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\begin{center}
\vskip 1cm{\LARGE\bf
Asymptotic Formulae for  \\
\vskip .1in
the $n$-th Perfect Power}
\vskip 1cm
\large
Rafael Jakimczuk\\
Divisi\'on Matem\'atica \\
Universidad Nacional de Luj\'an\\
Buenos Aires \\
Argentina\\ 
\href{mailto:jakimczu@mail.unlu.edu.ar}{\tt jakimczu@mail.unlu.edu.ar}\\
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\centerline{} 

\centerline{\it In memory of my sister Fedra Marina Jakimczuk (1970--2010)} 

\centerline{}

\vskip .2in

\begin{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
\[P_n=n^2-2n^{5/3}-2 n^{7/5} +\frac{13}{3}n^{4/3}-2n^{9/7}+2n^{6/5}-2n^{13/11}+o\left(n^{13/11}\right).\]
\end{abstract}
  
 
\section{Introduction}


A natural number of the form $m^n$ where $m$ is a positive integer and $n\geq 2$ is called a perfect power. The first few terms of the integer sequence of perfect powers are 
\[1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128 \ldots\]
and they form sequence \seqnum{A001597} in Sloane's {\it Encyclopedia}.

Let $P_n$ be the $n$-th perfect power. That is, $P_1=1, P_2=4, P_3=8,
P_4=9, \ldots$.

In this article we prove asymptotic formulae for $P_n$. For example,
\[P_n=n^2-2n^{5/3}-2 n^{7/5} +\frac{13}{3}n^{4/3}-2n^{9/7}+2n^{6/5}-2n^{13/11}+o\left(n^{13/11}\right).\]
This formula is a corollary of our main theorem (Theorem~\ref{T6}), 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:
\[\sum^{\infty}_{n=2}\frac{1}{P_n}=\sum^{\infty}_{k=2}\mu(k)\left(1-\zeta(k)\right)=0,87446\ldots\]
where $\mu(k)$ is the M\"{o}bius function and $\zeta(k)$ is the Riemann zeta function.

We also have the following theorem called the Goldbach-Euler theorem:
\[\sum^{\infty}_{n=2}\frac{1}{P_n-1}=1.\]
This result was first published by Euler in 1737. Euler attributed the
result to a letter (now lost) from Goldbach.

Mih\u{a}ilescu \cite{Mihailescu4, Mihailescu5, Mihailescu6} 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
\[\lim_{n\rightarrow \infty}(P_{n+1}-P_n)=\infty.\]
This is an unsolved problem.

There exist algorithms for detecting perfect powers
\cite{BachSorenson1, Bernstein2}.

Let $N(x)$ be the number of perfect powers not exceeding $x$. M. A. Nyblom \cite{Nyblom7} proved the following asymptotic formula
\[N(x)\sim \sqrt{x}.\]
M. A. Nyblom \cite{Nyblom8} also obtained a formula for the exact value of $N(x)$ using the inclusion-exclusion principle.

Let $p_h$ be the $h$-th prime. Consequently we have,
\[p_1=2, p_2=3, p_3=5, p_4=7, p_5=11, p_6=13,\ldots\]

Jakimczuk \cite{Jakimczuk3} proved the following theorem where more precise formulae for $N(x)$ are established. This theorem will be used later.

\begin{theorem} \label{T1}
Let $p_h$  be the $h$-th prime with $h\geq 2$, where $h$ is an arbitrary but fixed positive integer. Then  
\begin{equation} \label{E1}
N(x)=\sum^{h-1}_{k=1}(-1)^{k+1}\sum_{\stackrel{1\leq i_1<\cdots<i_k\leq h-1}{p_{i_1}\cdots p_{i_k}<p_h}  } x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}+ (1+o(1)) x^{1/p_h},
\end{equation} 
where the inner sum is taken over the $k$-element subsets $\left\{i_1,\ldots, i_k\right\}$ of the set $\left\{1, 2, \ldots, h-1\right\}$  such that the inequality $p_{i_1}\cdots p_{i_k}<p_h$ holds.
\end{theorem} 

If $h=5$ then Theorem~\ref{T1} becomes,
\begin{equation} \label{E2}
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}.
\end{equation}
Note that equation (\ref {E2}) include the cases $h=2, 3, 4$. In general,  equation (\ref {E1}) for a certain value of $h=k$ include the cases $h=2, 3,\ldots,k-1$. This fact is a direct consequence of equation (\ref {E1}). 



