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\begin{center}
\vskip 1cm{\LARGE\bf 
Some New Properties of Balancing Numbers \\
\vskip .1in
and Square Triangular Numbers
}
\vskip 1cm
\large
Refik Keskin and Olcay Karaatl\i  \\
Department of Mathematics \\
Faculty of Arts and Science \\
Sakarya University \\
54187 Serdivan \\
Sakarya \\
Turkey \\
\href{mailto:rkeskin@sakarya.edu.tr}{\tt rkeskin@sakarya.edu.tr}\\
\href{mailto:okaraatli@sakarya.edu.tr}{\tt okaraatli@sakarya.edu.tr}\\
\end{center}

\vskip .2 in

\begin{abstract}
A number $N$ is a {\it square} if it can be written as $N = n^2$ for some
natural number $n$;
it is a {\it triangular number} if it can be written as $N = n(n+1)/2$ for some
natural number $n$; and it is a {\it balancing number} if $8N^2 +1$ is a square.
In this paper, we study some
properties of balancing numbers and square triangular numbers.
\end{abstract}


\section{Introduction}

A triangular number is a number of the form $T_{n}=n(n+1)/2,$ where $n$ is a
natural number. So the first few triangular numbers are $1,\ 3,\ 6,\ 10,\
15,\ 21,\ 28,\ 36,45,$$\ldots$ (sequence \seqnum{A000217} in \cite{OEIS}). A well
known fact about the triangular numbers is that $x$ is a triangular number
if and only if $8x+1$ is a perfect square. Triangular numbers can be thought
of as the numbers of dots needed to make a triangle. In a similar way,
square numbers can be thought of as the numbers of dots that can be arranged
in the shape of a square. The $m$-th square number is formed using an outher
square whose sides have $m$ dots. Let us denote the expression for $m$-th
square number by $S_{m}=m^{2}$ \cite{STANDFOR}. Behera and Panda \cite{BEHERA}
introduced balancing numbers \ $m\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{+\text{ }}$as solutions of the equation%
\begin{equation}
1+2+\text{$\cdots$}+(m-1)=(m+1)+(m+2)+\text{$\cdots$}+(m+r),  \label{1.0}
\end{equation}%
calling $r\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{+},$ the balancer corresponding to the balancing number $m.$ For instance $%
6,35,$ and $204$ are balancing numbers with balancers $2,14,$ and $84,$
respectively. It is clear from (\ref{1.0}) that $m$ is a balancing number
with balancer $r$ if and only if 
\begin{equation*}
m^{2}=\dfrac{(m+r)(m+r+1)}{2},
\end{equation*}%
which when solved for $r$ gives%
\begin{equation}
r=\dfrac{-(2m+1)+\sqrt{8m^{2}+1}}{2}.  \label{1.56}
\end{equation}%
It follows from (\ref{1.56}) that $m$ is a balancing number if and only if $%
8m^{2}+1$ is a perfect square. Since $8\times 1^{2}+1=9$ is a perfect
square, we accept $1$ as a balancing number. In what follows, we introduce
cobalancing numbers in a way similar to the balancing numbers. By modifying (%
\ref{1.0}), we call $m\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{+},$ a cobalancing number if%
\begin{equation}
1+2+\text{$\cdots$}+(m-1)+m=(m+1)+(m+2)+\text{$\cdots$}+(m+r)  \label{1.1}
\end{equation}%
for some $r\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{+}$. Here, $r\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{+}$ is called a cobalancer corresponding to the cobalancing number $m$. A
few of the cobalancing numbers are $2,14,$ and $84$ with cobalancers$\ 6,\
35,$ and $204,$ respectively. It is clear from (\ref{1.1}) that $m$ is a
cobalancing number with cobalancer $r$ if and only if 
\begin{equation*}
m(m+1)=\dfrac{(m+r)(m+r+1)}{2},
\end{equation*}%
which when solved for $r$ gives 
\begin{equation}
r=\dfrac{-(2m+1)+\sqrt{8m^{2}+8m+1}}{2}.  \label{1.2}
\end{equation}%
It follows from (\ref{1.2}) that $m$ is a cobalancing number if and only if $%
8m^{2}+8m+1$ is a perfect square, that is, $m(m+1)$ is a triangular number.
Since $8\times 0^{2}+8\times 0+1=1$ is a perfect square, we accept $0$ is a
cobalancing number \cite{RAY, PANDA}. Also since $m(m+1)/2$ is known as a
triangular number by the very definition of triangular number, the above
discussion means that if $m$ is a cobalancing number, then both $m(m+1)$ and 
$m(m+1)/2$ are triangular numbers. Panda and Ray \cite{RAY} proved that
every cobalancing number is even. And also they showed that every balancer
is a cobalancing number and every cobalancer is a balancing number.

Oblong numbers are numbers of the form $O_{n}=n(n+1),$ where $n$ is a
positive integer. The $n$-th oblong number represents the number of points
in a rectangular array having $n$ columns and $n+1$ rows.\ The first few
oblong numbers are $2,6,12,20,30,42,56,72,90,110,$$\ldots$ (sequence 
\seqnum{A002378} in \cite{OEIS}). Since $2+4+6+$$\cdots$$+2n=2(1+2+3+$$\cdots$$%
+n)=2n(n+1)/2=n(n+1)=O_{n},$ the sum of the first $n$ even numbers equals
the $n$-th oblong number. Actually it is clear from the definition of oblong
numbers and triangular numbers that an oblong number is twice a triangular
number. After the definition of oblong numbers, we can say from (\ref{1.2})
that if $m$ is a cobalancing number, then $m(m+1)$ is both an oblong and
triangular number. Well then, what about the square triangular numbers?
Since triangular numbers are of the form $T_{n}=n(n+1)/2$ and square numbers
are of the form $S_{m}=m^{2},$ square triangular numbers are integer
solutions of the equation 
\begin{equation}
m^{2}=\dfrac{n(n+1)}{2}.  \label{1.3}
\end{equation}%
Eq.(\ref{1.3}) says something about the relation between balancing and
square triangular numbers. Behera and Panda \cite{BEHERA} proved that a
positive integer $m$ is a balancing number if and only if $m^{2}$ is a
triangular number, that is, $8m^{2}+1$ is a perfect square. Here, we will
get Eq.(\ref{1.3}) again using an amusing problem and we will see the
interesting relation between balancing and square triangular numbers by
means of this problem. In equation (\ref{1.0}), if we make the substitution $%
m+r=n$, then we get $1+2+$$\cdots$$+(m-1)=(m+1)+(m+2)+$$\cdots$$+n.$ Thus
this equation gives us a problem as follows.

