International Mathematical Forum, Vol. 12, 2017, no. 7, 331 - 338
HIKARI Ltd, www.m-hikari.com
https://doi.org/10.12988/imf.2017.7113
Two Topics in Number Theory:
Sum of Divisors of the Primorial and
Sum of Squarefree Parts
Rafael Jakimczuk
División Matemática, Universidad Nacional de Luján
Buenos Aires, Argentina
c 2017 Rafael Jakimczuk. This article is distributed under the Creative
Copyright Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
Let pn be the n-th prime and σ(n) the sum of the positive divisors
of n. Let us consider the primorial Pn = p1 .p2 . . . pn , the sum of its
Q
positive divisors is σ(Pn ) = ni=1 (1 + pi ). In the first section we prove
the following asymptotic formula
σ(Pn ) =
n
Y
(1 + pi ) =
i=1
6 γ
e Pn log pn + O (Pn ) .
π2
Let a(k) be the squarefree part of k, in the second section we prove the
formula
X
1≤k≤x
a(k) =
π2 2
x + o(x2 ).
30
We also study integers with restricted squarefree parts and generalize
these results to s-th free parts.
Mathematics Subject Classification: 11A99, 11B99
Keywords: Primorial, divisors, squarefree parts, s-th free parts, average
of arithmetical functions
332
1
Rafael Jakimczuk
Sum of Divisors of the Primorial
In this section p denotes a positive prime and pn denotes the n-th prime.
The following Mertens’s formulae are well-known (see [5, Chapter VI])
!
1
1
= log log x + M + O
,
log x
p≤x p
X
(1)
where M is called Mertens’s constant.
!
1
log 1 −
p
p≤x
X
!
1
= − log log x − γ + O
,
log x
(2)
where γ is the Euler’s constant.
1
1−
p
Y
p≤x

Y

p≤x
!
e−γ
1
=
+O
,
log x
log2 x
!
(3)

