)−HN(u))2du≤2
Z T+Y
T
y(u) X
m=N+1
bm mψ
u
m
2
du+O
Y k2(T)
.
Letη(T, m, n) = max T, y−1(m), y−1(n)
, then
2
Z T+Y
T
y(u) X
m=N+1
bm mψ
u
m
2
du= 2 y(T+Y)
X
m,n=N+1
bmbn mn
Z T+Y
η(T,m,n)
ψu m
ψu n
The Fourier series of ψ(u) =u− ⌊u⌋ −12, when u is not an integer, is given by
(8) ψ(u) =−1
so we obtain 2
Now, the integral above is equal to 1
For the first term we get
Z T+Y
Part (a) of Lemma 2.1 will now follow from the next three lemmas. Lemma 3.1. Let E = 4 +D2 as in Lemma 2.1. Then
Lemma 3.2. If D >0 satisfies condition (3), then
X
N <m,n≤X
In order to finish the proof of part (a) of Lemma 2.1 we just need to take
X=y(T +Y) in the previous lemmas. Hence
Z T+Y
T
(H(u)−HN(u))2 du≪ Y
N1−δ +y(T +Y) (logT) E
+ Y
k2(T). ✷
Before we prove the three lemmas above, we need the following technical result
Lemma 3.4. Let bn be a sequence satisfying condition (3). Then
X
n≤N
b2n≪NlogD2 N,
X
n≤N
|bn| ≪Nlog D
4 N,
X
n≤N b2
n
n ≪(logN)
1+D 2 ,
X
n≤N
|bn|
n ≪(logN)
1+D 4 , X
n>N b4n
n2τ(n)≪
1
N1−δ, for anyδ >0.
Proof. Follows from Cauchy’s inequality, partial summations and the fact
that, for anyǫ >0,τ(n) =O(nǫ).
Remark. IfH(x) can be expressed in the form (2), then (9) |H(x)| ≤ X
n≤y(x)
|bn|
n +O
1
k(x)
≪(logx)1+D4 .
Proof of Lemma 3.2 : Since the arithmetical mean is greater or equal to the geometrical mean, we have
X
m,n≤X
|bmbn| ∞
X
k,l=1
1
kl(kn+lm) ≤2
X
m,n≤X
|bmbn| ∞
X
k,l=1
1
kl√knlm
≪ X
m≤X
|bm|
√
m
2
≪ X
m≤X
1 X
M≤X b2
M M
≪X(logX)1+D2.
Proof of Lemma 3.3 : For the second sum, take d = (m, n), m =dα and
n =dβ. Since kn =lm, then α|k and β|l. Taking k =αγ, we also have
l=βγ. As (10)
∞
X
k,l=1 kn=lm
1
kl =
1
αβ ∞
X
γ=1
1
γ2 =
π2
6
(m, n)2
Then,
X
N <m,n≤X
|bmbn|
N <m,n≤X
|bmbn|(m, n)2
The next step is to estimate the inner sum using H¨older inequality
X
To complete the proof of Lemma 3.3 we use Cauchy’s inequality and Lemma 3.4. For anyδ >0,
X
N <m,n≤X
|bmbn|
N <m,n≤X
|bmbn|
Proof of Lemma 3.1 : This lemma is a generalization ofHilfssatz6 in [9] of A. Walfisz. Notice first that
Like in [9] we begin by separating the interior sum into four terms:
2 <lm<kn
kn<lm<2kn
For the first term, we use Lemma 3.4,
X
From the forth inequality of Lemma 3.4, we get
X
The estimation of the third term is more complicated and we have to use a different approach. In this case, 1l < 2knm,so that
2 <lm<kn
<2 X as in the first term. If there exists an integer lwith kn
m −1< l <
Notice that the fractional part of kn
m is at least
We are left with the estimation of
X
inequality. Take 1≤P ≤Q≤X, then,
X
P≤m<2P m|bm|
X
Q≤n<2Q kn≡a modm
|bn|
n ≪
P Q
X
P≤m<2P
|bm|
X
Q≤n<2Q kn≡a modm
|bn|.
