Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 2, Article 31, 2005
A NOTE ON SUMS OF POWERS WHICH HAVE A FIXED NUMBER OF PRIME
FACTORS
RAFAEL JAKIMCZUK
D EPARTMENT OF M ATHEMATICS
U NIVERSIDAD NACIONAL DE L UJÁN
A RGENTINA
jakimczu@mail.unlu.edu.ar
Received 05 October, 2004; accepted 16 February, 2005
Communicated by L. Tóth
A BSTRACT. Let us denote by cn,k the sequence of numbers which have in its factorization
k
Pn
prime factors (k ≥ 1), we obtain in short proofs asymptotic formulas for cn,k , i=1 cα
and
i,k
P
α
ci,k ≤ x ci,k . We generalize the work by T. Sálat y S. Znam when k = 1 (see reference [2]).
Key words and phrases: Sums of powers, Numbers with k prime factors.
2000 Mathematics Subject Classification. 11N25, 11N37.
Let πk (x) be the number of these numbers not exceeding x, it was proved by Landau [1] that
πk (x)
lim
(1)
k−1
x→∞ x(log log x)
(k−1)! log x
=1
Note that if k = 1 then π1 (x) = π(x), cn,1 = pn , and equation (1) is the prime number theorem.
Theorem 1. The following asymptotic formula holds:
cn,k ∼
(2)
(k − 1)!n log n
(log log n)k−1
.
Proof. If k = 1 the formula is true, since in this case (2) is the prime number theorem pn ∼
n log n. Suppose k ≥ 2. If we put x = cn,k and substitute into (1) we find that
(3)
lim
n→∞
(k − 1)!n log cn,k
cn,k (log log cn,k )k−1
= 1.
Writing
(4)
cn,k =
ISSN (electronic): 1443-5756
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182-04
(k − 1)!n log n
(log log n)k−1
f (n)
2
R AFAEL JAKIMCZUK
and substituting (4) into (3) we obtain
log cn,k (log log n)k−1
lim
(5)
n→∞
log n f (n) (log log cn,k )k−1
= 1.
From equation (1) we find that
πk (x)
= ∞.
x→∞ π(x)
lim
(6)
Assume that the inequalities cn,k ≥ pn have infinitely many solutions, then we have π(cn,k ) ≥
π(pn ) = n = πk (cn,k ), which contradicts (6). Hence for all sufficiently large n we have
cn,k < pn . On the other hand, clearly n ≤ cn,k . Therefore n ≤ cn,k ≤ pn , that is log n ≤
log cn,k ≤ log pn , and we find that
1≤
(7)
log cn,k
log pn
≤
.
log n
log n
From (7) and the prime number theorem pn ∼ n log n, we obtain
log cn,k
= 1.
n→∞ log n
lim
(8)
From (5) and (8) we find that
lim f (n) = 1.
(9)
n→∞
To finish, (9) and (4) give (2). The theorem is thus proved.
The following proposition is well known, we use it as a lemma
P
P∞
Lemma 2. Let ∞
lim abnn = 1. Then
i=1 ai and
i=1 bi be two series of positive terms such that n→∞
P∞
if i=1 bi is divergent, the following limit holds
Pn
ai
= 1.
lim Pi=1
n
n→∞
i=1 bi
Theorem 3. Let k ≥ 1 and let α be a positive number. The following asymptotic formula holds
n
X
(10)
cαi,k
∼
i=1
((k − 1)!)α nα+1 logα n
(α + 1) (log log n)α(k−1)
.
Proof. Let us consider the following two series:
∞
X
cαi,k
and
1+2+
i=1
Since the function
(11)
1+2+
(k−1)! t log t
(log log t)k−1
n
X
i=3
α
∞
X
(k − 1)! i log i
i=3
(log log i)k−1
!α
.
is increasing from a certain value of t, we find that
(k − 1)! i log i
(log log i)k−1
Z n
=
3
J. Inequal. Pure and Appl. Math., 6(2) Art. 31, 2005
!α
(k − 1)! t log t
(log log t)k−1
!α
dt + O
n log n
(log log n)k−1
!α !
.
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S UMS O F P OWERS
3
On the other hand, from the L’Hospital rule
!α
Z n
(k − 1)! t log t
((k − 1)!)α nα+1 logα n
(12)
.
dt
∼
(log log t)k−1
(α + 1) (log log n)α (k−1)
3
Equation (10) is an immediate consequence of (11), (12) and the lemma.
The theorem is thus proved.
Theorem 4. Let k ≥ 1 and let α be a positive number. The following asymptotic formula holds
X
xα+1 (log log x)k−1
(13)
cαi,k ∼
.
(α + 1) (k − 1)! log x
c ≤x
i,k
Proof. Equation (3) can be written in the form
n
lim
(14)
k−1
n→∞ cn,k (log log cn,k )
(k−1)! log cn,k
= 1.
From (8) we obtain
log log cn,k
= 1.
n→∞ log log n
lim
(15)
Substituting (14), (8) and (15) into (10) we find that
X
(16)
cαi,k ∼
ci,k ≤ cn,k
k−1
cα+1
n,k (log log cn,k )
.
(α + 1) (k − 1)! log cn,k
Equation (2) gives cn,k ∼ cn+1,k , therefore
X
(17)
cαi,k
ci,k ≤ cn,k
α+1
cn+1,k
(log log cn+1,k )k−1
∼
(α + 1) (k − 1)! log cn+1,k
k−1
cα+1
n,k (log log cn,k )
.
∼
(α + 1) (k − 1)! log cn,k
Since the function
xα+1 (log log x)k−1
(α + 1) (k − 1)! log x
is increasing from a certain value of x, we have for all n sufficiently large
P α
P
P
cαi,k
cαi,k
ci,k
(18)
ci,k ≤ cn,k
k−1
cα+1
n,k (log log cn,k )
(α+1) (k−1)! log cn,k
≤
ci,k ≤ x
xα+1 (log log x)k−1
(α+1) (k−1)! log x
6
ci,k ≤ cn,k
k−1
cα+1
n+1,k (log log cn+1,k )
(α+1) (k−1)! log cn+1,k
,
where cn,k ≤ x < cn+1,k .
To finish, (17) and (18) give (13). The theorem is proved.
Note. The case k = 1 was studied in the reference [2]. In this case (9) and (13) become
n
X
i=1
pαi ∼
nα+1 logα n
,
(α + 1)
X
pi ≤x
pαi ∼
xα+1
.
(α + 1) log x
Using equation (2) and the lemma, we can prove (as above) other theorems, for example the
following:
J. Inequal. Pure and Appl. Math., 6(2) Art. 31, 2005
http://jipam.vu.edu.au/
4
R AFAEL JAKIMCZUK
Theorem 5. The following asymptotic formulas holds
∞
X
X 1
1
(log log n)k
(log log x)k
∼
and
∼
.
c
k
!
c
k
!
n,k
n,k
n=1
c ≤x
n,k
When k = 1, this theorem is well known.
R EFERENCES
[1] G.H. HARDY AND E.M. WRIGHT, An Introduction to Number Theory, 4th Ed.1960. Chapter XXII.
[2] T. SÁLAT AND S. ZNAM, On the sums of prime powers, Acta Fac. Rer. Nat. Univ. Com. Math., 21
(1968), 21–25.
J. Inequal. Pure and Appl. Math., 6(2) Art. 31, 2005
http://jipam.vu.edu.au/