<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amscd}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics,amsmath,amssymb}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{latexsym}
\usepackage{epsf}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}

\newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}}

\newcommand {\R} {\ensuremath{\mathbb{R}}}
\newcommand {\N} {\ensuremath{\mathbb{N}}}
\DeclareMathOperator{\sinc}{sinc}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}

\begin{center}
\vskip 1cm{\LARGE\bf 
Kepler-Bouwkamp Radius of  \\
\vskip .1in
Combinatorial Sequences
}
\vskip 1cm
\large
Tomislav Do\v{s}li\'c \\
Faculty of Civil Engineering\\
University of Zagreb \\
Ka\v{c}i\'ceva 26 \\
10000 Zagreb \\
Croatia\\
\href{mailto:doslic@grad.hr}{\tt doslic@grad.hr} \\
\end{center}

\vskip .2 in

\begin{abstract}
The Kepler-Bouwkamp constant is defined as the limit of radii of a sequence
of concentric circles that are simultaneously inscribed in a regular $n$-gon
and circumscribed around a regular $(n+1)$-gon for $n \geq 3$. The outermost
circle, circumscribed around an equilateral triangle, has radius 1. We investigate
what happens when the number of sides of regular polygons from the definition
is given by a sequence different from the sequence of natural numbers.
\end{abstract}


\section{Introduction}

Take a unit circle and inscribe in it an equilateral triangle. Then inscribe
a circle in that triangle,
and inscribe an equilateral triangle in that circle. Continue the
procedure. It is a high-school exercise to show that the sequence of radii
of inscribed circles tends to zero as $\frac {1}{2^n}$. The conclusion
remains valid if we replace triangle with any other regular polygon, only the
rate of convergence is changed: For a regular $m$-gon, the radii tend to zero
as $\left ( \cos \frac {\pi }{m} \right ) ^n$. But what happens with the
limiting radius if the number of sides of inscribed polygons is not 
constant? What if it increases so that the number of sides of $n$-th 
polygon is given as the $n$-th element of a sequence $(a_n)$ of non-negative
integers that are all greater than two? For a given regular $n$-gon, the
ratio of radii of its inscribed and its circumscribed circle is equal to
$\cos \frac{\pi }{n}$. Hence the answer to our question will be given as an
infinite product $\prod _{a_n \geq 3}^{\infty} \cos \frac {\pi}{a_n}$.

When $a_n = n$, the answer is well known: The limiting radius is equal to the
Kepler-Bouwkamp constant $\rho = \prod _{n=3}^{\infty} \cos \frac {\pi}{n}
\doteq 0.1149420448$ (\cite[p.\ 428]{finch}; see also 
\cite[\seqnum{A085365}]{sloane}). To the best of my knowledge, there is 
only one other sequence, the
sequence of odd primes, for which the answer was sought. Kitson \cite{kitson}
computed the limiting radius as $\rho _P = 0.3128329295$. The quantities in
question were also computed for a few other sequences, mostly of the form
$n(n+1)$ for even and odd $n$, in a paper by Mathar \cite{mathar}, but they 
appear there as byproducts of some other computations. The goal of this note 
is to investigate for which sequences is that limit positive and
to compute the limiting radii for several classes of combinatorial sequences.
The results might be useful in further investigation of problems arising in
computational geometry \cite{mathar}, and they might shed additional light on 
properties of integrals of the type 
$\int _0 ^{\infty} \prod _{k=0}^n \frac {\sin (a_k x)}{a_k x} dx$ 
\cite{borwein}. They will also provide benchmarks
for testing numerical methods for efficient evaluation of slowly convergent
infinite products and series \cite{chamberland}.

In the rest of this paper we consider only (weakly) increasing sequences 
$(a_n)$ of non-negative integers. If a sequence has elements smaller than 3, 
we ignore them, along with the corresponding terms in any expressions
and formulas. We call such sequences {\it admissible}. Whenever referring to
a sequence that is not itself admissible, we mean its largest admissible
subsequence.

Let $(a_n)$ be an admissible sequence. Its {\it Kepler-Bouwkamp radius} is
denoted by $\kappa (a_n)$ and defined by 
$$\kappa (a_n)  = \prod _n^{\infty} \cos \frac {\pi}{a_n}.$$
Hence, $\kappa (n) = \rho$, $\kappa (p_n) = \rho _P$ and $\kappa (c) = 0$
for any constant sequence $a_n = c$. 