Next, we apply Cauchy’s inequality twice, first to the first sum on the right and afterwards to the second sum:
X
P≤m<2P
|bm|
X
Q≤n<2Q kn≡a modm
|bn|
!!2
≤ X
P≤M <2P
b2M X P≤m<2P
X
Q≤n<2Q kn≡a modm
|bn|
!2
≪PlogD2 P
X
P≤m<2P
X
Q≤n<2Q kn≡a modm
b2n
X
Q≤N <2Q kN≡a modm
1
!
≪PlogD2 P
X
P≤m<2P
1 + Qm
(k,m)
X
Q≤n<2Q kn≡a modm
b2n
!
.
Sincem≤2P ≤2Q, we have mQ ≥ 12. Using also (k, m)≤k, we obtain
1 + Qm
(k,m)
≤1 +Qk
m ≤
Q
m(k+ 2)≤3 Qk
m .
Therefore,
X
P≤m<2P
|bm|
X
Q≤n<2Q kn≡amodm
|bn|
!2
≪PlogD2 P
X
P≤m<2P 3Qk
m
X
Q≤n<2Q kn≡amodm
b2n
!
≪P3kQ P log
D 2 P
X
Q≤n<2Q
b2n X P≤m<2P
m|kn−a 1
!
≪kQ(logP)D2 X Q≤n<2Q
By a theorem of S. Ramanujan [7],P
n≤Xτ2(n)∼Xlog3Xand by another application of Cauchy inequality and condition (3), we get
X
Q≤n<2Q
b2n τ(kn−a)
2
≤ X
Q≤n<2Q b4n
X
Q≤n<2Q
τ2(kn−a)
≪QlogDQ X kQ−a≤N <2kQ−a
τ2(N)
≪kQ2logD+3X.
Therefore,
X
P≤m<2P
m|bm|
X
Q≤n<2Q kn≡a modm
|bn| n
≪ PQkQ(logP)D2
X
Q≤n<2Q
b2nτ(kn−a) 1 2
≪ PQkQ(logX)D2(kQ2logD+3X) 1 2
12
≪P k34(logX)1+ D
2.
The number of pairs of intervals of the form ([P,2P),[Q,2Q)) to be con-sidered is at most≪log2X, hence
X
a,k≤X 1
ak2 X
a<m≤X m|bm|
X
m<n≤X kn≡amodm
|bn|
n ≪
X
a,k≤X k34
ak2 X
P,Q
P(logX)1+D2
≪X(logX)4+D2 . The fourth term is treated as the third, hence
X
m≤n≤X
|bmbn| ∞
X
k,l=1 kn<lm<2kn
1
kl|kn−lm| ≪X(logX)
4+D 2 .
This completes the proof of Lemma 3.1.
4. Step II
In this section we prove part (b) of Lemma 2.1. From equation (8), we get
Z b
a
ψ(u) du= 1 2π2
∞
X
k=1
cos(2πkb)
k2 −
cos(2πka)
k2
Using the definition ofHN stated in (7), we obtain
After multiplying the terms inside the integral above, we obtain the follow-ing four terms that we will estimate below:
Notice that,
The third term is treated similarly to obtain
by Lemma 3.1. Similarly, obtained has some similarities with Lemma 3.3. We are going to use the same argument to prove:
(13) X
Sinced|mwe havem > hand so,h2P
h<d≤N
dP
m≤N d|m
|bm| m2
2
≪h1+δ. To estimate the first term we begin with H¨older inequality:
X
d≤h
d4 X m≤N d|m
|bm| m2
2
≤X
d≤h
d4 X m≤N d|m
|bm|4 m2
12 X
M≤N d|M
1
M2 32
.
The third sum isO d13
(similar to (11)). Then
X
d≤h
d4 X m≤N d|m
|bm|4 m2
12 X
M≤N d|M
1
M2 32
≪X
d≤h
d X
d≤m≤N d|m
|bm|4 m2
12
≪ X
d≤h d2
1 2h X
D≤h
X
D≤m≤N D|m
|bm|4 m2
i12
≪h32
X
m≤N
|bm|4 m2
X
D≤h D|m
1 1 2
≪h32 X m≤N
|bm|4 m2 τ(m)
12
.