\section{Some Lemmas}

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

\begin{lemma} \label{L2}
We have  
\[(1+x)^{\alpha}=1+(\alpha +o(1)) x \qquad (x\rightarrow 0),\]
\[(1+x)^{\alpha}=1+\alpha x +O(x^2) \qquad (x\rightarrow 0),\]
\[(1+x)^{\alpha}=1+\alpha x +\frac{\alpha (\alpha-1)}{2}x^2+O(x^3) \qquad (x\rightarrow 0).\]
\end{lemma} 

\begin{lemma} \label{L3}
Let $P_n$ be the $n$-th perfect power. We have
\[P_n \sim n^2.\]
\end{lemma}
\begin{proof} 
Equation (\ref {E2}) gives $N(x)\sim \sqrt{x}$. Consequently 	
$N(P_n)=n \sim \sqrt{P_n}$. Therefore $P_n\sim n^2.$ 
\end{proof}

\begin{lemma} \label{L4}
Let $p_h$ be the $h$-th prime. If $h\geq 3$ then we have
\[\frac{2}{p_{h-1}}-\frac{1}{3}<\frac{2}{p_h}.\]
\end{lemma}
\begin{proof} 
We have
\[\frac{2}{p_{h-1}}-\frac{1}{3}	<
\frac{2}{p_h}\Leftrightarrow \frac{2}{p_{h-1}}-\frac{2}{p_h}<\frac{1}{3}\Leftrightarrow 
\frac{1}{p_{h-1}}-\frac{1}{p_h}<\frac{1}{6}.\]
Clearly, the last inequality is true if $h\geq 5$ since $p_{h-1}\geq 7$.

On the other hand, we have
\[\frac{1}{p_2}-\frac{1}{p_3}=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}<\frac{1}{6}\]
\[\frac{1}{p_3}-\frac{1}{p_4}=\frac{1}{5}-\frac{1}{7}=\frac{2}{35}<\frac{1}{6}\]
\end{proof}

\section{The Fundamental Lemma}

The following lemma is a characterization of asymptotic formulae for $P_n$. The lemma prove the existence of asymptotic formulae for $P_n$.
 

\begin{lemma} \label{L5}
Let $p_h$ $(h\geq 3)$ be the $h$-th prime. We have 
\begin{equation} \label{E3} 
P_n=n^2-2n^{5/3}+\sum^{m}_{i=1}	d_in^{g_i}+(-2+o(1)) n^{1+\frac{2}{p_h}},
\end{equation} 
where $2>5/3>g_1>\cdots>g_m>1+\frac{2}{p_h}$, the $d_i$ are rational coefficients and in equation (\ref {E3}) appear the terms $-2 n^{1+\frac{2}{p_i}}$ $(i=2,\ldots, h-1)$. Besides the rational exponents $5/3$ and $g_i$ $(i=1, \ldots, m)$ 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 $p_{h-1}$.
\end{lemma} 
\begin{proof} 
We shall use mathematical induction. First, we shall prove that the lemma is true for $h=3$. 

