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\begin{center}
\vskip 1cm{\LARGE\bf
On the Distribution of Perfect Powers}
\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}\\
\end{center}

\centerline{} 

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

\centerline{}

\vskip .2in

\begin{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  $N(x)$.
\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 \textit{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 are sequence \seqnum{A001597} in Sloane's {\it Encylopedia}.
Let $N(x)$ be the number of perfect
powers that do not exceed $x$. M. A. Nyblom \cite{Nyblom3} proved the
following asymptotic formula, \[N(x)\sim \sqrt{x}.\] M. A. Nyblom
\cite{Nyblom4} 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
\[N(x)=\sqrt{x}+\sqrt[3]{x}+\sqrt[5]{x}-\sqrt[6]{x}+\sqrt[7]{x}+o\left( \sqrt[7]{x}\right).\]
Consequently
\[\sqrt{x}+\sqrt[3]{x}&lt;N(x)&lt;\sqrt{x}+\sqrt[3]{x}+\sqrt[5]{x}.\]




\section{Preliminary Results}

Let $A$ be a set. The number of elements in $A$ we  denote in the form $\left|A\right|$.
 
We  need the following results.
 
\begin{lemma} \label{L1}
(Inclusion-exclusion principle) Let us consider a given finite collection of sets $A_1, A_2, \ldots, A_n$. The number of elements in $\cup^{n}_{i=1}A_i$ is
\[\left|\cup^{n}_{i=1}A_i\right|=\sum^{n}_{k=1}(-1)^{k+1}\sum_{1\leq i_1&lt;\cdots&lt;i_k\leq n}
\left|A_{i_1}\cap \cdots \cap A_{i_k}\right|,\]
where the expression $1\leq i_1&lt;\cdots&lt;i_k\leq n$ indicates that the sum is taken over all the k-element subsets $\left\{i_1,\ldots, i_k\right\}$ of the set $\left\{1, 2, \ldots, n\right\}$.
\end{lemma}
\begin{proof} 
See for example either  \cite[page 233]{Hardy1} or \cite[page 84]{LeVeque2}.
\end{proof}
Let $A_n(x)$ $(n\geq 2)$ be the set $\left\{k^n : k \in N, k^n \leq x\right\}$, that is, the set of perfect powers whose exponent is $n$ that do not exceed $x$. 
 
\begin{lemma} \label{L2}
We have
\[\left|A_n(x)\right|=\left\lfloor \sqrt[n]{x}\right\rfloor,\]
where $\left\lfloor .\right\rfloor$ denotes the integer-part function.
\end{lemma}
\begin{proof}
We have
\[k^n\leq x \Leftrightarrow k\leq \sqrt[n]{x}.\]
\end{proof}
M. A. Nyblom \cite{Nyblom4} proved the following Lemma and the following Theorem.
 
\begin{lemma} \label{L3}
For any set consisting of $m\geq 2$ positive integers $\left\{n_1, \ldots,n_m \right\}$ all greater than unity, we have the set equality
\[\bigcap^{m}_{i=1}A_{n_i}(x)=A_{\left[n_1, \ldots,n_m\right]}(x),\]
where $\left[n_1, \ldots,n_m\right]$ denotes the least common multiple of the $m$ integers 
$n_1, \ldots,n_m$.
\end{lemma}

Let $p_n$ be the $n$-th prime. Consequently we have,
\[p_1=2, p_2=3, p_3=5, p_4=7, p_5=11, p_6=13,\ldots\]
 
\begin{theorem} \label{T4}
If $x\geq 4$ and $p_1, p_2,\ldots, p_m$ denote the prime numbers that do not exceed $\left\lfloor \log_2 x \right\rfloor$, then for $k\geq m$ the number of perfect powers that do not exceed $x$ is 
\[N(x)=\left|\bigcup^{k}_{i=1}A_{p_i}(x)\right|.\]
Besides $A_{p_i}(x)=\{1\}$ for $i\geq m+1$.
\end{theorem}

We also need the following two well-known results.

The binomial formula
\begin{equation}\label{E1}
(a+b)^n = \sum^{n}_{k=0}{{n}\choose{k}}	a^{n-k}b^k,
\end{equation}
and the following property of the absolute value
\begin{equation}\label{E2}
\left|a_1+a_2+\cdots+a_r\right|	\leq \left|a_1\right|+\left|a_2\right|+\cdots+\left|a_r\right|,
\end{equation}
where $a_1, a_2, \ldots, a_r$ are real numbers.



