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\begin{center}
\vskip 1cm{\LARGE\bf 
The Equation $(j+k+1)^{2}-4k=Qn^{2}$ and Related Dispersions
}
\vskip 1cm
\large
Clark Kimberling\\
Department of Mathematics\\
University of Evansville\\
1800 Lincoln Avenue\\
Evansville, IN 47722\\
USA\\
\href{mailto:ck6@evansville.edu}{\tt ck6@evansville.edu} \\
\end{center}

\vskip .2 in

\begin{abstract}
Suppose $Q$ is a positive nonsquare integer congruent to $0$ or $1$ mod $4$.
Then for every positive integer $n$, there exists a unique pair $(j,k)$ of
positive integers such that $(j+k+1)^{2}-4k=Qn^{2}$.  This representation
is used to generate the fixed-$j$ array for $Q$ and the fixed-$k$ array for 
$Q$.  These arrays are proved to be dispersions; i.e., each array contains
every positive integer exactly once and has certain compositional and
row-interspersion properties.
\end{abstract}

%\newtheorem{theorem}{Theorem}[section]
%\newtheorem{proposition}{Proposition}[section]
%\newtheorem{corollary}{Corollary}[section]
%\newtheorem{lemma}{Lemma}[section] 


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\section{Introduction}

The Pell-like equation $m^{2}-4k=Qn^{2},$ where $Q$ is a positive nonsquare
integer congruent to $0$ or $1$ mod $4$, can be written in the form

\begin{equation}
(j+k+1)^{2}-4k=Qn^{2},  \tag{1}
\end{equation}%
or equivalently%
\begin{equation}
(j+k-1)^{2}+4j=Qn^{2}.  \tag{2}
\end{equation}%
When so written, there is, for each $n,$ a unique solution $(j,k);$ here and
throughout this work, the symbols $n,j,k$ represent positive integers. \ In
section 2, the existence and uniqueness of such a solution is proved. \ In
section 3, the definition of dispersion is recalled, and in section 4, it is
proved that certain arrays associated with solutions of (2) are dispersions.
\ In section 5, another class of arrays are proved to be dispersions. \ In
sections 4 and 5, $Q$ is restricted to $0$-congruence mod\textit{\ }$4,$ and
in section 6, conjectures are given for $Q\equiv 1$\textit{\ }mod\textit{\ }$%
4.$ \ In section 7, various numerical sequences associated with the
dispersions in preceding sections are discussed.

Throughout, we abbreviate $\sqrt{Q}/2$ as $Q_{1}.$

\section{Unique Representations}

\textbf{Theorem 1.} \ \textit{Suppose that }$Q\equiv 0$\textit{\ mod }$4.$%
\textit{\ \ Then the unique solution of (1) is given by}%
\begin{eqnarray}
j &=&(Q_{1}n)^{2}-\left\lfloor Q_{1}n\right\rfloor ^{2},  \TCItag{3} \\
k &=&(1+\left\lfloor Q_{1}n\right\rfloor )^{2}-(Q_{1}n)^{2}.  \TCItag{4}
\end{eqnarray}%
\textbf{Proof:} \ The method of proof is to assume that (1) has a solution,
to find it, and then to observe that it is unique. \ Let $m=j+k+1.$ \
Clearly, in order for (1) to hold, $m$ must be even, so we write $m=2h$. \
The equation $m^{2}-4k=Qn^{2}$ then yields%
\begin{equation*}
k=h^{2}-Qn^{2}/4.
\end{equation*}%
Equation (2) can be written as $(m-2)^{2}+4j=Qn^{2},$ from which%
\begin{equation*}
j=Qn^{2}/4-(h-1)^{2},
\end{equation*}%
so that the requirement that $\dot{j}>0$ is equivalent to%
\begin{equation*}
h<1+Q_{1}n.
\end{equation*}%
Also, the requirement that $k>0$ is equivalent to%
\begin{equation*}
h>Q_{1}n.
\end{equation*}%
Clearly there is exactly one such integer:%
\begin{equation*}
h=1+\left\lfloor Q_{1}n\right\rfloor ,
\end{equation*}%
from which (3) and (4) follow. \ \ \ \ \ \ \ \ \ $\blacksquare $\bigskip 