If $h=2$ then Theorem~\ref{T1} becomes (see (\ref {E2}))
\[N(x)=\sqrt{x}+(1+o(1))\sqrt[3]{x}.\]
Substituting $x=P_n$ into this equation and using Lemma~\ref{L3} 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
\begin{eqnarray} \label{E4}
P_n&=&\left(n+(-1+o(1))n^{2/3}\right)^2\nonumber\\&=& n^2+2(-1+o(1))n^{5/3}+(-1+o(1))^2 n^{4/3}\nonumber\\&=&
n^2+(-2+o(1))n^{5/3}.	
\end{eqnarray}
If $h=3$ then Theorem~\ref{T1} becomes (see (\ref {E2}))
\[N(x)=\sqrt{x}+\sqrt[3]{x}+(1+o(1))\sqrt[5]{x}.\]
Substituting $x=P_n$ into this equation and using equation (\ref {E4}), Lemma~\ref{L3} and Lemma~\ref{L2} we obtain
\begin{eqnarray} 
N(P_n)&=&n=P_n^{1/2}+P_n^{1/3}+(1+o(1))n^{2/5}= P_n^{1/2}+\left(n^2+(-2+o(1))n^{5/3}\right)^{1/3}\nonumber\\&+&(1+o(1))n^{2/5}=
P_n^{1/2}+n^{2/3}\left(1+(-2+o(1))n^{-1/3}\right)^{1/3}+(1+o(1))n^{2/5}\nonumber\\&=&
P_n^{1/2}+n^{2/3}\left(1+((-2/3)+o(1))n^{-1/3}\right)\nonumber\\&+&(1+o(1))n^{2/5}=P_n^{1/2}+n^{2/3}+ (1+o(1))n^{2/5}.\nonumber
\end{eqnarray}
That is
\[P_n^{1/2}=n-n^{2/3}+(-1+o(1))n^{2/5}.	\]
Therefore 
\[P_n=\left(n-n^{2/3}+(-1+o(1))n^{2/5}\right)^2=n^2-2n^{5/3}+(-2+o(1))n^{7/5}. \]
That is
\begin{equation} \label{E5} 
P_n=n^2-2n^{5/3}+(-2+o(1))n^{7/5}.
\end{equation}
Equation (\ref {E5}) is Lemma~\ref{L5} for $h=3$. Consequently the lemma is true for $h=3$.

Suppose that the lemma is true for $h-1\geq 3$. We shall prove that the lemma is also true for $h\geq 4$.
 
We have (see (\ref {E1}))
\begin{eqnarray} \label{E6}
N(x)&=&\sum^{h-1}_{k=1}(-1)^{k+1}\sum_{\stackrel{1\leq i_1<\cdots<i_k\leq h-1}{p_{i_1}\cdots p_{i_k}<p_h}  } x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}+ (1+o(1)) x^{1/p_h}=x^{1/2}\nonumber\\&+&\sum^{h-1}_{i=2}x^{1/p_i}+\sum^{h-1}_{k=2}(-1)^{k+1}\sum_{\stackrel{1\leq i_1<\cdots<i_k\leq h-1}{p_{i_1}\cdots p_{i_k}<p_h}  } x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\nonumber\\&+& (1+o(1)) x^{1/p_h}\qquad (h\geq 4).
\end{eqnarray}
Substituting $x=P_n$ into (\ref {E6}) and using Lemma~\ref{L3} we obtain 
\begin{eqnarray} \label{E7}
n&=&P_n^{1/2}+\sum^{h-1}_{i=2}P_n^{1/p_i}+\sum^{h-1}_{k=2}(-1)^{k+1}\sum_{\stackrel{1\leq i_1<\cdots<i_k\leq h-1}{p_{i_1}\cdots p_{i_k}<p_h}  } P_n^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\nonumber\\&+& (1+o(1)) n^{2/p_h}\qquad (h\geq 4).
\end{eqnarray}
 

By inductive hypothesis we have
\begin{equation} \label{E8}
P_n=n^2-2n^{5/3}+\sum^{s}_{i=1}	a_in^{r_i}+(-2+o(1)) n^{1+\frac{2}{p_{h-1}}}\qquad (h\geq 4),
\end{equation}
where $2>5/3>r_1>\cdots>r_s>1+\frac{2}{p_{h-1}}$, the $a_i$ are rational coefficients and in equation (\ref {E8}) appear the terms $-2 n^{1+\frac{2}{p_i}}$ $(i=2,\ldots, h-2)$. Besides the rational exponents $5/3$ and $r_i$ $(i=1, \ldots, s)$ 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 (\ref {E8}) gives
\begin{equation}\label{E9}
P_n=n^2\left(1-2n^{-1/3}+\sum^{s}_{i=1}	a_i n^{-2+{r_i}}+(-2+o(1)) n^{-1+\frac{2}{p_{h-1}}}\right),
\end{equation}
where
\[-2n^{-1/3}+\sum^{s}_{i=1}	a_i n^{-2+{r_i}}+(-2+o(1)) n^{-1+\frac{2}{p_{h-1}}}\sim -2n^{-1/3}.\]
Consequently
\begin{equation}\label{E10}
-2n^{-1/3}+\sum^{s}_{i=1}	a_i n^{-2+{r_i}}+(-2+o(1)) n^{-1+\frac{2}{p_{h-1}}}=O\left(n^{-1/3}\right)=o(1).
\end{equation}