The last part of Lemma 3.4 implies P
m≤N |bm|4
m2 τ(m) =O(1), where the underlying constant doesn’t depend onN. Therefore, we obtain inequality
(13) and part (b) of Lemma 2.1, follows. ✷
5. A general Theorem
In this section, we prove Theorem 1.1, from which the main Theorems 1.2 and 1.3, will be deduced.
Proof of Theorem 1.1 : From the Main Lemma, we have, for all large T
and h≤min logT, k2(T)
,
Z 2T
T
Z t+h
t
H(u) du
2
dt≪T h32.
Assume #{n ≤ T : αH(n) ≺ 0} ≫ T. Let c > 0 be a constant and T
set of these subintervals. We have #(D) > cT /3h. Notice that any two members of Dare separated by a distance of at least 2h.
Let M be the number of subintervals K in D for which there exists an integer n in K such that αH(n) ≺ 0 and αH(m) ≤ 0 for every integer
m∈(n, n+ 2h), and let S be the set of the corresponding values of n. Lemma 5.1. For some absolute constant c1, we haveM ≤c1 T
h32.
Proof. Since H(x) =H(⌊x⌋)−α{x}+θ(x), then
αH(x)−αH(⌊x⌋) =−α2{x}+αθ(x).
So, ifx is sufficiently large and not an integer then
(14) −5
4α
2
{x}< αH(x)−αH(⌊x⌋)<−3
4α
2
{x}.
Let n1 be the smallest integer such that any non integer x > n1 satisfies
condition (14). If #{n∈ S :n≥n1} = 0 thenM ≤n1, so, the lemma is
clearly true for sufficiently large T. Otherwise,
#{n∈ S :n≥n1} ≥M−n1 ≫M.
Take n∈ S with n≥n1 and t∈[n, n+h]. For any integerm ∈[t, t+h],
αH(m)≤0. Now, for any 1≤j < h,
Z [t]+j+1
[t]+j
H(u) du=
Z [t]+j+1
[t]+j
(H(u)−H([t] +j)) du+
Z [t]+j+1
[t]+j
H([t] +j) du.
Therefore, by (14),
Z [t]+j+1
[t]+j
αH(u) du <
Z 1
0
−3
4α
2x
dx+αH([t] +j)
<−3
8α
2,
becauseαH([t] +j)≤0. Since [t]≥n, we also have
Z [t]+1
t
αH(u) du <
Z 1
{t}
−3
4α
2x
dx+αH([t]) (1− {t})<0 and
Z t+h
[t]+h
αH(u) du <
Z {t}
0
−34α2x
dx+αH([t] +h){t} ≤0.
Hence,
Z t+h
t
H(u) du
≥
3
Take an integerr=r(T) such that 2r>(logT)3+D2. Using (6) and (9), we obtain
Z 2T
0
Z t+h
t
H(u) du
2
dt=
Z T
2r
0
Z t+h
t
H(u) du
2
dt
+ r
X
j=0
Z T
2j−1
T 2j
Z t+h
t
H(u) du
2
dt
≪ T
2rh
2(logT)2+D 2 +h
3 2
r
X
j=0
T
2j
≪T h32, due to h≤logT. On the other hand,
Z 2T
0
Z t+h
t
H(u) du
2
dt≥X n∈S
Z n+h
n
Z t+h
t
H(u) du
2
dt
≥ X
n∈S n≥n1
Z n+h
n
3
8|α|(h−1)
2
dt
≫M h3.
HenceM ≤c1 T
h32
for some absolute constant c1.
Take c0 =c/6h. If h is a suitably large integer such that c1T < c0T h
3 2, then there are at leastc0T intervalsK inDsuch thatαH(n)≺0 for some
integer n ∈ K and αH(m) > 0 for some integer m lying in (n, n+ 2h). Now, suppose #{n≤T :αH(n) ≤0} ≫T. TakeT sufficiently large and take the order relation ‘≺’ to be ‘≤’. Therefore, we have c0T integers m
in the interval [1,2T], for which αH(m) is positive. In this case, #{n≤T :αH(n)>0} ≫T.
