International Mathematical Forum, Vol. 8, 2013, no. 19, 945 - 948
HIKARI Ltd, www.m-hikari.com
Asymptotic Formulae for the Square Root
of the n-th Perfect Power
Rafael Jakimczuk
División Matemática, Universidad Nacional de Luján
Buenos Aires, Argentina
jakimczu@mail.unlu.edu.ar
c 2013 Rafael Jakimczuk. This is an open access article distributed under
Copyright the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
Abstract
In this note we obtain asymptotic formulae for the square root of
the n-th perfect power.
Mathematics Subject Classification: 11A99, 11B99
Keywords: Asymptotic formulae, n-th perfect power, square root
1
Introduction and Lemmas
A natural number of the form mn where m is a positive integer and n ≥ 2 is
called a perfect power. The first few terms of the integer sequence of perfect
powers are
1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128 . . .
and they are sequence A001597 in Sloane’s Encyclopedia. In this note Pn
denotes the n-th perfect power.
A quadratfrei number is a number without square factors, a product of
different primes. The first few terms of the integer sequence of quadratfrei
numbers are
1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, . . .
On the other hand, the Möbius function µ(n) is defined as follows: µ(1) = 1,
if n is the product of r different primes, then µ(n) = (−1)r , if the square of a
R. Jakimczuk
946
prime divides n, then µ(n) = 0. In this note q denotes a quadratfrei number.
On the other hand (as usual) pn denotes the n-th prime number.
We shall need the following theorem.
Theorem 1.1 Let ph be the h-th prime with h ≥ 3, where h is an arbitrary
but fixed positive integer. We have the following asymptotic formula
⎛
⎞
⎜ 2⎟
13
32
1+ 2
Pn = n + n8/6 + n32/30 + ⎜
2µ(q)n1+ q ⎟
+ o n ph
⎝
⎠
3
15
2≤q≤ph
2
(1)
q=2,6,30
Note that 2 = 1 + 22 ,
8
6
2
6
=1+
32
30
and
=1+
2
.
30
Proof. See [1].
If h = 3 then Theorem 1.1 becomes
5
7
7
Pn = n2 − 2n 3 − 2n 5 + o n 5 .
(2)
If h = 4 then Theorem 1.1 becomes
5
7
Pn = n2 − 2n 3 − 2n 5 +
9
9
13 4
n 3 − 2n 7 + o n 7 .
3
(3)
If h = 5 then Theorem 1.1 becomes
13 5
7
9
6
13
13 4
Pn = n2 − 2n 3 − 2n 5 + n 3 − 2n 7 + 2n 5 − 2n 11 + o n 11 .
3
The following lemma is a immediate consequence of the binomial theorem.
Lemma 1.2 We have the following formulae
1
(1 + x)1/2 = 1 + x + O(x2 )
2
1
2
1
2
−1
(x → 0)
1
1
x2 +O(x3 ) = 1+ x− x2 +O(x3 )
2
2
8
√
In this note we obtain asymptotic formulae for Pn .
1
(1+x)1/2 = 1+ x+
2
2
(4)
(x → 0) (5)
Main Results
Theorem 2.1 Let ph be the h-th prime with h ≥ 3, where h is an arbitrary
but fixed positive integer. We have the following asymptotic formula
5
1
Pn = n + n2/6 + n2/30 +
µ(q)n2/q + o n2/ph
3
15
2≤q≤ph
q=2,6,30
(6)
Square rooth of perfect powers
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Proof. First, we shall prove the theorem if h = 3, that is ph = 5. We have (see
(2))
2
2
2
2
2
2
Pn = n2 − 2n1+ 3 − 2n1+ 5 + o n1+ 5 = n2 1 − 2n−1+ 3 − 2n−1+ 5 + o n−1+ 5
Therefore
2
2
2
Pn = n 1 − 2n−1+ 3 − 2n−1+ 5 + o n−1+ 5
where
2
2
2
1/2
= n (1 + x)1/2
(7)
x = −2n−1+ 3 − 2n−1+ 5 + o n−1+ 5 ∼ −2n−1/3
(8)
Equations (7), (8) and (4) give
Pn
2
2
2
1
= n 1+
−2n−1+ 3 − 2n−1+ 5 + o n−1+ 5 + O n−2/3
2
2/3
= n − n − n2/5 + o n2/5 + O n2/6 = n − n2/3 − n2/5 + o n2/5
That is, equation (6) if h = 3.
If h ≥ 4, that is ph ≥ 7, equation (1) can be written in the form (see (3))
2
2
Pn = n2 −2n1+ 3 −2n1+ 5 +
2
13 1+ 2 32 1+ 2
1+ 2
2µ(q)n1+ q +o n ph
n 6 + n 30 +
3
15
7≤q≤ph
(9)
q=30
Note that if 7 ≤ ph ≤ 29 equation (9) becomes
2
Pn = n − 2n
1+ 23
− 2n
1+ 25
2
2
13
1+ 2
+ n1+ 6 +
2µ(q)n1+ q + o n ph
3
7≤q≤ph
q=30
2
1+ p2
Since n1+ 30 = o n
h
.
Equation (9) gives
Pn = n2 (1 + x)
(10)
where
2
2
x = −2n−1+ 3 − 2n−1+ 5 +
−1+ p2
h
+ o n
2
13 −1+ 2 32 −1+ 2
6 +
30 +
n
n
2µ(q)n−1+ q
3
15
7≤q≤ph
q=30
∼ −2n−1/3
(11)
Equations (10) and (5) give
Pn
1
1
= n (1 + x) = n 1 + x − x2 + O(x3 )
2
8
1 2
1
= n + nx − nx + nO(x3 )
2
8
1/2
(12)
R. Jakimczuk
948
Equation (11) gives
nO(x3 ) = nO(n−1 ) = O(1) = o n2/ph
(13)
13
16
1
µ(q)n2/q + o n2/ph
nx = −n2/3 − n2/5 + n2/6 + n2/30 +
2
6
15
7≤q≤ph
(14)
q=30
1
− nx2
8
⎛
⎞2
2⎟
1 ⎜
13 −1+ 2 32 −1+ 2
−1+ 23
−1+ 25
6 +
30 +
= − n⎜
−2n
−
2n
+
2µ(q)n−1+ q ⎟
n
n
⎝
⎠
8
3
15
7≤q≤ph
2 2
q=30
2
1
1
+ o(1) = − n −2n−1+ 3 − n2 −2n−1+ 3
8
8
1 2/6
2/30
2/ph
= − n −n
+o n
2
2
−2n−1+ 5 + O(1) + o(1)
(15)
Substituting (13), (14) and (15) into (12) we find that
5
1
Pn = n − n2/3 − n2/5 + n2/6 + n2/30 +
µ(q)n2/q + o n2/ph
3
15
7≤q≤ph
q=30
5
1
= n + n2/6 + n2/30 +
3
15
µ(q)n2/q + o n2/ph
2≤q≤ph
q=2,6,30
That is, equation (6). The theorem is proved.
ACKNOWLEDGEMENTS. The author is very grateful to Universidad
Nacional de Luján.
References
[1] R. Jakimczuk, Asymptotic formulae for the n-th perfect power, J. Integer
Seq. 15 (2012), Article 12.5.5.
Received: March 29, 2013