%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%                                                               %%
%% This is the mc_template.tex file for the mc document class.   %%
%% It is used to prepare s manuscript for Mathematical 			 %%
%% Communications journal.                                       %%
%%                                                               %%
%% The mc.cls class works only with a pdflatex engine.           %%
%% The file newmc.cls should be placed where LaTeX 			     %%
%% can find it, e.g. in the current working directory.		     %%
%%                                                               %%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\documentclass{mc}

%%===============================================================%%
%% Please add here your own packages, macros and enviroments.    %%
%% It is not necessary to include ams* and graphicx packages     %%
%% since they are automatically included by the mc class.        %%
%% Avoid defining your own environments and use the already      %%
%% defined ones (e.g.~theorem, lemma etc.)                       %%
%%===============================================================%%

%\usepackage{enumerate}  % uncomment to use this package
%\newcommand{\E}{\mathbb{E}} % example of a macro

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


%%===============================================================%%
%% Journal info will be edited by the typesetter 				 %%
%% DO NOT CHANGE THIS PART						                 %%
%%===============================================================%%
\setcounter{page}{259}
\renewcommand\thisnumber{x}
\renewcommand\thisyear {2025}
\renewcommand\thismonth{xxx}
\renewcommand\thisvolume{30}
\renewcommand\datereceived{December 12, 2024}
\renewcommand\dateaccepted{June 12, 2025}
\renewcommand\doinum{10.1000/100}
%%===============================================================%%
%Date submitted: 12.12.2024.

%Date accepted: 12.06.2025.



\begin{document}
%\begin{linenumbers}
%%===============================================================%%
%% TITLE                                                         %%
%% Please add the title with \title[Short title]{Title}          %%
%% Short title is a running head apearing in the header.         %%
%%===============================================================%%
\title[Equidistant integer grids as integer sequence generators]	% at most 50 characters including spaces
		{Equidistant integer grids as integer sequence generators} 	% at most 150 characters including spaces ()
%%===============================================================%%


%%===============================================================%%
%% AUTHOR(S)                                                     %%
%%                                                               %%
%% Add author's details in the following format. For each author %%
%% provide the affiliation, address and Orcid identifier.		 %%
%% Mark the corresponding author with \comma\corrauth.			 %%
%%===============================================================%%


%% For single author use the following format. 				     %%
\author[B.~Kolarec]%
	{Biserka Kolarec \orcidnumber{0000-0003-1434-3533}
	}		 

\address{University of Zagreb, Faculty of Agriculture, Sveto\v simunska cesta 25, 10 000 Zagreb, Croatia} 

\emailsingle{
	\email{bkudelic@agr.hr}
}
%%===============================================================%%


%%===============================================================%%
%% ABSTRACT                                                      %%
%%===============================================================%%
\begin{abstract}
We investigate equidistant integer grids and semigrids as generators of integer sequences. Equidistant integer grids consist of arithmetic sequences arranged in rows and columns. Among other things, the partial sums of their diagonal elements result in polygonal and second polygonal numbers. Furthermore, sequences of row sums in integer semigrids fulfill certain recurrence relations. We analyze such sequences in detail for Pascal's semigrid obtained in two ways: by shifting columns only and by stretching and shifting columns.
\end{abstract}
%%===============================================================%%


%%===============================================================%%
%% KEYWORDS                                                      %%
%%===============================================================%%
\keywords{arithmetic sequence; integer grids; integer semigrids; recurrence relations}
%%===============================================================%%


%%===============================================================%%
%% AMS subject classification                                    %%
%%===============================================================%%
\ams{11B25, 11B37}
%%===============================================================%%


%%===============================================================%%
\maketitle
%%===============================================================%%





%%===============================================================%%
%% MAIN BODY                                                     %%
%%===============================================================%%

