Download Divisors of Numbers with k Prime Factors and Perfect

International Mathematical Forum, Vol. 10, 2015, no. 7, 339 - 347
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/imf.2015.5435
Divisors of Numbers with k
Prime Factors and Perfect Numbers
Rafael Jakimczuk
División Matemática, Universidad Nacional de Luján
Buenos Aires, Argentina
c 2015 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
In this article we study some arithmetical functions defined on numbers with a fixed number of primes in their prime factorization. These
arithmetical functions are relationed with the divisors of these numbers.
We also obtain some results on the distribution of perfect numbers, deficient numbers and abundant numbers with a fixed number of primes
in their prime factorization.
Mathematics Subject Classification: 11A99, 11B99
Keywords: Arithmetical functions on numbers with a fixed number of
primes in their prime factorization, divisors, average, perfect numbers, abundant numbers, deficient numbers
1
Notation and Preliminary Results
The following divisor functions are well-known. For a nonzero integer α we
define
σα (n) =
X
dα
(1)
d|n
the sum of the αth powers of the positive divisors of n.
Note that if h is a positive integer, we have
σ−h (n) =
X
d|n
d
−h
P
=
d|n d
nh
h
=
σh (n)
nh
(2)
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Rafael Jakimczuk
A number n is perfect if and only if
X
d=n
(3)
d|n
d6=n
A number n is abundant if and only if
X
d>n
d|n
d6=n
A number n is deficient if and only if
X
d<n
d|n
d6=n
A square-free number is a number without square factors, a product of
different primes. The first few terms of the integer sequence of square-free
numbers are
1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, . . .
We shall need the following well-known theorems.
Theorem 1.1 (Prime Number Theorem)Let π1 (x) be the number of primes
not exceeding x, we have
x
x
+o
π1 (x) =
log x
log x
!
Therefore there exist a constant C1 such that
π1 (x) ≤ C1
x
log x
(x ≥ 2)
Theorem 1.2 (Landau’s Theorem)Let πk (x) be the number of squarefree
with just k prime factors (k ≥ 2) not exceeding x, we have
x(log log x)k−1
x(log log x)k−2
πk (x) =
+O
(k − 1)! log x
log x
Therefore there exist a constant Ck such that
x(log log x)k−1
πk (x) ≤ Ck
(k − 1)! log x
(x ≥ 3)
!
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Divisors of numbers with k prime factors
Theorem 1.3 (Landau’s Theorem)Let νk (x) be the number of numbers with
just k prime factors (k ≥ 2) not exceeding x, we have
x(log log x)k−2
x(log log x)k−1
+O
νk (x) =
(k − 1)! log x
log x
!
Therefore there exist a constant Dk such that
νk (x) ≤ Dk
x(log log x)k−1
(k − 1)! log x
(x ≥ 3)
We also shall need the following theorem (see ([1], chapter XXII) )
Theorem 1.4 Let cn (n ≥ 1) a sequence of real numbers. Let us consider
the function
X
A(x) =
cn
n≤x
Suppose that cn = 0 for n < n1 and suppose that f (x) has a continuous
derivative f 0 (x) on the interval [n1 , ∞], then the following formula holds
X
cn f (n) = A(x)f (x) −
n≤x
2
Z x
A(t)f 0 (t) dt
n1
Square-Free with k prime factors
In this section we are interested in square-free numbers with just k prime
factors. A square-free with just k prime factors will be denoted qk . Note that
the number of divisors of a square-free with just k prime factors is 2k .
Theorem 2.1 Let h ≥ 1 and k ≥ 2 arbitrary but fixed positive integers.
The following asymptotic formula holds
x(log log x)k−2
x(log log x)k−1
+O
σ−h (qk ) =
(k − 1)! log x
log x
qk ≤x
!
X
(4)
Proof. We put qk = p1 p2 · · · pk , where p1 < p2 < · · · < pk are the different
prime factors in the prime factorization of qk .
We have (see (1))
X
σ−h (qk )
qk ≤x

