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\title{\bf Generalized Alcuin's Sequence}

% input author, affilliation, address and support information as follows;
% the address should include the country, and does not have to include
% the street address 

\author{Daniel Panario, Murat Sahin and Qiang Wang \thanks{The authors 
were supported by NSERC of Canada.}\\
\small School of Mathematics and Statistics\\[-0.8ex]
\small Carleton University\\[-0.8ex] 
\small Ottawa, K1S 5B6, Canada\\
\small\tt \{daniel,msahin,wang\}@math.carleton.ca
}

% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{}{}\\
\small Mathematics Subject Classifications: 11B50, 11P81, 05A15}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
% succinct statement of background followed by a listing of the
% principal new results that are to be found in the paper. The abstract
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\begin{abstract}
  We introduce a new family of sequences $\{t_k(n)\}_{n=-\infty}^{\infty}$
for given positive integer $k \ge 3$. We call these new sequences as
\emph{generalized Alcuin's sequences} because we get Alcuin's sequence which
has several interesting properties when $k=3$. Also, 
$\{t_k(n)\}_{n=0}^{\infty}$ counts the number of partitions of $n-k$
with parts being $k$, $\left(k-1\right)$, $2\left(k-1\right)$,
$3\left(k-1\right)$, $\ldots, \left(k-1\right)\left(k-1\right)$.
We find an explicit linear recurrence equation and
the generating function for $\{t_k(n)\}_{n=-\infty}^{\infty}$.
For the special case $k=4$ and $k=5$, we get a simpler formula for
$\{t_k(n)\}_{n=-\infty}^{\infty}$ and investigate the period of
$\{t_k(n)\}_{n=-\infty}^{\infty}$ modulo a fixed integer. Also, we get
a formula for $p_{5}\left(n\right)$ which is the number of partitions of
$n$ into exactly $5$ parts.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords: Alcuin's sequence, integer partition.} 
\end{abstract}


\section{Introduction}

Alcuin of York (c.~740-804) lived over four hundred years
before Fibonacci. Like Fibonacci, Alcuin has a sequence of
integers named after him. Alcuin's sequence
$\{t_3(n)\}_{n=0}^{\infty}$ has several interesting
properties and it can be defined as the number of incongruent
integer triangles of perimeter $n$. It is also related to
the solutions to the flask-sharing problem (see pages 150 to 165 of \cite{Olivastro})
and to $p_{3}\left(n\right)$, the number of partitions of $n$
into exactly $3$ parts. If $n$ is even then $t_{3}\left(n\right)
=p_{3}\left(\dfrac{n}{2}\right)$, else $t_{3}\left(n\right)
=t_{3}\left(n+3\right)$; see \cite{BindnerErickson}.
So we can write,
 \begin{equation*}
t_{3}\left(n\right)=\left \{ 
\begin{array}{lcl}
p_{3}\left(\dfrac{n}{2}\right), &  
    & \text{if }n\equiv 0\left( \text{mod } 2\right), \\ 
p_{3}\left(\dfrac{n+3}{2}\right), &  
    & \text{if }n\equiv 1\left( \text{mod } 2\right). \\ 
\end{array}
\right.
\end{equation*}

We now introduce a new sequence $\{t_k(n)\}_{n=0}^{\infty}$
for given positive integer $k \geq 3$ that is a generalization of the
Alcuin's sequence:

\begin{equation*}
t_{k}\left(n\right)=p_{k}\left(\dfrac{n+a_{i}}{k-1}\right),
\quad \text{if} \: n\equiv i\left( \text{mod } k-1\right),
\end{equation*}
where if $i=0$ then $a_{i}=0$ else $a_{i} = k\left(k-1-i\right)$
and $p_{k}\left(x\right)$ is the number of partitions of $x$ into
exactly $k$ parts. We call $\{t_k(n)\}_{n=0}^{\infty}$ a {\it generalized Alcuin's 
sequence}. Later on, we extend this definition so that $t_k(n)$ ranges from
$n=-\infty$ to $\infty$. 

In Section~\ref{section:generalization}, 
we find an explicit linear recurrence equation and
the generating function for $\{t_k(n)\}_{n=-\infty}^{\infty}$.
From the generating function, we can see that $t_k(n)$ counts the number of 
partitions of $n-k$ with parts being $k$, $(k-1)$, $2(k-1)$, $\ldots$, 
$(k-1)(k-1)$. 
In Sections~\ref{case:4} and~\ref{case:5}, we obtain simpler
formulae for $\lbrace t_{k}\left(n\right) \rbrace_{n=-\infty}^\infty$ with
$k=4$ and $k=5$, respectively, and using these formulae we study the period
of $\{t_k(n)\}_{n=-\infty}^{\infty}$ modulo a fixed integer for the special
cases $k=4$ and $k=5$. For $k=3$ the Alcuin sequence modulo $m$ has least 
period $12m$ \cite{BindnerErickson}. Motivated by this result we study the 
least period of generalized Alcuin's sequences for bigger $k$. 
For any integer $m\geq 2$, the sequence $\{t_4(n)\left(\text{mod} \: m \right)\}
_{n=-\infty}^\infty$ is periodic with the least period $36m$. Similarly,
for $k=5$, the least period is either $240m$  for odd $m$, or $480m$ for even $m$. 
As a by-product, we also obtain a formula for $p_{5}\left(n\right)$, the number
of partitions of $n$ into exactly $5$ parts. Conclusions and further
work are given in Section~\ref{concl}.

\section{Generalization of Alcuin's Sequence} \label{section:generalization}

\begin{definition}[Generalized Alcuin's Sequence]\label{dfn:GeneralAlcuin}
Let us define the sequence $\{t_k(n)\}_{n=0}^{\infty}$
\begin{equation*}
t_{k}\left(n\right)=p_{k}\left(\dfrac{n+a_{i}}{k-1}\right),
 \quad \text{if} \: \: n\equiv i\left( \text{mod } k-1\right),
\end{equation*}
where if $i=0$ then $a_{i}=0$ else $a_{i} = k\left(k-1-i\right)$
and $p_{k}\left(x\right)$ is the number of partitions of $x$ into
exactly $k$ parts.
\end{definition}

As we will see later, $t_{k}\left(n\right)$ is the number of partitions of $n-k$ such
that the parts are $k, \left(k-1\right),$
$2\left(k-1\right),3\left(k-1\right), \ldots , \left(k-1\right)\left(k-1\right)$.

If we take $k=3$ in Definition~\ref{dfn:GeneralAlcuin}, we get Alcuin's sequence :

\begin{equation*}
t_{3}\left(n\right)=\left \{ 
\begin{array}{lcl}
p_{3}\left(\dfrac{n}{2}\right), &  
    & \text{if }n\equiv 0\left( \text{mod } 2\right), \\ 
p_{3}\left(\dfrac{n+3}{2}\right), &  
    & \text{if }n\equiv 1\left( \text{mod } 2\right). \\ 
\end{array}
\right.
\end{equation*}

Obviously, we can get
\begin{eqnarray}
t_{k}\left(n\left(k-1\right)\right)=p_{k}\left(n\right)
\label{eq1}
\end{eqnarray}%
and, for $1 \leq i \leq k-2$,
\begin{eqnarray}
t_{k}\left(n\left(k-1\right)+i\right) &=& p_{k}\left(\dfrac{n\left(k-1\right)+i
+a_{i}}{k-1}\right)
=p_{k}\left(\dfrac{nk-n+i+k^2-k-ki}{k-1}\right) \notag \\
&=&p_{k}\left(\dfrac{\left(n+k-i\right)\left(k-1\right)}{k-1}\right)
=p_{k}\left(n+k-i\right).
\label{eq2}
\end{eqnarray}%
We use these facts in the following.

