Monatsh. Math. 145, 131–144 (2005)
DOI 10.1007/s00605-004-0288-6
On the Mean Value of the Index of Composition
of an Integer
By
J. M. De Koninck1; and I. Kátai2;y
2
1
Universite Laval, Quebec, Canada
E€
otv€
os Loránd University, Budapest, Hungary
Communicated by J. Schoissengeier
Received January 15, 2004; accepted in revised form July 26, 2004
Published online March 25, 2005 # Springer-Verlag 2005
n
Abstract. For each integer n 5 2, let ðnÞ ¼ loglogðnÞ
be the index of composition of n, where
Q
we write ð1ÞP
¼ ð1Þ ¼ 1. We obtain Psharp estimates for
ðnÞ ¼ pjn p. For convenience,
P
P
pffiffi
pffiffi
x 4 n 4 xþ x ðnÞ and
n 4 x ðnÞ, as well as for
x 4 n 4 xþ x 1=ðnÞ and
n 4 x 1=ðnÞ. Finally
we study the sum of running over shifted primes.
2000 Mathematics Subject Classification: 11A25, 11N25, 11N37
Key words: Arithmetic function, prime factorisation
1. Introduction
n
be the index of composition of n,
For each integer
n 5 2, let ðnÞ ¼ loglogðnÞ
Q
where ðnÞ ¼ pjn p. For convenience, we write ð1Þ ¼ ð1Þ ¼ 1. The index of
composition of an integer measures essentially the multiplicity of its prime factors.
It was shown by De Koninck and Doyon [1] that the average order of ðnÞ is 1 and
more precisely that
X
x
:
ð1Þ
ðnÞ ¼ x þ O
log x
n4x
P
1
A similar result was obtained for n 4 x ðnÞ
.
Here, we first prove a short interval version of (1) using
P a result of Filaseta
pffiffi
and
Trifonov
[2].
Then,
we
provide
estimates
for
x 4 n 4 xþ x ðnÞ and
x P
pffiffi 1=ðnÞ with an error term O
,
where
r
is
any
given
positive
x 4 n 4 xþ x
logrþ1 x
integer.
From these, we deduce a sharpening of the estimate (1) and similarly for
P
1
n 4 x ðnÞ. Finally we study the sum of running over shifted primes.
Research supported in part by a grant from NSERC.
y
Research supported by the Applied Number Theory Research Group of the Hungarian Academy of
Science and by a grant from OTKA.
132
J. M. De Koninck and I. Kátai
2. Main Results
Theorem 1. Let hðxÞ ¼ x1=5 log 3 x. Then
X
hðxÞ
ððnÞ 1Þ ¼ O
log x
x 4 n 4 xþhðxÞ
ð2Þ
and
1
hðxÞ
¼O
:
1
ðnÞ
log x
x 4 n 4 xþhðxÞ
X
ð3Þ
Remark. Observe that since ðnÞ 5 1, the summands on the left hand side of
(2) and (3) are non negative.
Theorem 2. Given any positive integer r, there exist computable constants
c1 ; c2 ; . . . ; cr ; c02 ; . . . ; c0r such that
X
1
1
1
1
1
pffiffiffi
þ c2
þ þ cr
þO
ðnÞ ¼ 1 þ c1
pffiffi
log x
log r x
x
log 2 x
log rþ1 x
x 4 n 4 xþ x
ð4Þ
and
X
x
x
x
þ c02
þ þ c0r
þO
ðnÞ ¼ x þ c1
2
log x
log r x
log x
n4x
x
:
log rþ1 x
ð5Þ
Theorem 3. Given any positive integer r, there exist computable constants
d1 ; d2 ; . . . ; dr ; d20 ; . . . ; dr0 such that
X
1
1
1
1
1
1
pffiffiffi
¼ 1 þ d1
þ d2
þO
þ þ dr
pffiffi ðnÞ
log x
log r x
x
log 2 x
log rþ1 x
x 4 n 4 xþ x
ð6Þ
and
X 1
x
x
x
x
0
0
¼ x þ d1
þ d2
þO
þ þ dr
:
ðnÞ
log x
log r x
log 2 x
log rþ1 x
n4x
ð7Þ
Theorem 4. Given any positive integer r, there exist computable constants
2 ; 3 ; . . . ; r such that
X
1
ðp þ 1Þ
ðx þ x2=3 Þ ðxÞ x < p 4 xþx2=3
1
1
1
1
ð8Þ
þO
¼ 1 þ 2
þ 3
þ þ r
log r x
log 2 x
log 3 x
log rþ1 x
Mean Value of the Index of Composition of an Integer
and
X
ðp þ 1Þ ¼
p4x
x
x
x
x
þ 2
þ 3
þ þ r
þO
2
3
log x
log r x
log x
log x
133
x
:
log rþ1 x
ð9Þ
3. Preliminary Results
Let N stand for the set of positive integers. A number n 2 N is said to be
squarefull (or powerful) if n ¼ 1 or if pjn ¼) p2 jn.