\textbf{Theorem 2.} \ \textit{Suppose that }$Q\equiv 1$\textit{\ mod }$4.$%
\textit{\ \ Then the unique solution of (1) is given by}%
\begin{eqnarray}
j &=&\left\{ 
\begin{tabular}{ll}
$(Q_{1}n)^{2}-(\left\lfloor 1/2+Q_{1}n\right\rfloor -1/2)^{2}$ & if $n$ is
odd, \\ 
$(Q_{1}n)^{2}-\left\lfloor Q_{1}n\right\rfloor ^{2}$ & if $n$ is even,%
\end{tabular}%
\right.  \TCItag{5} \\
k &=&\left\{ 
\begin{tabular}{ll}
$(\left\lfloor 1/2+Q_{1}n\right\rfloor +1/2)^{2}-(Q_{1}n)^{2}$ & if $n$ is
odd, \\ 
$(1+\left\lfloor Q_{1}n\right\rfloor )^{2}-(Q_{1}n)^{2}$ & if $n$ is even.%
\end{tabular}%
\right.  \TCItag{6}
\end{eqnarray}%
\textbf{Proof:} \ The method for even $n$ is the same as for Theorem 1. \ If 
$n$ is odd, then $m$ must be odd. \ Put $m=2h+1,$ and find the asserted
result using essentially the same method as for Theorem 1. \ \ \ \ \ \ \ \ \ 
$\blacksquare $\bigskip

\textbf{Theorem 3.} \ \textit{Suppose that }$Q\equiv 2$\textit{\ mod }$4$%
\textit{\ or }$Q\equiv 3$\textit{\ mod }$4$\textit{\ and that }$n$\textit{\
is even. \ Then the unique solution of (1) is given by (3) and (4). \ If }$n$%
\textit{\ is odd, then (1) has no solution.\medskip }

\textbf{Proof:} \ The method for even $n$ is the same as for Theorem 1. \
Now suppose $n$ is odd. \ Write $Q=r+4s,$ where $r=2$ or $r=3,$ and write $%
n=2h+1.$ \ Then $m^{2}=4k+(r+4s)(2h+1)^{2}\equiv r$ mod $4.$ \ But this is
contrary to the fact that the residues mod $4$ of the squares are all $0$ or 
$1.$ \ \ \ \ \ \ \ \ \ $\blacksquare $\bigskip 

\textbf{Corollary.} \ \textit{Suppose }$Q\equiv 2$\textit{\ mod }$4$\textit{%
\ or }$Q\equiv 3$\textit{\ mod }$4$\textit{\ and }$n$\textit{\ is a positive
integer. \ Then the unique solution of the equation}%
\begin{equation}
(j+k+1)^{2}-4k=4Qn^{2}  \tag{7}
\end{equation}%
\textit{is given by (3) and (4)}\medskip

\textbf{Proof:} \ The number $2n$ is even, so that Theorem 3 applies -- or,
as a second proof, apply Theorem 1, as $4Q\equiv 0$ mod $4$. \ \ \ \ \ \ \ \
\ $\blacksquare $

\section{Dispersions}

A dispersion is an array of positive integers in which each occurs exactly
once, and certain conditions (\cite{Kim1}, \cite{Slo2}) hold: \
specifically, an array $D=\{d(g,h)\}$ is \textit{the dispersion of a
strictly increasing sequence }$s(k)$ if the first column of $D$ is the
complement of $s(k)$ in increasing order, and%
\begin{eqnarray*}
d(1,2) &=&s(1)\geq 2; \\
d(1,h) &=&s(d(1,h-1))\text{ for all }h\geq 3; \\
d(g,h) &=&s(d(g,h-1))\text{ for all }h\geq 2,\text{ for all }g\geq 2.
\end{eqnarray*}%
In this section we introduce simple dispersions $D(r)$ and $E(r).$ \ In
later sections, related dispersions will be generated in connection with
equations (1) and (2).