I live on a street whose houses are numbered in order $1,2,3,$$\ldots$$,n-1,n
$; so the houses at the ends of the street are numbered $1$ and $n$. My own
house number is $m$ and of course $0<m<n$. One day, I add up the house
numbers of all the houses to the left of my house; then I do the same for
all the houses to the right of my house. I find that the sums are the same.
So how can we find $m$ and $n$ \cite{SHIRALI}? Since $1+2+3+$$\cdots$$%
+m-1=(m+1)+$$\cdots$$+(n-1)+n$, it follows that

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $%
\begin{equation*}
\frac{(m-1)m}{2}=\frac{n(n+1)}{2}-\frac{m(m+1)}{2}.
\end{equation*}%
Thus we get $m^{2}=n(n+1)/2$. Here, $m^{2}$ is both a triangular number and
a square number. That is, $m^{2}$ is a square triangular number. In Eq.(\ref%
{1.0}), since $m$ is a balancing number, it is easy to see that a balancing
number is the square root of a square triangular number. For more
information about triangular, square triangular and balancing numbers, one
can consult \cite{BEHERA, M.R. SCH, SUBRA 1, SUBRA 2, JAMES}.

\section{Preliminaries}

In this section, we introduce two kinds of sequences named generalized
Fibonacci and Lucas sequences $\left( U_{n}\right) $ and $\left(
V_{n}\right) $, respectively. Let $k$ and $t$ be two nonzero integers. The
generalized Fibonacci sequence is defined by $U_{0}=0,$ $U_{1}=1$ and $%
U_{n+1}=kU_{n}+tU_{n-1}$ for $n\geq 1$ and generalized Lucas sequence is
defined by $V_{0}=2,$ $V_{1}=k$ and $V_{n+1}=kV_{n}+tV_{n-1}$ for $n\geq 1,$
respectively. Also generalized Fibonacci and Lucas numbers for negative
subscript are defined as

\begin{equation}
U_{-n}=\dfrac{-U_{n}}{\left( -t\right) ^{n}}\text{ and }V_{-n}=\dfrac{V_{n}}{%
\left( -t\right) ^{n}}  \label{1.31}
\end{equation}%
for $n\geqslant 1$. For $k=t=1,$ the sequences $\left( U_{n}\right) $ and $%
\left( V_{n}\right) $ are called classic Fibonacci and Lucas sequences and
they are denoted as $\left( F_{n}\right) $ and $\left( L_{n}\right) ,$
respectively. The first Fibonacci numbers are $0,1,1,2,3,5,8,13,21,34,$%
$\ldots$ (sequence \seqnum{A000045} in \cite{OEIS}) and the first Lucas numbers are 
$2,1,3,4,7,11,18,29,47,76,$$\ldots$ (sequence \seqnum{A000032} in \cite{OEIS}). For 
$k=2$ and $t=1,$ the sequences $\left( U_{n}\right) $ and $\left(
V_{n}\right) $ are called Pell and Pell-Lucas sequences and they are denoted
as $\left( P_{n}\right) $ and $\left( Q_{n}\right) ,$ respectively. Thus $%
P_{0}=0,$ $P_{1}=1$ and $P_{n+1}=2P_{n}+P_{n-1}$ for $n\geqslant 1$ and $%
Q_{0}=2,$ $Q_{1}=2$ and $Q_{n+1}=2Q_{n}+Q_{n-1}$ for $n\geqslant 1.$ The
first few terms of Pell sequence are $0,1,2,5,12,29,70,169,408,985,$$\ldots$
(sequence \seqnum{A000129} in \cite{OEIS}) and the first few terms of Pell-Lucas
sequence are $2,2,6,14,34,82,198,478,1154,2786,$$\ldots$ (sequence \seqnum{A002203}
in \cite{OEIS}). Moreover, for $k=6$ and $t=-1,$ we represent $(U_{n})$ and $%
(V_{n})$ by $(u_{n})$ and $(v_{n}),$ respectively. Thus $u_{0}=0,$ $u_{1}=1$
and $u_{n+1}=6u_{n}-u_{n-1}$ and $v_{0}=2,$ $v_{1}=6$ and $%
v_{n+1}=6v_{n}-v_{n-1}$ for all $n\geqslant 1.$ The first few terms of the
sequence $(u_{n})$ are $0,1,6,35,204,$$\ldots$ (sequence \seqnum{A001109} in \cite%
{OEIS}) and the first few terms of the sequence $(v_{n})$ are $%
2,6,34,198,1154,$$\ldots$ (sequence \seqnum{A003499} in \cite{OEIS}). Furthermore,
from the equation (\ref{1.31}), it clearly follows that 
\begin{equation*}
u_{-n}=-u_{n}\text{ and }v_{-n}=v_{n}
\end{equation*}%
for all $n\geqslant 1.$ For more information about generalized Fibonacci and
Lucas sequences, one can consult \cite{KALMAN, OLCAY, MUSKAT, RABINOWITZ,
RIBENBOIM, ZFR}. Now we present some well known theorems and identities
regarding the sequences $(P_{n}),$ $(Q_{n}),$ $(u_{n}),$ and $(v_{n}),$
which will be useful during the proofs of the main theorems and the new
properties of the sequence $(y_{n}),$ where $y_{n}=(v_{n}-2)/4.$

\begin{theorem}
\label{t1.1}Let $\gamma $ and $\delta $ be the roots of the characteristic
equation $x^{2}-2x-1=0$. Then we have $P_{n}=\dfrac{\gamma ^{n}-\delta ^{n}}{%
2\sqrt{2}}$ and $Q_{n}=\gamma ^{n}+\delta ^{n}$ for all $n\geq 0.$
\end{theorem}

\begin{theorem}
\label{t1.2}Let $\alpha $ and $\beta $ be the roots of the characteristic
equation $x^{2}-6x+1=0$. Then 
\begin{equation}
u_{n}=\dfrac{\alpha ^{n}-\beta ^{n}}{4\sqrt{2}}  \label{1.44}
\end{equation}%
and 
\begin{equation*}
v_{n}=\alpha ^{n}+\beta ^{n}
\end{equation*}%
for all $n\geq 0.$
\end{theorem}

