Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 2, Article 33, 2004
ON THE SYMMETRY OF SQUARE-FREE SUPPORTED ARITHMETICAL
FUNCTIONS IN SHORT INTERVALS
GIOVANNI COPPOLA
DIIMA-U NIVERSITY OF S ALERNO
V IA P ONTE D ON M ELILLO
84084 F ISCIANO (SA) - ITALY.
gcoppola@diima.unisa.it
Received 17 March, 2003; accepted 02 April, 2004
Communicated by A. Fiorenza
A BSTRACT. We study the links between additive and multiplicative arithmetical functions, say
f , and their square-free supported counterparts, i.e. µ2 f (here µ2 is the square-free numbers
characteristic function), regarding the (upper bound) estimate of their symmetry around x in
almost all short intervals [x − h, x + h].
Key words and phrases: Symmetry, Square-free, Short intervals.
2000 Mathematics Subject Classification. 11N37, 11N36.
1. I NTRODUCTION AND S TATEMENT OF THE R ESULTS
In this paper we study the symmetry, in almost all short intervals, of square-free supported
arithmetical functions.
In our previous paper [3] we applied elementary methods, i.e. the Large Sieve, in order to
study the symmetry of distribution (around x) of the square-free numbers in "almost all" the
"short" intervals [x − h, x + h] (as usual, "almost all" means for all x ∈ [N, 2N ], except at most
o(N ) of them; "short" means that h = h(N ) and h → ∞, h = o(N ), as N → ∞).
As in [1], [2], [4], and [5] on (respectively) the prime-divisors function, von Mangoldt function, the divisor function and a wide class of arithmetical functions, we study the symmetry of
our arithmetical function f .
def
def
We define the "symmetry sum" of f as (here sgn(t) = t/|t|, sgn(0) = 0)
X
def
Sf± (x) =
f (n)sgn(n − x),
|n−x|≤h
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
The author wishes to thank Professor Saverio Salerno and Professor Alberto Perelli for friendly and helpful comments. Also, he wants to
express his sincere thanks to Professor Henryk Iwaniec, for his warm and familiar welcome during his stay in Rutgers University as a Visiting
Scholar (the present work was conceived and written during this period).
059-04
2
G IOVANNI C OPPOLA
and its mean-square as the "symmetry integral" of f :
2
X
X
def
.
If (N, h) =
f
(n)sgn(n
−
x)
x∼N |n−x|≤h
Here and hereafter x ∼ N stands for N < x ≤ 2N .
We will connect (in Theorem 1.1 and Theorem 1.2) If (N, h) and Iµ2 f (N, h), for suitable
f ; thus relating the symmetry of f to that of f on the square-free numbers (µ2 being their
characteristic function). Thus, we can estimate just one symmetry integral for two arithmetical
functions, whenever they agree on the square-free numbers.
As an example, for d(n) the divisor function, [4] estimates Id (N, h); then (using Theorem 1.5 to check the symmetry of d(n) in arithmetic progressions) in Theorem 1.3 we bound
Iµ2 d (N, h) = Iµ2 2Ω (N, h), and then obtain information on I2Ω (N, h) by Theorem 1.1 (here the
function 2Ω(n) is completely multiplicative, with 2Ω(p) = 2).
We denote with F the set of arithmetical functions f : N → C and with B the set of f ∈ F,
with |f | bounded (by an absolute constant); M denotes the multiplicative f ∈ F and A the
additive ones.
Also, we can define (∀α ∈]1, 2]) the set of "symmetric" arithmetical functions f as (where
we assume: ∀E > 0 supN |f | N E ):
N hα
def
cε
Sα = f ∈ F : sup If (N, h, k, q) 2 ε ∀k ≤ N , for some c, ε > 0
k N
q≤N cε
(the -constant is absolute, as well as c > 0), where we have set
2
X
X
x def
