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\begin{document}

%% Author, fill in article title here
\title{An interesting lemma for regular $C$-fractions}
%% Fill in author list here
\author{Kwang-Wu Chen
\\{\normalsize Department of International Business Management,} 
\\{\normalsize Ching Yun University,}
\\{\normalsize No. 229,  Jianshing Road, Jungli City,} 
\\{\normalsize Taoyuan, Taiwan 320, R.O.C.}
\\{\normalsize E-mail: kwchen@cyu.edu.tw} 
}
%
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\begin{abstract}  
\setlength{\baselineskip}{20pt}
In this short note we give an interesting lemma for regular $C$-fractions.
Applying this lemma we obtain some congruence properties of some classical numbers such as
the Springer numbers of even index, the median Euler numbers, the median Genocchi numbers, 
and the tangent numbers.
\end{abstract}

%%%% Authors begin text of article here %%%
\setlength{\baselineskip}{23pt}

\section{The interesting lemma}
A regular $C$-fraction is a continued fraction of the form 
\begin{eqnarray*}
  a_0+{\bf K}^\infty_{n=1}(a_nz/1) &=& a_0+\frac{a_1z}{1}\plus\frac{a_2z}{1}\plus\frac{a_3z}{1}\plus\ldots \\
  &=& a_0+\frac{a_1z}{1+\dfrac{a_2z}{1+\dfrac{a_3z}{\ddots}}},
\end{eqnarray*}
where $a_n\in \mathbb{C}$.

Let $f(z)=\sum^\infty_{n=0}c_nz^n\in\mathbb{C}[[z]]$ be a formal power series. It is known that 
there exists a one-to-one correspondence between regular $C$-fractions $a_0+{\bf K}^\infty_{n=1}(a_nz/1)$ 
and formal power series $\sum^\infty_{n=0}c_nz^n$ (ref. \cite[p.\ 252--p.\ 265]{Lisa}).

Now we assume that all coefficients are integral. The lemma we state here gives the division relation
between the integral coefficients of the regular $C$-fraction and the integral coefficients of 
its corresponding formal power series.

\begin{lemma}
\label{lma.2}
Let $f(z)=\sum^\infty_{n=0}a_nz^n\in\mathbb{Z}[[z]]$ be a integral formal power series. 
Assume the corresponding uniquely determined regular $C$-fraction is
\begin{equation}
\label{eq.32}
  \sum^\infty_{n=0}a_nz^n=\frac{b_0}1\plus\frac{bb_1z}1\plus\frac{bb_2z}1\plus\ldots,
\end{equation}
where $b$ and $(b_n)_{n\geq 0}$ are integral. Then $a_n$ is divisible by $(b_0b_1b^n)$ for $n\geq 1$.
\end{lemma}
\begin{proof}
Setting $z=y/b$, Equation \ref{eq.32} becomes
\begin{eqnarray*}
  \sum^\infty_{n=0}a_n(\frac yb)^n 
  &=& \frac{b_0}1\plus\frac{b_1y}1\plus\frac{b_2y}1\plus\frac{b_3y}1\plus\ldots \\
  &=& b_0-\frac{b_0b_1y}{1+b_1y}\plus\frac{b_2y}1\plus\frac{b_3y}1\plus\ldots.
\end{eqnarray*}
Since $a_0=b_0$, we have
$$
  \sum^\infty_{n=1}\frac{a_n}{b_0b_1b^n}y^n=\frac{-y}{1+b_1y}\plus\frac{b_2y}1\plus\frac{b_3y}1\plus\ldots.
$$
Since the right-hand side of the above identity can be uniquely expressed as a formal power series 
with integral coefficients, we conclude the proof.
\end{proof}

Let $f(t)=\sum_na_nt^n$ and $g(t)=\sum_nb_nt^n$
($n\geq 0$) be two formal power series with integral coefficients. For a non-negative integer $m$ we write
\begin{equation}
\label{eq.27}
  f(t)\equiv g(t)\ \pmod m\quad\mbox{iff}\quad a_n\equiv b_n\ \pmod m\quad \mbox{for all }n\geq 0.
\end{equation}

Applying Lemma \ref{lma.2} we could obtain some congruence properties of some classical numbers such as
the Springer numbers of even index, the median Euler numbers, the median Genocchi numbers, 
and the tangent numbers.