Let $t\geq 3$ be a positive integer. Equations (\ref {E9}), (\ref {E10}) and Lemma~\ref{L2} give
\begin{eqnarray}\label{E11}
P_n^{1/t}	&=&n^{2/t}\left(1-2n^{-1/3}+\sum^{s}_{i=1}	a_i n^{-2+{r_i}}+(-2+o(1)) n^{-1+\frac{2}{p_{h-1}}}\right)^{1/t}\nonumber\\&=&n^{2/t}\left(1+\frac{1}{t}\left(-2n^{-1/3}+\sum^{s}_{i=1}	a_i n^{-2+{r_i}}+(-2+o(1)) n^{-1+\frac{2}{p_{h-1}}}\right)\right)\nonumber\\&+&\left.O\left(n^{-2/3}\right)\right)=n^{2/t}-\frac{2}{t}n^{-\frac{1}{3}+\frac{2}{t}}+\sum^{s}_{i=1}\frac{1}{t}
a_i n^{-2+r_i+\frac{2}{t}}\nonumber\\&+&\frac{1}{t}(-2+o(1))n^{-1+\frac{2}{p_{h-1}}+\frac{2}{t}}+O\left(n^{-\frac{2}{3}+\frac{2}{t}}\right).
\end{eqnarray}
Note that if $t\geq 3$ then (see Lemma~\ref{L4})
\begin{equation}\label{E12}
\frac{1}{t}(-2+o(1)) n^{-1+\frac{2}{p_{h-1}}+\frac{2}{t}} =o\left(n^{2/p_h}\right),
\end{equation}
and if $t\geq 3$ then 
\begin{equation}\label{E13}
O\left(n^{-\frac{2}{3}+\frac{2}{t}}\right)=o\left(n^{2/p_h}\right).
\end{equation}
Consequently (\ref {E11}) becomes (see (\ref {E12}) and (\ref {E13}))
\begin{equation}\label{E14}
P_n^{1/t}=n^{2/t}-\frac{2}{t}n^{-\frac{1}{3}+\frac{2}{t}}+\sum^{s}_{i=1}\frac{1}{t}
a_i n^{-2+r_i+\frac{2}{t}}+o\left(n^{2/p_h}\right).
\end{equation}
Note that (see (\ref {E14})) if $t\geq 3$ 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 (\ref {E8}))

Substituting  (\ref {E14}) into (\ref {E7})  we find that
\begin{eqnarray}\label{E15}
n&=&P_n^{1/2}+\sum^{h-1}_{j=2}\left(n^{2/p_j}-\frac{2}{p_j}n^{-\frac{1}{3}+\frac{2}{p_j}}+\sum^{s}_{i=1}\frac{1}{p_j}
a_i n^{-2+r_i+\frac{2}{p_j}}\right)+\sum^{h-1}_{k=2}(-1)^{k+1}\nonumber\\&&\sum_{\stackrel{1\leq i_1<\cdots<i_k\leq h-1}{p_{i_1}\cdots p_{i_k}<p_h}  }\left(n^{\frac{2}{p_{i_1}\cdots p_{i_k}}}-\frac{2}{p_{i_1}\cdots p_{i_k}}n^{-\frac{1}{3}+\frac{2}{p_{i_1}\cdots p_{i_k}}}\right)\nonumber\\&+&\left.\sum^{s}_{i=1}\frac{1}{p_{i_1}\cdots p_{i_k}}
a_i n^{-2+r_i+\frac{2}{p_{i_1}\cdots p_{i_k}}} \right) + (1+o(1)) n^{2/p_h}
\nonumber\\&=&P_n^{1/2}+\sum^{l}_{i=1}b_in^{s_i}+(1+o(1))n^{2/p_h}\qquad (h\geq 4),
\end{eqnarray} 
where  $ 1> s_1> \cdots > s_l>\frac{2}{p_h}$. That is
\begin{equation}\label{E16}
P_n^{1/2}	=n-\sum^{l}_{i=1}b_i n^{s_i}+(-1+o(1))n^{2/p_h}.
\end{equation} 
 

