International Mathematical Forum, Vol. 11, 2016, no. 21, 1017 - 1026
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/imf.2016.68119
On the Quotient between the Greatest
Prime Factor and the Least Prime Factor
Rafael Jakimczuk
División Matemática, Universidad Nacional de Luján
Buenos Aires, Argentina
c 2016 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 b(n) be the greatest prime factor of n, a(n) the least prime factor
of n and m an arbitrary but fixed positive integer. In this article we
prove the asymptotic formulae
n
X
b(k)m
k=2
a(k)m
∼ Cm
nm+1
(m + 1) log n
where the constant Cm (depending of m) is defined in this article. In
particular if m = 1 we obtain
n
X
b(k)
k=2
a(k)
∼ C1
n2
2 log n
Mathematics Subject Classification: 11A41
Keywords: Greatest prime factor, least prime factor
1
Introduction and Preliminary Results
Let m be an arbitrary but fixed positive integer.
Let b(n) be the greatest prime factor in the prime factorization of n. In
previous articles [1][2], we proved the following asymptotic formula
n
X
ζ(m + 1) nm+1
b(i) ∼
,
m + 1 log n
i=2
m
1018
Rafael Jakimczuk
where ζ(s) is the Riemann’s Zeta Function. In articles [1][2] we use the notation
bm (i) = b(i)m .
Let a(n) be the least prime factor in the prime factorization of n. In a
previous article [2], we proved the following asymptotic formula
n
X
a(i)m ∼
i=2
1 nm+1
,
m + 1 log n
In article [2] we use the notation am (i) = a(i)m .
In this article we study the sum
n
X
i=2
b(i)
a(i)
!m
=
n
X
b(i)m
m
i=2 a(i)
The following lemma is well-known [3].
Lemma 1.1 Let m be a nonnegative integer and let Sm (x) be the sum of the
m-th powers of the primes not exceeding x. The following asymptotic formula
holds
X
1 xm+1
(1)
pm ∼
Sm (x) =
m + 1 log x
p≤x
where p denotes a positive prime.
2
Main Results
Theorem 2.1 The following asymptotic formula holds
n
X
b(k)m
nm+1
∼
C
m
m
(m + 1) log n
k=2 a(k)
(2)
where
Cm =
∞
X
1
h=2
a(h)m hm+1
(3)
Proof. Let k ≥ 2 be an arbitrary but fixed positive integer. Consider the
inequality
n
< p ≤ n,
(4)
k
where p denotes a positive prime number.
If n ≥ k 2 equation (4) gives p > k.
Consider the inequality
n
< p ≤ n.
2
1019
Greatest and least prime factors
The number of multiples of p not exceeding n is 1, namely p, since p ≤ n and
2p > n. Therefore p is the greatest prime factor in these multiples of p and p
is also the least prime factor in these multiples of p.
Consequently the sum of the quotients between the m-th powers of the
greatest prime factor and the m-th powers of the least prime factor in these
multiples of p not exceeding n will be
X
pm
1
=
pm
n
n
<p≤n
<p≤n
X
2
(5)
2
Consider the inequality
n
n
<p≤ .
3
2
The number of multiples of p not exceeding n is 2, namely p and 2p, since
2p ≤ n and 3p > n. Therefore p is the greatest prime factor in these multiples
of p and p and a(2) = 2 are the least prime factors in these multiples of p.
Consequently the sum of the quotients between the m-th powers of the
greatest prime factor and the m-th powers of the least prime factor in these
multiples of p not exceeding n will be
X
n
<p≤ n
3
2
pm
pm
+
pm a(2)m
!
=
X
n
<p≤ n
3
2
pm
1+
a(2)m
!
(6)
..
.
Consider the inequality
n
n
<p≤
.
k
k−1
The number of multiples of p not exceeding n is k−1, namely p, 2p, . . . , (k−1)p,
since (k − 1)p ≤ n and kp > n. Therefore p is the greatest prime factor in
these multiples of p. Since p > k (see above). On the other hand, p, a(2),
a(3), . . ., a(k − 1) are the least prime factors in these multiples of p.