\section{Main Results}

\begin{theorem} \label{T5}
Let $p_n$  be the $n$-th prime number with $n\geq 2$, where $n$ is an arbitrary but fixed positive integer. Then
\begin{equation}\label{E3} 
N(x)=\sum^{n-1}_{k=1}(-1)^{k+1}\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n-1}{p_{i_1}\cdots p_{i_k}&lt;p_n}} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}+ g(x) x^{\frac{1}{p_n}},
\end{equation} 
where $\lim_{x\rightarrow \infty}g(x)=1$ and the inner sum is taken over   the k-element subsets $\left\{i_1,\ldots, i_k\right\}$ of the set $\left\{1, 2, \ldots, n-1\right\}$ such that the inequality $p_{i_1}\cdots p_{i_k}&lt;p_n$ holds.
\end{theorem} 
\begin{proof} 
Let $k=\left\lfloor \log_2 x\right\rfloor+1=\left\lfloor \frac{\log x}{\log 2}\right\rfloor+1$, where $x\geq 4$. If $p_1, p_2,\ldots, p_m$ denote the prime numbers that do not exceed $\left\lfloor \log_2 x \right\rfloor$, then we have
\begin{equation}\label{E4}
p_1&lt;\cdots&lt;p_m\leq \left\lfloor \log_2 x \right\rfloor&lt;\left\lfloor \log_2 x \right\rfloor+1=k\leq p_{m+1}&lt;p_{m+2}&lt;\cdots
\end{equation}
Note that if $i\geq m+1$ we have $1&lt;x^{\frac{1}{p_i}}\leq x^{\frac{1}{k}}&lt;x^{\frac{1}{\frac{\log x}{\log 2}}}=2$. That is, $1&lt;x^{\frac{1}{p_i}}&lt;2$. Consequently if $i\geq m+1$ (see Lemma~\ref{L2}) $\left|A_{p_i}(x)\right|=1$. That is, $A_{p_i}(x)=\{1\}$.
Note also that $k$ and $m$ are increasing functions of $x$. On the other hand $n\geq 2$ is an arbitrary but fixed positive integer.

Equation (\ref {E4})  gives
\begin{equation}\label{E5}
p_m &lt; k.	
\end{equation}
On the other hand
\begin{equation}\label{E6}
m &lt; p_m.
\end{equation}
Therefore (\ref {E5}) and (\ref {E6}) give
\begin{equation}\label{E7}
m &lt; k,
\end{equation}
and consequently
\begin{equation}\label{E8}
p_m &lt; p_k.
\end{equation}
There exist three possible relations between $m$, $k$, $n$ and $n+1$ and consequently between $p_m$, $p_k$, $p_n$ and $p_{n+1}$.

First relation. 
\[m&lt;k\leq n &lt; n+1\]
and hence
\[p_m&lt;p_k\leq p_n &lt; p_{n+1}.\]
Second relation.
\[m\leq n&lt;n+1\leq k\]
and hence
\[p_m\leq p_n&lt;p_{n+1}\leq p_k.\]
Third relation.
\[n&lt;n+1\leq m&lt;k\]
and hence 
\[p_n&lt;p_{n+1}\leq p_m&lt;p_k.\]
If we define $S(x)=\max \{n+1, k\}$ then these three relations, Theorem~\ref{T4} and Lemma~\ref{L2} give us $(x\geq 4)$
\begin{eqnarray}\label{E9}
\left|\bigcup^{n}_{i=1}A_{p_i}(x)\right|\leq N(x)&amp;&lt;&amp;\left|\bigcup^{n}_{i=1}A_{p_i}(x)\right|+\sum^{S(x)}_{i=n+1}\left|A_{p_i}(x)\right|\nonumber\\&amp;=&amp;\left|\bigcup^{n}_{i=1}A_{p_i}(x)\right|+\sum^{S(x)}_{i=n+1}\left\lfloor x^{\frac{1}{p_i}}\right\rfloor \nonumber\\&amp;\leq&amp; \left|\bigcup^{n}_{i=1}A_{p_i}(x)\right| +(S(x)-n)\left\lfloor x^{\frac{1}{p_{n+1}}}\right\rfloor.
\end{eqnarray}
Note that there exists $x_0$ such that if $x\geq x_0$ the third relation holds.  