Note that all positive exponents in equation (\ref {E15}), that is, the positive exponents of the form
\[\frac{2}{p_j},\quad -\frac{1}{3}+\frac{2}{p_j}, \quad -2+r_i+\frac{2}{p_j}\]
\[\frac{2}{p_{i_1}\cdots p_{i_k}},\quad -\frac{1}{3}+\frac{2}{p_{i_1}\cdots p_{i_k}},\quad -2+r_i+\frac{2}{p_{i_1}\cdots p_{i_k}}\]
are (see (\ref {E8})) 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 $s_i$ $(i=1, \ldots, l)$ in (\ref {E16}) 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 (\ref {E16}) gives
\begin{eqnarray}\label{E17}
&&P_n=\left(	n-\sum^{l}_{i=1}b_in^{s_i}+(-1+o(1))n^{2/p_h}\right)^2=\left(	n-\sum^{l}_{i=1}b_in^{s_i}\right)^2 \nonumber\\&+&(-2+o(1))n^{1+\frac{2}{p_h}}=n^2-2n^{5/3}+\sum^{s}_{i=1}a_in^{r_i}-2 n^{1+\frac{2}{p_{h-1}}}+\sum^{q}_{i=1}c_in^{k_i}\nonumber\\&+&(-2+o(1))n^{1+\frac{2}{p_h}}\quad (h\geq 4),
\end{eqnarray}
where  $2>5/3>r_1>\cdots>r_s>1+\frac{2}{p_{h-1}}>k_1>\cdots>k_q>1+\frac{2}{p_h}$.
 
Note also that the first terms in equation  (\ref {E17}) are the terms of equation (\ref {E8}). On the other hand in equation (\ref {E17}) appear the term $ -2 n^{1+\frac{2}{p_{h-1}}}$ (see equation (\ref {E8})). We now prove these facts.

Equation (\ref {E17}) can be written in the form
\begin{equation}\label{E18}
P_n=Q(n)+\sum^{q}_{i=1}c_in^{k_i}+(-2+o(1))n^{1+\frac{2}{p_h}}=Q(n)+o\left(n^{1+\frac{2}{p_{h-1}}}\right),
\end{equation}
where $Q(n)$ is a sum of terms of the form $e_i n^{q_i}$ $\left(q_i\geq 1+\frac{2}{p_{h-1}}\right)$.

On the other hand, equation (\ref {E8}) can be written in the form
\begin{equation}\label{E19}
P_n=n^2-2n^{5/3}+\sum^{s}_{i=1}a_in^{r_i}-2 n^{1+\frac{2}{p_{h-1}}}	+o\left(n^{1+\frac{2}{p_{h-1}}}\right).	
\end{equation}
Equations (\ref {E18}) and (\ref {E19}) give
\[0=P_n-P_n=\left(Q(n)-\left(n^2-2n^{5/3}+\sum^{s}_{i=1}a_in^{r_i}-2 n^{1+\frac{2}{p_{h-1}}}\right)\right)+o\left(n^{1+\frac{2}{p_{h-1}}}\right).	\]
If 
\[Q(n)\neq n^2-2n^{5/3}+\sum^{s}_{i=1}a_in^{r_i}-2 n^{1+\frac{2}{p_{h-1}}}\]
then we obtain
\[0=(P_n-P_n)\sim a n^q \qquad (a\neq 0)\qquad \left(q\geq 1+\frac{2}{p_{h-1}}\right).\]
That is, an evident contradiction. Consequently
\begin{equation}\label{E20}
Q(n)= n^2-2n^{5/3}+\sum^{s}_{i=1}a_in^{r_i}-2 n^{1+\frac{2}{p_{h-1}}}.	
\end{equation}
Finally, equations (\ref {E18}) and (\ref {E20}) give (\ref {E17}). 
\end{proof}  

