Asymptotic Formulae for the n-th Perfect Power

Rafael Jakimczuk División Matemática Universidad Nacional de Luján Buenos Aires Argentina

                   jakimczu@mail.unlu.edu.ar

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

Abstract Let P_{n} be the n-th perfect power. In this article we prove asymptotic formulae for

P_{n}. For example, we prove the following formula

P_{n} = n^{2} − 2n^{5/3} − 2n^{7/5} + \frac{13}{3}n^{4/3} − 2n^{9/7} + 2n^{6/5} − 2n^{13/11} + o (n^{13/11}) .

1 Introduction

A natural number of the form m^{n} 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 form sequence \underline{A001597} in Sloane’s Encyclopedia.

Let P_{n} be the n-th perfect power. That is, P_{1} = 1, P_{2} = 4, P_{3} = 8, P_{4} = 9, . . ..

In this article we prove asymptotic formulae for P_{n}. For example,

P_{n} = n^{2} − 2n^{5/3} − 2n^{7/5} + \frac{13}{3}n^{4/3} − 2n^{9/7} + 2n^{6/5} − 2n^{13/11} + o (n^{13/11}) .

This formula is a corollary of our main theorem (Theorem 6), which can give as many terms in the expansion as desired.

There exist various theorems and conjectures on the sequence P_{n}. For example, the following theorem:

∑_{n=2}^{∞} \frac{1}{P_{n}} = ∑_{k=2}^{∞} μ(k) (1 − ζ(k)) = 0, 87446 …

where μ(k) is the Möbius function and ζ(k) is the Riemann zeta function.

We also have the following theorem called the Goldbach-Euler theorem:

∑_{n=2}^{∞} \frac{1}{P_{n} − 1} = 1.

This result was first published by Euler in 1737. Euler attributed the result to a letter (now lost) from Goldbach.

Mihăilescu [4, 5, 6] proved that the only pair of consecutive perfect powers is 8 and 9, thus proving Catalan’s conjecture.

The Pillai’s conjecture establish the following limit

lim_{n→∞}(P_{n+1} − P_{n}) = ∞.

This is an unsolved problem.

There exist algorithms for detecting perfect powers [1, 2].

Let N(x) be the number of perfect powers not exceeding x. M. A. Nyblom [7] proved the following asymptotic formula

N(x) ∼ \sqrt{x}.

M. A. Nyblom [8] also obtained a formula for the exact value of N(x) using the inclusionexclusion principle.

Let p_{h} be the h-th prime. Consequently we have,

p_{1} = 2, p_{2} = 3, p_{3} = 5, p_{4} = 7, p_{5} = 11, p_{6} = 13, …

Jakimczuk [3] proved the following theorem where more precise formulae for N(x) are established. This theorem will be used later.

Theorem 1. Let p_{h} be the h-th prime with h ≥ 2, where h is an arbitrary but fixed positive integer. Then

N(x) = ∑_{k=1}^{h−1}(−1)^{k+1} ∑_{p_{i_{1}}···p_{i_{k}} <p_{h}^{1≤i_{1}<···<i_{k}≤h−1}} x^{\frac{1}{p_{i_{1}} ···p_{i_{k}}}} + (1 + o(1))x^{1/p_{h}},  (1)

where the inner sum is taken over the k-element subsets {i_{1}, …, i_{k}} of the set {1, 2, …, h − 1} such that the inequality p_{i_{1}} ⋯ p_{i_{k}} < p_{h} holds.

If h = 5 then Theorem 1 becomes,

N(x) = \sqrt{x} + \sqrt[3]{x} + \sqrt[5]{x} − \sqrt[6]{x} + \sqrt[7]{x} − \sqrt[10]{x} + (1 + o(1)) \sqrt[11]{x}.  (2)

Note that equation (2) include the cases h = 2, 3, 4. In general, equation (1) for a certain value of h = k include the cases h = 2, 3, …, k − 1. This fact is a direct consequence of equation (1).