\section{Introduction}
To form an equidistant integer grid, we first place an arithmetic sequence $a_1, a_1+d_c, \ldots$ in the first column and an arithmetic sequence $a_1, a_1+d_r, \ldots$ in the first row. Then we distribute an integer grid by adding arithmetic sequences with the difference $d_c$ to columns (or with the difference $d_r$ to rows).
Here we study equidistant integer grids for $a_1=1$. Note that the entire equidistant integer grid is given by the upper left triangle
\[\begin{array}{ccc}
1&&1+d_r\\
1+d_c&&
\end{array}.\]
We define \emph{a grid characteristic $s$} as $s=2+d_c+d_r$, the sum of the elements on the first antidiagonal. The difference between consecutive elements in all diagonal sequences in an integer grid of characteristic $s$ is equal to $s-2$. We are interested in the partial sums of the elements of the diagonal sequences in an equidistant grid of characteristic $s$. In Section 2, we show that, depending on the position of the diagonal, they give $s$-gonal numbers, second $s$-gonal numbers and many other known integer sequences listed in The Online Encyclopedia of Integer Sequences, \cite{oe}. Section 3 is dedicated to integer semigrids, which consist of a single arithmetic sequence repeated column by column, each time shifted down by $k$. Depending on $k$, the row sums of the integer semigrid elements form sequences that fulfill certain linear recurrences. In the case of Pascal's semigrid, the row sums result in many known sequences. Moreover, the Fibonacci and Padovan sequences (\cite{pad}) are sums of integer semigrid elements lying on certain diagonals. Finally, in Section 4, we examine row sums in Pascal's semigrid transformed by shifting and stretching columns.

\section{Partial sums of diagonal sequences in equidistant integer grids}

For a positive integer $s$, up to symmetry with respect to the main diagonal, there are $\left \lfloor\frac{s}{2}\right\rfloor$ different equidistant grids of characteristic $s$, i.e. 
\[\begin{array}{ccc}
1&&s-p\\
p&&
\end{array}, \,\,\, p\in\{1,2,\ldots, \left \lfloor\frac{s}{2}\right\rfloor\}.\]

Some special values of $p$ lead to the special appearance of equidistant grids.
\begin{itemize}
\item For $p=1$, each diagonal sequence below the main diagonal is equal to the main diagonal sequence. The sequence  on the $l$th diagonal above the main diagonal is missing the first $l-1$ terms. Here, $l=1$ stands for the main diagonal.
\item If $p|s$, all sequences above the main diagonal also appear below it.
\item For even $s$ and $p=\frac{s}{2}$, the equidistant grid is symmetrical with respect to the main diagonal.
\end{itemize}

There are two possibilities when considering diagonal sequences. The first is to take $p\in \{1,2,\ldots, s-1\}$ and only consider sequences below the main diagonal in equidistant grids. The second option is to consider diagonal sequences below and above the main diagonal for $p\in\{1,2,\ldots, \left \lfloor\frac{s}{2}\right\rfloor\}$. We choose the second option and give formulas for partial sums of sequences on the $k$th diagonal below and the $l$th diagonal above the main diagonal. Here, $k=l=1$ stands for the main diagonal itself.

Let us fix $p\in\{1,2,\ldots, \left \lfloor\frac{s}{2}\right\rfloor\}$. The $k$th term of the first column is equal to $b_k=(k-1)p+(2-k)$. Since the difference of the diagonal elements is $s-2$, the $n$th element on the $k$th diagonal is $a_{n,k}=b_k +(n-1)(s-2)$. Therefore, the $n$th partial sums are
\[P_{k,n}^{s,p}=\frac{n}{2}\left[2b_k+(n-1)(s-2)\right]=\frac{n}{2}\left[2p(k-1)+2(2-k)+(n-1)(s-2)\right].\]
Here, $P_{1,n}^{s,p}=\frac{n}{2}\left[(s-2)n-(s-4)\right]$ is exactly the formula for the $n$th $s$-gonal number, see \cite{pol}.

Similarly, the first term of the $l$th diagonal above the main diagonal ($l=1$) is $c_l=1+ (l-1)(s-p-1)$. The $n$th partial sum of the elements on the $l$th diagonal above the main diagonal with $a_{n,l}=c_l + (n-1)(s-2)$ is
\[R_{l,n}^{s,p}=\frac{n}{2}\left[2c_l+(n-1)(s-2)\right]=\frac{n}{2}\left[2+2(l-1)(s-p-1)+(n-1)(s-2)\right].\]

The cases $l=1$ and $k=1$ result in partial sums of elements on the main diagonal and coincide. For $k=l=2$, the partial sums of the elements on the second diagonal below and above the main diagonal are
\[P_{2,n}^{s,p}=\frac{n}{2}\left[2p+(n-1)(s-2)\right], \,\,\, R_{2,n}^{s,p}=\frac{n}{2}\left[2(s-p)+(n-1)(s-2)\right], \]
 respectively. This reflects the symmetry of the grids
$\begin{array}{ccc}
1&&s-p\\
p&&
\end{array} \,\,\,\textup{and}\, \, \, \begin{array}{ccc}
1&&p\\
s-p&&
\end{array}.$ 

