On the Distribution of Perfect Powers

1 2 Journal of Integer Sequences, Vol. 14 (2011), 3 Article 11.8.5 47 6 23 11

Rafael Jakimczuk División Matemática Universidad Nacional de Luján Buenos Aires Argentina [email protected]

In memory of my sister Fedra Marina Jakimczuk (1970–2010)

Abstract Let N(x) be the number of perfect powers that do not exceed x. In this article we obtain asymptotic formulae for the counting function N(x).

1 Introduction

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 ar≥e 1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128,..., and they are sequence A001597 in Sloane’s Encylopedia. Let N(x) be the number of perfect powers that do not exceed x. M. A. Nyblom [3] proved the following asymptotic formula, N(x) √x. ∼ M. A. Nyblom [4] also obtained a formula for the exact value of N(x) using the inclusion- exclusion principle (also called the principle of cross-classiﬁcation). In this article we obtain more precise asymptotic formulae for the counting function N(x). For example, we prove N(x)= √x + √3 x + √5 x √6 x + √7 x + o √7 x . − Consequently √x + √3 x

1 2 Preliminary Results

Let A be a set. The number of elements in A we denote in the form A . We need the following results. | | Lemma 1. (Inclusion-exclusion principle) Let us consider a given ﬁnite collection of sets n A ,A ,...,An. The number of elements in Ai is 1 2 ∪i=1 n n k+1 Ai = ( 1) Ai Ai , |∪i=1 | − | 1 ∩···∩ k | Xk=1 1≤i1

n n Let An(x) (n 2) be the set k : k N, k x , that is, the set of perfect powers ≥ { ∈ ≤ } whose exponent is n that do not exceed x. Lemma 2. We have n An(x) = √x , | | where . denotes the integer-part function. ⌊ ⌋ Proof. We have kn x k √n x. ≤ ⇔ ≤

M. A. Nyblom [4] proved the following Lemma and the following Theorem.

Lemma 3. For any set consisting of m 2 positive integers n1,...,nm all greater than unity, we have the set equality ≥ { }

Ani (x)= A[n1,...,nm](x), i\=1 where [n1,...,nm] denotes the least common multiple of the m integers n1,...,nm.

Let pn be the n-th prime. Consequently we have,

p1 = 2,p2 = 3,p3 = 5,p4 = 7,p5 = 11,p6 = 13,...

Theorem 4. If x 4 and p1,p2,...,pm denote the prime numbers that do not exceed log x , then for k ≥ m the number of perfect powers that do not exceed x is ⌊ 2 ⌋ ≥ k N x A x . ( )= pi ( )

i[=1

Besides Ap (x)= 1 for i m + 1. i { } ≥ 2 We also need the following two well-known results. The binomial formula n n (a + b)n = an−kbk, (1) k Xk=0 and the following property of the absolute value

a + a + + ar a + a + + ar , (2) | 1 2 ··· | ≤ | 1| | 2| ··· | | where a1,a2,...,ar are real numbers.

3 Main Results

Theorem 5. Let pn be the n-th prime number with n 2, where n is an arbitrary but ﬁxed positive integer. Then ≥

n−1 1 1 k+1 p ···p N(x)= ( 1) x i1 ik + g(x)x pn , (3) − Xk=1 1≤i1<···X

p <

3 First relation. m < k n < n + 1 ≤ and hence pm

Note that there exists x such that if x x the third relation holds. 0 ≥ 0 Note also that S(x) n is either equal to 1 or k n < k 1= log x log x , and so in − − − log 2 ≤ log 2 either case j k log x S(x) n (10) − ≤ log 2 as x 4. Consequently (9) and (10) give ≥ n n log x 1 A x N x < A x x pn+1 . pi ( ) ( ) pi ( ) + (11) ≤ log 2 i[=1 i[=1

Lemma 1 gives

n n k+1 Ap (x) = ( 1) Ap (x) Ap (x) . (12) i i1 ik − ∩···∩ i[=1 Xk=1 1≤i1

On the other hand, Lemma 3 gives

Api (x) Api (x)= A p ,...,p (x)= Api ···pi (x). 1 ∩···∩ k [ i1 ik ] 1 k 4 Therefore (Lemma 2) we obtain