2 Some Lemmas

The following lemma is an immediate consequence of the binomial theorem.

Lemma 2. We have

(1 + x)^{α} = 1 + (α + o(1))x (x → 0),
(1 + x)^{α} = 1 + αx + O(x^{2}) (x → 0),

(1 + x)^{α} = 1 + αx + \frac{α(α − 1)}{2}x^{2} + O(x^{3}) (x → 0).

Lemma 3. Let P_{n} be the n-th perfect power. We have

P_{n} ∼ n^{2}.

Proof. Equation (2) gives N(x) ∼ \sqrt{x}. Consequently N(P_{n}) = n ∼ \sqrt{P_{n}}. Therefore P_{n} ∼ n^{2}.

Lemma 4. Let p_{h} be the h-th prime. If h ≥ 3 then we have

\frac{2}{p_{h−1}} − \frac{1}{3} < \frac{2}{p_{h}} .

Proof. We have

\frac{2}{p_{h−1}} − \frac{1}{3} < \frac{2}{p_{h}} ⇔ \frac{2}{p_{h−1}} − \frac{2}{p_{h}} < \frac{1}{3} ⇔ \frac{1}{p_{h−1}} − \frac{1}{p_{h}} < \frac{1}{6}.

Clearly, the last inequality is true if h ≥ 5 since p_{h−1} ≥ 7.

On the other hand, we have

\frac{1}{p_{2}} − \frac{1}{p_{3}} = \frac{1}{3} − \frac{1}{5} = \frac{2}{15} < \frac{1}{6}

\frac{1}{p_{3}} − \frac{1}{p_{4}} = \frac{1}{5} − \frac{1}{7} = \frac{2}{35} < \frac{1}{6}

3 The Fundamental Lemma

The following lemma is a characterization of asymptotic formulae for P_{n}. The lemma prove the existence of asymptotic formulae for P_{n}.

Lemma 5. Let p_{h} (h ≥ 3) be the h-th prime. We have

P_{n} = n^{2} − 2n^{5/3} + ∑_{i=1}^{m} d_{i}n^{g_{i}} + (−2 + o(1))n^{1+ \frac{2}{p_{h}}} ,  (3)

where 2 > 5/3 > g_{1} > ⋯ > g_{m} > 1 + \frac{2}{p_{h}}, the d_{i} are rational coefficients and in equation

(3) appear the terms −2n^{1+ \frac{2}{p_{i}}} (i = 2, …, h − 1). Besides the rational exponents 5/3 and g_{i} (i = 1, …, m) are of the form \frac{b_{i}}{c_{i}} where b_{i} and c_{i} are relatively prime and the c_{i} are squarefree integers with prime divisors bounded by p_{h−1}.

Proof. We shall use mathematical induction. First, we shall prove that the lemma is true for h = 3.

If h = 2 then Theorem 1 becomes (see (2))

N(x) = \sqrt{x} + (1 + o(1))\sqrt[3]{x}.

Substituting x = P_{n} into this equation and using Lemma 3 we obtain

N(P_{n}) = n = \sqrt{P_{n}} + (1 + o(1))\sqrt[3]{P_{n}} = \sqrt{P_{n}} + (1 + o(1))n^{2/3}.

That is

\sqrt{P_{n}} = n + (−1 + o(1))n^{2/3}.

Therefore

P_{n} = (n + (−1 + o(1))n^{2/3})^{2}
= n^{2} + 2(−1 + o(1))n^{5/3} + (−1 + o(1))^{2}n^{4/3}
= n^{2} + (−2 + o(1))n^{5/3}.  (4)

If h = 3 then Theorem 1 becomes (see (2))

N(x) = \sqrt{x} + \sqrt[3]{x} + (1 + o(1))\sqrt[5]{x}.