If (N, h, k, q) =
f (n)sgn n −
;
k x∼N |n−x/k|≤h/k
n≡0(q)
in the following, as here, we willl abbreviate n ≡ a(q) to mean n ≡ a(mod q).
def
We start giving a first link between f and µ2 f (in the sequel L = log N ):
Theorem 1.1.√Let N, h ∈ N, where h = h(N ), h/L2 → ∞ and h = o(N ) as N → ∞.
Assume J Lh , J → ∞ as N → ∞. Let kf k∞ := supN |f |.
If f is completely multiplicative then
X
N h
N h2
2
2
(i)
If (N, h) L max
d Iµ2 f
,
+
kf k2∞
2 d2
2
DJ
d
J
d∼D
and
(ii)
2
Iµ2 f (N, h) L max
DJ
X
2
d If
d∼D
N h
,
d2 d2
+
N h2
kf k2∞ .
J2
If f is completely additive then
X
√ 2
N h
N h2
2
2
(i)
If (N, h) L max
d Iµ2 f
,
+
+ N J hL kf k2∞
2 d2
2
DJ
d
J
d∼D
and
(ii)
2
Iµ2 f (N, h) L max
DJ
X
d∼D
2
d If
N h
,
d2 d2
+
N h2
2
+ N JL kf k2∞ .
2
J
We generalize Theorem 1.1 to additive and to multiplicative functions:
J. Inequal. Pure and Appl. Math., 5(2) Art. 33, 2004
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S YMMETRY OF S QUARE -F REE -S UPPORTED IN S HORT I NTERVALS
3
Theorem 1.2. Let f ∈ A ∪ M. Let N, h be natural numbers, with h = N θ (for 0 < θ < 1).
Assume that f is supported over the cube-free numbers and that ∀E > 0, kf k∞ N E , as
N → ∞. Choose ∀α ∈]1, 2] ε = θ(α−1)
> 0. Then
3
f ∈ Sα ⇔ µ 2 f ∈ Sα .
We give a concrete example: the function f (n) = 2Ω(n) (where Ω(n) is the total number of
prime divisors of n); in this case f ∈ Sα and µ2 f ∈ Sα ∀α > 23 , as we will prove directly, also
to detail the (more delicate) estimates
√ Theorem 1.3. Let N, h ∈ N, h = h(N ) ≥ L and h = o LN as N → ∞. Then
2
X X
Ω(n)
2
sgn(n − x) N h3/2 N ε
x∼N |n−x|≤h
and
2
X X
2
Ω(n)
N h3/2 N ε .
µ
(n)2
sgn(n
−
x)
x∼N |n−x|≤h
Remark 1.4. We explicitly remark that these bounds are non-optimal.
This result is obtained directly upon estimating the mean-square of the symmetry sum for the
divisor function over the arithmetic progressions:
√ Theorem 1.5. Let N, h ∈ N, with h = h(N ) → ∞ and h = o LN as N → ∞. Then,
uniformly ∀q ∈ N,
2
X X
d(n)sgn(n − x) N hL3 + N L2 log2 q,
|n−x|≤h
x∼N n≡0(q)
where the O-constant does not depend on q.
The paper is organized as follows
• In Section 2 we give the necessary lemmas;
• In Section 3 we prove our theorems.
2. L EMMAS
def
Lemma 2.1. Let f ∈ F be an arithmetical function, kf k∞ = supN |f (n)|.
Then, for N, h = h(N ) ∈ N and h → ∞, h = o(N ) as N → ∞:
2
X
X
X
x 2
a(d)
b(m)f (md )sgn m − 2 N hL2 kf k2∞ ,
√
d √
x∼N 2h<d≤ x+h
|m−dx2 |≤ dh2
uniformly ∀a, b ∈ B.
(Actually, for our purposes, kf k∞ =
J. Inequal. Pure and Appl. Math., 5(2) Art. 33, 2004
max
N −h≤n≤2N +h
|f (n)|).
http://jipam.vu.edu.au/
4
G IOVANNI C OPPOLA
Proof. Let Σ be the LHS. By a dyadic dissection and Cauchy inequality
2
X
X X
x
b(m)f (md2 )sgn m − 2 Σ L2 √ max√
a(d)
d
hD N
x∼N d∼D
|m− dx2 |≤ dh2
2
X X X
x
b(m)f (md2 )sgn m − 2 L2 √ max√ D
d
hD N
x∼N d∼D |m− x |≤ h
d2
d2
X
X
X
1.
kf k2∞ L2 √ max√ D
hD N
d∼D
N −h
N <x≤2N
≤m1 ,m2 ≤ 2N 2+h
d2
d
m1 d2 −h≤x≤m1 d2 +h
m2 d2 −h≤x≤m2 d2 +h
Clearly, the limitations on x imply m1 − 2h
≤ m2 ≤ m1 + 2h
(here we "reflect" the "sporadicd2
d2
√
2
ity") and this in turn, due to D h ⇒ d h, gives (∀m1 FIXED) O(1) possible values
to m2 . Hence Σ is bounded by
X
X
X
kf k2∞ hL2 √ max√ D
1 N hL2 kf k2∞ .
hD N
d∼D
N −h
≤m1 ≤ 2N 2+h
d2
d
|m2 −m1 |1
def
Lemma 2.2. Assume f ∈ F is completely additive and
√ kf k∞ = supN |f |. Let N, h ∈ N with
h = h(N ) → ∞, h = o(N ), as N → ∞. Then ∀J ≤ 2h
2
X X
X
x
a(d)
b(m)f (md2 )sgn m − 2 √
d x∼N d≤ 2h
|m− dx2 |≤ dh2
2