1 
= eγ log x + O(1).
1
1− p
(4)
Note that equation (4) is an immediate consequence of (3) if we use the formula
1
= 1 − x(1 + o(1))
1+x
(x → 0),
since then
!
1
1+O
1
log x
1
=1+O
.
log x
In this section we examine the sum p≤x log 1 + p1 and the product p≤x 1 +
Then, we apply the results to the sum of the divisors of the primorial. Our
main theorem is the following.
P
Q
Theorem 1.1 The following asymptotic formulae hold.
1
log 1 +
p
p≤x
!
X
Y
p≤x
!
6
1
= log log x + γ + log 2 + O
,
π
log x
1
1+
p
!
=
6 γ
e log x + O(1).
π2
(5)
(6)
1
p
.
333
Two topics in number theory: The primorial and squarefree parts
Proof. If we consider the formula
ex = 1 + x(1 + o(1))
(x → 0)
then equation (6) is an immediate consequence of equation (5). Since we have
1
exp O
log x
!!
!
1
=1+O
.
log x
Therefore we have that to prove equation (5). We have the equation
1
log(1 + x) = x − x2 (1 + o(1))
2
(x → 0)
Consequently
1
log 1 +
p
p≤x
!
X
=
1 1 X 1 + o(1)
−
.
2 p≤x
p2
p≤x p
X
(7)
Now, we have
1
1 + o(1) X 1 + o(1) X 1 + o(1) X 1 + o(1)
=
−
=
+O
2
2
2
2
p
p
p
p
x
p
p
p>x
p≤x
X
(8)
Note that we have used the well-known formula (see [1, Chapter 3])
1
1
=O
,
2
x
n>x n
X
where n denotes a positive integer. Substituting (1) and (8) into (7) we obtain
almost (5)
1
log 1 +
p
p≤x
!
X
!
1
= log log x + C + O
,
log x
where C is a constant. Therefore we have almost (6)
Y
p≤x
1
1+
p
!
= eC log x + O(1).
It is well-known the limit (Euler’s product formula)(see [1, Chapter 11])
lim
n→∞
2
n
Y
i=1
1
1− 2
pi
!
=
6
.
π2
(9)
Note that ζ(2) = π6 , where ζ(s) denotes the Riemann zeta function. Equations
(3) and (9) give eC = π62 eγ . The theorem is proved.
334
Rafael Jakimczuk
Theorem 1.2 Let us consider the primorial Pn = p1 .p2 . . . pn , the sum of
Q
its positive divisors is σ(Pn ) = ni=1 (1 + pi ). We have the following asymptotic
formula
σ(Pn ) =
n
Y
(1 + pi ) =
i=1
6 γ
e (p1 .p2 . . . pn ) log pn + O ((p1 .p2 . . . pn ))
π2
6 γ
=
e Pn log pn + O(Pn )
π2
(10)
Note that the primorial Pn = p1 .p2 . . . pn is its greatest divisor.
Proof. Put x = pn in equation (6). The theorem is proved.
It is well-known (see [2, Chapter XVIII]) that
lim sup
σ(n)
= eγ .
n log log n
We have
Theorem 1.3 The following limit holds
lim
n→∞
σ(Pn )
6
= 2 eγ .
Pn log log Pn
π
(11)
Proof. It is well-known that Pn = e(1+o(1))pn , therefore log log Pn ∼ log pn .
Now, limit (11) is an immediate consequence of (10). The theorem is proved.
2
Sum of Squarefree Parts and Generalization
Every positive integer k can be written in the form k = q.n2 where q is a
squarefree or quadratfrei number and n2 is the greatest square that divides k.
The positive integer q is called the squarefree part of k and it is denoted a(k)
(see [6]). Therefore if k is a square we have a(k) = 1, in contrary case a(k) is
the product of primes which have odd exponent in the prime factorization of
k. For exampe, if k = 24 .53 .112 .235 then a(k) = 5.23.
If q is relatively prime with n in the decomposition k = q.n2 , that is
(q, n2 ) = 1, we shall call q the relatively prime squarefree part of k and will
be denoted a0 (k), besides we shall say that k has a relatively prime squarefree
part.
The average of the arithmetic function a(k) is studied in the following
theorem.
335
Two topics in number theory: The primorial and squarefree parts
Theorem 2.1 The following formula holds
X
a(k) =
1≤k≤x
π2 2
x + o(x2 ).
30
(12)
Proof. The number of squarefree integers q not exceeding x (denoted Q(x)) is
well-known (see either [2, Chapter XVIII] or [4]). We have
Q(x) =
X
1=
q≤x
1
x + o(x),
ζ(2)
(13)
where ζ(s) denotes the Riemann zeta function. From here, we can obtain easily
(see [2, Chapter XXII]) that
Q1 (x) =
X
q=
q≤x
1
x2 + o(x2 ).
2ζ(2)
(14)
Let us consider the set Sn such that
Sn = q : q ≤
x
n2
(15)
Let us consider > 0. We choose the positive integer N such that
1
1
<
ζ(2) (N + 1)2
(16)
X 1
1
<
2ζ(2) n≥N +1 n4
(17)
and
Besides, for sake of simplicity, we put

B(x) =
X
√
N +1≤n≤ x

X

q≤ x2
n
q .
(18)

Now, we have (see (14), (15) and (18))

X
a(k) =
X
X

n2 ≤x
1≤k≤x
=
N
X
n=1
=
qSn
Q1


q =

X X

√
n≤ x
q≤
x
n2
q =

N
X
n=1


X

q≤
N
X
x
1
1
2
+ B(x) =
x
2
4
n
2ζ(2)
n=1 n
x
n2
q  + B(x)

!
+ o(x2 ) + B(x)
X 1
ζ(4) 2
x2
x −
+ o(1)x2 + B(x)
2ζ(2)
2ζ(2) n≥N +1 n4
336
Rafael Jakimczuk
That is
P
1≤k≤x
x2
a(k)
−
X 1
1
ζ(4)
B(x)
=−
+
o(1)
+
2ζ(2)
2ζ(2) n≥N +1 n4
x2
n
(19)
o
x
2
Note that SN +1 ∪ SN ∪ SN −1 ∪ · · · ⊆ q : q ≤ (N +1)
2 . Besides, since qn ≤ x,
the number of sets Si such that qSi isj less
than or equal to the number of
k
x
multiples of q not exceeding x, namely q , where b.c denotes the integer part
function. Hence (see (13) and (16))

0 ≤ B(x) =
√
N +1≤n≤ x
x
= xQ
(N + 1)2
!