\begin{theorem}
\label{GeneralGeneratingFunction} The generating function of the sequence
$\{t_k(n)\}_{n=0}^{\infty}$ is
\begin{equation*}
F_{k}\left( x\right) =\frac{x^{k}}{\left(1-x^k\right)
\prod_{i=1}^{k-1} \left(1-x^{i\left(k-1\right)} \right)}.
\end{equation*}
\end{theorem}

\begin{proof}
The generating function for 
$\{t_k(n)\}_{n=0}^{\infty}$ is 
\begin{equation*}
F_{k}\left(x\right) = \sum_{n=0}^{\infty }t_{k}\left(n\right)x^{n}.
\end{equation*}
Then we can write, using Equations ~(\ref{eq1}) and (\ref{eq2}),
\begin{eqnarray}
F_{k}\left(x\right)  &=&\sum_{n=0}^{\infty}
\sum_{i=0}^{k-2}t_{k}\left(n\left(k-1\right)+i\right)x^{n\left(k-1\right)+i} \notag \\
&=&\sum_{n=0}^{\infty}\left(p_{k}\left(n\right)x^{n\left(k-1\right)}+
\sum_{i=1}^{k-2}p_{k}\left(n+k-i\right)x^{n\left(k-1\right)+i} \right).
\label{gfeq1}
\end{eqnarray}
Formulas for the generating functions of the sequences $\{p_k(n)\}$ are well known
(see \cite{Biggs}, for example):
\begin{eqnarray*}
G_{k}\left(x\right)=\sum_{n=0}^{\infty}p_{k}\left(n\right)x^{n}
=\dfrac{x^{k}}{\prod_{i=1}^{k} \left(1-x^{i} \right)}.
\end{eqnarray*}%
Then, for $0 \le j \le k-1$, we derive
\begin{eqnarray*}
G_{k}^j\left(x\right)  &=&\sum_{n=0}^{\infty}p_{k}\left(n+j\right)x^{n}
=\dfrac{x^{k-j}}{\prod_{i=1}^{k} \left(1-x^{i} \right)}.
\end{eqnarray*}%
Substituting in Equation~(\ref{gfeq1}), we get
\begin{eqnarray*}
F_{k}\left( x\right) &= &G_{k}^{0}\left(x^{k-1}\right)+\sum_{i=1}^{k-2}x^{i}G_{k}^{k-i}\left(x^{k-1}\right) \\
& = & \frac{x^{k}\left(1+x^{k}+x^{2k}+\cdots+x^{\left( k-2 \right)k}\right)}
{\prod_{i=1}^{k} \left(1-x^{i\left(k-1\right)} \right)}.
\end{eqnarray*}%
The proof is complete using the fact that $$ 1-x^{k\left(k-1\right)}
=\left(1-x^{k}\right)\left(1+x^{k}+x^{2k}+\cdots+x^{\left( k-2 \right)k}\right).$$
\end{proof}

It is now clear that $t_{k}\left(n\right)$ counts the number of partitions of $n-k$
into parts $k, \left(k-1\right),$ $2\left(k-1\right),3\left(k-1\right),
\ldots , \left(k-1\right)\left(k-1\right)$.

Let $p\left(x\right)=\left(1-x^k\right)\prod_{i=1}^{k-1}
\left(1-x^{i\left(k-1\right)} \right)$ and $c_{i}$ be the
coefficient of the term $x^{i}$ in the expansion of $p\left(x\right)$.
Let $\ell$ be the degree of $p\left(x\right)$; then $\ell=\dfrac{k\left
(k^{2}-2k+3\right)}{2}$.

\begin{theorem}
\label{GeneralRecurrence} The sequence $\{t_k(n)\}_{n=0}^{\infty}$
is determined by the linear recurrence relation of order $\ell$
$$ t_{k}\left(n\right)=-\sum_{i=1}^{\ell}c_{i}t_{k}\left(n-i\right),$$
for $n \geq \ell$, where $c_{i}$'s are the coefficients of $p\left(x\right)$ defined
above, and the initial values of $t_{k}\left(n\right)$ for $0 \le n \le \ell-1$ are
obtained from Definition~\ref{dfn:GeneralAlcuin}.
\end{theorem}

\begin{proof}
By the generating function of the sequence
$\{t_k(n)\}_{n=0}^{\infty}$, we have
\begin{eqnarray*}
 x^{k}  &=& p\left(x\right)\sum_{n=0}^{\infty} t_{k}\left(n\right)x^{n} \\
 &=& \sum_{n=0}^{\infty} t_{k}\left(n\right)
 \left(x^{n}+c_1x^{n+1}+c_2x^{n+2}+\cdots+c_{\ell}x^{n+\ell}\right).
\end{eqnarray*}%
The linear recurrence relation of order $\ell$ can be read off
by equating coefficients of $x^{n}$ for $n \ge \ell$.
\end{proof}

Let us use the same recurrence relation in Theorem~\ref{GeneralRecurrence}
to define $t_k(n)$ for $n<0$, so this extends the ranges from $n=-\infty ~
to ~\infty$.

\begin{corollary}
If we take $k=3$ in the sequence $\{t_k(n)\}_{n=-\infty}^{\infty}$
we get as special case the Alcuin's sequence in \cite{BindnerErickson}.
Also, substituting $k=3$ in Theorem~\ref{GeneralGeneratingFunction}
and \ref{GeneralRecurrence}, we get Theorem 2 and Theorem 3 in \cite{BindnerErickson},
respectively.

\end{corollary}

\section{The case $k=4$} \label{case:4}

Let us take $k=4$ in the generalized Alcuin's sequences for
$\{t_k(n)\}_{n=0}^{\infty}$ given in Definition~\ref{dfn:GeneralAlcuin}.
We obtain the sequence $\{t_4(n)\}_{n=0}^{\infty}$
\begin{equation*}
t_{4}\left(n\right)=\left \{ 
\begin{array}{lcl}
p_{4}\left(\dfrac{n}{3}\right), &  
    & \text{if }n\equiv 0\left( \text{mod } 3\right), \\ 
p_{4}\left(\dfrac{n+8}{3}\right), &  
    & \text{if }n\equiv 1\left( \text{mod } 3\right), \\ 
p_{4}\left(\dfrac{n+4}{3}\right), &  
    & \text{if }n\equiv 2\left( \text{mod } 3\right). \\ 
\end{array}
\right.
\end{equation*}

If we substitute $k=4$ in Theorem~\ref{GeneralGeneratingFunction} and Theorem~\ref{GeneralRecurrence},
we get the generating function
\begin{equation} \label{GenaratingFunction:t4}
F_{4}\left( x\right) =\frac{x^{4}}{\left(1-x^4\right)
\left(1-x^3\right)\left(1-x^6\right)\left(1-x^9\right)}
\end{equation}
and the linear recurrence relation of order $22$
\begin{eqnarray} \label{Recurrence:t4}
t_{4}\left( n\right) &=& t_{4}\left( n-3\right)+t_{4}\left( n-4\right)+t_{4}\left( n-6\right)
-t_{4}\left( n-7\right)-t_{4}\left( n-10\right)-t_{4}\left( n-12\right)\\
& & -t_{4}\left( n-15\right)+t_{4}\left( n-16\right)+t_{4}\left( n-18\right)
+t_{4}\left( n-19\right)-t_{4}\left( n-22\right)  \notag 
\end{eqnarray}
for $n \geq 22$, together with the initial values $0, 0, 0, 0, 1, 0, 0,$ $1, 1, 0, 2,
1, 1, 3, 2, 1, 5, 3, 2, 6, 5, 3$ of $ t_{4}\left(n\right)$ for $0 \le n \le 21$.