If we don’t have #{n≤T :αH(n)≤0} ≫T, then #{n≤T :αH(n)>0}=T(1 +o(1)).
Hence, part 1 of Theorem 1.1 is proved. Next, we prove part 2. Take ‘≺’ to be ‘<’. Then, there exists a positive constant c0 and c0T disjoint
subintervals of [1, T], with each of them having at least two integers,mand
n, such thatH(m)>0 and H(n)<0. Therefore, in each of those intervals we have at least onel with either H(l) = 0 or H(l)H(l+ 1)<0. Whence,
6. A class of arithmetical functions
In this section, we consider arithmetical functions f(n), such that the sequencebnsatisfies conditions (3) and (4). We begin with some elementary results about this class of arithmetical functions. Using condition (4), we immediately obtain the following lemma:
Lemma 6.1. Let bn be a sequence of real numbers satisfying (4), for some constants B and A >1, then there exist constants γb and α such that
X
n≤x bn
n =Blogx+γb+O
1 logA−1x
,
(15)
∞
X
n=1
bn n2 =α,
(16)
X
n>x bn n2 =
B x +O
1
xlogA−1x
.
(17)
Next, we calculate the sum off(n) and describe the error termH(x). Lemma 6.2. Let bn be a sequence of real numbers as in Lemma 6.1, then
(18) X
n≤x
f(n) =X n≤x
X
d|n bd
d =αx−
Blog 2πx
2 −
γb
2 +H(x),
where,
(19) H(x) =− X n≤ x logC x
bn nψ
x
n
+O
1 logCx
+O
1 logA−C−1x
,
for any0< C < A−1. Proof. We have
X
n≤x
f(n) =X n≤x
X
d|n bd
d =
X
d≤x bd
d
X
n≤x d|n
1 = X m≤x
X
d≤x m
bd d.
We separate the double sum above in two parts. Let 0< C < A−1 and
y= logCx. Then
(20) X
m≤x
X
d≤x m
bd d =
X
n≤y
X
d≤x n
bd d +
X
d≤x y
bd d
X
y<n≤x d
In order to evaluate the first term on the right, we start with an application of formula (15):
X
Recall that by Stirling formula, P
n≤⌊y⌋logn=⌊y⌋log⌊y⌋ − ⌊y⌋+ log2⌊y⌋ +
For the second term, we get
X
O1y.Joining everything together, we obtain
Proof of Theorem 1.2 : We just have to show that H(x) satisfies the con-ditions of Theorem 1.1. From Lemma 6.2, for anyx
H(x)−H(⌊x⌋) =−α{x} −B
2 (log 2π⌊x⌋ −log 2πx) =−α{x}+B
2
{x}
x +O
1
x2
.
In Lemma 6.2, we also obtained
H(x) =− X n≤ x logC x
bn nψ
x
n
+O
1 logCx
+O
1 logA−C−1x
,
for any 0 < C < A− 1. Take C= 5 + D2, y(x) = x
logCx and k(x) = min logCx,logA−C−1x
. Since A > 6 + D/2, then C < A− 1 and
A−C −1 > 0. The first part of Theorem 1.2 now follows from Theo-rem 1.1.
Suppose that f(n) takes only rational values. In order to prove the second part of Theorem 1.2, we use the following result of A. Baker [1]. Proposition. Letα1, . . . , αnandβ0, . . . , βndenote nonzero algebraic num-bers. Then β0+β1logα1+· · ·+βnlogαn6= 0.
Using the result above, we obtain the next lemma.
Lemma 6.3. Let f(n) be a rational valued arithmetical function and sup-pose the sequencebnsatisfies condition (4) for some real B andA >1. Let r be a real number and suppose H(x) is given by (18). Then
(1) If B = 0 and α is irrational then#{n integer:H(n) =r} ≤1; (2) IfBis a nonzero algebraic number then#{ninteger:H(n) =r} ≤2; (3) If B is transcendental then there exists a constantC that depends on
r and on the function f(n), such that
#{n≤T, n integer:H(n) =r}<(logT)C.