Substituting x = P_{n} into this equation and using equation (4), Lemma 3 and Lemma 2 we obtain

N(P_{n}) = n = P_{n}^{1/2} + P_{n}^{1/3} + (1 + o(1))n^{2/5} = P_{n}^{1/2} + (n^{2} + (−2 + o(1))n^{5/3})^{1/3}
+ (1 + o(1))n^{2/5} = P_{n}^{1/2} + n^{2/3} (1 + (−2 + o(1))n^{−1/3})^{1/3} + (1 + o(1))n^{2/5}
= P_{n}^{1/2} + n^{2/3} (1 + ((−2/3) + o(1))n^{−1/3})
+ (1 + o(1))n^{2/5} = P_{n}^{1/2} + n^{2/3} + (1 + o(1))n^{2/5}.

That is

P_{n}^{1/2} = n − n^{2/3} + (−1 + o(1))n^{2/5}.

Therefore

P_{n} = (n − n^{2/3} + (−1 + o(1))n^{2/5})^{2} = n^{2} − 2n^{5/3} + (−2 + o(1))n^{7/5}.

That is

P_{n} = n^{2} − 2n^{5/3} + (−2 + o(1))n^{7/5}.  (5)

Equation (5) is Lemma 5 for h = 3. Consequently the lemma is true for h = 3.

Suppose that the lemma is true for h − 1 ≥ 3. We shall prove that the lemma is also true for h ≥ 4.

We have (see (1))

N(x) = ∑_{k=1}^{h−1}(−1)^{k+1} ∑_{p_{i_{1}}···p_{i_{k}} <p_{h}^{1≤i_{1}<···<i_{k}≤h−1}} x^{\frac{1}{p_{i_{1}} ···p_{i_{k}}}} + (1 + o(1))x^{1/p_{h}} = x^{1/2}

+ ∑_{i=2}^{h−1} x^{1/p_{i}} + ∑_{k=2}^{h−1}(−1)^{k+1} ∑_{p_{i_{1}}···p_{i_{k}} <p_{h}^{1≤i_{1}<···<i_{k}≤h−1}} x^{\frac{1}{p_{i_{1}} ···p_{i_{k}}}}

+ (1 + o(1))x^{1/p_{h}}  (h ≥ 4).  (6)

Substituting x = P_{n} into (6) and using Lemma 3 we obtain

n = P_{n}^{1/2} + ∑_{i=2}^{h−1} P_{n}^{1/p_{i}} + ∑_{k=2}^{h−1}(−1)^{k+1} ∑_{p_{i_{1}}···p_{i_{k}} <p_{h}^{1≤i_{1}<···<i_{k}≤h−1}} Pn^{\frac{1}{p_{i_{1}} ···p_{i_{k}}}}

+ (1 + o(1))n^{2/p_{h}}  (h ≥ 4).  (7)

By inductive hypothesis we have

P_{n} = n^{2} − 2n^{5/3} + ∑_{i=1}^{s} a_{i}n^{r_{i}} + (−2 + o(1))n^{1+ \frac{2}{p_{h−1}}}  (h ≥ 4),  (8)

where 2 > 5/3 > r_{1} > ⋯ > r_{s} > 1 + \frac{2}{p_{h−1}} , the a_{i} are rational coefficients and in equation

(8) appear the terms −2n^{1+ \frac{2}{p_{i}}} (i = 2, …, h − 2). Besides the rational exponents 5/3 and r_{i} (i = 1, …, s) are of the form \frac{l_{i}}{f_{i}} where l_{i} and f_{i} are relatively prime and the f_{i} are squarefree integers with prime divisors bounded by p_{h−2}.

Equation (8) gives

P_{n} = n^{2} (1 − 2n^{−1/3} + ∑_{i=1}^{s} a_{i}n^{−2+r_{i}} + (−2 + o(1))n^{−1+ \frac{2}{p_{h−1}}} ) ,  (9)

where

−2n^{−1/3} + ∑_{i=1}^{s} a_{i}n^{−2+r_{i}} + (−2 + o(1))n^{−1+ \frac{2}{p_{h−1}}} ∼ −2n^{−1/3}.