\begin{example}
From the generating function $F_{4}\left( x\right)$, $t_{4}\left( n\right)$
is the number of partitions of $n-4$ with parts $3,4,6 \: \text{and} \: \: 9$.
For example, $t_{4}\left( 16\right)=5$ and $12$ can be partitioned in five distinct ways:
$$ 3+3+3+3,\quad 4+4+4, \quad 3+3+6, \quad 6+6 \quad \text{and} \quad 3+9 .$$

\end{example}

\begin{theorem} \label{formula:p4}
Let $\Vert x \Vert $ be the nearest integer to $x$. Then for all $n \geq 0$,
\begin{equation*}
p_{4}\left(n\right)=\left \{ 
\begin{array}{lcl}
\left\Vert \dfrac{\left(n+1\right)^{3}}{144}-\dfrac{\left(n+1\right)}{48} \right\Vert, &  
    & \text{if } \: n \: \text{is even}, \\ 
    & & \\
\left\Vert \dfrac{\left(n+1\right)^{3}}{144}-\dfrac{\left(n+1\right)}{12} \right\Vert,&  
    & \text{if } \: n \: \text{is odd}. \\ 
\end{array}
\right.
\end{equation*}
\end{theorem}

\begin{proof}
The generating function of the numbers $p_{4}\left(n\right)$ is
\begin{equation*}
G_{4}\left( x\right) =\frac{x^{4}}{\left(1-x\right)
\left(1-x^2\right)\left(1-x^3\right)\left(1-x^4\right)}.
\end{equation*}
Now, by  partial fractions we obtain
\begin{eqnarray} \label{partialfraction:4}
G_{4}\left( x\right) & = & -\dfrac{13}{288\left(x-1\right)^2}
+\dfrac{1}{24\left(x-1\right)^3}+\dfrac{1}{32\left(x+1\right)^2}
+\dfrac{1}{24\left(x-1\right)^4}\\
& &+\dfrac{1}{8\left(x^2+1\right)}-\dfrac{1}{9\left(x^2+x+1\right)}. \notag 
\end{eqnarray}
Using the general binomial theorem (see \cite{Biggs}, for example),
the first four terms of $G_{4}\left( x\right)$
can be written as
% 
% \begin{eqnarray*}
% & & 
$$ -\dfrac{13}{288}\sum_{n=0}^{\infty}
\left(n+1\right)x^n
-\dfrac{1}{24}\sum_{n=0}^{\infty}\binom{n+2}{n}x^n
+\dfrac{1}{32}\sum_{n=0}^{\infty}\left(n+1\right)\left(-1\right)^{n}x^n
+\dfrac{1}{24}\sum_{n=0}^{\infty}\binom{n+3}{n}x^n. $$
% \end{eqnarray*}
Thus the first four terms give the following coefficient
of $x^n$ :
\begin{equation*}
\dfrac{2n^3+6n^2-9n+9n\left(-1\right)^n
+9\left(-1\right)^n-13}{288}.
\end{equation*}

The last two terms in Equation~(\ref{partialfraction:4})
can be handled by Maple using Taylor expansions and their
contribution to the coefficient of $x^n$ is $c/72$, where
$c$ is given in the following table.
$$
\begin{tabular}{c|cccccccccccc}
\rule[-1ex]{0pt}{2.5ex} $n$  \text{mod} 12 & 0 & 1 & 2 & 3 & 4 & 5
& 6 & 7 & 8 & 9 & 10 & 11 \\ 
\hline 
\rule[-1ex]{0pt}{2.5ex} $c$ & 1 & 8 & -9 & -8 & 17 & 0 & -17 & 8 & 9
& -8 & -1 & 0 \\ 
\end{tabular} 
$$
Combining all the contributions, we obtain the formula $p_{4}\left(n\right)=
\dfrac{n^3}{144}+\dfrac{n^2}{48}
-\dfrac{c-1}{72}$, for $n$ even, and $p_{4}\left(n\right)=
\dfrac{n^3}{144}+\dfrac{n^2}{48}-\dfrac{n}{16}
+\dfrac{2c-11}{144}$, for $n$ odd.
This can be represented as in the statement of the theorem.
\end{proof}

\begin{remark}
A search in internet \cite{internet} gives the previous
formula for $p_{4}\left(n\right)$, but we could not find
any proof of it. For the sake of completeness we give such
a proof in the previous theorem using generating functions.
\end{remark}

Plugging the formula for $p_{4}\left(n\right)$ into the definition
of $t_{4}\left(n\right)$ we obtain the following corollary.


\begin{corollary} \label{corollary:t4}
Let $C=1/3888$ and $\Vert x \Vert $ be the nearest integer to $x$. 
Then for all $n \geq 0$,
\begin{equation*}
t_{4}\left(n\right)=\left \{ 
\begin{array}{lcl}
\Vert C \left(n^3+9n^2-54\right) \Vert, &  
    & \text{if }n\equiv 0 \: \left( \text{mod } 6\right), \\ 
\Vert C \left(n^3+33n^2+255n+143\right) \Vert, & 
    & \text{if }n\equiv 1 \: \left( \text{mod } 6\right), \\
\Vert C \left(n^3+21n^2+120n+154\right) \Vert, & 
    & \text{if }n\equiv 2 \: \left( \text{mod } 6\right), \\ 
\Vert C \left(n^3+9n^2-81n-297\right) \Vert, & 
    & \text{if }n\equiv 3 \: \left( \text{mod } 6\right), \\ 
\Vert C \left(n^3+33n^2+336n+1034\right) \Vert, & 
    & \text{if }n\equiv 4 \: \left( \text{mod } 6\right), \\ 
\Vert C \left(n^3+21n^2+39n-413\right) \Vert, & 
    & \text{if }n\equiv 5 \: \left( \text{mod } 6\right). \\ 

\end{array}
\right.
\end{equation*}
\end{corollary}

Let us use the \emph{same} recurrence relation in (\ref{Recurrence:t4}) to define
$\{t_4(n)\}_{n=-\infty}^{\infty}$ for $ n <0$.
Then we get
\begin{eqnarray} \label{Recurrencenegative:t4}
t_{4}\left( n\right) &=& t_{4}\left( n+3\right)+t_{4}\left( n+4\right)+t_{4}\left( n+6\right)
-t_{4}\left( n+7\right)-t_{4}\left( n+10\right)-t_{4}\left( n+12\right)\\
& & -t_{4}\left( n+15\right)+t_{4}\left( n+16\right)+t_{4}\left( n+18\right)
+t_{4}\left( n+19\right)-t_{4}\left( n+22\right)  \notag 
\end{eqnarray}
together with the initial values of $ t_{4}\left(n\right)$ for $0 \le n \le 21$. Recall these initial
values are $0, 0, 0, 0, 1, 0, 0,$ $1, 1, 0, 2, 1, 1, 3, 2, 1, 5, 3, 2, 6, 5, 3.$ From this equation
we see that recurrence relation~(\ref{Recurrence:t4}) also holds for any integer $n$.