We can set up formulas for $P_{k,n}^{s,p}$ and $R_{l,n}^{s,p}$ and formulate the following theorem.
\begin{theorem} Let $s$ be a positive integer. For equidistant grids $\begin{array}{ccc}
1&&s-p\\
p&&
\end{array}$, $p\in\{1,2,\ldots, \left \lfloor\frac{s}{2}\right\rfloor\}$ of characteristic $s$, partial sums of their diagonal sequence elements on the $k$th diagonal below and the $l$th diagonal above the main diagonal are given by
\begin{eqnarray*}
P_{k,n}^{s,p}&=&\frac{n}{2}\left[(s-2)n +2k(p-1)+6-2p-s\right],\\
R_{l,n}^{s,p}&=&\frac{n}{2}\left[(s-2)n +2l(s-p-1)+6+2p-3s\right].
\end{eqnarray*}
\end{theorem}

Regardless of the specific diagonal position of the equidistant sequence in the equidistant grid of characteristic $s$, we can analyze the dependence of its partial sums on the first term alone. Let us denote such partial sums by $S_{n,m}^s$ and let $a_1=s-m$, the first term of the given sequence for $s\geq 5$ and $m\in\{1,2,\ldots, \left \lfloor\frac{s}{2}\right\rfloor\}$. The following theorem applies.
\begin{theorem} For an equidistant grid of characteristic $s$, $s\geq 5$ and $m\in\{1,2,\ldots, \left \lfloor\frac{s}{2}\right\rfloor\}$, the $n$-th partial sum of the equidistant sequences starting with $a_1=s-m$ is
\[S_{n,m}^s=\frac{n}{2}\left[(n+1)s-2(n +m-1)\right].\]
\end{theorem}
\begin{proof} The assertion follows from the standard formula for the sum of the first $n$ elements of the arithmetic sequence with the first term $a_1$ and the difference $d$, i.e. $s_n=\frac{n}{2}\left[2a_1+(n-1)d\right]$. Here, $d=s-2$.\end{proof}
A specific choice of $m=1,2,3$ leads to interesting relationships. Let us denote by $S_n$ the $n$th square number $S_n=n^2$ and by $T_n$ the $n$th triangular number $T_n= \frac{n(n+1)}{2}$.
\begin{corollary} For an equidistant grid of characteristic $s$, $s\geq 5$, with the first term $a_1=s-m$, we have the following formulas for $S_{n,m}^s$:
\begin{enumerate}
\item $S_{n,1}^s=sT_n-S_n,$
\item $S_{n,2}^s=(s-2)T_n,$
\item $S_{n,3}^s=\frac{n}{2}\left[(s-2)n +(s-4)\right]$, the $n$th second $s$-gonal number.
\end{enumerate}
\end{corollary}
\begin{proof} Proofs are achieved through simple calculations:
\begin{enumerate}
\item
$S_{n,1}^s=\frac{n}{2}\left[(n+1)s-2n\right]= \frac{n(n+1)}{2}s-n^2=sT_n-S_n. $
\item
$S_{n,2}^s=\frac{n}{2}\left[(n+1)s-2(n +1)\right]= \frac{n(n+1)}{2}\left(s-2\right)=(s-2)T_n.$

\item For $m=3$, \[S_{n,3}^s=\frac{n}{2}\left[(n+1)s-2(n +2)\right]=\frac{n}{2}\left[(s-2)n +(s-4)\right],\]
and this is exactly the formula for the second $s$-gonal numbers, cf. \cite{pol}.
\end{enumerate} \end{proof}
\section{Recurrence relations in equidistant semigrids}
The next recurrence relation applies to partial sums of elements of all arithmetic sequences, irrespective of their grid position.  
\begin{proposition} Let $(b_n)$ be any of partial sum sequences $(P_{k,n}^{s,p})$,  $(R_{l,n}^{s,p})$ or $(S_{n,m}^{s})$ of an equidistant grid of characteristic $s$. It satisfies the unique recurrence relation 
\[b_n=3b_{n-1}-3b_{n-2} +b_{n-3}.\]
\end{proposition}
\begin{proof}The difference between two consecutive elements in an arithmetic sequence is constant. Therefore,
\[(b_n-b_{n-1})-(b_{n-1}-b_{n-2})=(b_{n-1}-b_{n-2})-(b_{n-2}-b_{n-3}),\]
and the assertion follows. \end{proof}
The list of sequences satisfying linear recurrence relations of type $(3,-3,1)$ is the longest on the index page of linear recurrences of order three \cite{oer}. It contains almost 1500 sequences, but given the assertion of Proposition 1, it should be much longer.