Consequently

−2n^{−1/3} + ∑_{i=1}^{s} a_{i}n^{−2+r_{i}} + (−2 + o(1))n^{−1+ \frac{2}{p_{h−1}}} = O (n^{−1/3}) = o(1).  (10)

Let t ≥ 3 be a positive integer. Equations (9), (10) and Lemma 2 give

P_{n}^{1/t} = n^{2/t} (1 − 2n^{−1/3} + ∑_{i=1}^{s} a_{i}n^{−2+r_{i}} + (−2 + o(1))n^{−1+ \frac{2}{p_{h−1}}} )^{1/t}

= n^{2/t} (1 + \frac{1}{t} (−2n^{−1/3} + ∑_{i=1}^{s} a_{i}n^{−2+r_{i}} + (−2 + o(1))n^{−1+ \frac{2}{p_{h−1}}} ))

+ O (n^{−2/3})) = n^{2/t} − \frac{2}{t}n^{− \frac{1}{3} + \frac{2}{t}} + ∑_{i=1}^{s} \frac{1}{t}a_{i}n^{−2+r_{i}+ \frac{2}{t}}
+ \frac{1}{t}(−2 + o(1))n^{−1+ \frac{2}{p_{h−1}} + \frac{2}{t}} + O (n^{− \frac{2}{3} + \frac{2}{t}} ) .  (11)

Note that if t ≥ 3 then (see Lemma 4)

\frac{1}{t}(−2 + o(1))n^{−1+ \frac{2}{p_{h−1}} + \frac{2}{t}} = o (n^{2/p_{h}}) ,  (12)

and if t ≥ 3 then

O (n^{− \frac{2}{3} + \frac{2}{t}} ) = o (n^{2/p_{h}}) .  (13)

Consequently (11) becomes (see (12) and (13))

P_{n}^{1/t} = n^{2/t} − \frac{2}{t}n^{− \frac{1}{3} + \frac{2}{t}} + ∑_{i=1}^{s} \frac{1}{t}a_{i}n^{−2+r_{i}+ \frac{2}{t}} + o (n^{2/p_{h}}) .  (14)

Note that (see (14)) if t ≥ 3 the exponent \frac{2}{t} < 1 and consequently also −\frac{1}{3} + \frac{2}{t} < 1 and −2 + r_{i} + \frac{2}{t} < 1 since −2 + r_{i} < 0 (see (8))

Substituting (14) into (7) we find that

n = P_{n}^{1/2} + ∑_{j=2}^{h−1} (n^{2/p_{j}} − \frac{2}{p_{j}} n^{− \frac{1}{3} + \frac{2}{p_{j}}} + ∑_{i=1}^{s} \frac{1}{p_{j}} a_{i}n^{−2+r_{i}+ \frac{2}{p_{j}}} ) + ∑_{k=2}^{h−1}(−1)^{k+1}

∑_{p_{i_{1}}···p_{i_{k}} <p_{h}^{1≤i_{1}<···<i_{k}≤h−1}} (n^{\frac{2}{p_{i_{1}} ···p_{i_{k}}}} − \frac{2}{p_{i_{1}} ⋯ p_{i_{k}}} n^{− \frac{1}{3} +_{\frac{2}{p_{i_{1}} ···p_{i_{k}}}}} )

+ ∑_{i=1}^{s} \frac{1}{p_{i_{1}} ⋯ p_{i_{k}}} a_{i}n^{−2+r_{i}+_{\frac{2}{p_{i_{1}} ···p_{i_{k}}}}} ) + (1 + o(1))n^{2/p_{h}}

= P_{n}^{1/2} + ∑_{i=1}^{l} b_{i}n^{s_{i}} + (1 + o(1))n^{2/p_{h}}  (h ≥ 4),  (15)

where 1 > s_{1} > ⋯ > s_{l} > \frac{2}{p_{h}}. That is

P_{n}^{1/2} = n − ∑_{i=1}^{l} b_{i}n^{s_{i}} + (−1 + o(1))n^{2/p_{h}}.  (16)

