% EJC papers *must* begin with the following two lines. \documentclass[12pt]{article} \usepackage{e-jc} \specs{P55}{19(4)}{2012} \sloppy % Please remove all other commands that change parameters such as % margins or pagesizes. % only use standard LaTeX packages % only include packages that you actually need % we recommend these ams packages \usepackage{amsthm,amsmath,amssymb} % we recommend the graphicx package for importing figures \usepackage{graphicx} % use this command to create hyperlinks (optional and recommended) \usepackage[colorlinks=true,citecolor=black,linkcolor=black,urlcolor=blue]{hyperref} % use these commands for typesetting doi and arXiv references in the bibliography \newcommand{\doi}[1]{\href{http://dx.doi.org/#1}{\texttt{doi:#1}}} \newcommand{\arxiv}[1]{\href{http://arxiv.org/abs/#1}{\texttt{arXiv:#1}}} % all overfull boxes must be fixed; % i.e. there must be no text protruding into the margins % declare theorem-like environments \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{fact}[theorem]{Fact} \newtheorem{observation}[theorem]{Observation} \newtheorem{claim}[theorem]{Claim} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{open}[theorem]{Open Problem} \newtheorem{problem}[theorem]{Problem} \newtheorem{question}[theorem]{Question} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{note}[theorem]{Note} \newcommand{\rmv}[1]{} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % if needed include a line break (\\) at an appropriate place in the title \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{Oct 12, 2012}{Dec 18, 2012}{Dec 31, 2012}\\ \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 % should be informative, clear, and as complete as possible. Phrases % like "we investigate..." or "we study..." should be kept to a minimum % in favor of "we prove that..." or "we show that...". Do not % include equation numbers, unexpanded citations (such as "[23]"), or % any other references to things in the paper that are not defined in % the abstract. The abstract will be distributed without the rest of the % paper so it must be entirely self-contained. \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