Note also that $S(x)-n$ is either equal to 1 or $k-n&lt; k-1=\left\lfloor \frac{\log x}{\log 2}\right\rfloor\leq\frac{\log x}{\log 2} $, and so in either case 
\begin{equation}\label{E10}
S(x)-n\leq\frac{\log x}{\log 2} 
\end{equation}
as $x\geq 4$. Consequently (\ref {E9}) and (\ref {E10}) give 
\begin{equation}\label{E11}
\left|\bigcup^{n}_{i=1}A_{p_i}(x)\right|\leq N(x)&lt;\left|\bigcup^{n}_{i=1}A_{p_i}(x)\right|+ \frac{\log x}{\log 2 } x^{\frac{1}{p_{n+1}}}.
\end{equation}
Lemma~\ref{L1} gives 
\begin{equation}\label{E12}
\left|\bigcup^{n}_{i=1}A_{p_i}(x)\right|=\sum^{n}_{k=1}(-1)^{k+1}\sum_{1\leq i_1&lt;\cdots&lt;i_k\leq n}\left|A_{p_{i_1}}(x)\cap\cdots\cap A_{p_{i_k}}(x)\right|.
\end{equation}
On the other hand, Lemma~\ref{L3} gives 
\[A_{p_{i_1}}(x)\cap\cdots\cap A_{p_{i_k}}(x)=A_{\left[p_{i_1},\ldots ,p_{i_k}\right]}(x)=
A_{p_{i_1}\cdots p_{i_k}}(x).\]
Therefore (Lemma~\ref{L2}) we obtain
\begin{equation}\label{E13}
\left|A_{p_{i_1}}(x)\cap\cdots\cap A_{p_{i_k}}(x)\right|=\left|A_{p_{i_1}\cdots p_{i_k}}(x)\right|=\left\lfloor x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\right\rfloor.
\end{equation}
Substituting (\ref {E13}) into (\ref {E12}) we find that
\begin{eqnarray}\label{E14}
\left|\bigcup^{n}_{i=1}A_{p_i}(x)\right|&amp;=&amp;\sum^{n}_{k=1}(-1)^{k+1}\sum_{1\leq i_1&lt;\cdots&lt;i_k\leq n}\left\lfloor x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\right\rfloor \nonumber\\&amp;=&amp;\sum^{n}_{k=1}(-1)^{k+1}\sum_{1\leq i_1&lt;\cdots&lt;i_k\leq n} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\nonumber\\&amp;-&amp; \sum^{n}_{k=1}(-1)^{k+1}\sum_{1\leq i_1&lt;\cdots&lt;i_k\leq n} \left(x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}-\left\lfloor x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\right\rfloor\right).
\end{eqnarray}
Now
\begin{equation}\label{E15}
0\leq x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}-\left\lfloor x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\right\rfloor	&lt;1.
\end{equation}
Consequently (see (\ref {E1}) and (\ref {E2}))
\begin{eqnarray}\label{E16}
&amp;&amp;\left|\sum^{n}_{k=1}(-1)^{k+1}\sum_{1\leq i_1&lt;\cdots&lt;i_k\leq n} \left(x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}-\left\lfloor x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\right\rfloor\right)\right|\leq \sum^{n}_{k=1}\sum_{1\leq i_1&lt;\cdots&lt;i_k\leq n}1\nonumber\\&amp;=&amp;\sum^{n}_{k=1}{{n}\choose{k}}\leq	\sum^{n}_{k=0}{{n}\choose{k}}=(1+1)^n =2^n.
\end{eqnarray}
That is, 
\begin{equation}\label{E17}
\sum^{n}_{k=1}(-1)^{k+1}\sum_{1\leq i_1&lt;\cdots&lt;i_k\leq n} \left(x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}-\left\lfloor x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\right\rfloor\right)=O(1).	
\end{equation}
Equations (\ref {E14}) and (\ref {E17}) give
\begin{equation}\label{E18}
\left|\bigcup^{n}_{i=1}A_{p_i}(x)\right|=	\sum^{n}_{k=1}(-1)^{k+1}\sum_{1\leq i_1&lt;\cdots&lt;i_k\leq n} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}+O(1)
\end{equation}
If
\begin{equation}\label{E19}
B(x)=\sum^{n}_{k=1}(-1)^{k+1}\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n}{p_{i_1}\cdots p_{i_k}&lt;p_{n+1}}} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}
\end{equation}
and 
\begin{equation}\label{E20}