By using Equation~(\ref{Recurrencenegative:t4}) we can
construct the following table of values of $ t_{4}
\left(n\right)$ when $-22 \leq n \leq 8$:
$$
\begin{tabular}{c|ccccccccccccccccc}
\rule[-1ex]{0pt}{2.5ex} $n$ &-22 &-21 & -20 & -19 & -18 & -17 & -16 & -15 
& \ldots & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8  \\ 
\hline 
\rule[-1ex]{0pt}{2.5ex} $t_{4}\left(n\right)$ &-1 &-1 & 0 & 0 & -1 & 0 & 0& 0 & \ldots & 0 & 0
& 0 & 1 & 0 & 0& 1& 1\\ 
\end{tabular} 
$$
Obviously, the sequence is palindromic from the table
if we ignore signs; we can easily see that
\begin{eqnarray} \label{negative:t4}
t_{4}\left(-n\right)=-t_{4}\left(n-14\right).
\end{eqnarray}
Note that $\ell-2k =22-8=14$ for the sequence
$\{t_4(n)\}_{n=-\infty}^{\infty}$, where $k=4$ and $\ell=22$.
\begin{theorem}
For any integer $m \geq 2$, the sequence
$\{t_4(n) \left(\text{mod} \: m \right)\}_{n=-\infty}^{\infty}$ is
periodic with least period $36m$.
\end{theorem}

\begin{proof}
Let $L$ be the least period of $\{t_4(n)\left(\text{mod} \: m \right)\}
_{n=-\infty}^{\infty}$. First we prove
that $36m$ is a period of the sequence and thus
$L$ divides $36m$. That is, for each $m \ge 2$,
\begin{eqnarray} \label{periodequation:t4}
t_{4}\left(n+36m\right) \equiv t_{4}\left(n\right) \: \left(\text{mod} \: m\right)
\quad \text{for any integer} \: n.
\end{eqnarray}
In order to show this, we break it into 5 cases.
\begin{enumerate}% [(i)]
\item[(1)] $n \ge 0$ :
For $n \equiv 0 \: \left(\text{mod} \: 6\right)$, and by Corollary \ref{corollary:t4}, 
\begin{eqnarray*}
t_{4}\left(n+36m\right) & = & \left\Vert \dfrac{n^3}{3888}+\dfrac{n^2}{432}
-\dfrac{1}{72}\right\Vert+\dfrac{n^2}{36}m+\dfrac{n}{6}m+nm^2+12m^3+3m^2 \\
% \quad (\text{by Corollary} \: \ref{corollary:t4})\\
&  & \\
& \equiv & t_{4}\left(n\right)\: \left(\text{mod} \: m\right).
\end{eqnarray*}
In a similar way one shows that, Equation~(\ref{periodequation:t4}) is also 
satisfied for $n \equiv 1,2,3,4,5$  $\left(\text{mod} \: 6\right)$.

\item[(2)] $-14 < n < 0$ :
Equation~(\ref{periodequation:t4}) is also satisfied
for $n=-1,-2,-3,\ldots,-13$ by using Corollary
\ref{corollary:t4} and Equation~(\ref{negative:t4}).

\item[(3)] $-36m < n \leq -14$ : For $n \equiv 0 \: \left(\text{mod} \: 6\right)$,
and using again Corollary \ref{corollary:t4} and Equation~(\ref{negative:t4}),
\begin{eqnarray*}
t_{4}\left(n+36m\right) & = & -\left\Vert -\dfrac{n^3}{3888}-\dfrac{n^2}{432}
+\dfrac{1}{72}\right\Vert+\dfrac{n^2}{36}m+\dfrac{n}{6}m+nm^2+12m^3+3m^2 \\
% \quad (\text{by Corollary} \: \ref{corollary:t4})\\
&  & \\
& \equiv & -t_{4}\left(-n-14\right)\:  \left(\text{mod} \: m\right) \\
% \quad \left(\text{by} \: \text{Corollary} \: \ref{corollary:t4}\right) \\
& \equiv & t_{4}\left(n\right)\:  \left(\text{mod} \: m\right).
% \quad (\text{by Equation} \: (\ref{negative:t4})).
\end{eqnarray*}
Again, in a similar way, Equation~(\ref{periodequation:t4}) is also satisfied for
$n \equiv 1,2,3,4,5 \: \left(\text{mod} \: 6\right)$.

\item[(4)] $-36m-14 < n \leq -36m$ :
Since $n=-36m-a$ for $0 \leq a \leq 13$, we have
\begin{eqnarray*}
t_{4}\left(n+36m\right) & = & t_{4}\left(-36m-a+36m\right)
 =  t_{4}\left(-a\right) 
 \equiv  0\:  \left(\text{mod} \: m\right)
\end{eqnarray*}
and
\begin{eqnarray*}
t_{4}\left(n\right) & = & t_{4}\left(-36m-a\right)
 =  t_{4}\left(36m+a-14\right)
\quad (\text{by Equation} \: (\ref{negative:t4}))\\
& \equiv & 0\:  \left(\text{mod} \: m\right),
\end{eqnarray*}
so that $t_{4}\left(n+36m\right) \equiv t_{4}\left(n\right)
\: \left(\text{mod} \: m\right)$
for $-36m-14 < n \leq -36m$.

\item[(5)] $n \leq -36m-14$ : For $n \equiv 0 \: \left(\text{mod} \: 6\right)$,
and using Corollary \ref{corollary:t4} and Equation~(\ref{negative:t4}),
\begin{eqnarray*}
t_{4}\left(n+36m\right) & = & -t_{4}\left(-n-36m-14\right) \\
% \quad (\text{by Equation} \: (\ref{negative:t4})) \\ 
& = & -\left\Vert -\dfrac{n^3}{3888}-\dfrac{n^2}{432}
+\dfrac{1}{72}\right\Vert+\dfrac{n^2}{36}m+\dfrac{n}{6}m+nm^2+12m^3+3m^2 \\
% \quad (\text{by Corollary} \: \ref{corollary:t4})\\
& \equiv & -t_{4}\left(-n-14\right)\:  \left(\text{mod} \: m\right) \\
% \quad (\text{by} \: \text{Corollary} \: \ref{corollary:t4}) \\
& \equiv & t_{4}\left(n\right)\:  \left(\text{mod} \: m\right).
% \quad (\text{by Equation} \: (\ref{negative:t4})).
\end{eqnarray*}
As before, Equation~(\ref{periodequation:t4}) is also satisfied for
$n \equiv 1,2,3,4,5 \: \left(\text{mod}\: 6\right)$.

\end{enumerate}

Hence, Equation~(\ref{periodequation:t4}) is satisfied for
any integer $n$. Therefore, $36m$ is a period of the sequence and $L$ divides $36m$.

Secondly, we prove $L=36m$. This can be verified by a computer program
for small $m \leq 10$. Thus, here we prove $L=36m$ for only $m>10$. Let $\lambda=L,L-1$
or $L-2$ such that $\lambda \equiv 0 \: \left(\text{mod} \: 3 \right)$. First of all,
we assume that $\lambda$ is even. Since $t_{4}\left(-2\right)
=t_{4}\left(-1\right)=t_{4}\left(0\right)
=t_{4}\left(1\right)=t_{4}\left(2\right)=t_{4}\left(3\right)=0$, we have
$$t_{4}\left(\lambda\right)\equiv t_{4}\left(\lambda+1\right)
\equiv t_{4}\left(\lambda+2\right) \equiv t_{4}\left(\lambda+3\right)
\equiv 0 \: \left(\text{mod} \: m \right).$$
Hence, $m | M$ where $M=\left[t_{4}\left(\lambda+1\right)-t_{4}\left(\lambda+2\right)\right]
+\left[t_{4}\left(\lambda\right)-t_{4}\left(\lambda+3\right)\right]$.

We observe that $\{t_4(n) \left(\text{mod} \: m \right) \}_{n=-\infty}^{\infty}$
is an aperiodic pattern for $ -188 \le n \le 174$ with the aid of a computer program
for $m > 1495$. In addition to this, we can also show using a computer that $L \geq 362$
for $10 < m \leq 1495$. Therefore, we have $L \geq 362$ which means that $\lambda \geq 360 $.