The formulas given in the above theorems are known as Binet's formulas. Let $%
B_{n}$ denote the $n-$th balancing number. From \cite{PANDA}, we know that 
\begin{equation}
B_{n}=\dfrac{(3+\sqrt{8})^{n}-(3-\sqrt{8})^{n}}{2\sqrt{8}}.  \label{1.45}
\end{equation}%
From Theorems\ \ref{t1.1} and \ref{t1.2}, it is easily seen that $%
u_{n}=B_{n}=P_{2n}/2$ and $v_{n}=Q_{2n}$ for $n\geq 0.$ Moreover, from
identities (\ref{1.44}) and (\ref{1.45}), it is easily seen that $%
B_{n}=u_{n} $ for negative integer $n.$ Then well known identities for $%
\left( P_{n}\right) ,$ $\left( Q_{n}\right) ,$ $\left( B_{n}\right) $ and $%
\left( v_{n}\right) $ are

\begin{equation}
Q_{n}^{2}-8P_{n}^{2}=4(-1)^{n},  \label{1.4}
\end{equation}

\begin{equation}
v_{n}^{2}-32B_{n}^{2}=4,  \label{1.5}
\end{equation}

\begin{equation}
B_{n}^{2}-6B_{n}B_{n-1}+B_{n-1}^{2}=1,  \label{1.6}
\end{equation}%
\begin{equation}
Q_{n}^{2}=Q_{2n}+2(-1)^{n},  \label{1.7}
\end{equation}%
\begin{equation}
B_{2n}=B_{n}v_{n},  \label{1.8}
\end{equation}%
\begin{equation}
P_{2n}=P_{n}Q_{n},  \label{1.9}
\end{equation}

and\ 

\begin{equation}
v_{n}^{2}=v_{2n}+2.  \label{1.10}
\end{equation}

In order to see close relations between balancing numbers and square
triangular numbers, we can give the following well known theorem which
characterizes all square triangular numbers. We omit the proof of this
theorem due to Karaatl\i\ and Keskin \cite{OLCAY}.

\begin{theorem}
\label{t1.3}A natural number $x$ is a square triangular number if and only
if $x=B_{n}^{2}$ for some natural number $n.$
\end{theorem}

Since $y_{n}=(v_{n}-2)/4,$ it follows that 
\begin{equation*}
B_{n}^{2}=\dfrac{v_{n}^{2}-4}{32}=\frac{1}{2}\frac{(v_{n}-2)}{4}\left( \frac{%
(v_{n}-2)}{4}+1\right) =\dfrac{y_{n}(y_{n}+1)}{2}.
\end{equation*}%
Then, it is seen that $x^{2}=\dfrac{y(y+1)}{2}$ for some positive integers $%
x $ and $y$ if and only if $x=B_{n}$ and $y=y_{n}$ for some natural number $%
n.$ Now we prove the following lemma given in \cite{POTTER}.

\begin{lemma}
\label{L2.3} The sequence $(y_{n})$ satisfies the recurrence relation $%
y_{n+1}=6y_{n}-y_{n-1}+2$ for $n\geqslant 1$ where $y_{0}=0$ and $y_{1}=1.$
\end{lemma}

%TCIMACRO{\TeXButton{Proof}{\begin{proof}}}%
%BeginExpansion
\begin{proof}%
%EndExpansion
Using the fact that $y_{n}=\dfrac{v_{n}-2}{4}$, we get 
\begin{eqnarray*}
6y_{n}-y_{n-1}+2 &=&6(v_{n}-2)/4-(v_{n-1}-2)/4+2 \\
&=&(6v_{n}-v_{n-1}-2)/4=(v_{n+1}-2)/4 \\
&=&y_{n+1}.
\end{eqnarray*}%
%TCIMACRO{\TeXButton{End Proof}{\end{proof}}}%
%BeginExpansion
\end{proof}%
%EndExpansion

The first few terms of the sequence $(y_{n})$ are $0,1,8,49,288,$$\ldots$
(sequence \seqnum{A001108}
in \cite{OEIS}). For $n=1,2,$$\ldots$$,$ let $b_{n}$ be $%
n-$th cobalancing number and so let $(b_{n})$ denote the cobalancing number
sequence. Then, the cobalancing numbers satisfy the similar recurrence
relation given in Lemma \ref{L2.3}. That is, $b_{n+1}=6b_{n}-b_{n-1}+2$ for $%
n\geq 1$ where $b_{0}=0$ and $b_{1}=2$ (see \cite[p.\ 1191]{RAY}). The
first few terms of the cobalancing number sequence are $0,2,14,84,492,$%
$\ldots$ (sequence \seqnum{A053141} in \cite{OEIS}). Moreover, there is a close
relation between cobalancing numbers, balancing numbers and the sequence $%
(y_{n}).$ In order to see this relation, we can give the following lemma
without proof.

\begin{lemma}
\label{L2.2 copy(1)}For every $n\geqslant 1,$ $b_{n}=y_{n}+B_{n}$ and $%
b_{n}=y_{n+1}-B_{n+1}.$
\end{lemma}

\begin{lemma}
\label{L2.2}For every $n\geqslant 1,$ $y_{2n}=8B_{n}^{2}$ and $%
y_{2n+1}=8B_{n}B_{n+1}+1.$
\end{lemma}

