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\begin{center}
\vskip 1cm{\LARGE\bf Evaluations of Some Variant Euler Sums
}
\vskip 1cm
\large
Hongwei Chen\\
Department of Mathematics\\
Christopher Newport University\\
Newport News, VA 23606\\
\href{mailto:hchen@cnu.edu}{hchen@cnu.edu} \\
\end{center}

\vskip .2 in
\begin{abstract}
In this note we present some elementary methods for
the summation of certain Euler sums with terms involving $1 + 1/3 +
1/5 + \cdots + 1/(2k-1).$
\end{abstract}


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

\section{Introduction}

In the last decade, based on extensive experimentation with
computer algebraic systems, a large class of Euler sums have been
explicitly evaluated in terms of the Riemann zeta function
$\zeta(k)$. For example, let
$$H_{k} = 1 + \frac{1}{2} + \cdots + \frac{1}{k}.$$
Then
\begin{eqnarray*}
\sum_{k=1}^{\infty}\,\frac{1}{k^{2}}\,H_{k} & = &
 2\,\zeta(3),\\
\sum_{k=1}^{\infty}\,\frac{1}{2^{k}\,k^{2}}\,H_{k} & = & \zeta(3)
- \frac{\pi^{2}}{12}\,\ln 2,\\
\sum_{k=1}^{\infty}\,\frac{1}{k^{2}}\,H_{k}^{2} & = &
\frac{17}{4}\,\zeta(4),
\end{eqnarray*}
and $$ \sum_{k=1}^{\infty}\,\frac{(-1)^{k-1}}{k^{2}}\,H_{k} =
\frac{5}{8}\,\zeta(3).$$ More details can be found in \cite{Bailey,Borwein1, Borwein2, Borwein3}
In particular,  Borwein and Bradley
\cite{Borwein2} collected 32 beautiful proofs of the
first sum above.

Motivated by the above results, in this note, replacing $H_{k}$ by
$$h_{k} = H_{2k} - \frac{1}{2}\,H_{k} = 1 + \frac{1}{3} + \cdots + \frac{1}{2k - 1} ,\eqno(1)$$
we study the following variant Euler sums
$$\sum_{k=1}^{\infty}\,a_{k}\,h_{k}$$
where the $a_{k}$ are relatively simple function of $k$.\\

\section{The Main Results}

We begin to derive some series involving $h_{k}$. Since
$$-\,\ln(1 - x) = \int_{0}^{x}\,\frac{dt}{1 - t} = \sum_{k=1}^{\infty}\,\frac{x^{k}}{k},$$
replacing $x$ by $-x$ gives
$$\ln(1 + x) =  \sum_{k=1}^{\infty}\,\frac{(-1)^{k-1}x^{k}}{k}.$$
Averaging these two series gives us
$$\frac{1}{2}\,\ln\left(\frac{1+ x}{1 - x}\right)
= \sum_{k=1}^{\infty}\,\frac{1}{2k-1}\,x^{2k - 1}.\eqno(2)$$ In
term of the Cauchy product and partial fractions, we have
\begin{eqnarray*}
\frac{1}{4}\,\ln^{2}\left(\frac{1+ x}{1 - x}\right)  & = &
\sum_{k=1}^{\infty}\,\left(\frac{1}{(2k-1)\cdot 1} +
\frac{1}{(2k-3)\cdot 3} + \cdots +
\frac{1}{1\cdot(2k-1)}\right)x^{2k}\\
& = &
\sum_{k=1}^{\infty}\,\frac{1}{2k}\,\left[\left(\frac{1}{2k-1}
+ \frac{1}{1}\right) + \left( \frac{1}{2k-3} + \frac{1}{3}\right)
+ \cdots +
\left(\frac{1}{1} + \frac{1}{2k-1}\right)\right]x^{2k}\\
& = & \sum_{k=1}^{\infty}\,\left(1 + \frac{1}{3} +
\cdots + \frac{1}{2k -1}\right)\frac{x^{2k}}{k}.
\end{eqnarray*}
Noting that $h_{k}$ is given by (1), we have
$$ \sum_{k=1}^{\infty}\,\frac{h_{k}}{k}\,x^{2k}
= \frac{1}{4}\,\ln^{2}\left(\frac{1+ x}{1 - x}\right). \eqno(3)$$
This enables us to evaluate a wide variety of interesting series
via specialization, differentiation and integration.\\