By the definition of nearest integer function, we have
\begin{eqnarray}\label{property1}
\left\Vert x\mp y \right\Vert & = & \left\Vert x \right\Vert \mp \left\Vert y \right\Vert + \alpha,
\quad  \alpha = 0,1 \:\text{or}\: -1
\end{eqnarray}
and
\begin{eqnarray}\label{property2}
\left\Vert x \right\Vert + \left\Vert y \right\Vert & < & x+y+1
\end{eqnarray} for any $x$ and $y$.
From (\ref{property1}), we get
\begin{eqnarray*}
t_{4}\left(\lambda\right)+ t_{4}\left(\lambda+1\right)& = & \left\Vert \dfrac{2\lambda^3}{3888}+\dfrac{5\lambda^2}{432}+\dfrac{\lambda}{12}+\dfrac{7}{72}\right\Vert + \alpha_1, \quad -1 \leq \alpha_1 \leq 1 \notag \\ 
& = & \left\Vert\dfrac{2\lambda^3}{3888}\right\Vert + \left\Vert\dfrac{5\lambda^2}{432}\right\Vert 
+ \left\Vert\dfrac{\lambda}{12}\right\Vert
 + \left\Vert\dfrac{7}{72}\right\Vert + \alpha_2, \quad -4 \leq \alpha_2 \leq 4
\end{eqnarray*}
and
\begin{eqnarray*}
t_{4}\left(\lambda+2\right)+ t_{4}\left(\lambda+3\right)& = & \left\Vert \dfrac{2\lambda^3}{3888}+\dfrac{5\lambda^2}{432}+\dfrac{\lambda}{18}+\dfrac{1}{72}\right\Vert + \beta_1, \quad -1 \leq \beta_1 \leq 1 \notag \\ 
& = & \left\Vert\dfrac{2\lambda^3}{3888}\right\Vert + \left\Vert\dfrac{5\lambda^2}{432}\right\Vert 
+ \left\Vert\dfrac{\lambda}{18}\right\Vert
 + \left\Vert\dfrac{1}{72}\right\Vert + \beta_2, \quad -4 \leq \beta_2 \leq 4.
\end{eqnarray*}
Hence,
\begin{eqnarray*}
M & = &\left\Vert\dfrac{\lambda}{12}\right\Vert-\left\Vert\dfrac{\lambda}{18}\right\Vert+ \gamma_1,
\quad -8 \leq \gamma_1 \leq 8 \notag \\ 
& = & \left\Vert\dfrac{\lambda}{36}\right\Vert+\gamma_2, \quad -9 \leq \gamma_2 \leq 9
\end{eqnarray*}
Since $ \lambda \geq 360$, we have $M>1$.

Moreover, from (\ref{property2}), we have
\begin{eqnarray*}
t_{4}\left(\lambda+1\right)- t_{4}\left(\lambda+2\right)
<  \dfrac{\lambda^2}{432}+\dfrac{\lambda}{36}-\dfrac{1}{72}+1
\end{eqnarray*}
and
\begin{eqnarray*}
t_{4}\left(\lambda\right)-t_{4}\left(\lambda+3\right)
 <  -\dfrac{\lambda^2}{432}+\dfrac{7}{72}+1,
\end{eqnarray*}
so we get $ M < \dfrac{\lambda}{36}+\dfrac{6}{72}+2=\dfrac{\lambda+75}{36}$.
This means that $ m < \dfrac{\lambda+75}{36}$ since $m | M$ and $M \neq 0$,
and it follows that
\begin{equation} \label{ineq3}
36m < \lambda+75 \le L+75.
\end{equation}
Because $L$ is a divisor of $36m$, we conclude by inequality~(\ref{ineq3})
that $L=36m$ for any $m > 10$. Hence, $L=36m$ for $\lambda$ even. Similarly we can
also show $L=36m$ for $\lambda$ odd, and the proof is complete.

\end{proof}

\section{The Case $k=5$}\label{case:5}
Let us take $k=5$ in the generalized Alcuin's sequences for
$\{t_k(n)\}_{n=0}^{\infty}$ given in Definition~\ref{dfn:GeneralAlcuin}.
We obtain the sequence $\{t_5(n)\}_{n=0}^{\infty}$
\begin{equation*}
t_{5}\left(n\right)=\left \{ 
\begin{array}{lcl}
p_{5}\left(\dfrac{n}{4}\right), &  
    & \text{if }n\equiv 0\left( \text{mod } 4\right), \\ 
p_{5}\left(\dfrac{n+15}{4}\right), &  
    & \text{if }n\equiv 1\left( \text{mod } 4\right), \\ 
p_{5}\left(\dfrac{n+10}{4}\right), &  
    & \text{if }n\equiv 2\left( \text{mod } 4\right). \\ 
p_{5}\left(\dfrac{n+5}{4}\right), &  
    & \text{if }n\equiv 3\left( \text{mod } 4\right). \\ 
\end{array}
\right.
\end{equation*}

If we substitute $k=5$ in Theorem~\ref{GeneralGeneratingFunction} and
Theorem~\ref{GeneralRecurrence}, we get the generating function
\begin{equation} \label{GenaratingFunction:t5}
F_{5}\left( x\right) =\frac{x^{5}}{\left(1-x^5\right)
\left(1-x^4\right)\left(1-x^8\right)\left(1-x^{12}\right)\left(1-x^{16}\right)}
\end{equation}
and the linear recurrence relation of order $45$
\begin{eqnarray} \label{Recurrence:t5}
t_{5}\left( n\right) &=& t_{5}\left( n-4\right)+t_{5}\left( n-5\right)
+t_{5}\left( n-8\right)-t_{5}\left( n-9\right)-t_{5}\left( n-13\right)\\
& & -2t_{5}\left( n-20\right)+2t_{5}\left( n-25\right)+t_{5}\left( n-32\right)
+t_{5}\left( n-36\right)-t_{5}\left( n-37\right) \notag \\
& &  -t_{5}\left( n-40\right)-t_{5}\left( n-41\right)+t_{5}\left( n-45\right) \notag
\end{eqnarray}
for $n \geq 45$, together with the values of $ t_{5}\left(n\right)$ for $0 \le n \le 44$
that can be easily computed.

\begin{theorem}
Let $C=1/86400$. For all $n \geq 0$, we have
\begin{equation*}
p_{5}\left(n\right)=\left \{ 
\begin{array}{lcl}
\Vert C \left(30n^{4}+300n^{3}+300n^{2}-3600n-5224\right) \Vert, &  
    & \text{if } \: n \: \text{is even}, \\ 
\Vert C \left(30n^{4}+300n^{3}+300n^{2}-900n+1526\right) \Vert,&  
    & \text{if } \: n \: \text{is odd}. \\ 
\end{array}
\right.
\end{equation*}
\end{theorem}
\begin{proof}
The generating function of $p_{5}\left(n\right)$ is
\begin{equation*}
G_{5}\left( x\right) =\frac{x^{5}}{\left(1-x\right)
\left(1-x^2\right)\left(1-x^3\right)\left(1-x^4\right)
\left(1-x^5\right)}.
\end{equation*}
Now, by  partial fractions we obtain that $G_{5}\left( x\right)$ is 
\begin{eqnarray} \label{partialfraction:5}
% G_{5}\left( x\right) & = & 
& & -\dfrac{1}{64\left(x+1\right)^2}
    -\dfrac{3}{128\left(x+1\right)}-\dfrac{5}{288\left(x-1\right)^2}
    -\dfrac{431}{86400\left(x-1\right)}-\dfrac{1}{120\left(x-1\right)^5}\\
& & +\dfrac{5}{288\left(x-1\right)^3}-\dfrac{2x+1}{27\left(x^2+x+1\right)}
+\dfrac{x^3+2x^2+3x+4}{25\left(x^4+x^3+x^2+x+1\right)}
+\dfrac{x-1}{16\left(x^2+1\right)}. \notag 
\end{eqnarray}
Using the general binomial theorem, the first six terms of $G_{5}\left( x\right)$
can be written as

\begin{eqnarray*}
& & -\dfrac{1}{64}\sum_{n=0}^{\infty}
\left(n+1\right)\left(-1\right)^{n}x^n
-\dfrac{3}{128}\sum_{n=0}^{\infty}\left(-1\right)^{n}x^n
-\dfrac{5}{288}\sum_{n=0}^{\infty}\left(n+1\right)x^n
-\dfrac{431}{86400}\sum_{n=0}^{\infty}x^n \\
& & +\dfrac{1}{120}\sum_{n=0}^{\infty}\binom{n+4}{n}x^n
-\dfrac{5}{288}\sum_{n=0}^{\infty}\binom{n+2}{n}x^n.
\end{eqnarray*}
Thus the first six terms give the following coefficient
of $x^n$:
\begin{equation*}
\dfrac{30n^4+300n^3+300n^2-2250n-1350n\left(-1\right)^n
-3375\left(-1\right)^n-1849}{86400}.
\end{equation*}