X

X
q≤
x
n2

q
$ %
X
≤
q≤
q
x
(N +1)2
X
x
≤x
q
q≤ x
1
(N +1)2
!
=
1
1
+ o(1) x2 ≤ 2x2
ζ(2) (N + 1)2
That is,
0≤
B(x)
≤ 2.
x2
(20)
Therefore (see (19), (17) and (20)) we have
P
1≤k≤x a(k)
x2
ζ(4) −
≤ 4.
2ζ(2) (21)
Consequently equation (12) is proved, since can be arbitrarily small, ζ(2) =
4
π2
and ζ(4) = π90 (see [2, Chapter XVII]). The theorem is proved.
6
In the following theorem we examine the sum
X
a0 (k),
k≤x
where the sum run on the k that have a relatively prime squarefree part.
Suppose that the prime factorization of n2 is
n2 = ph1 1 ph2 2 · · · phs s ,
where p1 , p2 , . . . , ps are the different primes in the prime factorization and
h1 , h2 , . . . , hs are the even exponents.
We put
n0 = p1 p 2 · · · ps
n00 = (p1 + 1)(p2 + 1) · · · (ps + 1)
Two topics in number theory: The primorial and squarefree parts
337
If n2 = 1 then we put n0 = n00 = 1.
Then (see [3]) the number of squarefree relatively prime to n2 not exceeding
x, that is the number of squarefree relatively primes to n0 not exceeding x is
6 n0
x + o(x)
π 2 n00
(22)
6 n0 2
x + o(x2 )
2π 2 n00
(23)
Q2 (x) =
and consequently their sum is
Q3 (x) =
We have the following theorem
Theorem 2.2 The following asymptotic formula holds.
a0 (k) =
X
k≤x
1 6
Ax2 + o(x2 ),
2
2π
where
∞
X
n0 1
A=
00 4
n=1 n n
Proof. The proof is the same as the proof of Theorem 2.1. In this case we use
(22) and (23). The theorem is proved.
Let Q0 (x) be the number of positive integers not exceeding x with relatively
prime squarefree part. We have the following theorem.
Theorem 2.3 The following formula holds
Q0 (x) =
6
Cx + o(x),
π2
(24)
where
C=
∞
X
n0 1
.
00 2
n=1 n n
(25)
Proof. The proof is the same as the proof of Theorem 2.2. In this case we use
only (22). The theorem is proved.
Let s ≥ 2 a positive integer. A positive integer such that all prime in its
prime factorization has exponent less than or equal to s − 1 is called s-th
free number and will be denoted qs−1 . For example, if s = 2 we obtain the
squarefree numbers (q1 = q). The number of s-th free numbers not exceeding
x will be denoted Qs (x), for example, if s = 2 we have Q2 (x) = Q(x) (see
338
Rafael Jakimczuk
above, equation (13)). It is well-known the following generalization of equation
(13)(see, for example [4], for a simple proof)
Qs (x) =
1
x + o(x).
ζ(s)
Clearly, every positive integer k can be written in the form k = qs−1 .ns where
qs−1 is a s-th free number and ns is the greatest s-th power that divides k.
The positive integer qs−1 is called the s-th free part of k and will be denoted
as−1 (k). For example, if s = 2 then a1 (k) = a(k) (see above). We have the
following generalization of Theorem 2.1.
Theorem 2.4 The following formula holds
X
1≤k≤x
as−1 (k) =
1 ζ(2s) 2
x + o(x2 )
2 ζ(s)
Proof. The proof is the same as the proof of Theorem 2.1. The theorem is
proved.
Acknowledgements. The author is very grateful to Universidad Nacional
de Luján.
References
[1] T. M. Apostol, Introduction to Analytic Number Theory, Springer, 1976.
https://doi.org/10.1007/978-3-662-28579-4
[2] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers,
Oxford, 1960.
[3] R. Jakimczuk, On the distribution of certain subsets of quadratfrei numbers, International Mathematical Forum, 12 (2017), no 4, 185 - 194.
https://doi.org/10.12988/imf.2017.612176
[4] R. Jakimczuk, A simple proof that the square-free numbers have density
6/π 2 , Gulf Journal of Mathematics, 2 (2013), 189 - 192.
[5] W. J. LeVeque, Topics in Number Theory, Volume I, Addison-Wesley,
1958.
[6] N. J. A. Sloane, Sequence A007913, The On-Line Encyclopedia of Integer
Sequences.
Received: February 8, 2017; February 22, 2017