%============================================================================================

\section{Applications}
The Springer numbers (ref. \cite[p.\ 275]{Dumont}) are defined by 
\begin{equation}
\label{eq.10}
  S(x)=e^x\sech 2x=\sum^\infty_{n=0}\frac{S_nx^n}{n!}.
\end{equation}
The even (resp. odd) part of the Springer numbers is what Glaisher (ref. \cite[p.\ 276]{Dumont}) called the 
numbers $P_n$ (resp. $Q_n$).
That is to say,
\begin{equation}
\label{eq.11}
  \frac{\cosh x}{\cosh 2x}=\sum^\infty_{n=0}\frac{S_{2n}x^{2n}}{(2n)!},\qquad
  \frac{\sinh x}{\cosh 2x}=\sum^\infty_{n=0}\frac{S_{2n+1}x^{2n+1}}{(2n+1)!}.
\end{equation}
Springer introduced these numbers for a problem about root systems, and Arnold showed these
numbers as counting various types of snakes (ref. \cite[p.\ 6--p.\ 7]{Hoffman}).

Following the notation and the result in Corollary 3.3 of \cite{Dumont} we put
\begin{eqnarray}
\nonumber
  p(x) &=& \sum^\infty_{n=0}S_{2n}x^{2n+1}\ =\ x-3x^3+57x^5-\ldots \\
\nonumber
       &=& \frac x1\plus\frac{3x^2}1\plus\frac{16x^2}1\plus\frac{35x^2}1\plus\ldots \\
\label{eq.34}
       &&\quad    \plus\frac{16n^2x^2}1\plus\frac{(4n+1)(4n+3)x^2}1\plus\ldots.
\end{eqnarray}
Note that our definition of the Springer numbers $S_{2n}$ differs from that in \cite{Dumont}.
The unsign sequence $(-1)^nS_{2n}:$ $1$, $3$, $57$, $2763$, $250737$, $\ldots$, is the sequence 
A000281 in \cite{Sloane}.
Applying Lemma \ref{lma.2} we have $S_{2n}$ is divisible by $3$. Moreover, we have the following theorem.
\begin{theorem}
\label{thm.2}
For $n\geq 1$, the Springer number with even index $S_{2n}$ is divisible by $3$ and 
\begin{equation}
\label{eq.35}
  \frac{S_{2n}}3\equiv (-1)^n3^{n-1}\qquad \pmod{16}.
\end{equation}
\end{theorem}
\begin{proof}
Multiplying $x$ into $p(x)$ and setting $t=x^2$, we have
$$
  \sum^\infty_{n=0}S_{2n}t^{n+1}=t-3t^2+57t^3-\cdots=\frac t1\plus\frac{3t}1\plus\frac{16t}1\plus\ldots.
$$
Applying Lemma \ref{lma.2}, $S_{2n}$ is divisible by $3$ for $n\geq 1$. And
\begin{eqnarray}
\label{eq.51}
  \sum^\infty_{n=1}\frac{S_{2n}}3t^{n+1} &=& \frac{-t^2}{1+3t}\plus\frac{16t}1\plus\frac{35t}1\plus\ldots \\
  &\equiv & \frac{-t^2}{1+3t}\qquad \pmod{16}\nonumber\\
  &=& \sum^\infty_{n=1}(-1)^n3^{n-1}t^{n+1}.\nonumber
\end{eqnarray}
Comparing the coefficients of $t^{n+1}$, we have
$$
  \frac{S_{2n}}3\equiv (-1)^n3^{n-1}\qquad \pmod{16}, \qquad n\geq 1.
$$
\end{proof}

\begin{remark}
Now we write Equation \ref{eq.51} as
$$
  \frac{-t^2}{1+3t}\plus\frac{16t}1\plus\frac{35t}1\plus\ldots
  =\frac{-t^2}{1+3t}\plus\mathop{\bf K}^\infty_{n=1}\left(\frac{c_nt}1\right),
$$
where $c_{2n-1}=16n^2$ and $c_n=(4n+1)(4n+3)$, for $n\geq 1$.