%TCIMACRO{\TeXButton{Proof}{\begin{proof}} }%
%BeginExpansion
\begin{proof}
%EndExpansion
By identities (\ref{1.5}) and (\ref{1.10}), we get 
\begin{equation*}
B_{n}^{2}=(v_{n}^{2}-4)/32=(v_{2n}-2)/32=y_{2n}/8.
\end{equation*}%
Thus it follows that $y_{2n}=8B_{n}^{2}.$ Also since $%
y_{n+1}=6y_{n}-y_{n-1}+2,$ it is easy to see that $%
y_{n}=(y_{n+1}+y_{n-1}-2)/6.$ By using $y_{2n}=8B_{n}^{2},$ we find that

\begin{equation*}
y_{2n+1}=(y_{2n+2}+y_{2n}-2)/6=(8B_{n+1}^{2}+8B_{n}^{2}-2)/6=(8(B_{n+1}^{2}+B_{n}^{2})-2)/6
\end{equation*}%
Since $B_{n+1}^{2}+B_{n}^{2}=6B_{n}B_{n+1}+1$ by identity (\ref{1.6}), it
follows that

\begin{equation*}
y_{2n+1}=(8(B_{n+1}^{2}+B_{n}^{2})-2)/6=\left( 8(6B_{n}B_{n+1}+1)-2\right)
/6=8B_{n}B_{n+1}+1.
\end{equation*}%
This completes the proof.%
%TCIMACRO{\TeXButton{End Proof}{\end{proof}}}%
%BeginExpansion
\end{proof}%
%EndExpansion

Now we can give the following theorem. Since its proof is easy, we omit it.

\begin{theorem}
\label{t2.2} If $n$ is an odd natural number, then $y_{n}=Q_{n}^{2}/4$ and
if $n$ is an even natural number, then $y_{n}=Q_{n}^{2}/4-1.$
\end{theorem}

Since $y_{2n+1}=8B_{n}B_{n+1}+1$ and $y_{2n+1}=$ $Q_{2n+1}^{2}/4,$ it
follows that $B_{n}B_{n+1}$ is a triangular number. Moreover, it follows
from Lemma \ref{L2.2} that $y_{n}$ is odd if and only if $n$ is odd and $%
y_{n}$ is even if and only if $n$ is even.

\section{Main Theorems}

In the previous sections, we mentioned the well known elementary properties
about triangular, square triangular, balancing and cobalancing numbers. In
this chapter, by using the previous theorems, lemmas and identities we prove
some new properties concerning balancing numbers and square triangular
numbers. The principal question of our interest is whether the product of
two balancing numbers greater than $1$ is another balancing number. We will
show that the answer to this question is negative. Similarly, we will show
that the product of two square triangular numbers greater than $1$ is not a
triangular number. The product of two oblong numbers may be another oblong
number and similarly, the product of two triangular numbers may be another
one. For a simple example, $2$ and $6$ are two oblong numbers. The product
of them is $2\times 6=12$ and $12=3(3+1)$ is another oblong number.
Similarly, $3$ and $15$ are two triangular numbers. The product of $3$ and $%
15$ is $3\times 15=45$ and $45=\dfrac{9(9+1)}{2}$ is another triangular
number. Also it is obvious that the product of two consecutive oblong
numbers is another oblong:

\begin{equation*}
\left[ (x-1)x\right] \left[ x(x+1)\right] =(x^{2}-1)x^{2}.
\end{equation*}%
For solving the general problem, we need to solve the Diophantine equation 
\begin{equation}
x(x+1)y(y+1)=z(z+1).  \label{1.11}
\end{equation}%
In \cite{TRYGV}, Breiteig gave recursion formulae for the solutions $x,y,$
and $z$ satisfying the equation (\ref{1.11}). For more information about the
product of two oblong numbers, one can consult \cite{TRYGV}.

The question of when the product of two oblong numbers is another one
suggests an analogous question for balancing numbers. When is the product of
two balancing numbers another balancing number? Now before giving these
properties concerning balancing numbers and square triangular numbers, we
present some theorems which will be needed in the proof of the main
theorems. Since the following two theorems are given in \cite{ZFR}, we omit
their proofs.

\begin{theorem}
\label{t3.1}Let $n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
\cup \left\{ 0\right\} $ and $m,r\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$ Then
\end{theorem}

\begin{equation}
P_{2mn+r}\equiv (-1)^{(m+1)n}P_{r} \quad (\text{mod } Q_{m}),  \label{3.1}
\end{equation}

\begin{equation}
Q_{2mn+r}\equiv (-1)^{(m+1)n}Q_{r} \quad (\text{mod } Q_{m}),  \label{3.2}
\end{equation}

\begin{equation}
P_{2mn+r}\equiv (-1)^{mn}P_{r} \quad (\text{mod } P_{m}),  \label{3.3}
\end{equation}%
and

\begin{equation}
Q_{2mn+r}\equiv (-1)^{mn}Q_{r} \quad (\text{mod } P_{m}).  \label{3.4}
\end{equation}

\begin{theorem}
\label{t3.2} Let $n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
\cup \left\{ 0\right\} $ and $m,r\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$ Then
\end{theorem}

\begin{equation}
B_{2mn+r}\equiv B_{r} \quad (\text{\rm mod } B_m),  \label{3.5}
\end{equation}%
\begin{equation}
v_{2mn+r}\equiv v_{r} \quad (\text{\rm mod } u_{m}),  \label{3.6}
\end{equation}

\begin{equation}
B_{2mn+r}\equiv (-1)^{n}B_{r}\quad (\text{\rm mod } v_{m}),  \label{3.7}
\end{equation}%
and

\begin{equation}
v_{2mn+r}\equiv (-1)^{n}v_{r} \quad (\text{\rm mod } v_{m}).  \label{3.8}
\end{equation}