=
X
p1 p2 ···pk ≤x

1 +


X
d|(p1 p2 ···pk
d6=1


X
X
 X
1
1
=

(5)
1
+


h
h
d
d
p1 p2 ···pk ≤x
p1 p2 ···pk ≤x d|(p1 p2 ···p )
)
d6=1
k
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Rafael Jakimczuk
where (Theorem 1.2)
x(log log x)k−1
x(log log x)k−2
1 = πk (x) =
+O
(k − 1)! log x
log x
p1 p2 ···pk ≤x
!
X
(6)
and (Theorem 1.1 if k = 2 and Theorem 1.2 if k ≥ 3)


X
p1 p2 ···pk ≤x
X



d|(p1 p2 ···pk

X
X  X
1
1
1
 ≤ 2k
= 2k



h
d
p1 p2 ···pk ≤x p1
p1 ≤x p2 ···pk ≤ x p1
)
p1
d6=1
≤ 2k
1
x
πk−1
p1
p1 ≤x p1
!
= 2k
X
k
k−2
p1
k
k−2
≤ 2k Ck−1
1
x
πk−1
p1
p1
!
x log log px1
1
C
k−1
p21
(k − 2)! log x
X
p1 ≤ Px
X
p1 ≤ Px
≤ 2k

x (log log x)
(k − 2)! log x
X
p1 ≤ Px
k
1
1
p1
2
p1 1 − log
log x
(7)
where Pk is the product of the first k − 1 primes, thus
πk−1
x
p1
!
>0
Let us consider the nonnegative function of x.
X
p1 ≤ Px
k
1
1
p1
2
p1 1 − log
log x
(8)
Next, we shall prove that this function is bounded.
We shall denote sn the n-th prime.
If we put x = Pk sn into equation (8) then we obtain
X
si ≤sn
n
X
1
1
log si
2
si 1 − log(P
s
k n)
=
n
X
1
1
log si
2
i=1 si 1 − log(P s
≤
k n)
n
X
1
1
log si
2
i=1 si 1 − log(P s )
k i
n
n
1 log Pk + log si X
1
1 X
log si
=
+
<A
=
2
2
log Pk
log Pk i=1 s2i
i=1 si
i=1 si
where A is a positive constant, since the series
convergent.
On the interval Pk sn ≤ x < Pk sn+1 we have
X
p1 ≤ Px
k
P∞
1
i=1 s2
i
X 1
1
1
1
log p1 =
log si ≤ A
2
2
p1 1 − log x
si ≤sn si 1 − log x
and
P∞ log si
i=1
s2i
are
(9)
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Divisors of numbers with k prime factors
since if si ≤ sn then the function of x
1
1
si
2
si 1 − log
log x
is decreasing on the interval Pk sn ≤ x < Pk sn+1 .
Equations (7) and (9) give

X
0≤
p1 p2 ···pk ≤x




X
d|(p1 p2 ···pk
1
x (log log x)k−2
 ≤ 2k ACk−1
dh 
(k − 2)! log x
)
(10)
d6=1
That is