If we take the modulus $c_2=35$ instead of $c_1=16$ for Equation \ref{eq.51} in the above proof.
Then we have
\begin{eqnarray*}
  \sum^\infty_{n=1}\frac{S_{2n}}3t^{n+1}  &\equiv& \frac{-t^2}{1+19t} \qquad\pmod{35} \\
  &\equiv& \frac{-t^2}{1-16t}\qquad \pmod{35}\\
  &=& \sum^\infty_{n=1}(-16^{n-1})t^{n+1} .
\end{eqnarray*}
Comparing the coefficients of $t^{n+1}$, we have 
\begin{equation}
\label{eq.61}
  \frac{S_{2n}}3\equiv -16^{n-1}\qquad\pmod{35},\qquad n\geq 1.
\end{equation}
Since $16^3 \equiv 1$ (mod $35$), we can also write Equation \ref{eq.61} as follows: for $k\geq 1$, 
\begin{equation}
\label{eq.52}
  \frac{S_{2n}}3\equiv \left\{\begin{array}{ll} 34\ \ \pmod{35}, & \mbox{if\ }n=3k-2;\\
                19\ \ \pmod{35}, & \mbox{if\ }n=3k-1;\\
                24\ \ \pmod{35}, & \mbox{if\ }n=3k.\end{array}\right.
\end{equation}

Similarly, we take another $c_n$ as the modulus for Equation \ref{eq.51}, then
we can get the congruences for $S_{2n}/3$ under the modulus $c_n$. \qed
\end{remark}

Let us define the Euler numbers $E_n$ through the exponential generating function $E(x)$:
$$
  E(x)=\sech x+\tanh x=\sum^\infty_{n=0}\frac{E_nx^n}{n!}.
$$
We construct the Seidel matrix $(a_{n,m})_{n,m\ge 0}$ associated with the sequence
$(0,E_1,E_2,E_3,\ldots)$ as follows:
\begin{enumerate}
\item The first row $(a_{0,n})_{n\ge 0}$ of the matrix is the initial sequence
$(0$, $E_1$, $E_2$, $E_3$, $\ldots)$.
\item Each entry $a_{n,m}$ of the $n$-th row is the sum of the entry immediately above and 
of the entry above and to the right of it:
$$
  a_{n,m}=a_{n-1,m}+a_{n-1,m+1}.
$$
\end{enumerate}
The resulted Seidel matrix is
$$
\begin{array}{rrrrrrrr}
0 & 1 & -1 & -2 & 5 & 16 & -61 & \cdots \\
1 & 0 & -3 & 3 & 21 & -45 & \cdots \\
1 & -3 & 0 & 24 & -24 & \cdots \\
-2 & -3 & 24 & 0 & \cdots \\
-5 & 21 & 24 & \cdots \\
16 & 45 & \cdots \\
61 & \cdots \\
\cdots 
\end{array}
$$

The absolute values of the upper diagonal sequence $1$, $3$, $24$, $402$, $\ldots$ are called the median Euler 
numbers $R_n$ (ref. Section $4$ in \cite{Dumont} or the sequence A002832 in \cite{Sloane}). 
Using the same method as above, we have

\begin{theorem}
\label{thm.3}
For $n\geq 1$, the median Euler number $R_n$ is divisible by $3$ and 
\begin{equation}
\label{eq.36}
  \frac{R_n}3\equiv 3^{n-1}\qquad \pmod 5.
\end{equation}
\end{theorem}
\begin{proof}
Since the ordinary generating function of the median Euler numbers $R_n$ satisfies the continued fraction
representation (see Proposition 7 in \cite{Dumont}):
\begin{eqnarray}
\nonumber
  r(x) &=& \sum^\infty_{n=0}(-1)^nR_nx^{n+1}=x-3x^2+24x^3-402x^4+11616x^5-\cdots \\
\label{eq.37}
  &=& \frac x1\plus\frac{3x}1\plus\frac{5x}1\plus\frac{2\cdot 7x}1\plus\frac{2\cdot 9x}1\plus\ldots.
\end{eqnarray}
Applying Lemma \ref{lma.2}, $R_n$ is divisible by $3$ for $n\geq 1$. And
\begin{eqnarray*}
  \sum^\infty_{n=1}(-1)^n\frac{R_n}3x^{n+1} &=& \frac{-x^2}{1+3x}\plus\frac{5x}1\plus\frac{14x}1\plus\ldots \\
  &\equiv & \frac{-x^2}{1+3x}\qquad \pmod 5\\
  &=& \sum^\infty_{n=1}(-1)^n3^{n-1}x^{n+1}.
\end{eqnarray*}
Comparing the coefficients of $x^{n+1}$, we complete the proof.
\end{proof}