Lemma 1. Let be an arbitrary non negative integer. As y ! 1,
X
ðiÞ
k5y
k squarefull
log k
log y
pffiffiffi :
k
y
Moreover, given any real number > 12,
X
1
< þ1:
k
k squarefull
ðiiÞ
Proof. It is well known that
NðtÞ :¼
X
pffi
1 ¼ c t þ Oðt1=3 Þ;
n4t
n squarefull
c¼
ð3=2Þ
2:1732;
ð3Þ
where stands for the Riemann Zeta Function (see for instance Ivic and Shiu [4]).
Therefore, integration by parts yields
X log k ð 1 log t
¼
dNðtÞ
k
t
y
k5y
k squarefull
1 ð 1
ð1
log t
log t
log 1 t
NðtÞ þ
¼
NðtÞ
dt
NðtÞ dt
t
t2
t2
y
y
y
log y
log y
log y
log y
þ
O
¼ c pffiffiffi þ O
p
ffiffi
ffi
pffiffiffi ;
y
y
y
y2=3
which completes the proof of (i). For a proof of (ii), see De Koninck and Doyon [1].
For each positive integer k and each real number y, let
X
X
2 ðnÞ and MðyÞ ¼ Mðyj1Þ ¼
2 ðnÞ;
MðyjkÞ :¼
n4y
ðn;kÞ¼1
n4y
ð10Þ
134
J. M. De Koninck and I. Kátai
where stands for the Moebius function. For each positive integer k, let Bk be the
set of all positive integers
P all of whose prime factors divide k, and let 1 2 Bk .
Moreover, let ðnÞ :¼ p kn for each integer n 5 2 and set ð1Þ ¼ 0.
Walfisz [6] proved that there exists a positive constant A such that
pffiffiffi
6
MðxÞ ¼ 2 x þ TðxÞ;
TðxÞ x expfAð log xÞ3=5 ð log log xÞ1=5 g: ð11Þ
We shall use this result in the proof of the next lemma. But first, for each positive
integer k and real number > 0, let us set
Y
X ð1ÞðvÞ
Y
1 1
1 1
ðkÞ ¼
;
1þ
1
¼
ðkÞ
¼
: ð12Þ
p
p
v
v2B
pjk
pjk
k
Lemma 2. Let be a fixed positive integer. Then for each positive integer k,
P
(i) MðyjkÞ ¼ v 2 Bk ð1ÞðvÞ M yv ,
pffiffiffi
(ii) MðyjkÞ ¼ 62 ðkÞy þ Oð y expfð log yÞ3=5 ð log log yÞ1=5 gÞ uniformly
for k 4 ð log yÞ as y ! 1, for some positive constant .
Proof. (i) follows from the identity
Q 1
1
1
X
X
2 ðnÞ
2 ðnÞ Y
1
1
p 1 þ ps
1 s þ 2s ¼Q
¼
1
ns
ns pjk
p
p
pjk 1 þ ps
n¼1
n¼1
ðn;kÞ¼1
which is valid for all Rs > 1, and the uniqueness of Dirichlet series representation.