We continue with a few remarks.
\begin{enumerate}
\item There is no one-to-one correspondence between linear recurrence relations of type $(3,-3,1)$ and sequences of partial sums of arithmetic sequences. For example, the sequence of the centered $22$-gonal numbers $1,23,67,133, 221, 331, \ldots$ (A069173 in \cite{oe}) fulfills the recurrence. However, it is the partial sum sequence of the non-arithmetic sequence $1,22,44,66,88,\ldots$.

\item In general, for a positive integer $k$ and a sequence $1, k, 2k, 3k, \ldots$, the sequence of its partial sums $1, k+1, 3k+1, 6k+1, 10k+1,\ldots$ satisfies a linear recurrence relation of the type $(3,-3,1)$. In more general case:  For positive integers $b,k$ and an arithmetic sequence $b, b+k, b+2k, b+3k, \ldots$, the partial sum sequence $b, 2b+k, 3b+3k, 4b+6k, \ldots$ satisfies a linear recurrence of type $(3,-3,1)$.
\end{enumerate}

Not only equidistant grids generate recurrences, but also semigrids, which consist of an arithmetic sequence  continuously repeated in columns that are shifted downwards by a certain k.  A semigrid obtained by shifting by $k$ is called {\emph a $k$-semigrid}. For example, for an arithmetic sequence $a_1,a_2,a_3\ldots$ and $k=1,2,3$, respective $k$-semigrids are:
\[\begin{array}{ccccccccccc}
a_1&&&&&&&&&\\
a_2&&a_1&&&&&&&\\
a_3&&a_2&&a_1&&&&&\\
a_4&&a_3&&a_2&&a_1&&&\,,\,\,\,\\
a_5&&a_4&&a_3&&a_2&&\cdots\\
a_6&&a_5&&a_4&&a_3&&\cdots\\
a_7&&a_6&&a_5&&a_4&&\cdots\\
\vdots&&\vdots&&\vdots&&\vdots&&\ddots
\end{array}
\begin{array}{cccccccc}
a_1&&&&&&&\\
a_2&&&&&&&\\
a_3&&a_1&&&&&\\
a_4&&a_2&&&&&\,,\,\,\,\\
a_5&&a_3&&a_1&&\\
a_6&&a_4&&a_2&&\\
a_7&&a_5&&a_3&&a_1\\
\vdots&&\vdots&&\vdots&&\ddots
\end{array}
\begin{array}{ccccccc}
a_1&&&&&&\\
a_2&&&&&&\\
a_3&&&&&&\\
a_4&&a_1&&&\,.\\
a_5&&a_2&&&\\
a_6&&a_3&&&\\
a_7&&a_4&&a_1\\
\vdots&&\vdots&&\ddots
\end{array}\]

%% ____________________________________________________________________________________________%%
%% THE TEXT "By construction ..." SHOULD BE DISPLAYED BEFORE TABLE 1 AS GIVEN HERE, AND NOT AFTER IT !!!
%%_____________________________________________________________________________________________%%

By construction, sequences of row sums $(s_n)_{n\geq1}$ in such semigrids fulfill certain types of linear recurrence relations with constant coefficients. The general form of  recurrence relations becomes clear for $k\geq 4$, as can be seen in Table \ref{t:1}.


\begin{table}[h]
\centering
\begin{tabular}{cccccccccc}  
\hline
$k$&$s_n$&$s_{n-1}$&$s_{n-2}$&$s_{n-3}$&$s_{n-4}$&$s_{n-5}$&$s_{n-6}$&$s_{n-7}$&$s_{n-8}$\\
\hline
1&-1&3&-3&1&&&&&\\
2&-1&2&0&-2&1&&&&\\
3&-1&2&-1&1&-2&1&&&\\
4&-1&2&-1&0&1&-2&1&&\\
5&-1&2&-1&0&0&1&-2&1&\\
6&-1&2&-1&0&0&0&1&-2&1\\
\hline
\end{tabular}
\caption{Recurrence relations for row sum sequences depending on the shift $k$}
\label{t:1}
\end{table}

One could also consider shallow diagonal sums of elements. Note that shallow diagonal sums in the $k$-semigrid are equal to row sums in the $(k+1)$-semigrid.