\section{Growth rates and convergence}

In this section we investigate under what conditions a combinatorial sequence
has a positive Kepler-Bouwkamp radius. We start by some elementary 
observations.

\begin{proposition}
Let $(a_n)$ be an admissible sequence such that $\lim _{n \to \infty}
&lt; \infty$. Then $\kappa (a_n) = 0$.
\end{proposition}

\begin{proof}
If an admissible sequence is bounded, it must have a convergent subsequence.
The condition of integrality means that this subsequence must be constant.
Then the infinite product over that subsequence diverges toward zero and the
claim follows.
\end{proof}

Hence a sequence must grow without bounds to have a positive 
Kepler-Bouwkamp radius. Any sequence growing faster that the sequence of
natural numbers grows fast enough:

\begin{proposition}
Let an admissible sequence $(a_n)$ be a subsequence of $\mathbb N$. Then
$\kappa (a_n) &gt; \rho &gt; 0$.
\end{proposition}

What about sequences growing slower than $\mathbb N$? The following example
shows that linear growth, no matter how slow, still suffices for the
positivity of $\kappa (a_n)$.

\begin{proposition}
Let $(a_n) = \left \lfloor \frac {n}{k} \right \rfloor$ for a fixed $k \geq 2$,
$n \geq 3k$. Then $\kappa (a_n) = \rho ^k$.
\end{proposition}

\begin{proof}
It is enough to look at the case $k = 2$. Since each term in the product
is repeated twice, we have
$$\kappa \left (\left \lfloor \frac {n}{2} \right \rfloor \right ) = \prod _{n=3}^{\infty}
\left ( \cos \frac {\pi}{n} \right ) ^2 = \left ( \prod _{n=3}^{\infty} \cos \frac {\pi}{n} \right ) ^2 = \rho ^2.$$
\end{proof}

By the same reasoning we can conclude that the infinite product will 
converge for all admissible sequences in which the number of repetitions of
an element remains finite.

\begin{proposition}
Let $(a_n)$ be an admissible sequence and let there be a $k \in \N$ such that
no element of $\N$ appears in $(a_n)$ more than $k$ times. Then
$\kappa (a_n) &gt; \rho ^k &gt; 0$.
\end{proposition}

What happens when the number of repetitions of an element grows without 
bounds? We first consider a concrete example.

\begin{proposition}
$\kappa (\left \lfloor \sqrt n \right \rfloor) = 0$.
\end{proposition}

\begin{proof}
Let us look at the sequence $\left \lfloor \sqrt n \right \rfloor $, i.e., to
its admissible subsequence. It start with seven 3's, continues with nine 4's,
then eleven 5's and so on. In general, an integer $m$ appears in it exactly
$2m+1$ times. Hence, 
$$\kappa (\left \lfloor \sqrt n \right \rfloor) = \prod _{m=3}^{\infty} \left ( \cos \frac {\pi}{m} \right ) ^{2m+1} = \rho \left ( \prod _{m=3}^{\infty} \left ( \cos \frac {\pi}{m} \right )^m \right )^2.$$
Now, the above infinite product converges if and only if converges the
series of its logarithms $\sum _{m=3}^\infty m \ln \cos \frac {\pi}{m}$.
By expanding $\ln \cos \frac {\pi}{m}$ we obtain $\ln \cos \frac {\pi}{m}
\sim - \frac{\pi ^2}{2m^2}$, and then $m \ln \cos \frac {\pi}{m} \sim - 
\frac{\pi ^2}{2m}$. Hence the series diverges, and the infinite product goes 
to zero.
\end{proof}

From the above example we can conclude that the infinite product will 
converge even in cases when the number of repetitions grows without bounds,
as long as the growth is sublinear. The sequence 
$\left \lfloor \sqrt n \right \rfloor$ is a limiting case -- if a
sequence $(a_n)$ grows faster than $\left \lfloor \sqrt n \right \rfloor$,
its Kepler-Bouwkamp radius will be positive. The following result summarizes
our findings.

\begin{theorem}
$\kappa (a_n) &gt; 0$ if and only if $(a_n)$ grows faster than 
$\left \lfloor \sqrt n \right \rfloor$.
\end{theorem}

\section{Explicit formulas and (semi)numerical examples}

In this section we consider some admissible sequences and compute their
Kepler-Bouwkamp radii. We start with a class of sequences for which it is
possible to give explicit formulas. To the best of my knowledge, the class
is very narrow; it contains only integer multiples of the sequence of powers
of two. Hence, all sequences of this class are of the type $a_n = m \cdot 2^n$ 
for some integer $m$. The result follows from a classical infinite product
formula for $\sinc x$.