1 p ···p i1 ik Api (x) Api (x) = Api ···pi (x) = x . (13) 1 ∩···∩ k 1 k

Substituting (13) into (12) we ﬁnd that

n n 1 k+1 p ···p A x x i1 ik pi ( ) = ( 1) − i[=1 Xk=1 1≤i1

n 1 1 k+1 p ···p p ···p ( 1) x i1 ik x i1 ik . (14) − − − Xk=1 1≤i1

n 1 1 n k+1 pi ···pi pi ···pi ( 1) x 1 k x 1 k 1 − − ≤ Xk=1 1≤i1

n n 1 k+1 p ···p A x x i1 ik O pi ( ) = ( 1) + (1) (18) − i[=1 Xk=1 1≤i1

If n 1 k+1 p ···p B(x)= ( 1) x i1 ik (19) − Xk=1 1≤i1pn+1

5 then n 1 k+1 p ···p ( 1) x i1 ik = B(x)+ C(x). (21) − Xk=1 1≤i1

n A x B x C x O . pi ( ) = ( )+ ( )+ (1) (22)

i[=1

Equations (22) and (11) give

log x 1 B(x)+ C(x)+ O(1) N(x) < B(x)+ C(x)+ O(1) + x pn+1 . ≤ log 2 Therefore, log x 1 C(x)+ O(1) N(x) B(x)

N(x) B(x) 1 ǫ< − 1 < + ǫ (ǫ> 0). − log x x pn+1 log 2 That is, 1 N(x)= B(x)+ O log x x pn+1 . (24) Note that (see (19)) if k = 1,...,n then,

1 1 1 p ···p p ···p p ···p x i1 ik = x i1 ik + x i1 ik . (25)

1≤i1

1 1 p ···p p ···p x i1 ik = x i1 ik (27)

1≤i1

1 1 p x i1 = x pn (28)

1≤Xi1≤n pn≤pi1

6 Equations (19), (25), (26), (27) and (28) give

− n 1 1 k+1 p ···p B(x) = ( 1) x i1 ik − Xk=1 1≤i1<···X

That is,

− n 1 1 k+1 p ···p B(x) = ( 1) x i1 ik − Xk=1 1≤i1<···X

4 Examples

If n = 2 then Theorem 5 becomes

N(x)= √x + g(x)√3 x, (30) where limx→∞ g(x) = 1. If n = 3 then Theorem 5 becomes

N(x)= √x + √3 x + g(x)√5 x, where limx→∞ g(x) = 1. If n = 4 then Theorem 5 becomes

N(x)= √x + √3 x + √5 x √6 x + g(x)√7 x, − where limx→∞ g(x) = 1. Consequently

√x + √3 x

If n = 5 then Theorem 5 becomes

N(x)= √x + √3 x + √5 x √6 x + √7 x 10√x + g(x) 11√x, − − where limx→∞ g(x) = 1. To ﬁnish, we shall establish two simple theorems.

7 Theorem 6. Let us consider the n open intervals (0, 12), (12, 22),..., ((n 1)2, n2). Let S(n) be the number of these n open intervals that contain some perfect power.− Then

S(n) lim = 0. n→∞ n Therefore, almost all the open intervals are empty.

Proof. We have (Nyblom’s asymptotic formula)

N(x)= √x + f(x)√x, where limx→∞ f(x) = 0. Consequently

N(n2)= n + f(n2)n, where n are the n squares 12, 22,...,n2. Therefore

0 S(n) N(n2) n = f(n2)n. ≤ ≤ − That is S(n) 0 f(n2). ≤ n ≤

Using equation (30) we can obtain a more strong result.

Theorem 7. Let us consider the n open intervals (0, 12), (12, 22),..., ((n 1)2, n2). Let F (n) − 2 be the number of perfect powers in these n open intervals. Then F (n) n 3 . ∼ Proof. Equation (30) gives 2 2 2 N(n )= n + g(n )n 3 , where n are the n squares 12, 22,...,n2. Therefore

2 2 2 2 F (n)= N(n ) n = g(n )n 3 n 3 . − ∼

5 Acknowledgements

The author would like to thank the anonymous referee for his/her valuable comments and suggestions for improving the original version of this article. The author is also very grateful to Universidad Nacional de Luj´an.

8 References

[1] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford, 1960.

[2] W. J. LeVeque, Topics in Number Theory, Addison-Wesley, 1958.

[3] M. A. Nyblom, A counting function for the sequence of perfect powers, Austral. Math. Soc. Gaz. 33 (2006), 338–343.

[4] M. A. Nyblom, Counting the perfect powers, Math. Spectrum 41 (2008), 27–31.

2000 Mathematics Subject Classiﬁcation: Primary 11A99; Secondary 11B99. Keywords: Perfect powers, counting function, asymptotic formulae.

(Concerned with sequence A001597.)

Received May 28 2011; revised version received August 16 2011. Published in Journal of Integer Sequences, September 25 2011.

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