First, setting $x = 1/2$, we find
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{2^{2k}\,k} = \frac{1}{4}\,\ln^{2}3.\eqno(4)$$
For $x = \sqrt{2}/2$,
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{2^{k}\,k}
= \frac{1}{4}\,\ln^{2}(3 + 2\sqrt{2}).\eqno(5)$$
Putting $x = (\sqrt{5} -
1)/2 = \phi$, the golden ratio, we get
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{k}\,\phi^{2k}
= \frac{1}{4}\,\ln^{2}(2 + \sqrt{5}).\eqno(6)$$
Furthermore, for any
$\alpha \geq 2$, putting $x = (\sqrt{5} + 1)/2\alpha$ and $x =
(\sqrt{5} - 1)/2\alpha$ in (3) respectively, we get
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{\alpha^{2k}\,k}\,\left(\frac{\sqrt{5}
+ 1}{2}\right)^{2k} = \frac{1}{4}\,\ln^{2}\left(\frac{(2\alpha +1)
+ \sqrt{5}}{(2\alpha -1) - \sqrt{5}}\right)\eqno(7)$$ and
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{\alpha^{2k}\,k}\,\left(\frac{\sqrt{5}
- 1}{2}\right)^{2k} = \frac{1}{4}\,\ln^{2}\left(\frac{(2\alpha -1)
+ \sqrt{5}}{(2\alpha +1) - \sqrt{5}}\right).\eqno(8)$$
Recalling the Fibonacci numbers which are defined by
$$F_{1} = 1, F_{2} = 1, F_{k} = F_{k-1} + F_{k-2}\,\,\,\mbox{for $k \geq 2$}$$
and Binet's formula
$$F_{k} = \frac{1}{\sqrt{5}}\,\left(\left(\frac{\sqrt{5} + 1}{2}\right)^{k} -
\left(\frac{1- \sqrt{5} }{2}\right)^{k}\right),$$ combining (7)
and (8), we find
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{\alpha^{2k}\,k}\,F_{2k}
= \frac{\sqrt{5}}{20}\,\ln\left(\frac{\alpha^2 + \alpha
-1}{\alpha^{2} - \alpha -1}\right)\,\ln\left(\frac{\alpha^2 +
\alpha\sqrt{5} +1}{\alpha^{2} - \alpha\sqrt{5} +1}\right).\eqno(9)$$
In particular, for $\alpha = 2$
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{2^{2k}\,k}\,F_{2k}
= \frac{\sqrt{5}}{4}\,\ln 5\,\ln (9 + 4\sqrt{5}).\eqno(10)$$
Another step along this path is to change variables. Setting $x=
\cos\theta$ in (3) leads to
$$ \sum_{k=1}^{\infty}\,\frac{h_{k}}{k}\,\cos^{2k}\theta
= \ln^{2}\left(\cot(x/2)\right).\eqno(11)$$ Integrating both sides from $0$
to $\pi$, and using
$$\int_{0}^{\pi}\,\cos^{2k}\theta\,d\theta =
\frac{\pi}{2^{2k}}\,\left(\begin{array}{c} 2k \\k
\end{array}\right)$$
and
$$\int_{0}^{\pi}\,\ln^{2}\left(\cot(x/2)\right)\,d\theta =
\frac{\pi^{3}}{4},$$ we find
$$ \sum_{k=1}^{\infty}\,\frac{h_{k}}{2^{2k}\,k}\,\left(\begin{array}{c} 2k \\k
\end{array}\right) = \frac{\pi^{2}}{4}.\eqno(12)$$
This adds another interesting series to Lehmer's list \cite{Lehmer}.\\