The proofs of the following theorems can be given by using the above two
theorems. Also, we can find some of their proofs in \cite{HILTON}. Moreover,
some of them are given in \cite{RIBON} without proof.

\begin{theorem}
\label{t3.3} Let $m,n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ and $m\geqslant 2$. Then $P_{m}\mid P_{n}$ if and only if $m\mid n$.
\end{theorem}

\begin{theorem}
\label{t3.4}\ Let $m,n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ and $m\geqslant 2$. Then $Q_{m}\mid Q_{n}$ if and only if $m\mid n$ and $%
\dfrac{n}{m}$ is an odd integer.
\end{theorem}

\begin{theorem}
\label{t3.5} Let $m,n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ and $m\geqslant 2$. Then $Q_{m}\mid P_{n}$ if and only if $m\mid n$ and $%
\dfrac{n}{m}$ is an even integer.
\end{theorem}

Since $B_{n}=P_{2n}/2$ and $v_{n}=Q_{2n},$ the proofs of the following
theorems can be given by using the above theorems and identity (\ref{3.7}).

\begin{theorem}
\label{t3.6} Let $m,n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ and $m\geqslant 2$. Then $B_{m}\mid B_{n}$ if and only if $m\mid n$.
\end{theorem}

\begin{theorem}
\label{t3.7} Let $m,n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ and $m\geqslant 1$. Then $v_{m}\mid v_{n}$ if and only if $m\mid n$ and $%
\dfrac{n}{m}$ is an odd integer.
\end{theorem}

\begin{theorem}
\label{t3.8} Let $m,n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ and $m\geqslant 1$. Then $v_{m}\mid u_{n}$ if and only if $m\mid n$ and $%
\dfrac{n}{m}$ is an even integer.
\end{theorem}

The following theorem is a well known theorem (see \cite{PANDA, RIBON}).

\begin{theorem}
\label{t3.81}Let $m\geqslant 1$ and $n\geqslant 1$. Then $%
(B_{m},B_{n})=B_{(m,n)}$.
\end{theorem}

\begin{corollary}
\label{C3.1} Let $m\geqslant 1$ and $n\geqslant 1$. Then $%
(B_{m}^{2},B_{n}^{2})=B_{(m,n)}^{2}.$
\end{corollary}

Theorem \ref{t3.81} says that the greatest common divisor of any two
balancing numbers is again a balancing number. As a conclusion of this
theorem, Corollary \ref{C3.1} says that the greatest common divisor of any
two square triangular numbers is again a square triangular number. Now we
will discuss the least common multiple of any two balancing numbers. The
least common multiple of any two triangular numbers may be a triangular
number. For instance, $15$ and $21$ are two triangular numbers and $\left[
15,21\right] =105$ is again a triangular number. Note that $15\nmid 21$.
Similarly, the least common multiple of any two oblong numbers may be an
oblong number. For a simple example, $6$ and $15$ are two oblong numbers and 
$\left[ 6,15\right] =30$ is again an oblong number. But this is not true in
general for any two balancing numbers. This can be seen from the following
theorem.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ 

\begin{theorem}
\label{t3.10}Let $B_{n}>1,B_{m}>1$ and $B_{n}<B_{m}.$ Then $\left[
B_{n},B_{m}\right] $ is a balancing number if and only if $B_{n}\mid B_{m}.$
\end{theorem}

%TCIMACRO{\TeXButton{Proof}{\begin{proof}}}%
%BeginExpansion
\begin{proof}%
%EndExpansion
Assume that $B_{n}\mid B_{m}$. Then $\left[ B_{n},B_{m}\right] =B_{m}$ is
again a balancing number. Conversely, assume that $B_{n}>1,B_{m}>1$ and $%
B_{n}\nmid B_{m}.$ Then by Theorem \ref{t3.6}, $n\nmid m$. Let $d=(m,n)$.
Then by Theorem \ref{t3.81}, we get $(B_{n},B_{m})=B_{d}.$ Therefore

\begin{equation}
\left[ B_{n},B_{m}\right] =\frac{B_{n}B_{m}}{(B_{n},B_{m})}=\frac{B_{n}B_{m}%
}{B_{d}}.  \label{3.9}
\end{equation}%
Assume that $\left[ B_{n},B_{m}\right] $ is a balancing number. Thus $\left[
B_{n},B_{m}\right] =B_{r}$ for some natural number $r.$ Then by (\ref{3.9}),
we have $B_{n}B_{m}/B_{d}=B_{r}$. That is, $B_{n}B_{m}=B_{d}B_{r}$. Thus $%
\dfrac{B_{n}}{B_{d}}B_{m}=B_{r}$ and therefore $B_{m}\mid B_{r}$. This
implies that $r=mt$ for some natural number $t$ by Theorem \ref{t3.6}.
Assume that $t$ is an odd integer. Then $t=4q\mp 1$ for some $q\geqslant 1$.
Thus $B_{r}=B_{mt}=B_{4qm\mp m}=B_{2(2qm)\mp m}\equiv B_{\mp m}\ ({\rm mod\ } 
 B_{2m})$ by (\ref{3.5}).
This shows that $B_{r}\equiv \mp B_{m} \ ({\rm mod\ } B_{2m})$.
Since $B_{2m}=B_{m}v_{m}$ by (\ref{1.8}),
we see that $\dfrac{B_{n}}{B_{d}}B_{m}=B_{r}\equiv \mp B_{m}\ ({\rm mod}\ B_{m}v_{m})$.
Then it
follows that $\dfrac{B_{n}}{B_{d}}=\mp 1 \ ({\rm mod } v_{m})$. We assert
that $B_{n}\neq B_{d}$. On the contrary, assume that $B_{n}=B_{d}$. Then $%
n=d $ and this implies that $n\mid m$, which is impossible since $n\nmid m$.
Since $\dfrac{B_{n}}{B_{d}}\neq 1$ and $\dfrac{B_{n}}{B_{d}}\equiv \mp
1 \ ({\rm mod} \ v_{m})$,
it follows that $v_{m}\leqslant \dfrac{B_{n}}{B_{d}}\mp
1\leqslant \dfrac{B_{n}}{B_{d}}+1\leqslant B_{n}+1$. Since $B_{n}<B_{m}$, we
get $n<m$. This shows that $v_{n}<v_{m}\leqslant B_{n}+1$. On the other
hand, by identity (\ref{1.5}), we get $v_{n}>2B_{n}$. Therefore $%
2B_{n}<v_{n}<B_{n}+1$, which implies that $B_{n}<1$. But this is a
contradiction since $B_{n}>1$. Now assume that $t$ is an even integer. Then $%
t=2k$ and thus $r=mt=2mk$. Therefore $\dfrac{B_{n}}{B_{d}}%
B_{m}=B_{r}=B_{2km}=B_{km}v_{km}\geqslant B_{m}v_{m}$. This shows that $%
\dfrac{B_{n}}{B_{d}}\geqslant v_{m}$ and thus $v_{m}\leqslant \dfrac{B_{n}}{%
B_{d}}\leqslant B_{n}$. Since $n<m,$ we get $v_{n}<v_{m}\leqslant B_{n}$.
That is, $v_{n}<B_{n},$ which is impossible by identity (\ref{1.5}). This
completes the proof.%
%TCIMACRO{\TeXButton{End Proof}{\end{proof}}}%
%BeginExpansion
\end{proof}%
%EndExpansion