The Genocchi numbers $G_n$ (the sequence A036968 in \cite{Sloane}) are defined by
$$
  \frac{2x}{e^x+1}=\sum^\infty_{n=0}\frac{G_nx^n}{n!}.
$$
The median Genocchi numbers $H_{2n+1}$ (ref. \cite{Dumont}, \cite{Han}, or the sequence A005439 in \cite{Sloane})
can be defined by $H_1=1$ and 
$$
  H_{2n+1}=\sum^{\lfloor\frac{n-1}2\rfloor}_{k=0}{n\choose 2k+1}G_{2n-2k}, \qquad n\geq 1,
$$
where $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$.

\begin{theorem}
For $n\geq 1$, the median Genocchi number $H_{2n+3}$ is divisible by $2^n$ and 
\begin{equation}
\label{eq.53}
  \frac{H_{2n+3}}{2^n}\equiv\left\{\begin{array}{ll}
                         1\ \pmod{6}, & \mbox{if $n$ is odd;}\\
                         4\ \pmod{6}, & \mbox{if $n$ is even.}\end{array}\right.
\end{equation}
\end{theorem}
\begin{proof}
Since the ordinary generating function of the median Genocchi numbers $H_{2n+1}$ satisfies 
the continued fraction representation (see \cite[p.\ 295]{Dumont})
\begin{eqnarray}
   h(x) &=& \sum^\infty_{n=0}H_{2n+1}x^{n+1}\ =\ x-x^2+2x^3-8x^4+56x^5-\cdots \nonumber\\
        &=& \frac x1\plus\frac x1\plus\frac x1\plus\frac{2^2x}1\plus\frac{2^2x}1
            \plus\frac{3^2x}1\plus\frac{3^2x}1\plus\ldots.
\end{eqnarray}
Since (see Lemma 1 in \cite{Dumont})
\begin{eqnarray}
\lefteqn{\frac x1\plus\frac{c_1x}1\plus\frac{c_2x}1\plus\frac{c_3x}1\plus\ldots}\nonumber\\
    &=& x-\frac{c_1x^2}{1+(c_1+c_2)x}\minus\frac{c_2c_3x^2}{1+(c_3+c_4)x}
         \minus\frac{c_4c_5x^2}{1+(c_5+c_6)x}\minus\ldots \label{eq.55}\\
    &=& \frac{x}{1+c_1x}\minus\frac{c_1c_2x^2}{1+(c_2+c_3)x}
         \minus\frac{c_3c_4x^2}{1+(c_4+c_5)x}\minus\ldots.\label{eq.56}
\end{eqnarray}
Then we can rewrite the continued fraction representation of $h(x)$ as
$$
  h(x)=x-\frac{x^2}{1+2x}\minus\frac{2^2x^2}{1+2\cdot 2^2x}
       \minus\frac{2^2\cdot 3^2\cdot x^2}{1+2\cdot 3^2x}\minus\ldots
       \minus\frac{n^2(n+1)^2x^2}{1+2\cdot (n+1)^2x}\minus\ldots.
$$
Hence 
$$
  -\sum^\infty_{n=1}H_{2n+1}x^n=\frac{x}{1+2x}\minus\frac{2^2\cdot x^2}{1+2\cdot 2^2x}\minus\ldots.
$$
Now we apply Equation \ref{eq.56}, we transform the above equation to
$$
  -\sum^\infty_{n=0}H_{2n+3}x^{n+1}=\mathop{\bf K}^\infty_{n=0}\left(\frac{c_nx}1\right),
$$
where $c_0=1$, $c_{2n-1}=c_{2n}=n(n+1)$, for $n\geq 1$.