In order to prove (ii), we first observe that
log k
ðk 5 3Þ;
ðkÞ < exp 2
log log k
where ðkÞ stands for the number of positive divisors of k (see for instance Nicolas
and Robin [5]). It follows from this estimate that, given any > 0, 2 N, and
k 4 ð log yÞ
, as y becomes large, we have
log k
log log y
4 exp 2
:
ð13Þ
ðkÞ 4 ðkÞ 4 exp 2
log log k
log log log y
Now from (i) and (12), we have
0
1
0
1
B X 1C
B X y
C
6
C þ OB
MðyjkÞ ¼ 2 ðkÞy þ OB
y
T
C
@
A
@
v
v A
v4y
v 2 Bk
v>y
v 2 Bk
¼
6
ðkÞy þ OðE1 Þ þ OðE2 Þ;
2
say. First observe that
X 1
v>y
v 2 Bk
v
¼
X 1 1
1 X 1
ðkÞ
< 1
4 1 ;
1
v v
y
v
y
v>y
v>y
v 2 Bk
v 2 Bk
ð14Þ
Mean Value of the Index of Composition of an Integer
135
so that using (13), we get
E1 ðkÞy
y1=4 ;
ð15Þ
assuming that < 1=5, say.
To estimate E2 , we proceed as follows. If v 4 y1 for some fixed real number
> 0, then, in view of Walfisz’s result (11),
rffiffiffi
T y 4 y expfA ð log yÞ3=5 ð log log yÞ1=5 g
ð16Þ
v v
pffiffi
for some positive constant A , while the left hand side of (16) is 4 D yv if v 4 y,
for some constant D > 0, so that we may write
( pffiffi
3=5
y
ð log log yÞ1=5 g if v 4 y1 ;
ð log yÞ
v expfA
ffiffiy
p
E2 4
ð17Þ
D v
if v 4 y:
Now, using the definition of and then (13), we have
log log y
pffiffiffi X 1
pffiffiffi
pffiffiffi
p
ffiffi
ffi
:
y
4 y 1=2 ðkÞ 4 y exp 2
log log log y
v
v 4 y1
ð18Þ
v 2 Bk
On the other hand, if v > y1 , then
2
v ¼ v1 v > yð1Þ v ;
in which case, again using (13), we have
pffiffiffi X
y
1
pffiffiffi X 1
pffiffiffi <
y
2
=2
ð1Þ
=2
v
v
y
v 2 Bk
v > y1
v 2 Bk
2
1
4 y2ð1ð1Þ Þ
2
1
2ð1ð1Þ Þ
4y
=2 ðkÞ
exp 2
log log y
:
log log log y
ð19Þ
Choosing small and combining (18) with (19), we get from (17) that there exists
a positive constant such that
pffiffiffi
E2 y expfð log yÞ3=5 ð log log yÞ1=5 g:
ð20Þ
Gathering (15) and (20), (ii) follows, and Lemma 2 is proved.
For each positive integer k and each real number y, let
X
pþ1
ðyjkÞ :¼
2
k
p4y
p 1 ðmod kÞ
pþ1 k
;k ¼1
and further let ðx; D; ‘Þ be the number of primes p 4 x such that p ‘ ðmod DÞ.
136
J. M. De Koninck and I. Kátai
Lemma 3. Let k be a squarefull integer. Then
X X
ðÞðÞðy; k2 ; 1Þ:
ðyjkÞ ¼
jk ð;kÞ¼1
Proof. Since
P
djn
ðdÞ is always 0, unless n ¼ 1 and since
X
1 if n is squarefree;
ðdÞ ¼
0
otherwise;
2
d jn
it follows that
X
ðyjkÞ ¼
X
pþ1
p4y
p 1 ðmod kÞ
¼
X X
k
ðÞ
X
ðÞ
2 jpþ1
k
;k
X
ðÞðÞ
1;
p4y
p 1 ðmod k2 Þ
jk ð;kÞ¼1
which completes the proof of Lemma 3.
Let stand for Euler’s function and let liðyÞ :¼
Ðy
dt
2 log t.
Lemma 4. Let " > 0 be a small number. Let and r be fixed positive integers
7
and let y12þ" 4 H 4 y. Then, as y ! 1,
liðy þ HÞ liðyÞ Y
1
H
1
þO
ðy þ HjkÞ ðyjkÞ ¼
1
k
qðq 1Þ
k log rþ2 y
q6 j k
uniformly for k 4 ð log yÞ
.