\section{Recurrence relations in Pascal's semigrid}

Although not equidistant, it is interesting to study the row sums in Pascal's semigrid
\[\begin{array}{ccccccccccccc}
1&&&&&&&&&&\\
1&&1&&&&&&&&\\
1&&2&&1&&&&&&\\
1&&3&&3&&1&&&&\\
1&&4&&6&&4&&1&&\\
\vdots&&\vdots&&\vdots&&\vdots&&\ddots&&\ddots
\end{array}\,\, .\]
We consider the row sums of semigrids
whose columns are shifted downward by $k$, as well as the row sums of those semigrids
whose columns are both stretched and shifted. The first one is called Pascal’s $k$–semigrid.
Note that for $k = 0$, we have Pascal’s semigrid.

In Pascal's semigrid, the sum of the elements in the $n$th row is equal to $2^n$. The row sums in Pascal's $1$-semigrid are Fibonacci numbers. In Pascal's $2$-semigrid, the sums of the elements in the rows $1, 1, 1, 2, 3, 4, 6, 9, 13$, $19, 28,\ldots$, give Narayana's cows sequence (\cite{nar}) given by $N_0=N_1=N_2=1$, $N_n=N_{n-1}+N_{n-3}$, $n\geq 3$. In Pascal's $3$-semigrid, the sums are $1,1,1,1,2,3,4,5,7,10,14,19,\ldots$, which is the sequence $(s_n)_{n\geq0}$ given by the recurrence relation $s_0=s_1=s_2=s_3=1$, $s_n=s_{n-1}+s_{n-4}$, $n\geq 4$.
In general, we have the following result.
\begin{theorem} Sequences $(s_n)_{n\geq 0}$ of row sums in Pascal's $k$-semigrid satisfy the recurrence relation $s_n=s_{n-1}+s_{n-(k+1)}$ for $n\geq k+1$, with $s_0=s_1=\cdots=s_{k}=1$. In particular, \[s_n= \sum_{i=0}^{\lfloor \frac{n}{k+1}\rfloor} \left( \begin{array}{c}n-ki\\ i\end{array}\right). \]
\end{theorem}
\begin{proof} The first $k + 1$ rows in Pascal’s $k$–semigrid
each have a single term, which is $1$. The next $k + 1$ rows consist of two terms per row,
followed by $k + 1$ rows consisting of three terms per row, and this pattern continues. The elements in the $n$th row of  Pascal's $k$-semigrid are binomial coefficients
\[\left( \begin{array}{c}n\\ 0\end{array}\right),\,\left( \begin{array}{c}n-k\\ 1\end{array}\right),\, \left( \begin{array}{c}n-2k\\ 2\end{array}\right),\,\cdots,\, \left( \begin{array}{c}n-\lfloor \frac{n}{k+1}\rfloor k\\ \lfloor \frac{n}{k+1}\rfloor\end{array}\right).\]
Since $s_n$ is their sum, the statement follows. \end{proof}
One could also study row sums in semigrids where the columns are stretched by a certain $t\geq 1$. By stretching, we mean inserting $t$ zeros between every two consecutive elements in each column. The combination of stretching and shifting the columns of Pascal's semigrid results in a Padovan-like sequence of row sums. For example, a semigrid with   stretch $t=2$ and shift $k=1$ results in the row sum sequence: $1,0,1,1,1,2,2,3,4,6,12\ldots$. It fulfills the recurrence relation $P_n=P_{n-2}+P_{n-3}$, $n\geq 3$, with $P_0=1, P_1=0, P_2=1$. Similarly, $t=3$, $k=2$ results in the sequence $1,0,0,1,1,0,1,2,1,1,3,3\ldots$, which satisfies $P_n=P_{n-3}+P_{n-4}$, $n\geq 4$, with $P_0=1, P_1=P_2=0, P_3=1$.