Lemma~\ref{L5} 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 (\ref {E6}) is (see (\ref {E2}))
\begin{equation}\label{E21}
N(x)=x^{1/2}+x^{1/3}+x^{1/5}-x^{1/6}+(1+o(1))x^{1/7}.
\end{equation}
On the other hand (Lemma~\ref{L5}) equation (\ref {E8}) is (see (\ref {E5}))
\[P_n=n^2-2n^{5/3}+(-2+o(1)) n^{7/5}.\]
Consequently equation (\ref {E15}) is
\begin{eqnarray}
n&=&P_n^{1/2}+\left(n^{2/3}-\frac{2}{3}n^{-\frac{1}{3}+\frac{2}{3}}\right)+	
\left(n^{2/5}-\frac{2}{5}n^{-\frac{1}{3}+\frac{2}{5}}\right)-\left(n^{2/6}-\frac{2}{6}n^{-\frac{1}{3}+\frac{2}{6}}\right)\nonumber\\&+&(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}\nonumber\\&=&P_n^{1/2}+n^{2/3}+n^{2/5}-\frac{5}{3}n^{1/3}
+(1+o(1)) n^{2/7}.\nonumber
\end{eqnarray}
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 (\ref {E17})) 
\begin{eqnarray}
P_n &=& \left(n-n^{2/3}-n^{2/5}+\frac{5}{3}n^{1/3}\right)^2+(-2+o(1))n^{9/7}\nonumber\\&=&n^2-2n^{5/3}-2 n^{7/5} +\frac{13}{3}n^{8/6}+(-2+o(1))n^{9/7}.\nonumber
\end{eqnarray}
That is
\begin{equation}\label{E22}
P_n=n^2-2n^{5/3}-2 n^{7/5} +\frac{13}{3}n^{8/6}+(-2+o(1))n^{9/7}.
\end{equation}
  
\section{Main Result}

The following theorem is the main result of this article. In this theorem we obtain explicit formulae for $P_n$.

\begin{theorem}\label{T6} 
Let $p_h$ be the $h$-th prime with $h\geq 3$, where $h$ is an arbitrary but fixed positive integer. 
 
Let us consider the formula (see (\ref {E1}))
\begin{equation} \label{E23}
N(x)=\sum^{h-1}_{k=1}(-1)^{k+1}\sum_{\stackrel{1\leq i_1<\cdots<i_k\leq h-1}{p_{i_1}\cdots p_{i_k}<p_h}  } x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}+ (1+o(1)) x^{1/p_h}.
\end{equation}
We have 
\begin{eqnarray} \label{E24}
P_n&=&n^2+\frac{13}{3}n^{8/6}+\frac{32}{15}n^{32/30}+\sum^{h-1}_{k=1}(-1)^{k}\sum_{\stackrel{1\leq i_1<\cdots<i_k\leq h-1}{p_{i_1}\cdots p_{i_k}<p_h,\  p_{i_1}\cdots p_{i_k}\neq \ 2, \ 6, \ 30}  } 2 n^{1+\frac{2}{p_{i_1}\cdots p_{i_k}}}\nonumber\\&+& (-2+o(1)) n^{1+\frac{2}{p_h}}.
\end{eqnarray} 
\end{theorem}
\begin{proof} 
We shall see that everything relies on Theorem~\ref{T1}. The theorem is true for $h=3$ (see Lemma~\ref{L5}) and for $h=4$ (see (\ref {E21}) and (\ref {E22})). Suppose $h\geq 5$, that is $p_h\geq 11$. Equation (\ref {E23}) can be written in the form (see (\ref {E21}))
\begin{equation}\label{E25}
N(x)=x^{1/2}+	x^{1/3}+x^{1/5}-x^{1/6}+\sum^{s}_{i=1}(-1)^{1+a_i}x^{1/n_i}
+(1+o(1))x^{1/p_h},
\end{equation}
where $a_i$ is the number of different prime factors in $n_i$ and the exponents are in decreasing order,
\begin{equation}\label{E26}
\frac{1}{2}> \frac{1}{3}>\frac{1}{5}> \frac{1}{6}	>\frac{1}{n_1}>\cdots >\frac{1}{n_s}>\frac{1}{p_h}.
\end{equation}
For example, if $h=5$ then equation (\ref {E25}) becomes equation (\ref {E2}).