Proof. From Lemma 3 and Huxley’s Theorem for primes in arithmetic progressions (see Huxley [3]), we have that, given any number c0 > 0,
X X
ðy þ HjkÞ ðyjkÞ ¼
ðÞðÞððy þ H; k2 ; 1Þ ðy; k2 ; 1ÞÞ
jk ð;kÞ¼1
¼ ðliðy þ HÞ liðyÞÞ
0
BX
B
þ OB
@ jk
0
BX
B
þ OB
@ jk
X
ð;kÞ¼1
4 ðlog yÞc0
X
ð;kÞ¼1
> ðlog yÞc0
X X ðÞðÞ
ðk2 Þ
jk ð;kÞ¼1
1
pffiffiffiffiffiffiffiffiffiffi C
HjðÞj
C
expð log yÞC
2
A
ðk Þ
1
2H C
C
C:
k2 A
Mean Value of the Index of Composition of an Integer
137
Clearly this last error term is
X1
H
1
H
1
;
c0
k ð log yÞ jk k log rþ1 y
for k 4 ð log yÞ
, provided c0 is chosen sufficiently large, whereas the first error
term is
pffiffiffiffiffiffiffiffiffiffi
H expð log yÞ X jðÞj
H
1
:
ðkÞ
ðÞ
k log rþ1 y
jk
Gathering these estimates and observing that
X X ðÞðÞ
Y
1 X ðÞ X ðÞ
1
;
¼
¼
1
ðk2 Þ
ðkÞ jk ð;kÞ¼1 ð2 Þ q6 j k
qðq 1Þ
jk ð;kÞ¼1
the proof of Lemma 4 is complete.
Lemma 5. Let hðxÞ ¼ x1=5 log 3 x and let EðxÞ be the number of integers
n 2 ½x;
þ hðxÞ
which can be written as n ¼ p2 for some prime number
pxffiffiffiffiffiffiffiffi
ffi
p 5 hðxÞ. Then
EðxÞ hðxÞ
:
log x
Proof. This estimate is an immediate consequence of Theorem A, which is an
unpublished result of Michael Filaseta who kindly communicated it to the first author.
Theorem A (Filaseta). Let k be an integer 5 2. Let gðxÞ be a function satisfying 1 4 gðxÞ 4 log x for x sufficiently large, and set
h ¼ x1=ð2kþ1Þ gðxÞ3 :
Then the number of k-free numbers in the interval ðx; x þ h is
h
hð log xÞ
h
þO
:
þO
ðkÞ
gðxÞ
gðxÞ3
Proof. We modify the argument given in Section 4 of Filaseta and Trifonov [2].
Take z ¼ " log x, where " ¼ "ðkÞ > 0 is sufficiently small. A sieve-of-Eratosthenes
argument gives that the number of integers in ðx; x þ h free from divisors of the
form pk with p 4 z is
Y
Y
1
1
z
1 k þ Oð2 Þ ¼ h
1 k þ Oðh=zÞ þ Oð2z Þ
h
p
p
p4z
p
Y
1
h
þ Oðx" log 2 Þ
¼h
1 k þO
p
log
x
p
h
h
þO
:
¼
ðkÞ
gðxÞ
We now obtain an upper bound for the number of these integers that are divisible
by the kth power of a prime > z. If p > 2x1=k , then pk > 2k x > x þ h and, hence, pk
138
J. M. De Koninck and I. Kátai
does not divide integers in the interval ðx; x þ h. Let c be a sufficiently large
constant (for the purpose of having a condition in Theorem 7 of [2] hold later
in the argument). We break up our consideration of primes into two parts, those
primes p in I ¼ ðz; ch and those primes p in J ¼ ðch; 2x1=k . If Mu denotes the
number of integers in ðx; x þ h divisible by uk, then
X
X h
X h þ1 4
Mp 4
þ ðchÞ
k
p
pk
p5z
p2I
p2I
4
h
þ Oðh=log xÞ ¼ Oðh=log xÞ ¼ Oðh=gðxÞÞ:
z
Thus, it remains to show that
X
hð log xÞ
h
:
þO
Mp ¼ O
gðxÞ
gðxÞ3
p2J
P
It suffices to obtain the same bound for
u 2 J Mu (where u runs through the
integers in J). Since such u exceed h, we deduce that Mu 2 f0; 1g and Mu ¼ 1
precisely when there is a multiple of uk in ðx; x þ h. We use disjoint intervals Ji ,
with 1 4 i 4 r, of the form ðN; 2N satisfying
r
[
J
Ji ðch=2; 2x1=k :
i¼1
In particular, r log x. Fix Ji ¼ ðN; 2N. Note that
ch
4 N 4 x1=k :
2
As in [2], we obtain that
X
Mu jfu 2 ðN; 2N : k f ðuÞk < gj;
u 2 Ji
where f ðuÞ ¼ x=uk , ¼ hN k , and k f ðuÞk denotes the distance from f ðuÞ to the
nearest integer. We appeal to Theorem 7 from [2] (with s ¼ k and X ¼ x).