The shift $k$ can continue up to $k=t$ and beyond. Depending on the size of the stretch $t$ and the shift $k$, row sums of Pascal's semigrids are sequences with initial conditions and recurrence relations given in the following theorem. The places marked with a dot follow the start pattern and contain $0$ or $1$.
\begin{theorem} Fix the stretch size $t\geq 3$ of  the columns of Pascal's semigrid. Depending on the shift $k\geq 0$, sequences of row sums $(s_n)_{n\geq1}$ satisfy the initial conditions and recurrence relations given in Table \ref{t:2}.


\begin{table}[h]
\centering
\begin{tabular}{c|cccccccc|ccccccc} 
\hline
$k$&$s_0$&$s_1$&$s_2$&$s_3$&...&$s_t$&$s_{t+1}$&$s_{t+2}$&$s_n$&$s_{n-1}$&$s_{n-2}$&$s_{n-3}$&...&$s_{n-t}$&$s_{n-t-1}$\\
\hline
0&1&1&1&1&...&1&1&1&-1&1&0&0&...&0&1\\
1&1&0&1&0&...&.&.&.&-1&0&1&0&...&0&1\\
2&1&0&0&1&...&.&.&.&-1&0&0&1&...&0&1\\
$\vdots$&&&&&&&&&&&&&&\\
$t-1$&1&0&0&0&...&0&1&0&-1&0&0&0&...&1&1\\
$t$&1&0&0&0&...&0&0&1&-1&0&0&0&...&0&2\\
\hline
\end{tabular}
\caption{Pascal's $k$-semigrid row sum sequence recurrences with fixed stretch $t$}
\label{t:2}
\end{table}


\end{theorem}
\begin{proof} In   Pascal's semigrid stretched by $t$, the first $t+1$ row sums equal   $1$. The semigrid has a diagonal appearance: Each diagonal of the original Pascal's semigrid is followed by $t$ diagonals containing only zeros. The sequence $(s_n)_{n\geq0}$ of the row sums is therefore: $s_0=s_1=\ldots=s_{t}=1$ and, by construction, $s_n=s_{n-1}+s_{n-t-1}$, $n\geq t+1$. In   Pascal's $k$-semigrid with $k>0$, the initial conditions follow the pattern that a $1$  is followed by $k$ zeros. Therefore, for the stretch $t$, the shift $k$  and $n\geq t+1$, we have \[s_n=s_{n-k-1}+s_{n-t-1}.\] In the case $k=t$, this results in $s_n=2s_{n-t-1}$.
\end{proof}
\section {Conclusion} Integer grids and semigrids generate many integer sequences. They are obtained as partial sums of diagonal sequence elements and in integer semigrids as row sums. In the case of stretching and shifting columns in Pascal's semigrid, the row sum sequences fulfill certain linear recurrence relationships. Since all elements there are binomial coefficients, the recurrences apply directly to them. It would be interesting to consider the row sums of $n$-diagonal semigrids. They are constructed by placing the same arithmetic sequence on the main diagonal and $n-1$ diagonals below it, with zeros elsewhere.
In the special case that the sequence $1,2,3,\ldots$ lies on the diagonals of the $n$-diagonal semigrid, the row sums for $n=3$ form a sequence of coefficients in the expansion of $\frac{1+x+x^2}{(1-x)^2}$. Similarly, the row sums in a $4$-diagonal semigrid give the sequence of coefficients in the expansion of $\frac{1+x+x^2+x^3}{(1-x)^2}$. We assume that the row sums in an $n$-diagonal semigrid, $n\geq2$, generally correspond exactly to the coefficients in the expansion of $\frac{1+x+\cdots+x^{n-1}}{(1-x)^2}$. It would be interesting to form $n$-diagonal semigrids of other arithmetic sequences and analyze their row sums.

%%===============================================================%%
%% ACKNOWLEDGEMENTS                                              %%
%% Acknowledgements can be added here.                           %%
%%===============================================================%%
\section*{Acknowledgements}
%%===============================================================%%
The author thanks Laszlo Nemeth for his advice to study semigrids that generate sequences satisfying recurrence relations. She also thanks the reviewer(s) who, through their careful reading and valuable comments, helped to improve the mathematical rigor of the original manuscript.




%%===============================================================%%
%% REFERENCES                                                    %%
%% References should be provided in bibtex file.                 %%
%% We suggest using MR Lookup for finding bibtex entries.        %%
%%===============================================================%%
\bibliography{references}



%%===============================================================%%
%% APPENDICES	                                                 %%
%% Appendices can be added here.                                 %%
%%===============================================================%%
%\normalsize
%\begin{appendices}
%\section{Some Appendix}
%Appendices should be placed at the end of the manuscript, after the references list. 
%\end{appendices}
%%===============================================================%%

%\end{linenumbers}
\end{document} 