X
X
X
x 
2
2
L max D kf k∞
b(m)sgn m − 2 DJ
d d∼D x∼N |m− x |≤ h
d2
d2
2 
XX X
x  N h2
+
b(m)f (m)sgn m − 2  + 2 kf k2∞ ,
d J
d∼D x∼N |m− x |≤ h
d2
d2
uniformly ∀a, b ∈B (bounded arithmetical functions).
Proof. Let us call the left mean-square Σ. Then Σ is at most
2
X
X
X
x
N h2
2
2
b(m)f (md )sgn m − 2 + 2 kf k2∞ .
L max a(d)
DJ d J
x∼N
d∼D
|m− dx2 |≤ dh2
Since f is completely additive
2

X
X
X
x

Σ L2 max D 
b(m)f (m)sgn m − 2 DJ
d
x∼N d∼D |m− x |≤ h
d2
d2
J. Inequal. Pure and Appl. Math., 5(2) Art. 33, 2004
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S YMMETRY OF S QUARE -F REE -S UPPORTED IN S HORT I NTERVALS
5
2 
X
X
X
x
 N h2
b(m)sgn m − 2  + 2 kf k2∞ ,
+ kf k2∞
d J
x∼N d∼D |m− x |≤ h
d2
d2
by the Cauchy inequality. The lemma is thus proved.
Lemma 2.3. Let f be completely multiplicative.
Then, if N, h ∈ N, with h = h(N ) → ∞ and
√
h = o(N ) (as N → ∞ ), we have ∀J ≤ 2h
2
X X
X
x
a(d)
b(m)f (md2 )sgn m − 2 √
d x∼N d≤ 2h
|m− dx2 |≤ dh2
2