Observe that the condition there on holds as c is sufficiently large. We obtain
that
X
2
Mu x1=ð2kþ1Þ þ x1=ð6kþ3Þ N ð6k þk1Þ=ð6kþ3Þ
u 2 Ji
x1=ð2kþ1Þ þ hx1=ð6kþ3Þ N 1=3 :
Summing over i, we deduce
X
Mu x1=ð2kþ1Þ log x þ hx1=ð6kþ3Þ h1=3
u2J
x1=ð2kþ1Þ log x þ x1=ð2kþ1Þ gðxÞ2 :
The desired upper bound now follows, thus completing the proof of
Theorem A.
Mean Value of the Index of Composition of an Integer
139
4. Proof of Theorem 1
Each positive integer n can be written uniquely as
n ¼ km;
2 ðmÞ ¼ 1;
ðm; kÞ ¼ 1;
k squarefull;
so that
ðnÞ ¼ ðkÞm ¼
ðkÞ
ðkÞ
mk ¼
n
k
k
and therefore
ðnÞ ¼
log n
log n þ log
ðkÞ
k
¼
1
1
log ðk=ðkÞÞ
log n
:
ð21Þ
From this it follows in particular that there exists a positive constant B1 such that
1 4 ðnÞ 4 1 þ B1
log ðk=ðkÞÞ
log n
if k < x1=3 , say. Thus, writing kðnÞ for the squarefull part of n, we have
X
B1 X
ððnÞ 1Þ 4
ð log ðk=ðkÞÞÞ #fn 2 ½x; x þ hðxÞ : kðnÞ ¼ kg
log x k 4 hðxÞ
x 4 n 4 xþhðxÞ
X
ððnÞ 1Þ
þ
x 4 n 4 xþhðxÞ
kðnÞ > hðxÞ
¼
B1
S1 þ S2 ;
log x
ð22Þ
say. Using Lemma 1, we have that
X
hðxÞ
4 B2 hðxÞ
S1 4 2
ð log ðk=ðkÞÞÞ
k
k 4 hðxÞ
ð23Þ
k squarefull
for some absolute constant B2 > 0, since the series over k converges by (ii) of
Lemma 2.
It remains to estimate S2 . Let U ¼ fn 2 ½x; x þ hðxÞ : kðnÞ > hðxÞg and let
U1 U be the subset of those n 2 U for which there is some squarefull divisor
tjkðnÞ such that log 4 x < t < hðxÞ. It is clear that, using (i) of Lemma 1 with ¼ 0
and since ðnÞ log x for n 4 2x,
X
X 1
log x
hðxÞ
:
ð24Þ
ððnÞ 1Þ log x hðxÞ hðxÞ ¼
2
t
log x
log x
n 2 U1
t squarefull
t > log4 x
Let U2 ¼ U nU1 . Given n 2 U2 , let t ¼ 1 1 ‘ ‘ , with 1 < < ‘ primes, be
the smallest squarefull divisor of kðnÞ which is larger than hðxÞ. Since kðnÞ is such
140
J. M. De Koninck and I. Kátai
a candidate, then such a divisor t must exist. Furthermore, for every j 2 ½1; ‘, either
hðxÞ
4
t
t
t
2
2
j or 2j is squarefull, and therefore 2j 4 log x, so that j 5 log4 x. Since j jkðnÞ
2
and n 2
= U1 , it follows that j > hðxÞ, so that the conditions of Lemma 5 are
satisfied. Observe furthermore that if n 2 U2 , then
2x
1
log x þ Oð log log xÞ
log ðk=ðkÞÞ < log pffiffiffiffiffiffiffiffiffi < 1 10
hðxÞ
and therefore, using (21),
1
ðnÞ <
1
9
10log x þ Oðlog log xÞ
log x
<
1
10
þO
1
log log x ;
log x
and so by Lemma 5, we obtain that
X
hðxÞ
:
ððnÞ 1Þ log x
n 2 U2
ð25Þ
Combining (22), (23), (24) and (25), we obtain (2). Since the proof of (3) is
similar, we shall omit it.