C(x)=\sum^{n}_{k=1}(-1)^{k+1}\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n}{p_{i_1}\cdots p_{i_k}&gt;p_{n+1}}} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}=o\left(x^{\frac{1}{p_{n+1}}}\right)
\end{equation}
then 
\begin{equation}\label{E21}
\sum^{n}_{k=1}(-1)^{k+1}\sum_{1\leq i_1&lt;\cdots&lt;i_k\leq n} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}=B(x)+C(x).
\end{equation}
Equations (\ref {E18}) and (\ref {E21}) give
\begin{equation}\label{E22}
\left|\bigcup^{n}_{i=1}A_{p_i}(x)\right|=B(x)+C(x)+O(1).
\end{equation}
Equations (\ref {E22}) and (\ref {E11}) give
\[B(x)+C(x)+O(1)\leq N(x)&lt;B(x)+C(x)+O(1)+ \frac{\log x}{\log 2 } x^{\frac{1}{p_{n+1}}}.\]
Therefore, 
\begin{equation}\label{E23}
C(x)+O(1)\leq N(x)-B(x)&lt;C(x)+O(1)+ \frac{\log x}{\log 2 } x^{\frac{1}{p_{n+1}}}.
\end{equation}
Equations (\ref {E23}) and (\ref {E20}) give
\[-\epsilon &lt; \frac{N(x)-B(x)}{\log x\  x^{\frac{1}{p_{n+1}}}}&lt;\frac{1}{\log 2}+\epsilon \qquad (\epsilon&gt;0).\]
That is,
\begin{equation}\label{E24} 
N(x)=B(x)+O\left(\log x\  x^{\frac{1}{p_{n+1}}}\right).
\end{equation}
Note that (see (\ref {E19})) if $k=1,\ldots,n$ then,
\begin{equation}\label{E25}
\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n}{p_{i_1}\cdots p_{i_k}&lt;p_{n+1}}} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}=	\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n}{p_{i_1}\cdots p_{i_k}&lt;p_n}  } x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}+\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n}{p_n\leq p_{i_1}\cdots p_{i_k}&lt;p_{n+1}}} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}.
\end{equation}
Now, if $k=1,\ldots,n-1$ then
\begin{equation}\label{E26}
\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n}{p_{i_1}\cdots p_{i_k}&lt;p_n}} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}=	\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n-1}{p_{i_1}\cdots p_{i_k}&lt;p_n}} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}},
\end{equation}
and if $k=2,\ldots,n$ then
\begin{equation}\label{E27}
\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n}{p_n\leq p_{i_1}\cdots p_{i_k}&lt;p_{n+1}}} x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}=\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n}{p_n&lt; p_{i_1}\cdots p_{i_k}&lt;p_{n+1}} } x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}
\end{equation} 
On the other hand, if $k=1$ then
\begin{equation}\label{E28}
\sum_{\stackrel{1\leq i_1 \leq n}{p_n\leq p_{i_1}&lt;p_{n+1}} }x^{\frac{1}{p_{i_1}}}=x^{\frac{1}{p_n}}	
\end{equation}
Equations (\ref {E19}), (\ref {E25}), (\ref {E26}), (\ref {E27}) and (\ref {E28}) give
\begin{eqnarray}
B(x)&amp;=&amp;\sum^{n-1}_{k=1}(-1)^{k+1}\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n-1}{p_{i_1}\cdots p_{i_k}&lt;p_n}  } x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\nonumber\\&amp;+&amp;x^{\frac{1}{p_n}}+
\sum^{n}_{k=2}(-1)^{k+1}\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n}{p_n&lt; p_{i_1}\cdots p_{i_k}&lt;p_{n+1}} } x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\nonumber
\end{eqnarray}
That is, 
\begin{eqnarray}\label{E29}
B(x)&amp;=&amp;\sum^{n-1}_{k=1}(-1)^{k+1}\sum_{\stackrel{1\leq i_1&lt;\cdots&lt;i_k\leq n-1}{p_{i_1}\cdots p_{i_k}&lt;p_n}  } x^{\frac{1}{p_{i_1}\cdots p_{i_k}}}\nonumber\\&amp;+&amp;x^{\frac{1}{p_n}}+o\left(x^{\frac{1}{p_n}}\right)
\end{eqnarray}
Finally, equations (\ref {E29}) and (\ref {E24}) give (\ref {E3}). 
\end{proof}
 