Note that all positive exponents in equation (15), that is, the positive exponents of the form

\frac{2}{p_{j}} , −\frac{1}{3} + \frac{2}{p_{j}} , −2 + r_{i} + \frac{2}{p_{j}}

\frac{2}{p_{i_{1}} ⋯ p_{i_{k}}} , −\frac{1}{3} + \frac{2}{p_{i_{1}} ⋯ p_{i_{k}}} , −2 + r_{i} + \frac{2}{p_{i_{1}} ⋯ p_{i_{k}}}

are (see (8)) of the form \frac{m_{i}}{n_{i}} where m_{i} and n_{i} are relatively prime and the n_{i} are squarefree integers with prime divisors bounded by p_{h−1}. Therefore these exponents are different from \frac{2}{p_{h}} and consequently the exponents s_{i} (i = 1, …, l) in (16) are of this same form.

Note that 1 + \frac{2}{p_{h}} > 2 \frac{2}{p_{h}}, since \frac{2}{p_{h}} < 1. Consequently equation (16) gives

P_{n} = (n − ∑_{i=1}^{l} b_{i}n^{s_{i}} + (−1 + o(1))n^{2/p_{h}})^{2} = (n − ∑_{i=1}^{l} b_{i}n^{s_{i}})^{2}

+ (−2 + o(1))n^{1+ \frac{2}{p_{h}}} = n^{2} − 2n^{5/3} + ∑_{i=1}^{s} a_{i}n^{r_{i}} − 2n^{1+ \frac{2}{p_{h−1}}} + ∑_{i=1}^{q} c_{i}n^{k_{i}}
+ (−2 + o(1))n^{1+ \frac{2}{p_{h}}} (h ≥ 4),  (17)

where 2 > 5/3 > r_{1} > ⋯ > r_{s} > 1 + \frac{2}{p_{h−1}} > k_{1} > ⋯ > k_{q} > 1 + \frac{2}{p_{h}}.

Note also that the first terms in equation (17) are the terms of equation (8). On the other hand in equation (17) appear the term −2n^{1+ \frac{2}{p_{h−1}}} (see equation (8)). We now prove these facts.

Equation (17) can be written in the form

P_{n} = Q(n) + ∑_{i=1}^{q} c_{i}n^{k_{i}} + (−2 + o(1))n^{1+ \frac{2}{p_{h}}} = Q(n) + o (n^{1+ \frac{2}{p_{h−1}}} ) ,  (18)

where Q(n) is a sum of terms of the form e_{i}n^{q_{i}} (q_{i} ≥ 1 + \frac{2}{p_{h−1}} ).

On the other hand, equation (8) can be written in the form

P_{n} = n^{2} − 2n^{5/3} + ∑_{i=1}^{s} a_{i}n^{r_{i}} − 2n^{1+ \frac{2}{p_{h−1}}} + o (n^{1+ \frac{2}{p_{h−1}}} ) .  (19)

Equations (18) and (19) give

0 = P_{n} − P_{n} = (Q(n) − (n^{2} − 2n^{5/3} + ∑_{i=1}^{s} a_{i}n^{r_{i}} − 2n^{1+ \frac{2}{p_{h−1}}} )) + o (n^{1+ \frac{2}{p_{h−1}}} ) .

If

Q(n) ≠ n^{2} − 2n^{5/3} + ∑_{i=1}^{s} a_{i}n^{r_{i}} − 2n^{1+ \frac{2}{p_{h−1}}}

then we obtain

0 = (P_{n} − P_{n}) ∼ an^{q}  (a ≠ 0)  (q ≥ 1 + \frac{2}{p_{h−1}} ) .

That is, an evident contradiction. Consequently

Q(n) = n^{2} − 2n^{5/3} + ∑_{i=1}^{s} a_{i}n^{r_{i}} − 2n^{1+ \frac{2}{p_{h−1}}} .  (20)

Finally, equations (18) and (20) give (17).

Lemma 5 is constructive, we can build the next formula using the former formula. Next, we build the formula that correspond to h = 4. We shall need this formula.