X
p1 p2 ···pk ≤x




X
d|(p1 p2 ···pk
1
x (log log x)k−2
=O
dh 
log x
)
!
(11)
d6=1
Equations (5), (6) and (11) give (4). The theorem is proved.
Theorem 2.2 Let h ≥ 1 and k ≥ 2 arbitrary but fixed positive integers.
The following asymptotic formula holds.
xh+1 (log log x)k−1
xh+1 (log log x)k−2
σh (qk ) =
+O
(h + 1)(k − 1)! log x
log x
qk ≤x
!
X
(12)
Proof. We use Theorem 1.4. We put cn = 0 if n is not a square-free with just
k prime factors and cn = σ−h (n) if n is a square-free with just k prime factors.
Therefore we have (see (2) and (4))
σh (qk )
x(log log x)k−1
x(log log x)k−2
A(x) =
+
O
=
qkh
(k − 1)! log x
log x
qk ≤x
!
X
(13)
If we put f (x) = xh , f 0 (x) = hxh−1 , then we obtain (Theorem 1.4)
xh+1 (log log x)k−1
xh+1 (log log x)k−2
σh (qk ) =
+O
(k − 1)! log x
log x
qk ≤x
!
X
−
=
+
+
Z x
Z x
h
(log log t)k−1
th (log log t)k−2
th
dt +
O
(k − 1)! Pk
log t
log t
Pk
!
xh+1 (log log x)k−1
xh+1 (log log x)k−2
+O
(h + 1)(k − 1)! log x
log x
!
Z x h
k−2
h+1
t (log log t)
x (log log x)k−1
dt =
O
log t
(h + 1)(k − 1)! log x
Pk
!
xh+1 (log log x)k−2
O
log x
!
dt
(14)
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Rafael Jakimczuk
That is (12). Note that in this theorem Pk is the product of the first k primes.
We have used the formula (Integration by parts)
Z x
Pk
−
Z x
Pk
(log log t)k−1
xh+1 (log log x)k−1
dt =
+ O(1)
log t
h+1
log x
th (k − 1)(log log t)k−2 − (log log t)k−1
dt
h+1
log2 t
th
and the limits (L’Hospital’s rule)
Z x h
t (log log t)k−2
log2 t
Pk
Z x h
t (log log t)k−1
log2 t
Pk
xh+1 (log log x)k−2
dt = o
log x
!
xh+1 (log log x)k−2
dt = o
log x
!
Z x h
t (log log t)k−2
log t
Pk
dt ∼
xh+1 (log log x)k−2
(h + 1) log x
The theorem is proved.
Lemma 2.3 Let h ≥ 1 and k ≥ 2 arbitrary but fixed positive integers. The
following asymptotic formula holds.
xh+1 (log log x)k−1
xh+1 (log log x)k−2
+O
qkh =
(h + 1)(k − 1)! log x
log x
qk ≤x
!
X
(15)
Proof. We use Theorem 1.4. We put cn = 0 if n is not a square-free with
just k prime factors and cn = 1 if n is a square-free with just k prime factors.
Therefore the proof is identical to Theorem 2.2, since (compare with (13))
x(log log x)k−1
x(log log x)k−2
A(x) = πk (x) =
1=
+O
(k − 1)! log x
log x
qk ≤x
!
X
If we put f (x) = xh , f 0 (x) = hxh−1 , then we obtain (15) (see (14)). The lemma
is proved.
Theorem 2.4 Let h ≥ 1 and k ≥ 2 arbitrary but fixed positive integers.
The following asymptotic formula holds.
X
qk ≤x
X
σh (qk ) ∼
qk ≤x
Note that qk is the greatest divisor of qk .
qkh
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Divisors of numbers with k prime factors
Proof. It is a immediate consequence of Theorem 2.2 and Lemma 2.3. The
theorem is proved.
Let us consider the squarefree with k prime factors not exceeding x, the number
of numbers in this set is πk (x). Let 0 < ≤ 1 and consider the squarefree with
k prime factors not exceeding x such that
X
d|(p1 p2 ···pk
1
1
=
d
p1 p2 · · · pk
)
d6=1
X
d≥
(16)
d|(p1 p2 ···pk )
d6=p1 p2 ···pk
Let αk (x) the number of these numbers not exceeding x.
Theorem 2.5 The following limit holds
lim
x→∞
αk (x)
=0
πk (x)
(17)
Proof. we have (see (16) and (10) with h = 1)

αk (x) ≤
X
p1 p2 ···pk ≤x




X
d|(p1 p2 ···pk
1
x (log log x)k−2
 ≤ 2k ACk−1
d
(k − 2)! log x
)
d6=1
Therefore