\section{Examples}

If $n=2$ then Theorem~\ref{T5} becomes
\begin{equation}\label{E30}
N(x)=\sqrt{x}+g(x)\sqrt[3]{x},
\end{equation}
where $\lim_{x\rightarrow \infty}g(x)=1$.
 
If $n=3$ then Theorem~\ref{T5} becomes
\[N(x)=\sqrt{x}+\sqrt[3]{x}+g(x)\sqrt[5]{x},\]
where $\lim_{x\rightarrow \infty}g(x)=1$.

If $n=4$ then Theorem~\ref{T5} becomes
\[N(x)=\sqrt{x}+\sqrt[3]{x}+\sqrt[5]{x}-\sqrt[6]{x}+g(x)\sqrt[7]{x},\]
where $\lim_{x\rightarrow \infty}g(x)=1$. Consequently
\[\sqrt{x}+\sqrt[3]{x}&lt;N(x)&lt;\sqrt{x}+\sqrt[3]{x}+\sqrt[5]{x}\]

If $n=5$ then Theorem~\ref{T5} becomes
\[N(x)=\sqrt{x}+\sqrt[3]{x}+\sqrt[5]{x}-\sqrt[6]{x}+\sqrt[7]{x}-\sqrt[10]{x}+g(x)\sqrt[11]{x},\]
where $\lim_{x\rightarrow \infty}g(x)=1$. 

To finish, we shall establish two simple theorems. 

\begin{theorem} \label{T6}
Let us consider the $n$ open intervals $(0, 1^2), (1^2, 2^2),\ldots, ((n-1)^2, n^2)$. Let $S(n)$ be the number of these $n$ open intervals that contain some perfect power. Then
\[\lim_{n\rightarrow \infty}\frac{S(n)}{n}=0.\]
Therefore, almost all the open intervals are empty.
\end{theorem}
\begin{proof} 
We have (Nyblom's asymptotic formula)
\[N(x)=\sqrt{x}+f(x)\sqrt{x},\]
where $\lim_{x\rightarrow \infty}f(x)=0$. Consequently
\[N(n^2)=n+f(n^2)n,\]
where $n$ are the $n$ squares $1^2, 2^2,\ldots, n^2$.
Therefore
\[0\leq S(n)\leq N(n^2)-n=f(n^2)n.\]
That is
\[0\leq\frac{S(n)}{n}\leq f(n^2).\]
\end{proof}
Using equation (\ref {E30}) we can obtain a more strong result.
\begin{theorem}\label{T7}
Let us consider the $n$ open intervals $(0, 1^2), (1^2, 2^2),\ldots, ((n-1)^2, n^2)$. Let $F(n)$ be the number of perfect powers in these $n$ open intervals.
Then $F(n)\sim n^{\frac{2}{3}}$.
\end{theorem}
\begin{proof} 
Equation (\ref {E30}) gives
\[N(n^2)=n+g(n^2)n^{\frac{2}{3}},\]
where $n$ are the $n$ squares $1^2, 2^2,\ldots, n^2$. Therefore
\[F(n)=N(n^2)-n=g(n^2)n^{\frac{2}{3}}\sim n^{\frac{2}{3}}.\]
\end{proof}

 

\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.

\begin{thebibliography}{9}
 
\bibitem{Hardy1}
G. H. Hardy and E. M. Wright, \textit{An Introduction to the Theory of Numbers}, Oxford, 1960. 

\bibitem{LeVeque2}
W. J. LeVeque, \textit{Topics in Number Theory}, Addison-Wesley, 1958.

\bibitem{Nyblom3}
M. A. Nyblom, A counting function for the sequence of perfect powers, \textit{Austral. Math. Soc. Gaz.} \textbf{33} (2006), 338--343. 

\bibitem{Nyblom4}
M. A. Nyblom, Counting the perfect powers, \textit{Math. Spectrum} \textbf{41} (2008), 27--31.


\end{thebibliography}

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

\noindent \emph{Keywords: } 
Perfect powers, counting function, asymptotic formulae.

\bigskip
\hrule
\bigskip

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

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  May 28 2011;
revised version received  August 16 2011.
Published in {\it Journal of Integer Sequences}, September 25 2011.

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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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