If h = 4 equation (6) is (see (2))

N(x) = x^{1/2} + x^{1/3} + x^{1/5} − x^{1/6} + (1 + o(1))x^{1/7}.  (21)

On the other hand (Lemma 5) equation (8) is (see (5))

P_{n} = n^{2} − 2n^{5/3} + (−2 + o(1))n^{7/5}.

Consequently equation (15) is

n = P_{n}^{1/2} + (n^{2/3} − \frac{2}{3}n^{− \frac{1}{3} + \frac{2}{3}} ) + (n^{2/5} − \frac{2}{5}n^{− \frac{1}{3} + \frac{2}{5}} ) − (n^{2/6} − \frac{2}{6}n^{− \frac{1}{3} + \frac{2}{6}} )
+ (1 + o(1))n^{2/7} = P_{n}^{1/2} + n^{2/3} − \frac{2}{3}n^{1/3} + n^{2/5} − \frac{2}{5}n^{1/15} − n^{1/3} + \frac{1}{3} + (1 + o(1))n^{2/7}
= P_{n}^{1/2} + n^{2/3} + n^{2/5} − \frac{5}{3}n^{1/3} + (1 + o(1))n^{2/7}.

Therefore

P_{n}^{1/2} = n − n^{2/3} − n^{2/5} + \frac{5}{3}n^{1/3} + (−1 + o(1))n^{2/7}.

Consequently (see (17))

P_{n} = (n − n^{2/3} − n^{2/5} + \frac{5}{3}n^{1/3})^{2} + (−2 + o(1))n^{9/7}
= n^{2} − 2n^{5/3} − 2n^{7/5} + \frac{13}{3}n^{8/6} + (−2 + o(1))n^{9/7}.

That is

P_{n} = n^{2} − 2n^{5/3} − 2n^{7/5} + \frac{13}{3}n^{8/6} + (−2 + o(1))n^{9/7}.  (22)

4 Main Result

The following theorem is the main result of this article. In this theorem we obtain explicit formulae for P_{n}.

Theorem 6. Let p_{h} be the h-th prime with h ≥ 3, where h is an arbitrary but fixed positive integer.

Let us consider the formula (see (1))

N(x) = ∑_{k=1}^{h−1}(−1)^{k+1} ∑_{p_{i_{1}}···p_{i_{k}} <p_{h}^{1≤i_{1}<···<i_{k}≤h−1}} x^{\frac{1}{p_{i_{1}} ···p_{i_{k}}}} + (1 + o(1))x^{1/p_{h}}.  (23)

We have

P_{n} = n^{2} + \frac{13}{3}n^{8/6} + \frac{32}{15}n^{32/30} + ∑_{k=1}^{h−1}(−1)^{k} ∑_{p_{i_{1}}···p_{i_{k}} <p_{h}, p_{i_{1}}···p_{i_{k}} ≠ 2, 6, 30^{1≤i_{1}<···<i_{k}≤h−1}} 2n^{1+_{\frac{2}{p_{i_{1}} ···p_{i_{k}}}}}

+ (−2 + o(1))n^{1+ \frac{2}{p_{h}}} .  (24)

Proof. We shall see that everything relies on Theorem 1. The theorem is true for h = 3 (see Lemma 5) and for h = 4 (see (21) and (22)). Suppose h ≥ 5, that is p_{h} ≥ 11. Equation (23) can be written in the form (see (21))

N(x) = x^{1/2} + x^{1/3} + x^{1/5} − x^{1/6} + ∑_{i=1}^{s}(−1)^{1+a_{i}}x^{1/n_{i}} + (1 + o(1))x^{1/p_{h}},  (25)

where a_{i} is the number of different prime factors in n_{i} and the exponents are in decreasing order,

\frac{1}{2} > \frac{1}{3} > \frac{1}{5} > \frac{1}{6} > \frac{1}{n_{1}} > ⋯ > \frac{1}{n_{s}} > \frac{1}{p_{h}} .  (26)

For example, if h = 5 then equation (25) becomes equation (2).