αk (x) ≤
X
p1 p2 ···pk ≤x




X
d|(p1 p2 ···pk
1
2k ACk−1 x (log log x)k−2
≤
d
(k − 2)! log x
)
(18)
d6=1
Equation (18) and Theorem 1.2 give (17). The theorem is proved.
If = 1 (see (16) and the first section) we obtain the perfect numbers and
abundant numbers, squarefree with k prime factors not exceeding x. We shall
denote χk (x) the number of these perfect and abundant numbers not exceeding
x. Theorem 2.5 becomes.
Theorem 2.6 The following limit holds
χk (x)
=0
x→∞ π (x)
k
lim
Therefore, in the sequence qk almost all numbers are deficient (see the first
section).
346
3
Rafael Jakimczuk
Numbers with k prime factors
In this section we are interested in numbers with just k prime factors. A
number with just k prime factors will be denoted ak . Note that the number of
divisors of a number with just k prime factors is bounded.
If the prime factorization of ak is
ak = q1r1 q2r2 · · · qsrs
(r1 + r2 + · · · + rs = k)
where q1 , q2 , . . . , qs are the different primes in the prime factorization then el
number of divisors of ak is
(r1 + 1)(r2 + 1) · · · (rs + 1) ≤ (k + 1)k
The number of numbers with just k prime factors (k ≥ 2) not exceeding x is
denoted νk (x) (see Theorem 1.3 and compare with Theorem 1.2). Note that
the first number with just k prime factors is 2k . The proofs of the following
theorems are the same as in the former section.
Theorem 3.1 Let h ≥ 1 and k ≥ 2 arbitrary but fixed positive integers.
The following asymptotic formula holds
x(log log x)k−2
x(log log x)k−1
+O
σ−h (ak ) =
(k − 1)! log x
log x
ak ≤x
!
X
Theorem 3.2 Let h ≥ 1 and k ≥ 2 arbitrary but fixed positive integers.
The following asymptotic formula holds.
xh+1 (log log x)k−1
xh+1 (log log x)k−2
σh (ak ) =
+O
(h + 1)(k − 1)! log x
log x
ak ≤x
!
X
Lemma 3.3 Let h ≥ 1 and k ≥ 2 arbitrary but fixed positive integers. The
following asymptotic formula holds.
X
ak ≤x
ahk
xh+1 (log log x)k−1
xh+1 (log log x)k−2
=
+O
(h + 1)(k − 1)! log x
log x
!
Theorem 3.4 Let h ≥ 1 and k ≥ 2 arbitrary but fixed positive integers.
The following asymptotic formula holds.
X
ak ≤x
X
σh (ak ) ∼
ak ≤x
Note that ak is the greatest divisor of ak .
ahk
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Divisors of numbers with k prime factors
Let us consider the numbers with k prime factors not exceeding x, the
number of numbers in this set is νk (x). Let 0 < ≤ 1 and consider the
numbers with k prime factors not exceeding x such that
X
d|(p1 p2 ···pk
1
1
=
d
p1 p2 · · · pk
)
d6=1
X
d≥
d|(p1 p2 ···pk )
d6=p1 p2 ···pk
Let βk (x) the number of these numbers not exceeding x.
Theorem 3.5 The following limit holds
βk (x)
=0
x→∞ ν (x)
k
lim
(19)
If = 1 we obtain the perfect numbers and abundant numbers with k prime
factors not exceeding x. We shall denote δk (x) the number of these perfect
and abundant numbers not exceeding x. Theorem 3.5 becomes.
Theorem 3.6 The following limit holds
δk (x)
=0
x→∞ ν (x)
k
lim
Therefore, in the sequence ak almost all numbers are deficient (see the first
section).
Acknowledgements. The author is very grateful to Universidad Nacional
de Luján.
References
[1] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers,
Oxford, 1960.
Received: May 9, 2015; Published: June 12, 2015