Next, for $0 < x < 1$, differentiating (3), then multiplying both
sides by $x$, we obtain
$$\sum_{k=1}^{\infty}\,h_{k}x^{2k} = \frac{x}{2(1-x^{2})}\,\ln\left(\frac{1+ x}{1 -
x}\right).\eqno(13)$$ Setting $x = 1/2$, we get
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{2^{2k}} = \frac{1}{3}\,\ln3.\eqno(14)$$
For $x = \sqrt{2}/2$,
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{2^{k}} = \frac{\sqrt{2}}{2}\,
\ln (3 + 2\sqrt{2}).\eqno(15)$$
Similar to (10), we have
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{2^{2k}}\,F_{2k} = \frac{\sqrt{5}}{50}\,
(10\ln(5 + 2\sqrt{5}) + 3 \sqrt{5}\ln 5 - 5 \ln5).\eqno(16)$$

Finally, for $ 0 < x \leq 1$, dividing both sides of (3) by $x$
and integrating from $0$ to $x$, we obtain
$$ \sum_{k=1}^{\infty}\,\frac{h_{k}}{k^{2}}\,x^{2k} = \frac{1}{2}\,\int_{0}^{x}\,
\frac{1}{t}\ln^{2}\left(\frac{1+ t}{1 - t}\right)\,dt.\eqno(17)$$
Using the substitution $u = (1-x)/(1 + x)$ and integration by
parts, we get
\begin{eqnarray*}
\sum_{k=1}^{\infty}\,\frac{h_{k}}{k^{2}}\,x^{2k}
& = & \int_{(1-x)/(1 + x)}^{1}\,\frac{\ln^{2}u}{1 - u^{2}}\,du\\
& = & \frac{1}{2}\,\ln x\ln^{2}\left(\frac{1-x}{1 + x}\right) +
\int_{(1-x)/(1 + x)}^{1}\,\frac{\ln u}{u}\,\ln\left(\frac{1-u}{1 +
u}\right)du.
\end{eqnarray*}
In view of (2), we have
\begin{eqnarray*}
\int_{(1-x)/(1 + x)}^{1}\,\frac{\ln u}{u}\,\ln\left(\frac{1-u}{1 +
u}\right)du & = & -2\,\sum_{k=0}^{\infty}\,\frac{1}{2k +
1}\,\int_{(1-x)/(1 + x)}^{1}\,u^{2k}\ln u\,du.
\end{eqnarray*}
Since
$$\int\,u^{2k}\ln u\,du = \frac{1}{2k +1}\,u^{2k+1}\ln u -
\frac{1}{(2k+1)^{2}}\,u^{2k+1} + C,$$
we find
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{k^{2}}\,x^{2k}
=  \frac{1}{2}\,\ln x\ln^{2}\left(\frac{1-x}{1 + x}\right)+
2\ln\left(\frac{1-x}{1 + x}\right)\,
\sum_{k=0}^{\infty}\,\frac{1}{(2k+1)^{2}}\,\left(\frac{1-x}{1 +
x}\right)^{2k+1}$$
$$+ 2\,\sum_{k=0}^{\infty}\,\frac{1}{(2k+1)^{3}} -
2\,\sum_{k=0}^{\infty}\,\frac{1}{(2k+1)^{3}}\,\left(\frac{1-x}{1 +
x}\right)^{2k+1}.\eqno(18)$$ 
In terms of the polylogarithm
function \cite{Lewin}
$$Li_{n}(x) = \sum_{k=1}^{\infty}\,\frac{x^{n}}{k^{n}},$$
and noting that
$$\sum_{k=0}^{\infty}\,\frac{x^{n}}{(2k+1)^{n}} = \frac{1}{2}\,(Li_{n}(x) -
Li_{n}(-x))$$ and
$$\sum_{k=0}^{\infty}\,\frac{1}{(2k + 1)^{3}} =
\sum_{k=0}^{\infty}\,\frac{1}{k^{3}} -
\sum_{k=0}^{\infty}\,\frac{1}{(2k)^{3}} = \frac{7}{8}\,\zeta(3),$$