The sinc function is defined by
$$\sinc x = 
\begin{cases}
\frac {\sin x}{x}, &amp;\mbox{if } x \neq 0; \\ 
1, &amp;\mbox{if } x =0; \end{cases} .$$
The following infinite product representation of $\sinc x$ was known already
to Vi\`ete:
$$\sinc x = \prod _{n = 1}^{\infty} \cos \frac {x}{2^n}.$$
It immediately yields formulas for the Kepler-Bouwkamp radius of an integer
multiple of the sequence of powers of two.

\begin{theorem}
$\kappa (m \cdot 2^n) = \sinc \frac{\pi }{m}.$
\end{theorem}

Here are the values for some small $m$.

\begin{corollary}
\begin{align*}
\kappa (2 \cdot 2^n) &amp; =  \frac {2}{\pi}; &amp;
\kappa (3 \cdot 2^n) &amp; =  \frac {3 \sqrt{3}}{2\pi}; &amp;
\kappa (4 \cdot 2^n) &amp; =  \frac {3 \sqrt{2}}{\pi}; \\
\kappa (5 \cdot 2^n) &amp; =  \frac {5 \sqrt{5-\sqrt{5}}}{2\sqrt{2}\pi}; &amp;
\kappa (6 \cdot 2^n) &amp; =  \frac {3}{\pi}. 
\end{align*}
\end{corollary}

Let us now look at a general admissible sequence $(a_n)$ with $\kappa (a_n) &gt; 
0$. We follow, with minor modifications, the approach outlined by Kitson 
\cite{kitson}. First we take the logarithm of both sides in the expression for
$\kappa (a_n)$,
$$ \ln \kappa (a_n) = \sum _n \ln \cos \frac{\pi}{a_n}.$$
The summand $\ln \cos \frac{\pi}{a_n}$ can be expanded into a series
$$\ln \cos \frac{\pi}{a_n} = - \sum _{k=1}^{\infty} 
\frac{2^{2k} (2^{2k}-1)}{2k (2k)!} |B_{2k}| \left ( \frac{\pi}{a_n} \right )^{2k}$$
\cite[Formula 4.3.72]{abramowitz}. Here $B_k$ denote the Bernoulli
numbers. Their exponential generating function is given by
$$\sum _{k=0}^{\infty} \frac{B_k x^k}{k!} = \frac {x}{e^x -1}.$$
We have the following representation of even Bernoulli numbers:
$$B_{2k} = (-1)^{k-1} \frac {2(2k)!}{(2\pi )^{2k}} \zeta (2k)$$
\cite[Formula 23.1.18]{abramowitz}. 
It can be obtained by
expanding Bernoulli polynomials $B_n(x)$ into a cosine Fourier series
$$B_{2k}(x) = (-1)^{k-1} \frac{2(2k)!}{(2\pi )^{2k}} \sum _{l=1}^{\infty}
\frac {\cos 2l\pi x}{l^{2k}}$$
and substituting $x = 0$. When this is plugged into expansion of 
$\ln \cos \frac{\pi}{a_n}$ a lot of cancellation occurs and we end with
$$\ln \cos \frac{\pi}{a_n} = - \sum _{k=1}^{\infty} \frac{2^{2k}-1}{k}
\zeta (2k) \frac{1}{a_n^{2k}}.$$
Now we can exchange the order of summation in the double sum for
$\ln \kappa (a_n)$. We obtain
$$\ln \kappa (a_n) = - \sum _{k=1}^{\infty} \frac{2^{2k}-1}{k} \zeta (2k) 
\sum _n \frac{1}{a_n^{2k}}.$$
Here the interior sum runs over all $n$ such that $a_n \geq 3$. Hence, we have
proved the following result.

\begin{theorem}\label{main}
$$\kappa (a_n) = \exp \left ( -\sum _{k=1}^{\infty} \frac{2^{2k}-1}{k} \zeta (2k) \sum _{a_n \geq 3} a_n ^{-2k} \right ).$$
\end{theorem}

We see that all dependence on the sequence $(a_n)$ is well isolated and 
contained in the sum $\sum _{a_n \geq 3} a_n ^{-2k}$. For some sequences
that sum can be expressed in closed formulas in terms of $k$. By plugging them
into the formula from the above theorem, we obtain rapidly converging 
expressions for $\kappa (a_n)$; hence the (semi)numerical in the section title.