Now as a result of the above theorem, we can give the following corollary
which says something about the least common multiple of any two square
triangular numbers. The proof of the following corollary is straightforward,
using the fact that 
\begin{equation*}
\left[ a^{2},b^{2}\right] =\dfrac{a^{2}b^{2}}{(a^{2},b^{2})}=\dfrac{%
a^{2}b^{2}}{(a,b)^{2}}=\left( \dfrac{ab}{(a,b)}\right) ^{2}=\left[ a,b\right]
^{2}
\end{equation*}%
where $a$ and $b$ are positive integers.

\begin{corollary}
\label{C3.10}Let $B_{n}>1,B_{m}>1$ and $B_{n}<B_{m}$. Then $\left[
B_{n}^{2},B_{m}^{2}\right] $ is a triangular number if and only if $%
B_{n}^{2}\mid B_{m}^{2}.$ \ 
\end{corollary}

In order to answer the main question which is about the product of two
balancing numbers, we give the following theorem. This theorem says
something more than the above theorem.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 

\begin{theorem}
\label{t3.11} Let $n>1,m>1$ and $m\geqslant n$. Then there is no integer $r$
such that $B_{n}B_{m}=B_{r}$.
\end{theorem}

%TCIMACRO{\TeXButton{Proof}{\begin{proof}} }%
%BeginExpansion
\begin{proof}
%EndExpansion
Assume that $m>1,n>1$ and $B_{n}B_{m}=B_{r}$ for some $r>1$. Then $B_{m}\mid
B_{r}$ and therefore $m\mid r$ by Theorem \ref{t3.6}. Thus $r=mt$ for some
positive integer $t$. Assume that $t$ is an even integer. Then $t=2k$ and
therefore $r=mt=2mk$. Thus 
\begin{equation*}
B_{n}B_{m}=B_{r}=B_{2km}=B_{km}v_{km}
\end{equation*}%
by identity (\ref{1.8}). This shows that $B_{n}=\dfrac{B_{km}}{B_{m}}v_{km}$
and therefore $v_{km}\mid B_{n}$. By Theorem \ref{t3.8}, we get $km\mid n$
and $n/km=2s$ for some integer $s$. Then $n=2kms$. Since $n=2kms$ and $r=2km$%
, we get $n=rs$. Thus $r\mid n$. On the other hand, since $B_{n}B_{m}=B_{r},$
it follows that $B_{n}\mid B_{r}$ and therefore $n\mid r$ by Theorem \ref%
{t3.6}. This implies that $n=r$ and $B_{n}=B_{r}$. Since $B_{n}B_{m}=B_{r}$,
we get $B_{m}=1$, which is a contradiction. Now assume that $t$ is an odd
integer. Then $t=4q\mp 1$ for some positive integer $q$. Thus $r=mt=4qm\mp m$
and therefore 
\begin{equation*}
B_{r}=B_{4qm\mp m}=B_{2(2qm)\mp m}\equiv B_{\mp m} \ ({\rm mod}\text{ }B_{2m})
\end{equation*}%
by (\ref{3.5}). This shows that $B_{m}B_{n}\equiv \mp B_{m}\ ({\rm mod} \ B_{2m})$.
Since $B_{2m}=B_{m}v_{m}$, we get $B_{m}B_{n}\equiv \mp B_{m}\ 
({\rm mod}\ B_{m}v_{m})$,
which implies that $B_{n}\equiv \mp 1 \ ({\rm mod}\ 
v_{m})$. Therefore $v_{m}\mid B_{n}\mp 1$ and thus $v_{m}\leqslant B_{n}\mp
1 $. Since $v_{n}>2B_{n}$ and $m\geq n,$ we get $B_{n}+1\geqslant B_{n}\mp
1\geqslant v_{m}\geqslant v_{n}>2B_{n}$. This implies that $B_{n}+1>2B_{n}$.
Then $B_{n}<1$, which is a contradiction. This completes the proof.%
%TCIMACRO{\TeXButton{End Proof}{\end{proof}}}%
%BeginExpansion
\end{proof}%
%EndExpansion
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ 

Since square triangular numbers are square of the balancing numbers, the
above theorem says that the product of two square triangular numbers greater
than one is not a triangular number. Now we can give the following corollary
easily.

\begin{corollary}
\label{C3.2} The only positive integer solution of the system of Diophantine
equations $2u^{2}=x(x+1),$ $2v^{2}=$ $y(y+1)$ and $2u^{2}v^{2}=z(z+1)$ is
given by $(x,y,u,v,z)=(1,1,1,1,1).$
\end{corollary}