Suppose $r>2$ is an irrational number, and let $D(r)$ denote the dispersion
of the sequence $\{\left\lfloor rn\right\rfloor \}$. \ In order to construct 
$D(r),$ let $r^{\prime }$ be the number given by%
\begin{equation*}
1/r+1/r^{\prime }=1,
\end{equation*}%
so that the Beatty sequences $\{\left\lfloor rn\right\rfloor \}$ and $%
\{\left\lfloor r^{\prime }n\right\rfloor \}$ partition the set of positive
integers. \ Writing $d(g,h)$ for the general term of $D(r),$ we have%
\begin{eqnarray*}
d(g,1) &=&\left\lfloor r^{\prime }g\right\rfloor \text{ for all }g\geq 1, \\
d(g,h) &=&\left\lfloor rd(g,h-1)\right\rfloor \text{ for all }g\geq 1\text{
and }h\geq 2.
\end{eqnarray*}%
Thus, all the terms of the sequence $\{\left\lfloor rn\right\rfloor \}$ are
dispersed in the columns of $D(r)$ excluding the first.\pagebreak \medskip

\textbf{Example 1.} \ The dispersion $D(r)$ for $r=3+2\sqrt{2}$ is
represented here and as \seqnum{A120858} in \cite{Slo1}.
Column 1 is the sequence $
d(g,1)=\left\lfloor r^{\prime }g\right\rfloor $ for $r^{\prime }=(1+\sqrt{2}%
)/2.$%
\begin{equation*}
\begin{tabular}{cccccc}
$1$ & $5$ & $29$ & $169$ & $985$ & $\cdots $ \\ 
$2$ & $11$ & $64$ & $373$ & $2174$ &  \\ 
$3$ & $17$ & $99$ & $577$ & $3363$ &  \\ 
$4$ & $23$ & $134$ & $781$ & $4552$ &  \\ 
$6$ & $34$ & $198$ & $1154$ & $6726$ &  \\ 
$7$ & $40$ & $233$ & $1358$ & $7915$ &  \\ 
$8$ & $46$ & $268$ & $1562$ & $9104$ &  \\ 
$9$ & $52$ & $303$ & $1766$ & $10293$ &  \\ 
$10$ & $58$ & $338$ & $1970$ & $11482$ &  \\ 
$12$ & $69$ & $402$ & $2343$ & $13656$ &  \\ 
$13$ & $75$ & $437$ & $2547$ & $14845$ &  \\ 
$\vdots $ &  &  &  &  & 
\end{tabular}%
\end{equation*}

Now suppose $r>1$ is an irrational number, and define%
\begin{equation*}
E(1,1)=1,\text{ \ \ \ \ }E(1,h)=\left\lfloor rE(1,h-1)\right\rfloor +1\text{
\ for }h\geq 2,
\end{equation*}%
and inductively define, for $g\geq 2,$%
\begin{eqnarray*}
E(g,1) &=&\text{least positive integer not among }E(i,h)\text{ for }1\leq
i\leq g-1,\text{ }h\geq 1; \\
E(g,h) &=&\left\lfloor rE(1,h-1)\right\rfloor +1\text{ \ for }h\geq 2.
\end{eqnarray*}%
Then $E=\{E(g,h)\}$ is clearly the dispersion of the sequence $s(k)$ given by%
\begin{equation*}
s(k)=\left\lfloor rk\right\rfloor +1,
\end{equation*}%
and the complement $\{s^{\prime }(k)\}$ of $\{s(k)\}$ is given by%
\begin{equation*}
s^{\prime }(k)=\left\lfloor r^{\prime }(k-1)\right\rfloor +1,\text{ \ \
where }1/r+1/r^{\prime }=1.
\end{equation*}