5. Proofs of Theorems 2 and 3
We start with a general remark concerning the main idea used in the upcoming
proofs. As we shall see, the contribution of k ¼ kðnÞ, the squarefull part of n, in
the estimates of Theorems 2 and 3 (or of k ¼ kð p þ 1Þ in Theorem 4) may be
restricted to k 4 log x where is a large generic positive constant. This follows
by trivial estimation and (i) of Lemma 1. Therefore, in the representation (21) for
ðnÞ, one can replace log n in the denominator by log x with negligible error.
Hence this procedure permits, in view of the unique representation n ¼
km ¼ kðnÞmðnÞ, ðk; mÞ ¼ 1, to split all sums in the following fashion: inner sum
over m is performed using (ii) of Lemma 2, noting that partial summation permits
one to evaluate (in terms
P of asymptotic expansion in decreasing powers of the
logarithm) the quantity m 4 x;ðm;kÞ¼1 log j m for j ¼ 0; 1; 2; . . . , where as always
m denotes a generic squarefree number.
5.1. Regarding Theorems 2 and 3, it is enough to prove the short interval
version, that is (4) and (6). Indeed, in each case, the ‘‘long interval’’ version,
namely (5) or (7), follows by integrating with respect to x, as we will show in
Section 5.2. Moreover, we shall only prove (4), the proof of (6) being very similar.
We start as in the proof of Theorem 1 by writing
1
ðnÞ ¼
;
log ðk=ðkÞÞ
1 log n
from which it follows that
ðnÞ ¼
1
þ
O
;
x1=3
1 log ðk=ðkÞÞ
1
log x
pffiffiffi
whenever n 2 ½x; x þ x and kðnÞ < ð log xÞ
, being an arbitrary positive integer.
Mean Value of the Index of Composition of an Integer
141
First, we shall obtain an upper bound for the number of integers
pffiffiffi n 2 M, namely
2
those integerspwhich
can
be
written
in
the
form
n
¼
p
a
2
½x;
x
þ
x for some
ffiffiffi
pffiffiffi
pffiffiffi prime
number p > x. Observe that if p2 a 2 ½x; x þ x , then p2 ða þ 1Þ > x þ x, so that
for each prime number p, there exists at most one positive integer
pffiffi a with this property.
x
2
We first count
those
integers
n
¼
p
a
for
which
p
<
. Their contribution
pffiffi
pffiffi
logR x
x
is at most C logRþ1 x for some constant C > 0. On the other hand, if p > logRxx, then
pffiffiffi
ðx þ xÞð log 2R xÞ
< 2 log 2R x:
a<
x
h
i
pffiffix pffiffixqffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffi
p1ffiffi , and since the length
;
1
þ
Since au2 2 ½x; x þ x implies that u 2
a
a
x
of this interval is bounded, no more than C log 2R x such integers exist.
Gathering these estimates, we obtain that
pffiffiffi
C x
þ C log 2R x:
#M 4
log 2R x
pffiffiffi
Now let M1 be the set ofpffiffithose
integers n ¼ p2 a 2 ½x; x þ x for which
ffi
p2 jkðnÞ, where ð log xÞ
< p2 < x. Then clearly
pffiffiffi
pffiffiffi
X
x
x
:
#M1 4
2
pffiffi p
ð log xÞ
=2
ðlog xÞ
< p2 < x
We now estimate
SðxÞ :¼ #fn 2 ½x; x þ
pffiffiffi
x : kðnÞ > ð log xÞ
1 g ¼ S1 ðxÞ þ S2 ðxÞ;
n for which there is a squarefull divisor tjkðnÞ
where in S1 ðxÞ we count
pffiffithose
ffi
belonging to ½ð log xÞ
1 ; x , the others being counted by S2 ðxÞ. Repeating the
argument used in the proof of Theorem 1 and using Lemma 1, we obtain that
pffiffiffi
x
SðxÞ 4
;
Rþ2
log
x
provided 1 is taken large enough.