Consequently the sum of the quotients between the m-th powers of the
greatest prime factor and the m-th powers of the least prime factor in these
multiples of p not exceeding n will be
X
n
n
<p≤ k−1
k
pm
pm
pm
1+
+
+
·
·
·
+
a(2)m a(3)m
a(k − 1)m
!
(7)
If we put
A(n) =
X
2≤i≤n
b(i)> n
k
b(i)m
a(i)m
(8)
1020
Rafael Jakimczuk
then equations (5), (6), . . ., (7) give (see (1))
A(n) =
X
1+
n
<p≤n
k
+
1
a(k − 1)m
1
a(2)m
pm +
X
n
<p≤ n
k
2
X
1
a(3)m
pm = π(n) − π
X
pm + · · ·
n
<p≤ n
k
3
n
n
1
n
+
Sm
− Sm
m
k
a(2)
2
k
n
n
<p≤ k−1
k
1
n
n
+
Sm
− Sm
+ ···
m
a(3)
3
k
n
n
1
S
−
S
+
m
m
m
a(k − 1)
k−1
k
(9)
where π(x) ∼ logx x is the prime counting function.
Note that (see (1))
lim
n→∞
Sm
n
h
Sm (n)
=
1
(10)
hm+1
Consequently (see (9) and (10))
lim
n→∞
=
k−1
X
1
A(n)
=
Sm (n) h=2 a(h)m
1
hm+1
−
1
k m+1
k−1
X
X
1
1 k−1
1
−
m m+1
k m+1 h=2 a(h)m
h=2 a(h) h
(11)
Hence (see (11) and (1))
A(n) =
k−1
X
X
1
1 k−1
1
−
m
m+1
m+1
m
k
h=2 a(h) h
h=2 a(h)
+ rk (n)
!
nm+1
(m + 1) log n
nm+1
(m + 1) log n
(12)
where rk (n) → 0.
Note that
0≤
1
k m+1
k−1
X
1
k
1
≤ m+1 = m
m
k
k
h=2 a(h)
(13)
and note that (see (1)) there exists t > 1 such that from a certain value of x
we have
Sm−1 (x) =
X
p≤x
pm−1 ≤ t
xm
m log x
(14)
1021
Greatest and least prime factors
If we put
X
B(n) =
2≤i≤n
b(i)m
a(i)m
(15)
2≤b(i)≤ n
k
Then we have (see (14))
m
B(n) ≤
X
2≤p≤ n
k
m
n
$ %
p
n
k
≤n
pm−1 ≤ nt
m
n
1
p
n
m
log
2≤p≤
X
k
k
m+1
=
≤
t(m + 1)
1
n
log k
m
mk
1 − log n (m + 1) log n
t(m + 1) 1
+
mk m
k
!
nm+1
(m + 1) log n
(16)
where b.c denotes the integer part function.
Consequently if we put
B(n) = g(n)
nm+1
(m + 1) log n
(17)
then from a certain value of n we have
0 ≤ g(n) ≤
t(m + 1) 1
+
mk m
k
(18)
nm+1
nm+1
+ f (n)
(m + 1) log n
(m + 1) log n
(19)
We can write
∞
X
b(k)m
1
=
m
m
m+1
k=2 a(k)
h=2 a(h) h
n
X
!
Now (see (12) and (17))
n
X
b(k)m
= A(n) + B(n)
m
k=2 a(k)
=
k−1
X
X
1 k−1
1
1
−
m m+1
k m+1 h=2 a(h)m
h=2 a(h) h
!
nm+1
(m + 1) log n
nm+1
nm+1
+ g(n)
(m + 1) log n
(m + 1) log n
!