X X X
N h2
x 
kf k2∞ L2 maxD
b(m)f (m)sgn m − 2 + 2 
DJ
d J
d∼Dx∼N |m− x |≤ h
d2
d2
uniformly ∀a, b ∈B.
Proof. Let us call the left mean-square Σ. Then
2
X
X
X
N h2
x 2
2
Σ
L max a(d)
b(m)f (md )sgn m − 2 + 2 kf k2∞ ,
DJ d J
x∼N
d∼D
|m− dx2 |≤ dh2
and being f completely multiplicative we get
2
X
X
X
N h2
x
Σ kf k2∞ L2 max D
b(m)f (m)sgn m − 2 + 2 kf k2∞ ,
DJ
d J
d∼D x∼N |m− x |≤ h
d2
d2
by the Cauchy inequality. The lemma is thus proved.
√
Lemma 2.4. Let N, h, J and D be as in Lemma 2.2, with D = o( h). Then
2
X X X
x f (m)sgn m − 2 d d∼D x∼N |m− x |≤ h
2
2
d
d
2
2
X X X
h
2
d
f (m)sgn(m − y) +
+ N D kf k2∞ .
D
d∼D
y∼ N |m−y|≤h/d2
d2
2
Proof.
Write
r (0 ≤ r < d2 ) and let Σ be the left mean-square; since we have
P
P x = yd +
2
x∼N =
y∼ N2 +O (d ), then
d
2
X X X X
r h2
Σ
f (m)sgn m − y − 2 +
kf k2∞
d
D
d∼D 0≤r<d2 y∼ N |m−y− r |≤ h
d2
d2
d2
J. Inequal. Pure and Appl. Math., 5(2) Art. 33, 2004
http://jipam.vu.edu.au/
6
G IOVANNI C OPPOLA
2
(thus hD is due to x-range remainders); then correcting O(1) values of the m-sum gives as a
remainder (due to h-range)
!
X N
O
d2 2 kf k2∞ = O N D kf k2∞ .
d
d∼D
Gathering the estimates we then obtain the lemma.
3. P ROOF OF THE T HEOREMS
We start by proving Theorem 1.1.
Proof. In both cases (f completely additive or completely multiplicative) we use the hypothesis
on f to "separate variables" after having expressed the symmetry of f by that of µ2 f (for i),
say) and the symmetry of µ2 f by that of f (for ii), say). Thus, to prove i) it will suffice to
remember that each natural number n = md2 , where m and d are natural and µ2 (m) = 1, i.e.
m is square-free:
X
X
X
x
2
2
f (n)sgn(n − x) =
µ (m)f (md )sgn m − 2 .
d
√
x
h
|n−x|≤h
d≤ x+h |m− 2 |≤ 2
d
d
Instead, to prove ii) we simply use the following formula (see [7]):
X
µ2 (n) =
µ(d)
∀n ∈ N
d2 |n
to get
X
X
2
µ (n)f (n)sgn(n − x) =
µ(d)
√
|n−x|≤h
d≤ x+h
X
|m− dx2 |≤ dh2
x
f (md )sgn m − 2 .
d
2
As for the additional terms in the completely additive case, they come from the estimate of
the square-free symmetry sum as in [3].
Putting together Lemmas 2.1, 2.2, 2.3 and 2.4, the theorem is proved.
We now come to the proof of Theorem 1.2.
Proof. We first prove that f ∈ S ⇒ µ2 f ∈ S.
As before, we split at D (to be chosen); say (here [a, b] is the l.c.m. of a, b)
def
Σ=
X
d≤D
X
µ(d)
|
f (n)sgn(n − x)
|
n− x ≤ h
k
k
n≡0([q,d2 ])
=
X
d≤D
J. Inequal. Pure and Appl. Math., 5(2) Art. 33, 2004
µ(d)
X
X
t|[q,d2 ]
m− x ≤ h
kt2 g kt2 g
g=[q,d2 ]/t (m,g)=1
x
f (mt g)sgn m − 2
kt g
2
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S YMMETRY OF S QUARE -F REE -S UPPORTED IN S HORT I NTERVALS
7
and observe that, since f is supported over the cube-free numbers, Σ is
X
X
X
X
x
2
µ(j)
f (t g)
f (m)sgn m − 2
µ(d)
kt g
2
d≤D
t|[q,d ]
j|g
m− x ≤ h
kt2 g kt2 g
g=[q,d2 ]/t
m≡0(j)
X1
X
x
δ
kf k∞ N
d max2 f (m)sgn m −
,
2
d j,t≤qd kt[q, d ] d≤D
m− x 2 ≤ h 2
kt[q,d ] kt[q,d ]
m≡0(j)
by (see [7]) the estimate ∀δ > 0 d(n) nδ ; using the hypothesis f ∈ Sα we get, by Cauchy
inequality
α
X
X 1 X
N hα
2 Nh
d
|Σ|2 kf k2∞ N 2δ
d2 d≤D k 2 d4 N ε
k2N ε
x∼N
d≤D
Hence, it remains to prove that the mean-square of, say
X
X
def
f (n)sgn(n − x)
Σ0 =
µ(d)
√
h
x
D<d≤ x+h
|n− k |≤ k
n≡0([q,d2 ])
is
X
|Σ0 |2 x∼N
N hα
.
k2N ε
By the Cauchy inequality and a "sporadicity" argument as in the proof of Lemma 2.1,

2
X
X X
h

|Σ0 |2 kf k2∞
+1 

2
kd
√
x∼N
x∼N
h
k
D<d≤
2

+ kf k2∞ L2 √ max √ J
h
J
k
δ
N N
h2
h
+
k2 D2 k
N
XX

d∼J x∼N
+ N δ √ max √ J
h
J
k
N
X
m− k[dx2 ,q] ≤ k[dh2 ,q]