For $a_n = n$ the sum can be expressed in terms of zeta function, taking into
account the corrections for terms smaller than 3: $\sum _{a_n \geq 3} n ^{-2k}
= \frac{4^k(\zeta (2k) -1)-1}{4^k}$. Similarly, Kitson used the prime zeta
function to express the analogous sum in \cite{kitson}. Let us denote by
$Z(a_n)$ the sum $\sum _{a_n \geq 3} a_n ^{-2k}$ for the sequence $(a_n)$. In
the following proposition we present the values of $Z(a_n)$ 
for several sequences that yield to this approach.

\begin{proposition}
\begin{align*}
Z(2n) &amp;= 2^{-2k}(\zeta (2k)-1);  \\
Z(2n+1) &amp;= 2^{-2k} \zeta (2k,\frac{3}{2});  \\
Z(m \cdot n + p) &amp;= m^{-2k} \zeta (2k, \frac{m+p}{m})\qquad \mbox{ for \,\,} m \geq 3;  \\
Z(n^m) &amp;= \zeta (2mk) -1 ; \\
Z(m^n) &amp;= \frac {1}{m^{2k}-1} \qquad \mbox{ for \,\,} m \geq 3.
\end{align*}
(Here $\zeta (2k,q)$ is the Hurwitz zeta function.)
\end{proposition}

The Kepler-Bouwkamp radii of several sequences now follow by plugging the
above expressions into formula of Theorem \ref{main} and evaluating the resulting sums.

\begin{corollary}
\begin{align*}
\kappa (2n) &amp; =  0.4297802164;  \\
\kappa (2n+1) &amp; =  0.2674437781;  \\
\kappa (n^2) &amp; =  0.6402929927; \\
\kappa (3^n) &amp; =  0.4662745790.
\end{align*}
\end{corollary}

Here are the Kepler-Bouwkamp radii of some sequences as reported in Mathar's 
paper \cite{mathar}.

\begin{proposition} 
\begin{align*}
\kappa (n(n+1)) &amp; = 0.8154881209; \\
\kappa ((2n+1)(2n+2)) &amp; =  0.8373758680; \\
\kappa (p_n p_{n+1}) &amp; =  0.9729664541 \qquad \mbox{for odd primes}.
\end{align*}
\end{proposition}

We conclude the paper by reporting the numerical values of Kepler-Bouwkamp 
radii for several interesting combinatorial sequences.

\begin{proposition}
\begin{align*}
\kappa (n!) &amp; = 0.8583138700;  \\
\kappa ((2n-1)!!) &amp; = 0.4888521829; \\
\kappa ((2n)!!) &amp; =  0.9218702724; \\
\kappa (n^n) &amp; =  0.7022723378.
\end{align*}
\end{proposition}

The class of sequences that allow the seminumerical approach is probably 
much wider than reported here, but I am not aware of any simple way
to decide if a given sequence belongs to it.

\section{Acknowledgments}

Partial support of the Ministry of Science, Education and Sport of the
Republic of Croatia (Grants No.\ 037-0000000-2779 and 177-0000000-0884)
is gratefully acknowledged. I am thankful to the referee for careful reading
and correcting several mistakes.

\begin{thebibliography}{9}

\bibitem{abramowitz}
M. Abramowitz and I. A. Stegun, eds., 
\newblock {\em Handbook of Mathematical Functions}, 
National Bureau of Standards, Applied Math.\ Series {\bf 55}, 1972.

\bibitem{borwein}
D. Borwein and J. M. Borwein,
\newblock Some remarkable properties of sinc and related integrals,
{\em Ramanujan J.} {\bf 5} (2001) 73--89.

\bibitem{chamberland}
M. Chamberland and A. Straub,
\newblock On gamma quotients and infinite products, preprint,
\url{http://arxiv.org/abs/1309.3455}, 2013.

\bibitem{finch}
S. R. Finch,
\newblock {\em Mathematical Constants},
Cambridge University Press, Cambridge, 2003.

\bibitem{kitson}
A. R. Kitson,
\newblock The prime analog of the Kepler-Bouwkamp constant,
{\em Math. Gazette} {\bf 92} (2008) 293--294.

\bibitem{mathar}
R. J. Mathar,
\newblock Tightly circumscribed regular polygons, preprint, \newline
\url{http://arxiv.org/abs/1301.6293}, 2013.

\bibitem{sloane}
N. J. A. Sloane,
\newblock {\em On-Line Encyclopedia of Integer Sequences},
available electronically at \url{http://oeis.org}.

\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11Z05; Secondary: 51M04, 51M25

\noindent \emph{Keywords: } 
Kepler-Bouwkamp constant, infinite product.
\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000027},
\seqnum{A000040},
\seqnum{A000079},
\seqnum{A000142},
\seqnum{A000244},
\seqnum{A000290},
\seqnum{A000714},
\seqnum{A002378},
\seqnum{A005408},
\seqnum{A005843},
\seqnum{A007283}, and
\seqnum{A085365}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 30 2014;
revised versions received November 4 2014; November 5 2014.
Published in {\it Journal of Integer Sequences}, November 7 2014.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                

</pre></body></html>