\textbf{Example 2.} \ The dispersion $E(3+2\sqrt{2})$ is represented here
and as \seqnum{A120859} in \cite{Slo1}:%
\begin{equation*}
\begin{tabular}{cccccc}
$1$ & $6$ & $35$ & $204$ & $1189$ & $\cdots $ \\ 
$2$ & $12$ & $70$ & $408$ & $2378$ &  \\ 
$3$ & $18$ & $105$ & $612$ & $4756$ &  \\ 
$4$ & $24$ & $140$ & $816$ & $4348$ &  \\ 
$5$ & $30$ & $175$ & $1020$ & $5945$ &  \\ 
$7$ & $41$ & $239$ & $1393$ & $8119$ &  \\ 
$8$ & $47$ & $274$ & $1597$ & $9308$ &  \\ 
$9$ & $53$ & $309$ & $1801$ & $10497$ &  \\ 
$10$ & $59$ & $344$ & $2005$ & $11686$ &  \\ 
$11$ & $65$ & $379$ & $2209$ & $12875$ &  \\ 
$13$ & $76$ & $443$ & $2582$ & $15049$ &  \\ 
$\vdots $ &  &  &  &  & 
\end{tabular}%
\end{equation*}

Next, suppose that $r$ is a quadratic irrational number: \ $r=a+b\sqrt{Q},$
where $Q$ is not a square. \ By Theorem 4 in $\cite{Kim1}$, the rows of $D(r)
$ and also the rows of $E(r)$ satisfy the recurrence%
\begin{equation*}
x_{n}=2ax_{n-1}+(b^{2}Q-a^{2})x_{n-2}.
\end{equation*}%
For example, the row recurrence for the dispersions in Examples 1 and 2 is%
\begin{equation*}
x_{n}=6x_{n-1}-x_{n-2}.
\end{equation*}

\section{Fixed-$j$ Dispersions for $Q\equiv 0$ mod $4$}

Suppose that $Q\equiv 0$ mod $4$ and that (2) holds for a triple $(j,k,n)$.
\ Keeping $j$ fixed, we ask what other pairs $(k_{1},n_{1})$ there are for
which $(j,k_{1},n_{1})$ is another solution of (1). \ This question is
answered by Theorem 4, which, with Theorems 5 and 6, shows that for given $j$%
, the set of such $n_{1}$ forms a row, or in some cases, several rows, of a
dispersion.\bigskip 

\textbf{Theorem 4.} \ \textit{Suppose that }$Q\equiv 0$\textit{\ mod }$4$%
\textit{\ and that }$n,j,k$\textit{\ satisfy (2). \ Let }$u$\textit{\ be the
least positive integer }$x$\textit{\ such that}%
\begin{equation*}
x^{2}-1=Qy^{2}/4
\end{equation*}%
\textit{\ for some positive integer }$y,$\textit{\ and let}%
\begin{eqnarray}
n_{1} &=&un+y\left\lfloor Q_{1}n\right\rfloor ,  \TCItag{8} \\
k_{1} &=&(1+\left\lfloor Q_{1}n_{1}\right\rfloor )^{2}-(Q_{1}n_{1})^{2}. 
\TCItag{9}
\end{eqnarray}%
\textit{Then}%
\begin{equation}
(j+k_{1}-1)^{2}+4j=Qn_{1}^{2}.  \tag{2A}
\end{equation}%
\bigskip 