It remains to consider the sum
pffiffiffi X
1
x þ x x S0 ðxÞ :¼
M
k :
k M
log ðk=ðkÞÞ
k
k
k squarefull 1 log x
k < ðlog xÞ
1
By (ii) of Lemma 2, we have
pffiffiffi
X
x
1
6 ðkÞ pffiffiffi
1 þ1
x
þ
O
ð
log
xÞ
:
S0 ðxÞ ¼
R
log ðk=ðkÞÞ 2 k
x
log
k squarefull 1 log x
k < ðlog xÞ
1
ð26Þ
pffiffi
x
provided we choose R 5 ð
1 þ 1Þ þ r þ 1.
The error term can be bounded by logrþ1
x
On the other hand, the main sum in (26) can be rewritten as
pffiffiffi
#1 ðxÞ
#r ðxÞ
#rþ1 ðxÞ
þ þ
;
þO
x #0 ðxÞ þ
log x
log r x
log rþ1 x
142
J. M. De Koninck and I. Kátai
where
#j ðxÞ ¼
6
2
X
ð log ðk=ðkÞÞÞ j
k squarefull
k < ðlog xÞ
1
ðkÞ
k
ð j ¼ 0; 1; 2; . . . ; rÞ:
Set
#j :¼ lim #j ðxÞ;
x!1
observing that indeed the limit exists and in fact that, using Lemma 1, we have
#j ðxÞ #j ð log log xÞ j
ð log xÞ
1 =2
:
It is easy to see that #1 ¼ 1. Hence, we obtain that (4) holds with ci ¼ #i for
i ¼ 2; 3; . . . ; r. The proofs of Theorems 2 and 3 are thus complete.
Observe incidentally that
6 X ðkÞ log ðk=ðkÞÞ X log p
¼
0:75536;
c1 ¼ 2
k squarefull
k
pðp 1Þ
p
as in De Koninck and Doyon [1], and that one can easily show that d1 ¼ c1 .
5.2. P
We now show how (5) follows from (4). For this purpose, let
EðxÞ :¼ n 4 x ðnÞ, so that
pffiffiffi
r
X
X
Eðx þ xÞ EðxÞ
1
ci
1
pffiffiffi
ð27Þ
ðnÞ ¼
þ
O
¼ pffiffiffi
i
x
x x < n 4 xþpffiffix
log rþ1 x
i¼0 log x
with c0 ¼ 1, and therefore
pffiffiffi
ðX
ðX
r
X
Eðx þ xÞ EðxÞ
dx
X
pffiffiffi
:
ð28Þ
ci
dx ¼
i þO
x
log rþ1 X
2
2 log x
i¼0
pffiffiffi
pffiffiffi pffiffiffi
pffiffiffi
On the other hand, since for x 4 n 4 x þ x, we have n ¼ xð1 þ Oð1= xÞÞ,
so that
X
X
1
ðnÞ
pffiffiffi
pffiffiffi þ Oðx1=2 log xÞ:
ðnÞ ¼
ð29Þ
pffiffi
x x < n 4 xþpffiffix
n
x 4 n 4 xþ x
pffiffiffi pffiffiffi
pffiffiffi
Now, again since x ¼ n þ Oð1Þ for x 4 n 4 x þ x, we have
ðX
ð
X
X
ðnÞ
ðnÞ n
pffiffiffi dx ¼
pffiffiffi
dx
pffiffi
pffiffiffi
pffiffi
n
n
2 x 4 n 4 xþ x
2 4 n 4 Xþ X
n nþOð1Þ
X
ðnÞ pffiffiffi
pffiffiffi ð n þ Oð1ÞÞ
pffiffiffi
n
2 4 n 4 Xþ X
X
pffiffiffiffi
ðnÞ þ Oð X log XÞ:
¼
¼
24n4X
ð30Þ
Mean Value of the Index of Composition of an Integer
143
Finally, by partial integration, one easily obtains that, for each integer
i 2 ½1; r,
ðX
dx
X
X
X
X
þ iði þ 1Þ
þ þ O
: ð31Þ
þi
i ¼
log i X
log iþ1 X
log iþ2 X
log rþ1 X
e log x
Gathering all estimates (27) through (31), estimate (5) follows, thus completing
the proof of Theorem 2.