Applying Lemma \ref{lma.2}, $H_{2n+3}$ is divisible by $2^n$ for $n\geq 1$, and 
\begin{eqnarray}
\label{eq.57}
   \sum^\infty_{n=1}\frac{H_{2n+3}}{2^n}x^n &=& 
   \frac{x}{1+x}\plus\frac x1\plus\frac{3x}1\plus\frac{3x}1\plus\frac{6x}1\plus\frac{6x}1\plus\ldots \\
   &\equiv& \frac{x}{1+x}\plus\frac x1\plus\frac{3x}{1+3x}\qquad\pmod{6}\nonumber\\
   &\equiv& \frac{x}{3x^2+2x+1}\qquad\pmod{6}\nonumber\\
   &\equiv& \frac{x}{3x^2-4x+1}\qquad\pmod{6}\nonumber\\
   &=& \frac{x}{(3x-1)(x-1)}\ =\ \frac 12\cdot \frac 1{1-3x}-\frac 12\cdot \frac 1{1-x} \nonumber\\
   &=& \sum^\infty_{n=0} \left(\frac{3^n-1}2\right)x^n.\nonumber
\end{eqnarray}
Comparing the coefficients of $x^n$, we have
\begin{eqnarray*}
  \frac{H_{2n+3}}{2^n} &\equiv& \frac{3^n-1}2\qquad\pmod{6} \\
  &=& 3^{n-1}+3^{n-2}+\cdots+3+3^0.
\end{eqnarray*}
Since $3^n \equiv 3$ (mod $6$), for $n\geq 1$, we have
\begin{equation}
\label{eq.62}
  \frac{H_{2n+3}}{2^n}\ \equiv\ (n-1)\cdot 3+1\ \equiv\ 3n-2\qquad\pmod{6}.
\end{equation}
If $n=2k-1$, for $k\geq 1$, then 
$$
  \frac{H_{2n+3}}{2^n}\equiv 3(2k-1)-2\equiv 1\quad\pmod{6}.
$$
If $n=2k$, for $k\geq 1$, then
$$
  \frac{H_{2n+3}}{2^n}\equiv 3(2k)-2\equiv 4\quad\pmod{6}.
$$
Hence we complete our proof.
\end{proof}
Using the similar method, we could get the Barsky's result
(ref. Theorem 1 in \cite{Han}): for $n\geq 1$,
\begin{equation}
\label{eq.58}
  \frac{H_{2n+3}}{2^n}\equiv\left\{\begin{array}{ll}
     3\ \pmod{4}, & \mbox{if $n$ is odd;}\\
     2\ \pmod{4}, & \mbox{if $n$ is even.}\end{array}\right.
\end{equation}

The tangent numbers $T_n$ are defined by 
$$
  1+\tanh x=\sum^\infty_{n=0}\frac{T_nx^n}{n!}.
$$
The unsign tangent numbers are the sequence A009006 in \cite{Sloane}.
The tangent numbers $T_n$
are closely related to the Bernoulli numbers:
\begin{equation}
\label{eq.40}
  T_{2n-1}=2^{2n}(2^{2n}-1)B_{2n}/2n.
\end{equation}
\begin{theorem}
\label{thm.4}
For $n\geq 1$, the tangent number $T_{2n+1}$ is divisible by $2^n$ and
\begin{equation}
\label{eq.38}
 \frac{T_{2n+1}}{2^n}\equiv (-1)^n4^{n-1}\qquad \pmod 6.
\end{equation}
\end{theorem}
\begin{proof}
We use the classical continued fraction representation for the ordinary generating function
of the tangent numbers $T_n$ (see Corollary 3.1 in \cite{Dumont})
\begin{eqnarray}
\nonumber
  \sum^\infty_{n=0}T_nx^{n+1} 
  &=& x+x^2-2x^4+16x^6-272x^8+\ldots \\
\label{eq.39}
  &=& x+\frac{x^2}1\plus\frac{2x^2}1\plus\frac{6x^2}1\plus\ldots\plus\frac{n(n+1)x^2}1\plus\ldots.
\end{eqnarray}
Changing the variable $x^2$ as $t$ we have
$$
  t+\sum^\infty_{n=1}T_{2n+1}t^{n+1}=\frac t1\plus\frac{2t}1\plus\frac{6t}1
  \plus\ldots\plus\frac{n(n+1)t}1\plus\ldots.
$$
Applying Lemma \ref{lma.2}, $T_{2n+1}$ is divisible by $2^n$ for $n\geq 1$. And
\begin{eqnarray*}
  \sum^\infty_{n=1}\frac{T_{2n+1}}{2^n}t^{n+1} &=& \frac{-t^2}{1+t}\plus\frac{3t}1\plus\frac{6t}1\plus\ldots \\
  &\equiv & \frac{-t^2}{1+t+3t}\qquad \pmod 6\\
  &=& \sum^\infty_{n=1}(-1)^n4^{n-1}t^{n+1}.
\end{eqnarray*}
Comparing the coefficients of $t^{n+1}$, we complete the proof.
\end{proof}