The following theorem gives a new property of the sequence $(y_{n}).$ It is
about the product of any two elements of the sequence $(y_{n})$ greater than 
$1.$

\begin{theorem}
\label{t3.12}Let $n>1$ and $m>1$. Then there is no integer $r$ such that $%
y_{n}y_{m}=y_{r}.$
\end{theorem}

%TCIMACRO{\TeXButton{Proof}{\begin{proof}}}%
%BeginExpansion
\begin{proof}%
%EndExpansion
Assume that $y_{n}y_{m}=y_{r}$. Since $y_{k}$ is odd if and only if $k$ is
odd and $y_{k}$ is even if and only if $k$ is even, we see that $m,n,$ and $%
r $ are odd or $r$ and at least one of the numbers $n$ and $m$ are even.
Assume that $n$ and $r$ are even. Then $n=2k$ and $r=2t$ for some positive
integers $k$ and $t$. By Lemma \ref{L2.2}, we have $y_{n}=y_{2k}=8B_{k}^{2}$
and $y_{r}=y_{2t}=8B_{t}^{2}$. Therefore 
\begin{equation*}
y_{m}=\dfrac{y_{r}}{y_{n}}=\dfrac{8B_{t}^{2}}{8B_{k}^{2}}=\left( \dfrac{B_{t}%
}{B_{k}}\right) ^{2}.
\end{equation*}%
If $m$ is even, then $m=2l$ for some positive integer $l$. This implies that 
$y_{m}=y_{2l}=8B_{l}^{2}$ and therefore $\left( \dfrac{B_{t}}{B_{u}}\right)
^{2}=8B_{l}^{2},$ which is impossible. So $m$ is odd. Then, by Theorem \ref%
{t2.2}, it follows that $y_{m}=\dfrac{Q_{m}^{2}}{4}$. Thus, we get $\dfrac{%
Q_{m}}{2}=\dfrac{B_{t}}{B_{k}}$ $=\dfrac{P_{2t}/2}{P_{2k}/2}=\dfrac{P_{2t}}{%
P_{2k}}=\dfrac{P_{r}}{P_{n}}$. This shows that $2P_{r}=P_{n}Q_{m}$. Since $n$
is even, $P_{n}$ is even. Also, since $P_{r}=P_{n}\dfrac{Q_{m}}{2}$, we see
that $P_{n}\mid P_{r}$ and $Q_{m}\mid P_{r}$. By Theorem \ref{t3.3} and
Theorem \ref{t3.5}, we get $r=nu$ and $r=2ms$ for some natural numbers $u$
and $s$. Since $2P_{r}=P_{n}Q_{m}$, we have 
\begin{equation*}
P_{n}Q_{m}=2P_{r}=2P_{2ms}=2P_{ms}Q_{ms}>P_{ms}Q_{ms}>P_{ms}Q_{m}.
\end{equation*}%
Therefore $P_{n}>P_{ms}$ and this implies that $n>ms$. Then $2n>2ms$ and
thus $2n>r=nu$. This shows that $u<2$. That is, $u=1$. Since $u=1$, we get $%
r=nu=n,$ which is impossible. Assume that $m,n$ and $r$ are odd integers.
Since $y_{n}y_{m}=y_{r},$ we get $\dfrac{Q_{n}^{2}}{4}\dfrac{Q_{m}^{2}}{4}=%
\dfrac{Q_{r}^{2}}{4}$ by Theorem \ref{t2.2}. Therefore $Q_{n}Q_{m}=2Q_{r}$.
This shows that $Q_{n}\mid Q_{r}$ and $Q_{m}\mid Q_{r}$. Then $r=nt$ and $%
r=mk$ for some odd natural numbers $t$ and $k$ by Theorem \ref{t3.4}. Since $%
t$ is an odd integer, $t=4q\mp 1$ for some $q\geqslant 1$. Thus 
\begin{equation*}
Q_{r}=Q_{nt}=Q_{4qn\mp n}=Q_{2(2qn)\mp n}\equiv \mp Q_{n}\ ({\rm mod}\text{ }
Q_{2n})
\end{equation*}%
by the congruence (\ref{3.2}). That is, $Q_{r}\equiv \mp Q_{n}\ ({\rm mod}\ 
Q_{2n})$. This implies that $2Q_{r}\equiv \mp 2Q_{n}\ ({\rm mod}\ 
Q_{2n}) $ and thus $Q_{n}Q_{m}\equiv \mp 2Q_{n}\ ({\rm mod}\ Q_{2n})$.
Since $Q_{2n}=Q_{n}^{2}+2,$ when $n$ is odd, by identity (\ref{1.7}), we
clearly have that $(Q_{n},Q_{2n})=2$. Then the congruence 
\begin{equation*}
Q_{n}Q_{m}\equiv \mp 2Q_{n} \quad ({\rm mod}\ Q_{2n})
\end{equation*}%
implies that 
\begin{equation*}
\dfrac{Q_{n}}{2}Q_{m}\equiv \mp 2\dfrac{Q_{n}}{2} \quad \left ({\rm mod}\ 
\dfrac{Q_{2n}}{2} \right) .
\end{equation*}%
This shows that $Q_{m}\equiv \mp 2\ ({\rm mod}\ Q_{2n}/2)$. Since $m\geq
2,$ we get $Q_{m}>2$ and therefore $Q_{2n}/2\leqslant Q_{m}\mp 2$. Therefore 
$Q_{2n}\leqslant 2Q_{m}\mp 4<2Q_{m}+4$. Similarly, by using $r=mk,$ it is
seen that $Q_{2m}<2Q_{n}+4$. Then it follows that $%
Q_{2n}+Q_{2m}<2Q_{m}+2Q_{n}+8$. Since $n$ and $m$ are odd integers, $%
Q_{2n}=Q_{n}^{2}+2$ and $Q_{2m}=Q_{m}^{2}+2$ by identity (\ref{1.7}). Thus,
we get 
\begin{equation*}
Q_{n}^{2}+2+Q_{m}^{2}+2<2Q_{m}+2Q_{n}+8.
\end{equation*}%
Since $Q_{n}^{2}+2+Q_{m}^{2}+2<2Q_{m}+2Q_{n}+8$, it follows that $%
Q_{n}^{2}+Q_{m}^{2}<2Q_{m}+2Q_{n}+4$. This implies that $%
Q_{n}^{2}-2Q_{n}+Q_{m}^{2}-2Q_{m}<4$. Then we get 
\begin{equation*}
Q_{n}(Q_{n}-2)+Q_{m}(Q_{m}-2)<4,
\end{equation*}%
which implies that $Q_{n}+Q_{m}<4$. This is a contradiction since $m>1$ and $%
n>1$. This completes the proof.%
%TCIMACRO{\TeXButton{End Proof}{\end{proof}}}%
%BeginExpansion
\end{proof}%
%EndExpansion