∞
m+1
X
1
n
m m+1
(m + 1) log n
h=2 a(h) h
+ rk (n)
=
+
∞
X
X
1
1 k−1
1
rk (n) + g(n) −
−
m
m+1
m+1
m
k
h=k a(h) h
h=2 a(h)
!
nm+1
(20)
(m + 1) log n
1022
Rafael Jakimczuk
Therefore equations (19) and (20) give
f (n) = rk (n) + g(n) −
∞
X
X
1
1 k−1
1
−
m
m+1
m+1
m
k
h=k a(h) h
h=2 a(h)
(21)
On the other hand, if we take n and k sufficiently large then (see (12), (18),
(13))
∞
X
1
1 k−1
X
1 −
<
,
<
(22)
|rk (n)| < , |g(n)| < , −
m m+1 4
4
4 k m+1 h=2 a(h)m 4
h=k a(h) h
and consequently from a certain value of n (see (21) and (22))
|f (n)| < That is
lim f (n) = 0
(23)
n→∞
since > 0 is arbitrarily small. Equations (19) and (23) give (2). The theorem
is proved.
Note that equation (12) can be written in the form
k−1
X
X
1
1 k−1
1
A(n) ∼
−
m
m+1
m+1
m
k
h=2 a(h) h
h=2 a(h)
!
nm+1
(m + 1) log n
(24)
Equations (2), (3) and (24) give
∞
X
X
1
1 k−1
1
B(n) ∼
+
m
m+1
m+1
m
k
h=k a(h) h
h=2 a(h)
!
nm+1
(m + 1) log n
since (see (20))
n
X
b(k)m
= A(n) + B(n)
m
k=2 a(k)
If m = 1 then equation (2) becomes
n
X
b(k)
n2
∼ C1
2 log n
k=2 a(k)
where
C1 =
∞
X
1
2
h=2 a(h)h
(25)
1023
Greatest and least prime factors
If M (n) denotes the mean quotient between the greatest prime factor and the
least prime factor, that is
b(k)
k=2 a(k)
Pn
M (n) =
n
We have (see (25))
M (n) ∼ C1
n
2 log n
and consequently M (n) → ∞.
We recall the definition of function of slow increase (see [5])
Definition 2.2 Let f (x) be a function defined on interval [a, ∞) such that
f (x) > 0, limx→∞ f (x) = ∞ and with continuous derivative f 0 (x) > 0 . The
function f (x) is of slow increase if and only if the following condition holds
xf 0 (x)
=0
x→∞ f (x)
lim
Typical functions of slow increase are f (x) = log x, f (x) = logα x (α > 0),
f (x) = log log x, f (x) = log log log x, f (x) = logloglogx x , etc.
Theorem 2.3 Let f (x) be a function of slow increase on the interval [k, ∞],
b(i)
where k is a positive integer. Let us consider the first n − 1 quotients a(i)
(i =
b(i)
2, 3, . . . , n) and let n0 be the number of these quotients such that a(i)
≥ iflog(i)i .
Then limn→∞ nn0 = 0. Consequently if n1 is the number of these quotients such
b(i)
that a(i)
< iflog(i)i , then limn→∞ nn1 = 1, since k − 2 + n0 + n1 = n − 1.
Proof. Equation (25) can be written in the form
n
X
b(i)
n2
= h(n)C1
2 log n
i=2 a(i)
(26)
where h(n) → 1. Note that (L’Hospital’s rule and ( ))
R x tf (t)
k log t dt
x2 f (x)
x→∞
2 log x
lim
xf (x)
= lim
x→∞ d
dx
log2 x
x f (x)
2 log x
= lim
x→∞
1
1+
xf 0 (x)
2f (x)
−
1
2 log x
=1
That is, we have
Z x
k
tf (t)
x2 f (x)
dt = g(x)
log t
2 log x
(27)
1024
Rafael Jakimczuk
where limx→∞ g(x) = 1.
(x)
Note that there exists a positive integer k such that the function xf
is
log x
x
positive and strictly increasing in the interval [k, ∞), since the functions log x
and f (x) are positive and strictly increasing in the interval [k, ∞).
Suppose that limit (26) is not fulfilled. Then there exist α > 0 such that
for infinite values of n we have nn0 ≥ α. Consequently for these infinite values
of n we have (see (28))
n
X
b(i)
≥
i=2 a(i)
X
k≤i≤n
b(i)
if (i)
≥ log i
a(i)
b(i)
≥
a(i)
X
k≤i≤n
b(i)
if (i)
≥ log i
a(i)
0 −1
X
if (i) k+n
if (i)
≥
log i
log i
i=k
Z k+n0 −1
Z k+αn−1
xf (x)
xf (x)
dx ≥
dx
log x
log x
k
k
(k + αn − 1)2 f (k + αn − 1)
= g(k + αn − 1)
2 log (k + αn − 1)
α 2 n2
f (k + αn − 1)
= h(n)
2 log n
≥
(28)
where h(n) → 1.