1
X
X
d∼J
N −h
<m≤ 2N2+h
k[d2 ,q]
k[d ,q]
h.
Hence
N hα
|Σ | 2 ε
k N
x∼N
X
0 2
N δ+ε h2−α
1−α δ+ε
+h N k .
D2
In order to obtain the above required estimate we need ε ≤
θ(α−1)
3
(for the II term in brackets)
4−α
and, comparing the mean-squares of Σ and of Σ , we come to the choice D = N 2(α−1) ε (I term).
This proves the first implication.
As for the reverse implication µ2 f ∈ S ⇒ f ∈ S we do not need the hypothesis on the
support of f and we use the same method (but using n = md2 instead of the identity for µ2 ).
This finally proves Theorem 1.2.
0
We now prove Theorem 1.5.
J. Inequal. Pure and Appl. Math., 5(2) Art. 33, 2004
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8
G IOVANNI C OPPOLA
Proof. First of all, let us call Iq (N, h) the mean-square to evaluate.
We will closely follow the proof of Theorem 1 in [4].
In fact, we start from the "flipping" property to write:


X
1X X
h
 X 
d(n)sgn(n − x) =
eq (rn) 2
1 sgn(n − x) + O √ + 1 ,
q r≤q
N
|n−x|≤h
d|n
|n−x|≤h
n≡0(q)
√
d≤ n
having used the orthogonality of the additive characters (see [7]). By our hypothesis on h (see
[4] for the details)
X
2X X X
d(n)sgn(n − x) =
eq (rn)sgn(n − x) + O(1)
q r≤q √ |n−x|≤h
|n−x|≤h
d≤ x
n≡0(q)
n≡0(d)
(here the constant is independent of q, like all the others following).
Next, write n − x = s to get (again by orthogonality)
X
X
eq (rn)sgn(n − x) = eq (rx)
eq (rs)sgn(s)
|n−x|≤h
n≡0(d)
|s|≤h
s≡−x(d)
X
eq (rx) X
ed (jx)
eq (rs)ed (js)sgn(s)
d j≤d
|s|≤h
X
= eq (rx)
cj,d (q, r)ed (jx),
=
j≤d
say, where
def 2i
cj,d (q, r) =
d
X
s≤h
sin 2πs
r j
+
q d
.
Here (w.r.t. the quoted [4, Theorem 1]) we have the dependence of the Fourier coefficients on q
and r; also, while cd,d = 0 there, here (by the estimate in of [6, Chap. 25])
cd,d (q, r) =
2i X
2πsr
q
sin
.
d s≤h
q
rd
Hence, this term’s contribute to the mean-square Iq (N, h) is:
2
!2
X X1
X 1 X
X
eq (rx)
cd,d (q, r)ed (jx) L N L2 log2 q
q
r
√
r≤q
x∼N r≤q
x∼N
d≤ x
(that is why we have this additional remainder, here!).
Henceforth, we can rely upon the proof of [4, Theorem 1], the only difference being the r, s
dependence:
(*)
2
2
X 1 X
XX
XXX X
1
e
(rx)
c
(q,
r)e
(jx)
c
(q,
r)e
(jx)
q
j,d
d
j,d
d
q
√
q
√
r≤q
r≤q x∼N d≤ x j<d
x∼N
d≤ x j<d
(we have used the Cauchy inequality).
J. Inequal. Pure and Appl. Math., 5(2) Art. 33, 2004
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S YMMETRY OF S QUARE -F REE -S UPPORTED IN S HORT I NTERVALS
9
We apply, then, exactly the same estimates; while there we get (we are quoting inequalities
to ease comparison)
X
X
2h
|cj,d |2 ≤
|cj,d |2 ≤
,
d
j<d
j≤d
here we have (the constant c > 0 is ininfluent)
X 1 X
r j
|cj,d (q, r)| = c 2
sgn(s1 )sgn(s2 )
e (s1 − s2 )
+
d
q
d
j≤d
j≤d
|s1 |,|s2 |≤h
X
c X
=
sgn(s1 )
sgn(s2 )eq (r(s1 − s2 )),
d
|s |≤h
X
2
|s1 |≤h
2
s2 ≡s1 (d)
whence, by (*), we get (see [4, Theorem 1]), ignoring the remainder O(N L2 log2 q):
Iq (N, h) X
X 1 X
1X
sgn(s1 )
sgn(s2 )eq (r(s1 − s2 ))
N L2
d
q r≤q
√
|s |≤h
|s1 |≤h
d≤ 2N
2
s2 ≡s1 (d)
X
X 1 X
sgn(s1 )
sgn(s2 )
d
√
|s |≤h
= N L2
d≤ 2N
|s1 |≤h
2
s2 ≡s1 (d)
s2 ≡s1 (q)