\textbf{Proof:} \ In view of (2) and (2A), it suffices to prove that%
\begin{equation}
(j+k-1)^{2}-Qn^{2}=(j+k_{1}-1)^{2}-Qn_{1}^{2}.  \tag{10}
\end{equation}%
Using (3) and (4), we have%
\begin{equation*}
j+k-1=2\left\lfloor Q_{1}n\right\rfloor ,
\end{equation*}%
which, along with (3) and (8), leads to the following equivalent of (10):%
\begin{eqnarray}
(Q_{1}^{2}n^{2}-\left\lfloor Q_{1}n\right\rfloor ^{2}+k_{1}-1)^{2}
&=&4\left\lfloor Q_{1}n\right\rfloor ^{2}-Qn^{2}+Qu^{2}n^{2}  \notag \\
&&+2Quny\left\lfloor Q_{1}n\right\rfloor +Qy^{2}\left\lfloor
Q_{1}n\right\rfloor ^{2}.  \TCItag{11}
\end{eqnarray}%
Next using%
\begin{equation}
u^{2}-1=Q_{1}^{2}y^{2}  \tag{12}
\end{equation}%
and $Q_{1}^{2}=Q/4$, we find, after simplifications, that (11) is equivalent
to%
\begin{equation}
(Q_{1}^{2}n^{2}-\left\lfloor Q_{1}n\right\rfloor
^{2}+k_{1}-1)^{2}=(2u\left\lfloor Q_{1}n\right\rfloor +2nyQ_{1}^{2})^{2}. 
\tag{13}
\end{equation}%
Taking square roots leads to%
\begin{equation}
k_{1}=\left\lfloor Q_{1}n\right\rfloor ^{2}-Q_{1}^{2}n^{2}+2u\left\lfloor
Q_{1}n\right\rfloor +2ynQ_{1}^{2}+1  \tag{14}
\end{equation}%
as an equivalent of (13). \ Thus, using (9), what we must prove is that%
\begin{eqnarray}
(1+\left\lfloor Q_{1}n_{1}\right\rfloor )^{2} &=&\left\lfloor
Q_{1}n\right\rfloor ^{2}-Q_{1}^{2}n^{2}+2u\left\lfloor Q_{1}n\right\rfloor
+2ynQ_{1}^{2}+1  \notag \\
&&+Q_{1}^{2}(un+y\left\lfloor Q_{1}n\right\rfloor )^{2}.  \TCItag{15}
\end{eqnarray}%
Expanding the right side of (15) and again using (12), we find that the
right-hand side of (15) is a square, and taking square roots leaves the
following equivalent of (10):%
\begin{equation*}
1+\left\lfloor Q_{1}n_{1}\right\rfloor =u\left\lfloor Q_{1}n\right\rfloor
+ynQ_{1}^{2}+1,
\end{equation*}%
so that, using (8), what we must prove has now been reduced to%
\begin{equation}
\left\lfloor Q_{1}(un+y\left\lfloor Q_{1}n\right\rfloor )\right\rfloor
=u\left\lfloor Q_{1}n\right\rfloor +ynQ_{1}^{2}.  \tag{16}
\end{equation}

To prove (16), we begin by noting from (12) that%
\begin{equation}
(u-yQ_{1})(u+yQ_{1})=1,  \tag{17}
\end{equation}%
showing that%
\begin{equation*}
u-yQ_{1}>0.
\end{equation*}%
Let $\epsilon =Q_{1}n-\left\lfloor Q_{1}n\right\rfloor ,$ the fractional
part of $Q_{1}n.$ \ The fact that $(u-yQ_{1})\epsilon >0$ readily renders%
\begin{equation}
u\left\lfloor Q_{1}n\right\rfloor +ynQ_{1}^{2}<Q_{1}(un+y\left\lfloor
Q_{1}n\right\rfloor ).  \tag{18}
\end{equation}%
Also, since $u+yQ_{1}>1,$ we have from (17)%
\begin{equation}
u-yQ_{1}<1,  \tag{19}
\end{equation}%
so that $(u-yQ_{1})\epsilon <1,$ and consequently,%
\begin{equation*}
Q_{1}(un+y\left\lfloor Q_{1}n\right\rfloor )<u\left\lfloor
Q_{1}n\right\rfloor +ynQ_{1}^{2}+1.
\end{equation*}%
This and (18) imply (16) and hence (10). \ \ \ \ \ \ \ \ \ $\blacksquare $%
\bigskip 