6. Proof of Theorem 4
2=3
Assume that p 2 ½x; x þ x
p þ 1 ¼ km;
and write
ðm; kÞ ¼ 1;
2 ðmÞ ¼ 1;
k squarefull:
Then, using the same approach as in the proof of Theorem 1, we can write
ðp þ 1Þ ¼
1
1
log ðk=ðkÞÞ
log ðpþ1Þ
:
Now let be a large constant. We shall first bound the set of those shifted
primes p þ 1 2 ½x; x þ x2=3 which have a squarefull divisor t > ð log xÞ. The
contribution of those primes p with a corresponding t < x2=3 can be estimated
using a sieve approach. Indeed, letting PðxÞ stand for the number of these
primes p, we have
X
PðxÞ ððx þ x2=3 ; t; 1Þ ðx; t; 1ÞÞ
ffiffi
p
ðlog xÞ < t 4 x
X x þ x2=3 x þ
pffiffi
t
t
2=3
x<t<x
x2=3
1
þ Oðx2=31=4 Þ:
log x ð log xÞ=2
It remains to estimate those shifted primes p þ 1 2 ½x; x þ x2=3 for which there is a
prime number q such that
p þ 1 ¼ aq2
with q2 > x2=3 :
ð32Þ
To each prime q, there
pffiffiffiffiffi can only correspond one integer a. Moreover,
pffiffiffi it follows
from (32) that q < 2x and therefore that there can be at most x such shifted
primes p þ 1.
We now consider the set F consisting of the shifted primes p þ 1 2 ½x; x þ x2=3 for which the corresponding squarefull number k satisfies k 4 ð log xÞ . Given
p 2 F, we have
1
1
ðp þ 1Þ ¼
þ O 1=4 ;
x
1 log ðk=ðkÞÞ
log x
144
J. M. De Koninck and I. Kátai: Mean Value of the Index of Composition of an Integer
say, and therefore
X
ðp þ 1Þ ¼
F :¼
p2F
X
1
k squarefull 1
log ðk=ðkÞÞ
log x
ððx þ x2=3 jkÞ ðxjkÞÞ
k 4 ðlog xÞ
þ Oðx2=31=4 Þ:
Hence, by using Lemma 4, we get that
X 1Y
1
1
F ¼ ðliðx þ x2=3 Þ liðxÞÞ
1
log
k
qðq 1Þ 1 ðk=ðkÞÞ
k squarefull q6 j k
log x
k 4 ðlog xÞ
2=3 log log x
:
þO x
ð log xÞrþ2
From here on, the proof is similar to that of Theorem 2 and we shall therefore
omit it.
Incidentally, observe that 2 can easily be computed and in fact that
X log ðk=ðkÞÞ Y 1
0:069:
2 ¼
1
k2
pðp 1Þ
k squarefull
p6 j k
7. Final Remarks
A key element in the proofs of Theorems 2 and 3 is the estimate (ii) of Lemma
x 2.
But our results used only the fact that the error term in this estimate is O
log R x
for any fixed R > 0. However, it should be mentioned that using the full force of
the error term and at the cost of some additional computations, one can show that
estimates (4) and (6) still hold when the p
indicated
sums run over shorter intervals,
ffiffiffi
such as for instance the interval ½x; x þ x logr x.
Acknowledgements. The authors are indebted to Professor Michael Filaseta for allowing them to
include here an important unpublished result (Theorem A). The authors are also grateful to the referees
for valuable suggestions which helped to improve the final version of this paper.
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139: 151–167.
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Number Theory. Proc London Math Soc 73: 241–278
Huxley M (1972) On the difference between consecutive primes. Invent Math 15: 164–170
Ivic A, Shiu P (1982) The distribution of powerful integers. Illinois J Math 26: 576–590
Nicolas JL, Robin G (1983) Majorations explicites pour le nombre de diviseurs de n. Bull Can Math
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Walfisz A (1963) Weylsche Exponentialsummen in der neueren Zahlentheorie. Berlin: Springer
Authors’ addresses: Jean-Marie De Koninck, Departement de mathematiques, Universite Laval,
Quebec G1K 7P4, Canada, e-mail: [email protected]; Imre Kátai, Computer Algebra Department,
E€otv€os Loránd University, 1117 Budapest, Pázmány Peter Setány I=C, Hungary, e-mail: katai@
compalg.inf.elte.hu