 X 1

NL 
h+
 h d
X
2
d≤
L
[d,q]≤ h
L
h
<d≤
L
√
2N
1
d

h2

+ h .
d

Thus
Iq (N, h) N hL3 + N L2 log2 q.
We now prove Theorem 1.3.
Proof. We first show the second estimate.
First of all, we observe that µ2 (n)2Ω(n) = µ2 (n)d(n), ∀n ∈ N; here we will apply the flipping
property of the divisor function as in [4].
Then, we will try to link our symmetry integral (for µ2 2Ω ) with that of d(n).
Writing µ2 (n) as before
X
X
X
µ(d)
d(n)sgn(n − x).
µ2 (n)d(n)sgn(n − x) =
√
d≤ x+h
|n−x|≤h
Splitting the range at D = D(x) ≤
def
Σ1 (x) =
√
x + h (to be chosen later), we treat, say
X
d≤D
J. Inequal. Pure and Appl. Math., 5(2) Art. 33, 2004
|n−x|≤h
n≡0(d2 )
µ(d)
X
d(n)sgn(n − x)
|n−x|≤h
n≡0(d2 )
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10
G IOVANNI C OPPOLA
by the Cauchy inequality and Theorem 1.5 to get
2
X
X X X
2
|Σ1 (x)| D
d(n)sgn(n − x)
x∼N
d≤D x∼N |n−x|≤h
2
n≡0(d )
2
3
N D L (h + L) N D2 hL3 ,
by our hypothesis on h. It remains to bound the mean-square of, say
X
X
def
µ(d)
d(n)sgn(n − x).
Σ2 (x) =
√
D<d≤ x+h
|n−x|≤h
n≡0(d2 )
√
We split again at 2h (to distinguish non-sporadic and sporadic terms).
Since by the classical estimate d(n) nε (see [7]; here ε > 0 will not be the same at each
occurrence) we estimate trivially (the non-sporadic terms)
X
X
X hN ε
N h2 ε
d(n)sgn(n − x) N
µ(d)
2
2
d
D
√
√
|n−x|≤h
D<d≤ 2h
D<d≤ 2h
n≡0(d2 )
we get, together with (the sporadic terms, treated by Lemma 2.1)
2
X
X
X
µ(d)
d(n)sgn(n − x) N hN ε ,
√
√
|n−x|≤h
x∼N 2h<d≤ x+h
2
n≡0(d )
that
X
2
|Σ2 (x)| x∼N
N h2
+ N h N ε.
D2
Thus, comparing the mean-squares of Σ1 (x) and Σ2 (x) we make the best choice D = h1/4 ,
finally proving the second estimate.
Writing I2Ω for the symmetry integral of 2Ω , we apply Theorem 1.1 to this function; then, i)
gives us
X N h3/2
N h2
I2Ω (N, h) L2 max
d2 2 3 N ε + 2 N ε N h3/2 N ε ,
DJ
d d
J
d∼D
√
by the choice J = h. This gives the first estimate, hence finally proving Theorem 1.3.
R EFERENCES
[1] G. COPPOLA, On the symmetry of distribution of the prime-divisors function in almost all short
intervals, to appear.
[2] G. COPPOLA, On the symmetry of primes in almost all short intervals, Ricerche Mat., 52(1) (2003),
21–29.
[3] G. COPPOLA, On the symmetry of the square-free numbers in almost all short intervals, submitted.
[4] G. COPPOLA AND S. SALERNO, On the symmetry of the divisor function in almost all short
intervals, to appear in Acta Arithmetica.
[5] G. COPPOLA AND S. SALERNO, On the symmetry of arithmetical functions in almost all short
intervals, submitted.
[6] H. DAVENPORT, Multiplicative Number Theory, Springer Verlag, New York 1980.
J. Inequal. Pure and Appl. Math., 5(2) Art. 33, 2004
http://jipam.vu.edu.au/
S YMMETRY OF S QUARE -F REE -S UPPORTED IN S HORT I NTERVALS
11
[7] G. TENENBAUM, Introduction to Analytic and Probabilistic Number Theory, Cambridge Studies
in Advanced Mathematics, 46, Cambridge University Press, 1995.
J. Inequal. Pure and Appl. Math., 5(2) Art. 33, 2004
http://jipam.vu.edu.au/