As in the case of Theorem~\ref{formula:p4}, the last three terms in
Equation~(\ref{partialfraction:5}) can be handled by Maple using Taylor
expansions and their contribution to the coefficient of $x^n$ is $c/10800$, where
$c$ follows now a pattern modulo $60$ by taking the values 
$c=653,-157,1043,-1507$, $3203,-307,2003$ and $1853$.
Combining all contributions, we get the desired result.
\end{proof}

Plugging the formula for $p_{5}\left(n\right)$ into the definition
of $t_{5}\left(n\right)$ we obtain the following corollary.

\begin{corollary} \label{corollary:t5}
Let $C=1/11059200$. For all $n \geq 0$,
\begin{equation*}
t_{5}\left(n\right)=\left \{ 
\begin{array}{lcl}

\Vert C\left(15n^4+600n^3+2400n^2-115200n-668672\right) \Vert, &  
    & \text{if }n\equiv 0 \: \left( \text{mod } 8\right), \\ 
\Vert C\left(15n^4+1500n^3+49650n^2+564300n+927703\right) \Vert, & 
    & \text{if }n\equiv 1 \: \left( \text{mod } 8\right), \\
\Vert C\left(15n^4+1200n^3+29400n^2+259200n+897328\right) \Vert, & 
    & \text{if }n\equiv 2 \: \left( \text{mod } 8\right), \\ 
\Vert C\left(15n^4+900n^3+13650n^2-38700n-1100297\right) \Vert, & 
    & \text{if }n\equiv 3 \: \left( \text{mod } 8\right), \\ 
\Vert C\left(15n^4+600n^3+2400n^2-28800n+195328\right) \Vert, & 
    & \text{if }n\equiv 4 \: \left( \text{mod } 8\right), \\ 
\Vert C\left(15n^4+1500n^3+49650n^2+650700n+3087703\right) \Vert, & 
    & \text{if }n\equiv 5 \: \left( \text{mod } 8\right), \\ 
\Vert C\left(15n^4+1200n^3+29400n^2+172800n-830672\right) \Vert, &  
    & \text{if }n\equiv 6 \: \left( \text{mod } 8\right), \\ 
\Vert C\left(15n^4+900n^3+13650n^2+47700n+195703\right) \Vert, &  
    & \text{if }n\equiv 7 \: \left( \text{mod } 8\right). \\ 
\end{array}
\right.
\end{equation*}
\end{corollary}

Similar to Section~\ref{case:4}, we use the \emph{same} recurrence
relation in (\ref{Recurrence:t5}) to define $\{t_5(n)\}_{n=-\infty}^{\infty}$
for $ n <0$, obtaining
\begin{eqnarray} \label{negative:t5}
t_{5}\left(-n\right)=t_{5}\left(n-35\right).
\end{eqnarray}
We note that $\ell-2k=45-10=35$ for the sequence $\{t_5(n)\}_{n=-\infty}^{\infty}$
where $k=5$ and $\ell=45$.

\begin{lemma} \label{lemma}
For any integer $n$,
$4n^3 \pm 15n^2+5n \equiv 0 \: \left(\text{mod} \:\: 6\right)$
and $4n^3+5n \equiv 0 \: \left(\text{mod} \:\: 3\right)$.
\end{lemma}
\begin{proof}
Since $$4n^3 \pm 15n^2+5n \equiv n^2-n \equiv n\left(n-1\right) 
\equiv 0 \: \left(\text{mod} \: 2\right)$$ and
$$4n^3 \pm 15n^2+5n \equiv n^3-n \equiv n\left(n-1\right)\left(n-2\right)
\equiv 0 \: \left(\text{mod} \: 3\right),$$ we get
$$4n^3 \pm 15n^2+5n \equiv 0 \: \left(\text{mod} \: 6\right).$$ Similarly,
$$4n^3+5n \equiv n^3-n \equiv n\left(n-1\right)\left(n-2\right)
\equiv 0 \: \left(\text{mod} \: 3\right).$$
\end{proof}

\begin{theorem}
For any odd integer $m \geq 3$, the sequence
$\{t_5(n) \left(\text{mod} \: m\right)\}_{n=-\infty}^{\infty}$ is
periodic with least period $240m$. For any even integer $m \geq 2$, 
the sequence $\{t_5(n) \left(\text{mod} \: m\right)\}_{n=-\infty}^{\infty}$
is periodic with least period $480m$.
\end{theorem}

\begin{proof}
Let $L$ be the least period of $\{t_5(n) \left(\text{mod} \: m\right)\}_{n=-\infty}^{\infty}$.
First we prove that if $m$ is odd then $240m$ is a period of the sequence
else $480m$ is a period of the sequence so that if $m$ is odd then
$L$ divides $240m$ else $L$ divides $480m$. That is for each $m \ge 2$
and any integer $n$, if $m$ is odd then
\begin{eqnarray} \label{periodequation1:t5}
t_{5}\left(n+240m\right) \equiv t_{5}\left(n\right) \: \left(\text{mod} \: m\right),
\end{eqnarray}
and else
\begin{eqnarray} \label{periodequation2:t5}
t_{5}\left(n+480m\right) \equiv t_{5}\left(n\right) \: \left(\text{mod} \: m\right).
\end{eqnarray}
In order to prove this, as before, we break it into 5 cases. For simplicity,
we only show the case $m$ is odd; the even case is analogous.
\begin{enumerate} % [(i)]
\item[(1)] $n \ge 0$ : For $n \equiv 0 \: \left(\text{mod}\: 8\right)$,
using Corollary \ref{corollary:t5} and $m$ odd, we have
$$\begin{array}{rll}
   && t_{5}\left(n+240m\right) \\
& = & \left\Vert \dfrac{n^4}{737280}+\dfrac{n^3}{18432}
+\dfrac{n^2}{4608}-\dfrac{n}{96}-\dfrac{653}{10800}\right\Vert+\dfrac{n^3}{768}m
+\dfrac{5n^2}{128}m+\dfrac{5n}{48}m\\
& & \\
& & +\dfrac{15n^2}{32}m^2+\dfrac{5\left(5m-1\right)}{2}m +\dfrac{75n}{8}m^2+75nm^3+750m^3+4500m^4\\
&=& t_{5}\left(n\right)+ \dfrac{4t^3}{6}m +\dfrac{15t^2}{6}m +\dfrac{5t}{6}m
+30t^2m^2 +\dfrac{5\left(5m-1\right)}{2}m +75tm^2 \\
& & \\
& & +600tm^3+750m^3+4500m^4 \quad (\text{by substituting} \: n=8t)\\
& \equiv & t_{5}\left(n\right)\: \left(\text{mod} \: m\right)
\quad (\text{by Lemma~\ref{lemma}} \: \text{and} \: 
5\left(5m-1\right) \equiv 0 \:\left(\text{mod} \: 2\right)).\\
\end{array}$$
In a similar way, we show that Equation~(\ref{periodequation1:t5}) is also satisfied for
$n \equiv 1,2,3,4,5,6,7$  $\left(\text{mod}\: 8\right)$.