Proof. Suppose thatB = 0 andαis irrational. Suppose also that there are two integers, sayM 6=N, such that H(M) =H(N). Then
X
n≤M
f(n)−αM +γb 2 =
X
n≤N
f(n)−αN+γb 2. But this implies thatα is rational, a contradiction.
Next, supposeB6= 0 is algebraic number and that there areM > N > Q
integers, satisfyingH(M) =H(N) =H(Q). We have
X
n≤M
f(n)−αM +Blog 2πM
2 +
γb 2 =
X
n≤N
f(n)−αN+Blog 2πN
2 +
which implies
α= B
M−N log
M N
+ 1
M−N
X
N <n≤M f(n).
Consequently
Blog
M N
M−N1
M Q
M1−Q
=
1
M−Q
X
Q<n≤M
f(n)− 1
M−N
X
N <n≤M f(n).
We are going to prove that (21)
M N
M1−N
6
=
M Q
M1−Q
.
SinceB is a nonzero algebraic number and the values off(n) are rational, for any integern, the proposition implies
Blog
M N
M−N1
M Q
M1−Q
6=
1
M−Q
X
Q<n≤M
f(n)− 1
M−N
X
N <n≤M f(n),
and so we get a contradiction, which implies #{ninteger :H(n) =r} ≤2, for any real number r. In fact, instead of proving (21), we are going to prove that
(22) MN−QQM−N < NM−Q,
for any positive integers M > N > Q. Clearly, this implies (21). The inequality (22) is just a particular case of the geometric mean-analytic mean inequality
(23)
n
Y
i=1
ui
!n1
≤ n1
n
X
i=1
ui,
where equality happens only if u1 = u2 = · · · = un. In fact, if we take n=M−Q,ui =M for 1≤i≤N−Qandui =QforN−Q < i≤M−Q, we derive
MN−QQM−NM−Q1
< 1
M−Q((N−Q)M+ (M−N)Q) =N.
Hence, we obtain (22) and part 2 of the Lemma.
Let Q < N < M be the three smallest positive integers in the above set, then
06=Blog
M N
M−N1
M Q
M1−Q
=
1
M−Q
X
Q<n≤M
f(n)− 1
M−N
X
N <n≤M f(n).
SupposeL is such thatH(L) =r. ThenL > N > Q, and as in part 2:
06=Blog
L N
L−N1
L Q
1
L−Q
=
1
L−Q
X
Q<n≤L
f(n)− 1
L−N
X
N <n≤L f(n).
After we cross multiply the two expressions above, we obtain
log
M N
M−N1
M Q
1
M−Q
=r1log
L N
L−N1
L Q
1
L−Q
,
for some rational r1. Therefore, there are four rational numbers r2, r3, r4
and r5, such that
Lr2 =Mr3Nr4Qr5.
Now, any prime dividingLmust divideM N Q. Notice that, ifpis a prime,
kis an integer andpk≤xthenk≤ logx
logp. Therefore, the number of integers smaller than x, which have all prime divisors smaller than M is smaller than (logx)π(M). This finishes the proof.
Except whenα= 0, orB= 0 andαis rational, we cannot havezH(T)≫ T. Hence, we obtain the second part of Theorem 1.2.
Example. We finish this section by proving that Theorem 1.2 is valid for the arithmetical function φ(nn).
Notice that
n φ(n) =
Y
p|n
1 + 1
p−1
=X
d|n µ2(d)
φ(d) .