Note that in (29) the sum
k+n
0 −1
X
i=k
if (i)
log i
is a sum of rectangles of basis 1 and height iflog(i)i .
Now, (27) and (29) are an evident contradiction since
f (k + αn − 1) → ∞
and consequently limit (26) is fulfilled. The theorem is proved.
We also have the following analogous three theorems.
Theorem 2.4 Let f (x) be a function of slow increase on the interval [k, ∞],
where k is a positive integer. Let us consider the first n − 1 greatest prime
factors b(i)(i = 2, 3, . . . , n) and let n0 be the number of these greatest prime
factors such that b(i) ≥ iflog(i)i . Then limn→∞ nn0 = 0. Consequently if n1 is the
number of these greatest prime factors such that b(i) <
1, since k − 2 + n0 + n1 = n − 1.
if (i)
,
log i
then limn→∞
n1
n
=
Proof. The proof is the same as Theorem 2.2 since we have (see the introduction)
n
X
ζ(2) n2
b(i) ∼
2 log n
i=2
1025
Greatest and least prime factors
The theorem is proved.
Theorem 2.5 Let f (x) be a function of slow increase on the interval [k, ∞],
where k is a positive integer. Let us consider the first n − 1 least prime factors
a(i)(i = 2, 3, . . . , n) and let n0 be the number of these least prime factors such
that a(i) ≥ iflog(i)i . Then limn→∞ nn0 = 0. Consequently if n1 is the number of
these least prime factors such that a(i) <
k − 2 + n0 + n1 = n − 1.
if (i)
,
log i
then limn→∞
n1
n
= 1, since
Proof. The proof is the same as Theorem 2.2 since we have (see the introduction)
n
X
1 n2
a(i) ∼
2 log n
i=2
The theorem is proved.
Let c(n) be the sum of the prime factors of n.
Theorem 2.6 Let f (x) be a function of slow increase on the interval [k, ∞],
where k is a positive integer. Let us consider the first n − 1 values of c(i)
c(i)(i = 2, 3, . . . , n) and let n0 be the number of these c(i) such that c(i) ≥ iflog(i)i .
Then limn→∞ nn0 = 0. Consequently if n1 is the number of these c(i) such that
c(i) < iflog(i)i , then limn→∞ nn1 = 1, since k − 2 + n0 + n1 = n − 1.
Proof. The proof is the same as Theorem 2.2 since we have (see [4])
n
X
i=2
c(i) ∼
ζ(2) n2
2 log n
The theorem is proved.
In [4] a weaker theorem was proved where f (x) = log x and 0 < < 1.
Acknowledgements. The author is very grateful to Universidad Nacional de
Luján.
References
[1] R. Jakimczuk, Sums of greatest prime factors, International Mathematical
Forum, 8 (2013), no. 19, 937 - 943.
http://dx.doi.org/10.12988/imf.2013.13100
1026
Rafael Jakimczuk
[2] R. Jakimczuk, A note on sums of greatest (least) prime factors, International Journal of Contemporary Mathematical Sciences, 8 (2013), no. 9,
423 - 432. http://dx.doi.org/10.12988/ijcms.2013.13042
[3] R. Jakimczuk, A note on sums of powers which have a fixed number of
prime factors, Journal of Inequalities in Pure and Applied Mathematics,
6 (2005), Article 31.
[4] R. Jakimczuk, Sums of prime factors in the prime factorization of an
integer, International Mathematical Forum, 7 (2012), no. 53, 2617 - 2621.
[5] R. Jakimczuk, Functions of slow increase and integer sequences, Journal
of Integer Sequences, 13 (2010), Article 10.1.1.
Received: September 6, 2016; Published: October 14, 2016