On the other hand, we have (Lemma~\ref{L5} and equation (\ref {E22}))
\begin{equation}\label{E27}
P_n=n^{2}	-2n^{5/3}-2n^{7/5}+\frac{13}{3}n^{8/6}+\sum^{t}_{i=1}d_i n^{r_i}+(-2+o(1))n^{1+\frac{2}{p_h}},
\end{equation}
where the exponents are in decreasing order,
\begin{equation}\label{E28}
2> \frac{5}{3}> \frac{7}{5}>\frac{8}{6}	>r_1>\cdots > r_t>1+\frac{2}{p_h}.
\end{equation} 
Equation (\ref {E27}) gives
\begin{eqnarray}\label{E29}
P_n=n^{2}\left(	1-2n^{-1/3}-2n^{-3/5}+\frac{13}{3}n^{-4/6}+\sum^{t}_{i=1}d_i n^{r_i-2}+(-2+o(1))n^{-1+\frac{2}{p_h}}\right)	
\end{eqnarray}
where
\[-2n^{-1/3}-2n^{-3/5}+\frac{13}{3}n^{-4/6}+\sum^{t}_{i=1}d_i n^{r_i-2}+(-2+o(1))n^{-1+\frac{2}{p_h}}\sim -2 n^{-1/3},\]
since (see (\ref {E28}))
\begin{equation}\label{E30}
-\frac{1}{3}>-\frac{3}{5}	>-\frac{4}{6}>r_1-2>\cdots>r_t-2>-1+\frac{2}{p_h}.
\end{equation}  
Consequently
\begin{eqnarray}\label{E31}
A_n&=&-2n^{-1/3}-2n^{-3/5}+\frac{13}{3}n^{-4/6}+\sum^{t}_{i=1}d_i n^{r_i-2}+(-2+o(1))n^{-1+\frac{2}{p_h}}\nonumber\\&=&O\left( n^{-1/3}\right)=o(1).
\end{eqnarray}
Besides
\begin{eqnarray}\label{E32}
B_n&=&\left(-2n^{-1/3}-2n^{-3/5}+\frac{13}{3}n^{-4/6}+\sum^{t}_{i=1}d_i n^{r_i-2}+(-2+o(1))n^{-1+\frac{2}{p_h}}\right)^2\nonumber\\&=&	
\left(-2n^{-1/3}-2n^{-3/5}+\left(\frac{13}{3}+o(1)\right)n^{-4/6}\right)^2\nonumber\\&=&4n^{-2/3}+8n^{-14/15}+O\left(n^{-1}\right).
\end{eqnarray}
Substituting $x=P_n$ into equation (\ref {E25}) and using Lemma~\ref{L3} we obtain
\begin{equation}\label{E33}
n=P_n^{1/2}+	P_n^{1/3}+P_n^{1/5}-P_n^{1/6}+\sum^{s}_{i=1}(-1)^{1+a_i}P_n^{1/n_i}
+n^{2/p_h}+o\left(n^{2/p_h}\right).
\end{equation}
Equations (\ref {E29}), (\ref {E31}), (\ref {E32}) and Lemma~\ref{L2} give
\begin{eqnarray}\label{E34}
P^{1/2}_{n}&=& n\left(1+\frac{1}{2}A_n+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2}B_n +O\left(n^{-1}\right)\right)
\nonumber\\&=&n-n^{2/3}-n^{2/5}+\frac{13}{6}n^{2/6}	+\sum^{t}_{i=1}\frac{d_i}{2}n^{r_i-1}\nonumber\\&-&n^{2/p_h}+o\left(n^{2/p_h}\right)
-\frac{1}{2}n^{2/6}-n^{2/30}+O(1).
\end{eqnarray}
Equations (\ref {E29}), (\ref {E31}), (\ref {E30}) and Lemma~\ref{L2} give
\begin{equation}\label{E35}
P_n^{1/3}	=n^{2/3}\left(1+\frac{1}{3}A_n +O\left(n^{-2/3}\right)\right)
= n^{2/3}-\frac{2}{3}n^{2/6}-\frac{2}{3}n^{2/30}+O(1).
\end{equation}  
 