Letbn= µ 2(n)n
φ(n) , thenf(n) = P
d|n bd
d. In [8], R. Sitaramachandrarao proved that
X
n≤x
µ2(n)n
φ(n) =x+O
x12
so condition (4) is satisfied for anyAand withB = 1. By Merten’s Theorem
Q
p|n
1−1p−1 ≤Q
p≤n
1−1p−1 ∼eγlogn, hence
X
n≤x
b4n=X n≤x
µ2(n) n
4
φ4(n) = X
n≤x
O log4n
=O xlog4x
,
and condition (3) is satisfied forD≥4. In this case,
α= ζ(2)ζ(3)
ζ(6) , γb =γ+
X
p
logp p(p−1) and
H(x) =X n≤x
n φ(n) −
ζ(2)ζ(3)
ζ(6) x+ logx
2 +
log 2π+γ+P
p
logp p(p−1)
2 .
Since B = 1 we can apply Theorem 6.3, and so z(T) ≤ 2. Therefore, if #{n≤T :αH(n)<0} ≫T, then NH(T)≫T.
7. Second class of arithmetical functions
Given a sequence of real numbersbn, and a complex numbers, we define the Dirichlet series B(s) = P∞
n=1nbns. In this section, we consider arith-metical functions f(n), such that the sequence bn satisfies conditions (3) and (5) for someD >0, β real and a function g(s) with a Dirichlet series expansion absolutely convergent forσ >1−λ, for someλ >0.
U. Balakrishnan and Y.-F. S. P´etermann [2] proved that:
Proposition. Letf(n)be a complex valued arithmetical function satisfying ∞
X
n=1
f(n)
ns =ζ(s)ζ
β(s+ 1)g(s+ 1),
for a complex numberβ, andg(s) having a Dirichlet series expansion g(s) =
∞
X
n=1
cn ns,
which is absolutely convergent in the half planeσ >1−λfor some λ >0. Let β0 be the real part of β. If
ζβ(s)g(s) = ∞
X
n=1
then there is a real number b, 0 < b < 1/2, and constants Bj, such that, taking y(x) =xexp −(logx)b
andα =ζβ(2)g(2),
X
n≤x
f(n) =
(
αx−P
n≤y(x)bnnψ x n
+o(1) ifβ0<0,
αx+P[β0]
j=0Bj(logx)β−j−Pn≤y(x) bnnψ x n
+o(1) ifβ0≥0,
The real version of the previous proposition allows us to prove Theo-rem 1.3:
Proof. Notice that, for any c >0, logc⌊x⌋= logcx−c{x}x logc−1x+O 1x
.
So,H(x) =H(⌊x⌋)−α{x}+o(1). From the previous proposition, there is an increasing functionk(x), with limx→∞k(x) =∞, such that
H(x) =− X n≤y(x)
bn nψ
x
n
+O
1
k(x)
,
where y(x) =xexp −(logx)b
, for some 0 < b < 1/2. Hence, the result
follows from Theorem 1.1.
Acknowledgements: I would like to thank my advisor Andrew Granville for his guidance and encouragement in this research and to the Referee for his useful comments.
References
[1] A. Baker, Linear forms in the logarithms of algebraic numbers (III). Mathematika 14
(1967), 220–228.
[2] U. Balakrishnan, Y.-F. S. P´etermann,The Dirichlet series ofζ(s)ζα(s+ 1)f(s+ 1): On an error term associated with its coefficients. Acta Arithmetica75(1996), 39–69.
[3] S. Chowla,Contributions to the analytic theory of numbers. Math. Zeit.35(1932), 279–
299.
[4] G. H. Hardy, E. M. Wright,An Introduction to the Theory of Numbers. 5th ed. Oxford 1979.
[5] Y.-K. Lau,Sign changes of error terms related to the Euler function. Mathematika 46
(1999), 391–395.
[6] Y.-F. S. P´etermann,On the distribution of values of an error term related to the Euler function. Th´eorie des Nombres, (Quebec, PQ, 1987), 785–797.
[7] S. Ramanujan,Collected papers. Cambridge, 1927, 133–135.
[8] R. Sitaramachandrarao,On an error term of Landau - II. Rocky Mount. J. of Math.
152 (1985), 579–588.
[9] A. Walfisz,Teilerprobleme II. Math. Zeit.34(1931), 448–472.
Paulo J.Almeida
Departamento de Matem´atica Universidade de Aveiro
Campus Universit´ario de Santiago 3810-193 Aveiro, Portugal