Regarding $u$ and $y$ in Theorem 4 as sequences with terms $u(n)$ and $y(n),$
where $n=Q/4,$ we note that $u$ and $y$ are registered in \cite{Slo1} as
\seqnum{A033313} and \seqnum{A033317}. \ Several initial terms are shown here:%
\begin{equation*}
\begin{tabular}{||c||c|c|c|c|c|c|c|c|c|}
\hline
$Q$ & $8$ & $12$ & $20$ & $24$ & $28$ & $32$ & $40$ & $44$ & $48$ \\ \hline
$u$ & $3$ & $2$ & $9$ & $5$ & $8$ & $3$ & $19$ & $10$ & $7$ \\ \hline
$y$ & $2$ & $1$ & $4$ & $2$ & $3$ & $1$ & $6$ & $3$ & $4$ \\ \hline
\end{tabular}%
\end{equation*}%
\textbf{Theorem 5.}\textit{\ \ Suppose that }$Q,j,k,u,y,n,$\textit{\ and }$%
n_{1}$\textit{\ are as in Theorem 4, and let}%
\begin{equation}
n_{2}=un_{1}+y\left\lfloor Q_{1}n_{1}\right\rfloor .  \tag{20}
\end{equation}%
\textit{\ Then}%
\begin{equation}
n_{2}=2un_{1}-n.  \tag{21}
\end{equation}%
\textbf{Proof:} \ It suffices to prove that the right-hand sides of (21) and
(22) are equal, or equivalently, that%
\begin{equation}
n=un_{1}-y\left\lfloor Q_{1}n_{1}\right\rfloor .  \tag{22}
\end{equation}%
Using (8) to substitute for $n$ shows that (22) is equivalent to%
\begin{equation}
\left\lfloor unQ_{1}+yQ_{1}\left\lfloor Q_{1}n_{1}\right\rfloor
\right\rfloor =nyQ_{1}^{2}+u\left\lfloor Q_{1}n\right\rfloor .  \tag{23}
\end{equation}%
Now since $\left\lfloor nQ_{1}\right\rfloor <nQ_{1},$ we have%
\begin{equation}
nyQ_{1}^{2}+u\left\lfloor Q_{1}n\right\rfloor <unQ_{1}+yQ_{1}\left\lfloor
Q_{1}n_{1}\right\rfloor .  \tag{24}
\end{equation}%
Also, in connection with (17),%
\begin{equation*}
Q_{1}n-\left\lfloor Q_{1}n\right\rfloor <1/(u-yQ_{1}),
\end{equation*}%
which implies%
\begin{equation*}
unQ_{1}+yQ_{1}\left\lfloor Q_{1}n_{1}\right\rfloor
<nyQ_{1}^{2}+u\left\lfloor Q_{1}n\right\rfloor +1,
\end{equation*}%
which with (24) establishes (23) and hence (20). \ \ \ \ \ \ \ \ \ $%
\blacksquare $\bigskip 

We turn now to the construction of the \textit{fixed-}$j$\textit{\ array of }%
$Q$. \ By Theorem 1, every $n$ has a unique representation of a certain form
depending on a pair $(j,k)$ which we shall now write as $(j_{n},k_{n}).$ \
By Theorem 4, for each $j_{n},$ there are infinitely many pairs $(n^{\prime
},k^{\prime })$ such that (1) and (2) hold; that is,%
\begin{equation}
(j_{n}+k_{n^{\prime }}-1)^{2}+4j_{n}=Q(n^{\prime })^{2}.  \tag{25}
\end{equation}