we finally obtain
$$\sum_{k=1}^{\infty}\,\frac{h_{k}}{k^{2}}\,x^{2k}
=  \frac{7}{4}\,\zeta(3) + \frac{1}{2}\,\ln
x\ln^{2}\left(\frac{1-x}{1 + x}\right)$$
$$+ \ln\left(\frac{1-x}{1 + x}\right)\,\left(Li_{2}\left(\frac{1-x}{1 + x}\right) -
Li_{2}\left(\frac{x-1}{1 + x}\right)\right) -
\left(Li_{3}\left(\frac{1-x}{1 + x}\right) -
Li_{3}\left(\frac{x-1}{1 + x}\right)\right).
$$
Setting $x = 1$, we get
$$ \sum_{k=1}^{\infty}\,\frac{h_{k}}{k^{2}}  =
\frac{7}{4}\,\zeta(3).\eqno(19)$$ For $x = 1/3$,
\begin{eqnarray*}
\sum_{k=1}^{\infty}\,\frac{h_{k}}{3^{2k}\,k^{2}} & = &
\frac{7}{8}\,\zeta(3) - \frac{1}{2}\,\ln 3\ln^{3}2\\
& + &\frac{1}{3}\,\ln^{3}2 + \ln 2\,Li_{2}(-1/2) + Li_{3}(-1/2),
\end{eqnarray*}
where we have used
\begin{eqnarray*}
Li_{2}(1/2) & = & \frac{\pi^{2}}{12} - \frac{1}{2}\,\ln^{2}2;\\
 Li_{3}(1/2) & = &
\frac{7}{8}\,\zeta(3) + \frac{1}{6}\,\ln^{3}2 -
\frac{\pi^{2}}{12}\,\ln 2 .
\end{eqnarray*}
Moreover, noting that
$$h_{k} = \sum_{i=1}^{k}\,\int_{0}^{1}\, x^{2(i-1)}\,dt = \int_{0}^{1}\, (\sum_{i=1}^{k}\,x^{2(i-1)})\,dt
= \int_{0}^{1}\,\frac{1 - x^{2k}}{1 - x^{2}}\,dx$$ and rewriting
(8) as
$$ \sum_{k=1}^{\infty}\,\frac{h_{k}}{k^{2}}\,(1-x^{2k}) = \frac{1}{2}\,\int_{x}^{1}\,
\frac{1}{t}\ln^{2}\left(\frac{1+ t}{1 - t}\right)\,dt, $$ we have
\begin{eqnarray*}
\sum_{k=1}^{\infty}\,\frac{h_{k}^{2}}{k^{2}} & = &
\sum_{k=1}^{\infty}\,\frac{h_{k}}{k^{2}}\,\int_{0}^{1}\,\frac{1 - x^{2k}}{1 - x^{2}}\,dx\\
& = & \frac{1}{2}\,\int_{0}^{1}\,\left(\frac{1}{1 - x^{2}}\,\int_{x}^{1}\,
\frac{1}{t}\ln^{2}\left(\frac{1+ t}{1 - t}\right)\,dt\right)\,dx.
\end{eqnarray*}
Exchanging the order of the integration, we get
\begin{eqnarray*}
\sum_{k=1}^{\infty}\,\frac{h_{k}^{2}}{k^{2}} & = &
\frac{1}{2}\,\int_{0}^{1}\,\left(\frac{1}{t}\ln^{2}\left(\frac{1+
t}{1 - t}\right)\,\int_{0}^{t}\,\frac{1}{1 -
x^{2}}\,dx\right)\,dt.\\
& = & \frac{1}{4}\,\int_{0}^{1}\,\frac{1}{t}\ln^{3}\left(\frac{1+
t}{1 - t}\right)\,dt.
\end{eqnarray*}
Using the substitution $x = (1-t)/(1 + t)$ and the well-known fact
that
$$\int_{0}^{1}\,x^{k}\,\ln^{3}x\,dx = -\,\frac{6}{(k + 1)^{3}},$$
we find
$$
\sum_{k=1}^{\infty}\,\frac{h_{k}^{2}}{k^{2}}  =  -
\frac{1}{2}\,\int_{0}^{1}\,\frac{\ln^{3}x}{1 - x^{2}}\,dx$$
$$ =  -
\frac{1}{2}\,\sum_{k=0}^{\infty}\,\int_{0}^{1}\,x^{2k}\ln^{3}x\,dx\\
 =  3\,\sum_{k=0}^{\infty}\,\frac{1}{(2k+1)^{4}} =
\frac{45}{16}\,\zeta(4)\eqno(20)$$