\item[(2)] $-35 < n < 0$ : If $m$ is odd then
Equation~(\ref{periodequation1:t5}) is immediately satisfied for
this range by using Corollary \ref{corollary:t5} and Equation~(\ref{negative:t5}).

\item[(3)] $-240m < n \leq -35$ : For $n \equiv 0 \: \left(\text{mod}\: 8\right)$,
by using Corollary \ref{corollary:t5}, Equation~(\ref{negative:t5}) and $m$ is odd, we have
$$
\begin{array}{rll}
   && t_{5}\left(n+240m\right) \\
& = & \left\Vert \dfrac{n^4}{737280}+\dfrac{n^3}{18432}
+\dfrac{n^2}{4608}-\dfrac{n}{96}-\dfrac{653}{10800}\right\Vert+\dfrac{n^3}{768}m
+\dfrac{5n^2}{128}m+\dfrac{5n}{48}m\\
& & \\
& & +\dfrac{15n^2}{32}m^2+\dfrac{5\left(5m-1\right)}{2}m +\dfrac{75n}{8}m^2+75nm^3+750m^3+4500m^4 \\
& & \\
&=& t_{5}\left(-n-35\right)+ \dfrac{4t^3}{6}m +\dfrac{15t^2}{6}m +\dfrac{5t}{6}m
+30t^2m^2 +\dfrac{5\left(5m-1\right)}{2}m +75tm^2 \\
& & \\
& & +600tm^3+750m^3+4500m^4 \quad (\text{by substituting} \: n=8t)\\
& \equiv & t_{5}\left(n\right)\: \left(\text{mod} \: m\right)
\quad (\text{by Lemma~\ref{lemma}} \: \text{and} \: 
5\left(5m-1\right) \equiv 0 \:\left(\text{mod} \: 2\right)).\\
\end{array}$$
Again, Equation~(\ref{periodequation1:t5}) is also satisfied for
$n \equiv 1,2,3,4,5,6,7 \: \left(\text{mod}\: 8\right)$.

\item[(4)] $-240m-35 < n \leq -240m$ :

That is, $n=-240m-a$ for $0 \leq a \leq 34$ then
\begin{eqnarray*}
t_{5}\left(n+240m\right) & = & t_{5}\left(-240m-a+240m\right)
 =  t_{5}\left(-a\right)
 \equiv  0\:  \left(\text{mod} \: m\right)
\end{eqnarray*}
and
\begin{eqnarray*}
t_{5}\left(n\right) & = & t_{5}\left(-240m-a\right) 
 =  t_{5}\left(240m+a-35\right) 
\quad (\text{by Equation~(\ref{negative:t5})})\\
& \equiv & 0\:  \left(\text{mod} \: m\right),
\end{eqnarray*}

and so $t_{5}\left(n+240m\right) \equiv t_{5}\left(n\right)
\: \left(\text{mod} \: m\right)$.

\item[(5)] $n \leq -240m-35$ : For $n \equiv 0 \: \left(\text{mod}\: 8\right)$,
using Corollary \ref{corollary:t5}, Equation~(\ref{negative:t5})
and $m$ odd, we have
\begin{eqnarray*}
   && t_{5}\left(n+240m\right) \\ 
& = & t_{5}\left(-n-240m-35\right)  \\ 
& = & \left\Vert \dfrac{n^4}{737280}+\dfrac{n^3}{18432}
+\dfrac{n^2}{4608}-\dfrac{n}{96}-\dfrac{653}{10800}\right\Vert+\dfrac{n^3}{768}m
+\dfrac{5n^2}{128}m+\dfrac{5n}{48}m\\
& & \\
& & +\dfrac{15n^2}{32}m^2+\dfrac{5\left(5m-1\right)}{2}m +\dfrac{75n}{8}m^2+75nm^3+750m^3+4500m^4 \\
& & \\
&=& t_{5}\left(-n-35\right)+ \dfrac{4t^3}{6}m +\dfrac{15t^2}{6}m +\dfrac{5t}{6}m
+30t^2m^2 +\dfrac{5\left(5m-1\right)}{2}m +75tm^2 \\
& & \\
& & +600tm^3+750m^3+4500m^4 \quad (\text{by substituting} \: n=8t)\\
& \equiv & t_{5}\left(n\right)\: \left(\text{mod} \: m\right)
\quad (\text{by Lemma~\ref{lemma}} \: \text{and} \: 
5\left(5m-1\right) \equiv 0 \:\left(\text{mod} \: 2\right)).
\end{eqnarray*}
As before, Equation~(\ref{periodequation1:t5}) is also satisfied for
$n \equiv 1,2,3,4,5,6,7 \: \left(\text{mod}\: 8\right)$.

\end{enumerate}

If $m$ is odd then Equation~(\ref{periodequation1:t5})
is satisfied for any integer $n$ from (1), (2), (3),
(4) and (5). Similarly, one can show that if $m$ is even
then Equation~(\ref{periodequation2:t5}) is satisfied for any
integer $n$ by using Lemma~\ref{lemma}, Corollary \ref{corollary:t5}
and Equation~(\ref{negative:t5}). Therefore, if $m$ is odd
$240m$ is a period of the sequence and $L$ divides $240m$, else $480m$ is a
period of the sequence and $L$ divides $480m$.

Secondly, we prove that if $m$ is odd then $L=240m$ else $L=480m$.
This can be verified by a computer program for small $m \leq 10$.
So here, we prove $L=36m$ for only $m>10$.
Let $\lambda=L,L-1,L-2,L-3,L-4,L-5,L-6$ or $L-7$ so that 
$\lambda \equiv 0 \: \left(\text{mod} \: 8 \right)$.

Since $t_{5}\left(-7\right)=t_{5}\left(-6\right)=\cdots=
t_{5}\left(4\right)=0$, we have
$$t_{5}\left(\lambda\right)\equiv t_{5}\left(\lambda+1\right)
\equiv t_{5}\left(\lambda+2\right) \equiv t_{5}\left(\lambda+4\right)
\equiv 0 \: \left( \text{mod} \: m \right).$$
So, we get $m | M$ where $$M=\left[t_{5}\left(\lambda+4\right)-t_{5}\left(\lambda\right)\right]+
\left[t_{5}\left(\lambda+1\right)-t_{5}\left(\lambda+2\right)\right]+
\left[t_{5}\left(\lambda+4\right)-t_{5}\left(\lambda+2\right)\right].$$

$$t_{5}\left(\lambda\right)= \left\Vert\dfrac{\lambda^4}{737280}
+\dfrac{\lambda^3}{18432}+\dfrac{\lambda^2}{4608}-\dfrac{\lambda}{96}-\dfrac{653}{10800}\right\Vert$$

$$t_{5}\left(\lambda+1\right)=\left\Vert\dfrac{\lambda^4}{737280}
+\dfrac{13\lambda^3}{92160}+\dfrac{113\lambda^2}{23040}+\dfrac{29\lambda}{480}+\dfrac{1507}{10800}\right\Vert$$

$$t_{5}\left(\lambda+2\right)=\left\Vert\dfrac{\lambda^4}{737280}
+\dfrac{11\lambda^3}{92160}+\dfrac{77\lambda^2}{23040}+\dfrac{17\lambda}{480}+\dfrac{1507}{10800}\right\Vert$$

$$t_{5}\left(\lambda+4\right)=\left\Vert\dfrac{\lambda^4}{737280}
+\dfrac{7\lambda^3}{92160}+\dfrac{23\lambda^2}{23040}+\dfrac{\lambda}{480}+\dfrac{157}{10800}\right\Vert$$