We easily obtain the following corollary.

\begin{corollary}
\label{C3.3}The only positive integer solution of the system of Diophantine
equations $x(x+1)=2u^{2},$ $y(y+1)=2v^{2},$ and $xy(xy+1)=2z^{2}$ is given
by $(x,y,u,v,z)=(1,1,1,1,1)$.
\end{corollary}

Balancing numbers and cobalancing numbers are related to the solutions of \
some Diophantine equations. Solutions of some of the Diophantine equations
are given in \cite{OLCAY}. Now we give four of them from \cite{OLCAY}.

\begin{theorem}
All positive integer solutions of the equation $x^{2}=y(y+1)/2$ are given by 
$(x,y)=(B_{n},y_{n})$ with $n\geq 1.$
\end{theorem}

\begin{theorem}
All positive integer solutions of the equation $(x+y-1)^{2}=8xy$ are given
by $(x,y)=(y_{n},y_{n+1})$ with $n\geq 1.$
\end{theorem}

\begin{theorem}
All positive integer solutions of the equation $x^{2}-6xy+y^{2}-1=0$ are
given by $(x,y)=(B_{n},B_{n+1})$ with $n\geq 1.$
\end{theorem}

\begin{theorem}
\label{t3.16}All positive integer solutions of the equation $%
(x+y-1)^{2}=8xy+1$ are given by $(x,y)=(b_{n},b_{n+1})$ with $n\geq 1.$
\end{theorem}

From Theorem \ref{t3.16}, it follows that $b_{n}b_{n+1}$ is a triangular
number for every natural number $n.$

Moreover, we can easily state the following theorems.

\begin{theorem}
All positive integer solutions of the equation $x^{2}-y^{2}+2xy+x-y=0$ are
given by $(x,y)=(B_{n},b_{n})$ with $n\geq 1.$
\end{theorem}

\begin{theorem}
All positive integer solutions of the equation $x^{2}+2y^{2}-4xy-x=0$ are
given by $(x,y)=(y_{n},b_{n})$ with $n\geq 1.$
\end{theorem}

\section{Concluding Remarks}

The sum of two triangular numbers may be a triangular number. For instance, $%
6$ and $15$ are triangular numbers and $6+15=21$ is again a triangular
number. Similarly, the sum of two oblong numbers may be another oblong
number. For instance, $12$ and $30$ are oblong numbers and $12+30=42$ is
again an oblong number. But we think that the sum of two square triangular
numbers is not a square triangular number. That is, $%
B_{n}^{2}+B_{m}^{2}=B_{r}^{2}$ has no solution if $n\geqslant 1$ and $%
m\geqslant 1$. We also think that there is no integer $r$ such that $%
B_{n}+B_{m}=B_{r}$ and $b_{n}+b_{m}=b_{r}$ for $n\geqslant 1$ and $%
m\geqslant 1.$ On the other hand, we think that the product of any two
cobalancing numbers greater than $1$ is not a cobalancing number. That is,
there is no integer $r$ such that $b_{n}b_{m}=b_{r}$ for $n\geqslant 1$ and $%
m\geqslant 1.$ Moreover, we think that there is no solution of the equation $%
y_{n}+y_{m}=y_{r}$ if $n\geqslant 1$ and $m\geqslant 1.$

\section{Acknowledgements}

The authors would like to thank the anonymous referee for careful reading
and suggestions that improved the clarity of this manuscript.

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\bibitem{KALMAN} D. Kalman and R. Mena, \textrm{The Fibonacci numbers-exposed%
}, \textit{Math. Mag.} \textbf{76 }(2003), 167--181.

\bibitem{OLCAY} O. Karaatl\i\ and R. Keskin, \textrm{On some Diophantine
equations related to square triangular and balancing numbers,}\emph{\ }%
\textit{J. Algebra, Number Theory:\ Adv. Appl.} \textbf{4 }(2010), 71--89.

\bibitem{MUSKAT} J. B. Muskat, \textrm{Generalized Fibonacci and Lucas
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\bibitem{RAY} G. K. Panda and P. K. Ray, \textrm{Cobalancing numbers and
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\bibitem{PANDA} G. K. Panda, \textrm{Some fascinating properties of
balancing numbers}, in \textit{Proc. Eleventh Internat. Conference on
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\bibitem{SHIRALI} S. A. Sh{\i}ral\i, \textrm{Fun with triangular numbers}, \hfil\newline
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\bibitem{STANDFOR} Stanford Math Circle, \textrm{Sunday, May 9, 2010,
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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11D25; Secondary 11D41, 11D72.

\noindent \emph{Keywords: } 
triangular number, square triangular number, balancing number, cobalancing number.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A000129},
\seqnum{A000217},
\seqnum{A001108},
\seqnum{A001109},
\seqnum{A002203},
\seqnum{A002378},
\seqnum{A003499}, and
\seqnum{A053141}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received June 13 2011;
revised version received  December 2 2011.
Published in {\it Journal of Integer Sequences}, December 27 2011.

\bigskip
\hrule
\bigskip

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


\end{document}