The result that $T_{2n+1}$ is divisible by $2^n$ is not new. 
Howard (see Theorem 8 in \cite{Howard}) proved in an elementary way that for every 
$n\geq 1$ the number $(2^{n+1}(1-2^{2n})/2n)B_{2n}$ is an integer. That is to say,
$T_{2n-1}$ is divisible by $2^{n-1}$. Ramanujan (see \cite[p.\ 7]{Hardy}) proved 
some similar congruence properties, such as
$$
  \frac{2(2^{4n+2}-1)}{2n+1}B_{4n+2}, \quad\mbox{and}\quad
  \frac{-2(2^{8n+4}-1)}{2n+1}B_{8n+4}
$$
are integers of the form $30k+1$, for $n\geq 0$. And it means that
$T_{4n+1}$, $T_{8n+3}$ are divisible by $2^{4n}$, $2^{8n+1}$, respectively,
and 
$$
  \frac{T_{4n+1}}{2^{4n}} \equiv \frac{-T_{8n+3}}{2^{8n+1}} \equiv 1 \ \pmod{30}.
$$

%============================================================================================

\noindent{\bf Acknowledgements.}
The author would like to thank the referee for some useful comments and suggestions.

%============================================================================================

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D. Dumont,
Further triangles of Seidel-Arnold type and continued fractions related to Euler and Springer numbers,
\textit{Adv. in Appl. Math.}
\textbf{16}, No. 3 (1995),
275--296.

\bibitem{Han}
G.-N. Han, J. Zeng,
On a $q$-sequence that generalizes the median Genocchi numbers,
\textit{Ann. Sci. Math. Qu\'{e}bec}
\textbf{23}, No. 1 (1999),
63--72.

\bibitem{Hardy}
G. H. Hardy, P. V. Seshu Aiyar, and B. M. Wilson,
\textit{Collected Papers of Srinivasa Ramanujan},
Chelsea Pub. Co., 1962.

\bibitem{Hoffman}
M. E. Hoffman,
Derivative polynomials, Euler polynomials, and associated integer sequence,
\textit{Electron. J. Combin.}
\textbf{6} (1999), \#R21, 13 pp.

\bibitem{Howard}
F. T. Howard,
Applications of a recurrence for the Bernoulli Numbers,
\textit{J. Number Theory}
\textbf{52}, No. 1 (1995),
157--172.

\bibitem{Lisa}
L. Lorentzen, H. Waadeland,
\textit{Continued Fractions with Applications},
North-Holland, Netherlands, 1992.

\bibitem{Sloane}
N. J. A. Sloane, editor (2003),
The On-Line Encyclopedia of Integer Sequences, 
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\end{thebibliography}

%
%============================================================================================
%

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
11A55, 11B68.\ \

\noindent \emph{Keywords:}
Continued fractions, Springer numbers, Euler numbers, Genocchi numbers, Tangent numbers.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences A000281, A002832, A005439, A009006, and A036968.)

\end{document}