Another path out of (3) is to bring in complex variables. Since
$$ \frac{1}{i}\tan^{-1}(iz) = \tanh^{-1} z = \frac{1}{2} \ln
\left(\frac{1+z}{1 - z}\right)$$ Replacing $x$ by $ix$ in (3), we
obtain
$$ \sum_{k=1}^{\infty}\,\frac{(-1)^{k-1}h_{k}}{k}\,x^{2k}
= (\tan^{-1}x)^{2}.  \eqno(21)$$
This series may be evaluated at
values such as $x = 2- \sqrt{3}, \sqrt{3}/3, 1$ explicitly:
$$\sum_{k=1}^{\infty}\,\frac{(-1)^{k-1}h_{k}(2 -
\sqrt{3})^{2k}}{k} =   \frac{\pi^{2}}{144} =
\frac{3}{72}\,\zeta(2),\eqno(22)$$
$$\sum_{k=1}^{\infty}\,\frac{(-1)^{k-1}h_{k}}{3^{k}\,k} =
\frac{\pi^{2}}{36} = \frac{1}{6}\,\zeta(2),\eqno(23)$$
$$\sum_{k=1}^{\infty}\,\frac{(-1)^{k-1}h_{k}}{k} =
\frac{\pi^{2}}{16} = \frac{3}{8}\,\zeta(2).\eqno(24)$$ Similarly,
applying differentiation and integration to (21), we deduce the
corresponding formulas
$$\sum_{k=1}^{\infty}\,(-1)^{k-1}h_{k}\,x^{2k}
=  \frac{x}{1 + x^{2}}\,\tan^{-1}x,\eqno(25)$$
$$\sum_{k=1}^{\infty}\,\frac{(-1)^{k-1}h_{k}}{k^{2}}\,x^{2k}  =
2\,\int_{0}^{x}\,\frac{(\tan^{-1}t)^{2}}{t}\,dt.\eqno(26)$$ In
particular, we find $$
\sum_{k=1}^{\infty}\,\frac{(-1)^{k-1}h_{k}}{3^{k}} =
\frac{\sqrt{3}}{24}\,\pi,\eqno(27)$$
$$\sum_{k=1}^{\infty}\,\frac{(-1)^{k-1}h_{k}}{k^{2}}  =   G\,\pi -
\frac{7}{4}\,\zeta(3),\eqno(28)$$ where $G$ is the Catalan's
constant which is defined by
$$ G = \sum_{k=0}^{\infty}\,\frac{(-1)^{k}}{(2k + 1)^{2}}.$$
\vspace{0.1in}