\begin{equation}\label{E36}
P_n^{1/5}	=n^{2/5}\left(1+\frac{1}{5}A_n +O\left(n^{-2/3}\right)\right)
= n^{2/5}-\frac{2}{5}n^{2/30}+o(1).
\end{equation}

\begin{equation}\label{E37}
P_n^{1/6}	=n^{2/6}\left(1+\frac{1}{6}A_n +O\left(n^{-2/3}\right)\right)
= n^{2/6}+O(1).
\end{equation}

\begin{equation}\label{E38}
P_n^{1/n_i}	=n^{2/n_i}\left(1+\frac{1}{n_i}A_n +O\left(n^{-2/3}\right)\right)
= n^{2/n_i}+o(1)\qquad (i=1, \ldots, s).
\end{equation} 

Substituting equations (\ref {E34}), (\ref {E35}), (\ref {E36}), (\ref {E37}) and (\ref {E38}) into equation (\ref {E33}) we find that
\begin{equation}\label{E39}
0= \sum^{t}_{i=1}	\frac{d_i}{2}n^{r_i-1}+\sum^{s}_{i=1}(-1)^{1+a_i}n^{2/n_i}-\frac{31}{15}n^{2/30}+o\left(n^{2
/p_h}\right).
\end{equation}
Note that (see (\ref {E28}) and (\ref {E26})) $r_i-1>\frac{2}{p_h}$ and $\frac{2}{n_i}>\frac{2}{p_h}$.

If $p_h\leq 29$ then $-\frac{31}{15}n^{2/30}=o\left(n^{2/p_h}\right)$. Consequently we have
\[\sum^{t}_{i=1}	\frac{d_i}{2}n^{r_i-1}=\sum^{s}_{i=1}(-1)^{a_i}n^{2/n_i},\]
where $t=s$, $d_i=(-1)^{a_i}2$ $(i=1, \ldots, s)$ and $r_i=1+\frac{2}{n_i}$ $(i=1, \ldots, s)$. Since in contrary case we have $0\sim a n^b$ where $a\neq 0$ and $b>\frac{2}{p_h}$, an evident contradiction. Substituting these values into (\ref {E27}) we obtain (\ref {E24}) (see (\ref {E25})).
 
If $p_h\geq 31$ then $\frac{2}{30}> \frac{2}{p_h}$ and there exists $k$ such that $n_k=30=2.3.5$ (see (\ref {E23})). Consequently we have
\[\sum^{t}_{i=1}	\frac{d_i}{2}n^{r_i-1}=\sum^{s}_{i=1}(-1)^{a_i}n^{2/n_i}+\frac{31}{15}n^{2/30},\]
where $t=s$, $d_i=(-1)^{a_i}2$ $(i\neq k)$, $d_k=2(-1+\frac{31}{15})=\frac{32}{15}$ and 
$r_i=1+\frac{2}{n_i}$ $(i=1, \ldots, s)$. Since in contrary case we have $0\sim a n^b$ where $a\neq 0$ and $b>\frac{2}{p_h}$, an evident contradiction. Substituting these values into (\ref {E27}) we obtain (\ref {E24}) (see (\ref {E25})). 
\end{proof} 
 
\begin{example}
If $h=5$ equation (\ref {E23})  is (see (\ref {E2}))
\[N(x)=x^{1/2}+x^{1/3}+x^{1/5}-x^{1/6}+x^{1/7}-x^{1/10}+(1+o(1))x^{1/11}.\]
Consequently Theorem~\ref{T6} gives
\[P_n=n^2-2n^{5/3}-2 n^{7/5} +\frac{13}{3}n^{4/3}-2n^{9/7}+2n^{6/5}+(-2+o(1)) n^{13/11}\]
\end{example}
  
\section{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. 

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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11A99; Secondary 11B99. 

\noindent \emph{Keywords: } 
$n$-th perfect power, asymptotic formula.

\bigskip
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\noindent (Concerned with sequence
\seqnum{A001597}.)

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\vspace*{+.1in}
\noindent
Received October 1 2011;
revised versions received  January 14 2012; March 20 2012; May 29 2012.
Published in {\it Journal of Integer Sequences}, May 29 2012.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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