On the other hand, we have (Lemma 5 and equation (22))

P_{n} = n^{2} − 2n^{5/3} − 2n^{7/5} + \frac{13}{3}n^{8/6} + ∑_{i=1}^{t} d_{i}n^{r_{i}} + (−2 + o(1))n^{1+ \frac{2}{p_{h}}} ,  (27)

where the exponents are in decreasing order,

2 > \frac{5}{3} > \frac{7}{5} > \frac{8}{6} > r_{1} > ⋯ > r_{t} > 1 + \frac{2}{p_{h}} .  (28)

Equation (27) gives

P_{n} = n^{2} (1 − 2n^{−1/3} − 2n^{−3/5} + \frac{13}{3}n^{−4/6} + ∑_{i=1}^{t} d_{i}n^{r_{i}−2} + (−2 + o(1))n^{−1+ \frac{2}{p_{h}}} )  (29)

where

−2n^{−1/3} − 2n^{−3/5} + \frac{13}{3}n^{−4/6} + ∑_{i=1}^{t} d_{i}n^{r_{i}−2} + (−2 + o(1))n^{−1+ \frac{2}{p_{h}}} ∼ −2n^{−1/3},

since (see (28))

−\frac{1}{3} > −\frac{3}{5} > −\frac{4}{6} > r_{1} − 2 > ⋯ > r_{t} − 2 > −1 + \frac{2}{p_{h}} .  (30)

Consequently

A_{n} = −2n^{−1/3} − 2n^{−3/5} + \frac{13}{3}n^{−4/6} + ∑_{i=1}^{t} d_{i}n^{r_{i}−2} + (−2 + o(1))n^{−1+ \frac{2}{p_{h}}}
= O (n^{−1/3}) = o(1).  (31)

Besides

B_{n} = (−2n^{−1/3} − 2n^{−3/5} + \frac{13}{3}n^{−4/6} + ∑_{i=1}^{t} d_{i}n^{r_{i}−2} + (−2 + o(1))n^{−1+ \frac{2}{p_{h}}} )^{2}

= (−2n^{−1/3} − 2n^{−3/5} + (\frac{13}{3} + o(1)) n^{−4/6})^{2}
= 4n^{−2/3} + 8n^{−14/15} + O (n^{−1}) .  (32)

Substituting x = P_{n} into equation (25) and using Lemma 3 we obtain

n = P_{n}^{1/2} + P_{n}^{1/3} + P_{n}^{1/5} − P_{n}^{1/6} + ∑_{i=1}^{s}(−1)^{1+a_{i}}P_{n}^{1/n_{i}} + n^{2/p_{h}} + o (n^{2/p_{h}}) .  (33)

Equations (29), (31), (32) and Lemma 2 give

P_{n}^{1/2} = n (1 + \frac{1}{2}A_{n} + \frac{\frac{1}{2} (\frac{1}{2} − 1)}{2}B_{n} + O (n^{−1}))

= n − n^{2/3} − n^{2/5} + \frac{13}{6}n^{2/6} + ∑_{i=1}^{t} \frac{d_{i}}{2} n^{r_{i}−1}
− n^{2/p_{h}} + o (n^{2/p_{h}}) − \frac{1}{2}n^{2/6} − n^{2/30} + O(1).  (34)

Equations (29), (31), (30) and Lemma 2 give

P_{n}^{1/3} = n^{2/3} (1 + \frac{1}{3}A_{n} + O (n^{−2/3})) = n^{2/3} − \frac{2}{3}n^{2/6} − \frac{2}{3}n^{2/30} + O(1).  (35)

P_{n}^{1/5} = n^{2/5} (1 + \frac{1}{5}A_{n} + O (n^{−2/3})) = n^{2/5} − \frac{2}{5}n^{2/30} + o(1).  (36)

P_{n}^{1/6} = n^{2/6} (1 + \frac{1}{6}A_{n} + O (n^{−2/3})) = n^{2/6} + O(1).  (37)