Hence, from Equation (\ref{property2}), we can write
$$t_{5}\left(\lambda+4\right)-t_{5}\left(\lambda\right)<\dfrac{\lambda^3}{46080}
+\dfrac{\lambda^2}{1280}+\dfrac{\lambda}{80}+\dfrac{3}{40}+1$$
$$t_{5}\left(\lambda+1\right)-t_{5}\left(\lambda+2\right)<\dfrac{\lambda^3}{46080}
+\dfrac{\lambda^2}{640}+\dfrac{\lambda}{40}+1$$
$$t_{5}\left(\lambda+4\right)-t_{5}\left(\lambda+2\right)<-\dfrac{\lambda^3}{23040}
-\dfrac{3\lambda^2}{1280}-\dfrac{\lambda}{30}-\dfrac{1}{80}+1,$$
and it follows that
$$M=\left[t_{5}\left(\lambda+4\right)-t_{5}\left(\lambda\right)\right]+
\left[t_{5}\left(\lambda+1\right)-t_{5}\left(\lambda+2\right)\right]+
\left[t_{5}\left(\lambda+4\right)-t_{5}\left(\lambda+2\right)\right]
<\dfrac{\lambda}{240}-\dfrac{1}{20}+3.$$
Similar to Section~\ref{case:4}, we can conclude that $L \geq 2880$  for $m > 10$
by observing an aperiodic pattern for $\{t_5(n) \left(\text{mod}\: m\right)\}_{n=-\infty}^{\infty}$
with the aid of a computer program. That is, we have $L \geq 2880$ so that $\lambda  \geq 2880 $.
From (\ref{property1}), we get
\begin{eqnarray*}
t_{5}\left(\lambda+1\right)+ t_{5}\left(\lambda\right)& = & \left\Vert \dfrac{8\lambda^3}{92160}+\dfrac{108\lambda^2}{23040}+\dfrac{34\lambda}{480}+\dfrac{2160}{10800}\right\Vert + \alpha_1, \quad -1 \leq \alpha_1 \leq 1 \notag \\ 
& = & \left\Vert\dfrac{8\lambda^3}{92160}\right\Vert + \left\Vert\dfrac{108\lambda^2}{23040}\right\Vert 
+ \left\Vert\dfrac{34\lambda}{480}\right\Vert
 + \left\Vert\dfrac{2160}{10800}\right\Vert + \alpha_2, \quad -4 \leq \alpha_2 \leq 4
\end{eqnarray*}
and
\begin{eqnarray*}
  & & 2\left(t_{5}\left(\lambda+2\right)+ t_{5}\left(\lambda+3\right)\right) \\
& = & 2 \left(\left\Vert \dfrac{4\lambda^3}{92160}+\dfrac{54\lambda^2}{23040}+\dfrac{16\lambda}{480}+\dfrac{1350}{10800}\right\Vert + \beta_1 \right), \quad -1 \leq \beta_1 \leq 1 \notag \\ 
& = & \left\Vert\dfrac{8\lambda^3}{92160} + \dfrac{108\lambda^2}{23040} + \dfrac{32\lambda}{480}
 + \dfrac{2700}{10800}\right\Vert + \beta_2, \quad -3 \leq \beta_2 \leq 3 \notag \\
 & = & \left\Vert\dfrac{8\lambda^3}{92160}\right\Vert + \left\Vert\dfrac{108\lambda^2}{23040}\right\Vert 
+ \left\Vert\dfrac{32\lambda}{480}\right\Vert
 + \left\Vert\dfrac{2700}{10800}\right\Vert + \beta_3, \quad -6 \leq \beta_3 \leq 6.
\end{eqnarray*}
Hence,
\begin{eqnarray*}
M & = &\left\Vert\dfrac{34\lambda}{480}\right\Vert-\left\Vert\dfrac{32\lambda}{480}\right\Vert+ \gamma_1,
\quad -10 \leq \gamma_1 \leq 10 \notag \\ 
& = & \left\Vert\dfrac{\lambda}{240}\right\Vert+\gamma_2, \quad -11 \leq \gamma_2 \leq 11
\end{eqnarray*}
Since $ \lambda \geq 2880$, we have $M>1$. 
We have $ M < \dfrac{\lambda}{240}-\dfrac{1}{20}+3 =\dfrac{\lambda+708}{240}$.
This means that $ m < \dfrac{\lambda+708}{240}
\left( \text{since} \: m | M \: \text{and} \: M \neq 0 \right)$,
and it follows that
\begin{equation} \label{ineq5}
240m < \lambda+708 \le L+708.
\end{equation}
If $m$ is odd then $L$ is a divisor of $240m$, and so $L=240m$ by
Inequality~(\ref{ineq5}) for any $m > 10$. Hence, if $m$
is odd then $L=240m$. If $m$ is even then $L$ is a divisor of $480m$, so
$L=240m$ or $480m$. However, we can get
\begin{eqnarray*}
t_{5}\left(240m\right) &=&\left\Vert 4500m^4+750m^3
+\dfrac{25m^2}{2}-\dfrac{5m}{2}-\dfrac{653}{10800}\right\Vert \\
&=& 4500m^4+750m^3+\dfrac{5m}{2}\left(5m-1\right)\\
& \equiv & \dfrac{5m}{2} \: \left(\text{mod} \: m \right)\\
& \not\equiv & 0 \: \left(\text{mod} \: m \right),
\end{eqnarray*}
when $m$ is even. Therefore, if $m$ is even then $L$
can not be $240m$. This means that if $m$ is even
then $L=480m$ and the theorem is proved.
\end{proof}

\section{Conclusion}\label{concl}
In this paper we introduce a new sequence $\{t_k(n)\}_{n=-\infty}^{\infty}$
for any given positive integer $k\ge 3$ that is a generalization of Alcuin's sequence.
We find explicitly a linear recurrence equation and
the generating function for $\{t_k(n)\}_{n=-\infty}^{\infty}$.
The case $k=3$ is Alcuin's sequence.
For the special case $k=4$ and $k=5$, we get simpler formulas for
$\{t_k(n)\}_{n=-\infty}^{\infty}$ and investigate the period of
$\{t_k(n)\}_{n=-\infty}^{\infty}$ modulo a fixed integer.
Also, we get a formula for $p_{5}\left(n\right)$, the number of
partitions of $n$ into exactly $5$ parts.

In \cite{Andrews},  Andrews  discussed the geometric interpretation of the sequence
$\{t_3(n)\}_{n=0}^{\infty}$ corresponding to Alcuin's sequence. For $k=4$, let $a,b,c$ and $d$
be integers satisfying the condition
\begin{equation} \label{cond:equilateral}
a+b+c+d=3n,\: a+b+c>2n,\: a+b+d>2n,\: b+c+d>2n \: \hbox{and} \: a+c+d>2n.
\end{equation}
If this is case $a,b,c,d<n$, and since $$t_{4}\left(3n\right)
=p_{4}\left(n\right)$$ via the bijection
$$\left(a,b,c,d\right)\longleftrightarrow \left(n-a,n-b,n-c,n-d\right),$$
we conclude that $t_{4}\left(3n\right)$ counts the number of different
$a,b,c,d$ integers which satisfy Condition~(\ref{cond:equilateral}).
We wonder what kind of geometric interpretation does $\{t_4(n)\}_{n=-\infty}^\infty$, or more generally $\{t_k(n)\}_{n=-\infty}^\infty$ for $k\geq 4$,  represent?



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\bibitem{Biggs} N. Biggs. 
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\newblock Oxford University Press, 2002.

\bibitem{BindnerErickson} D. J. Bindner and M. Erickson. 
\newblock Alcuin's Sequence.
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\bibitem{Olivastro} D. Olivastro. 
\newblock {\em Ancient Puzzles}, 
\newblock Bantam Books, New York, NY, 1993.

\bibitem{internet} 
\newblock http://mathworld.wolfram.com/PartitionFunctionP.html  
\newblock (as seen on October 11, 2012).

\end{thebibliography}

\end{document}