Finally, following the excellent suggestion of an anonymous referee,
recalling that
$$h_{k} = H_{2k} - \frac{1}{2}\,H_{k},\eqno(29)$$
we find from (19)
$$\sum_{k=1}^{\infty}\,\frac{1}{k^{2}}\,H_{2k} = \sum_{k=1}^{\infty}\,\frac{1}{k^{2}}\,h_{k}
+ \frac{1}{2}\,\sum_{k=1}^{\infty}\,\frac{1}{k^{2}}\,H_{k} =
\frac{11}{4}\,\zeta(3).\eqno(30)$$ Furthermore, in terms of the
multiple series \cite{Zagier} 
$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\,\frac{1}{ij(i + j)} =
2\zeta(3),\,\,\,\,\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\,\frac{(-1)^{i
+ j}}{ij(i + j)} = \frac{1}{4}\,\zeta(3),$$ the difference gives
$$\sum_{i, j> 1, i + j = \mbox{odd}}\,\frac{1}{ij(i + j)} =
\frac{7}{8}\,\zeta(3).$$ Setting $i + j = 2k + 1$ and using partail
fractions, we have
\begin{eqnarray*}
\sum_{i, j> 1, i + j = \mbox{odd}}\,\frac{1}{ij(i + j)} & = &
\sum_{k=1}^{\infty}\,\sum_{j=1}^{2k}\,\frac{1}{j(2k+1-j)(2k + 1)}\\
& = & \sum_{k=1}^{\infty}\,\frac{1}{(2k+ 1)^{2}}\sum_{j=1}^{2k}\,\left(\frac{1}{j} + \frac{1}{2k+1-j}\right)\\
& = & \sum_{k=1}^{\infty}\,\frac{1}{(2k+ 1)^{2}}\,2H_{2k}.
\end{eqnarray*}
Thus,
$$\sum_{k=1}^{\infty}\,\frac{1}{(2k+ 1)^{2}}\,H_{2k} =
\frac{7}{16}\,\zeta(3).\eqno(31)$$  Subsequently, we have
$$\sum_{k=1}^{\infty}\,\frac{1}{(2k-1)^{2}}\,H_{2k}
  =  \sum_{k=1}^{\infty}\,\frac{1}{(2k+ 1)^{2}}\,H_{2k} + $$
$$\sum_{k=1}^{\infty}\,\frac{1}{(2k- 1)^{3}}
 + \sum_{k=1}^{\infty}\,\frac{1}{2k(2k- 1)^{2}}
 = \frac{21}{16}\,\zeta(3) +
 \frac{1}{8}\,(\pi^{2} - 8\ln 2).\eqno(32)$$
From this and the known result \cite{Bailey}
$$\sum_{k=1}^{\infty}\,\frac{1}{(2k- 1)^{2}}\,H_{k} =
\frac{1}{4}\,(\pi^{2} - \pi^{2}\ln 2 - 8\ln 2 + 7\zeta(3)),$$ we
finally get
$$\sum_{k=1}^{\infty}\,\frac{1}{(2k- 1)^{2}}\,h_{k} =
\frac{7}{16}\,\zeta(3) + \frac{3}{4}\,\zeta(2)\ln 2. \eqno(33)$$

\section{Acknowledgments}

The author would like to thank the referee
and the editor for their thoughtful comments and suggestions for
improving the original version of the manuscript.

\begin{thebibliography}{9}
\bibliographystyle{plain}

\bibitem{Bailey} D.\ Bailey, J.\ Borwein and R.\ Girgensohn, Experimental
evaluation of Euler sums, {\it Experimental Math.}, {\bf
3} (1994), 17--30.

\bibitem{Borwein1} J.\ Borwein, D.\ Bailey and R.\ Girgensohn, {\it
Experimentation in Mathematics}, A. K. Peters, 2004.

\bibitem{Borwein2} J.\ Borwein and D.\ V. Bradley, Thirty-two Goldbach
variations, preprint. Available at
\href{http://users.cs.dal.ca/~jborwein/32goldbach.pdf}{\tt http://users.cs.dal.ca/\symbol{126}jborwein/32goldbach.pdf}.

\bibitem{Borwein3} J.\ Borwein and I.\ J. Zucker and J.\ Boersma, The
evaluation of character Euler double sums, preprint. Available at
\href{http://users.cs.dal.ca/~jborwein/bzb7.pdf}{\tt http://users.cs.dal.ca/\symbol{126}jborwein/bzb7.pdf}.

\bibitem{Lewin} L.\ Lewin, {\it Polylogarithms and Associated Functions},
Elsevier North Holland, New York-Amsterdam, 1981.

\bibitem{Lehmer} D.\ H.\ Lehmer, Interesting series involving the central
binomial coefficient, {\it Amer.\ Math.\ Monthly}, {\bf 92} (1985),
449--457.

\bibitem{Zagier} D.\ Zagier, Values of zeta functions and their applications,
{\it First European Congress of Mathematics}, Vol.\  2, Paris,
Birkhauser, 1994, pp.\ 497--512.
\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 40C15; Secondary 40A25, 11M41, 11M06.

\noindent \emph{Keywords: } Euler sums, Riemann zeta function, closed
forms.


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000045},
\seqnum{A000984},
\seqnum{A001008}, and
\seqnum{A005408}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received February 9 2006;
revised version received April 21 2006. 
Published in {\it Journal of Integer Sequences}, May 18 2006.

\bigskip
\hrule
\bigskip

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


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