P_{n}^{1/n_{i}} = n^{2/n_{i}} (1 + \frac{1}{n_{i}} A_{n} + O (n^{−2/3})) = n^{2/n_{i}} + o(1) (i = 1, …, s).  (38)

Substituting equations (34), (35), (36), (37) and (38) into equation (33) we find that

0 = ∑_{i=1}^{t} \frac{d_{i}}{2} n^{r_{i}−1} + ∑_{i=1}^{s}(−1)^{1+a_{i}}n^{2/n_{i}} − \frac{31}{15}n^{2/30} + o (n^{2/p_{h}}) .  (39)

Note that (see (28) and (26)) r_{i} − 1 > \frac{2}{p_{h}} and \frac{2}{n_{i}} > \frac{2}{p_{h}}.

If p_{h} ≤ 29 then −\frac{31}{15}n^{2/30} = o (n^{2/p_{h}}). Consequently we have

∑_{i=1}^{t} \frac{d_{i}}{2} n^{r_{i}−1} = ∑_{i=1}^{s}(−1)^{a_{i}}n^{2/n_{i}},

where t = s, d_{i} = (−1)^{a_{i}}2 (i = 1, …, s) and r_{i} = 1 + \frac{2}{n_{i}} (i = 1, …, s). Since in contrary case we have 0 ∼ an^{b} where a ≠ 0 and b > \frac{2}{p_{h}}, an evident contradiction. Substituting these values into (27) we obtain (24) (see (25)).

If p_{h} ≥ 31 then \frac{2}{30} > \frac{2}{p_{h}} and there exists k such that n_{k} = 30 = 2.3.5 (see (23)). Consequently we have

∑_{i=1}^{t} \frac{d_{i}}{2} n^{r_{i}−1} = ∑_{i=1}^{s}(−1)^{a_{i}}n^{2/n_{i}} + \frac{31}{15}n^{2/30},

where t = s, d_{i} = (−1)^{a_{i}}2 (i ≠ k), d_{k} = 2(−1 + \frac{31}{15}) = \frac{32}{15} and r_{i} = 1 + \frac{2}{n_{i}} (i = 1, …, s). Since in contrary case we have 0 ∼ an^{b} where a ≠ 0 and b > \frac{2}{p_{h}}, an evident contradiction. Substituting these values into (27) we obtain (24) (see (25)).

Example 7. If h = 5 equation (23) is (see (2))

N(x) = x^{1/2} + x^{1/3} + x^{1/5} − x^{1/6} + x^{1/7} − x^{1/10} + (1 + o(1))x^{1/11}.

Consequently Theorem 6 gives

P_{n} = n^{2} − 2n^{5/3} − 2n^{7/5} + \frac{13}{3}n^{4/3} − 2n^{9/7} + 2n^{6/5} + (−2 + o(1))n^{13/11}

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 Luján.

References

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[2] D. Bernstein, Detecting perfect powers in essentially linear time, Math. Comp. 67 (1998), 1253–1283.

[3] R. Jakimczuk, On the distribution of perfect powers, J. Integer Seq. 14 (2011), Article 11.8.5.

[4] P. Mihăilescu, A class number free criterion for Catalan’s conjecture, J. Number Theory 99 (2003), 225–231.

[5] P. Mihăilescu, Primary cyclotomic units and a proof of Catalan’s conjecture, J. Reine Angew. Math 572 (2004), 167–195.

[6] P. Mihăilescu, On the class groups of cyclotomic extensions in presence of a solution to Catalan’s equation, J. Number Theory 118 (2006), 123–144.

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

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

2000 Mathematics Subject Classification: Primary 11A99; Secondary 11B99. Keywords: n-th perfect power, asymptotic formula.

(Concerned with sequence \underline{A001597}.)

Received October 1 2011; revised versions received January 14 2012; March 20 2012; May 29 2012. Published in Journal of Integer Sequences, May 29 2012.

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