Let $S_{1}$ be the set of all $n^{\prime }$ for which (1A) holds for $n=1$
and some $k_{n^{\prime }}.$ \ By Theorem 4, $n_{1}=u+y\left\lfloor
Q_{1}\right\rfloor \in S_{1}$. \ (Note that in (25), 
\begin{equation*}
j_{1}=Q/4-\left\lfloor \sqrt{Q}/2\right\rfloor ,
\end{equation*}%
this being the \textquotedblleft fixed $j$\textquotedblright\ used to define 
$S_{1}).$ \ By Theorem 5, the numbers given by the recurrence relation%
\begin{equation*}
x_{i}=2ux_{i-1}-x_{i-1},
\end{equation*}%
with initial values $x_{1}=1$ and $x_{2}=n_{1},$ all lie in $S_{1}.$ \
Possibly they comprise all of $S_{1},$ but if not, let $w$ be the least
number in $S_{1}$ that is not an $x_{i}.$ \ By Theorems 4 and 5, we obtain
another recurrence sequence $\{w_{i}\}$ lying in $S_{1}.$ \ If $S_{1}$
contains a number not in $\{x_{i}\}$ and not in $\{w_{i}\},$ the production
of recurrence sequences with all terms in $S_{1}$ can be continued
inductively, so that $S_{1}$ is partitioned into a set of such sequences.

Next, let $\upsilon $ be the least positive integer not in $S_{1},$ and let $%
S_{2}$ be the set of all $\upsilon ^{\prime }$ for which (25) holds for $%
n=\upsilon $ and some $k_{\upsilon ^{\prime }}.$ \ Again Theorems 4 and 5
apply, so that $S_{2}$ is partitioned into a set of recurrence sequences.

Continuing in this manner, the set of all positive integers is partitioned
into sets $S_{i},$ which are themselves partitioned into sets of recurrence
sequences. \ Next, arrange these sequences so that their first terms form an
increasing sequence, thus producing an array of all the positive integers,
with increasing first column, and whose rows all satisfy the recurrence
relation $x_{i}=2ux_{i-1}-x_{i-1}.$\bigskip

\textbf{Example 3.} \ The fixed-$j$ array for $Q=8.$ \ (This is \seqnum{A120860} in 
\cite{Slo1}.)%
\begin{equation*}
\begin{tabular}{cccccc}
$1$ & $5$ & $29$ & $169$ & $985$ & $\cdots $ \\ 
$2$ & $10$ & $58$ & $338$ & $1970$ &  \\ 
$3$ & $17$ & $99$ & $577$ & $3363$ &  \\ 
$4$ & $22$ & $128$ & $746$ & $4348$ &  \\ 
$6$ & $34$ & $198$ & $1154$ & $6726$ &  \\ 
$7$ & $39$ & $227$ & $1323$ & $7711$ &  \\ 
$8$ & $46$ & $268$ & $1562$ & $9104$ &  \\ 
$9$ & $51$ & $297$ & $1731$ & $10089$ &  \\ 
$11$ & $63$ & $367$ & $2139$ & $12467$ &  \\ 
$12$ & $68$ & $396$ & $2308$ & $13452$ &  \\ 
$13$ & $75$ & $437$ & $2547$ & $14845$ &  \\ 
$\vdots $ &  &  &  &  & 
\end{tabular}%
\end{equation*}

We prove below that the array in Example 3 is the dispersion of the ordered
sequence $s$ of numbers not in its first column; this sequence appears to be
\seqnum{A098021}, described in \cite{Slo1} as \textquotedblleft Irrational rotation
of the square root of $2$ as an implicit sequence with an uneven Cantor
cartoon.\textquotedblright\ \ (A Cantor cartoon is a kind of geometric
fractal, so that \seqnum{A098021} indicates a connection between geometric fractals
and fractal integer sequences as defined just before Example 15. 
The sequence $s$ and \seqnum{A098021} agree for 30
terms and no later discrepancies were detected.) 

In order to prove that the fixed-$j$ array of $Q$ is actually a dispersion,
we begin with a lemma which applies to a wider class of arrays all of whose
row sequences satisfy a common second-order recurrence.\bigskip

\textbf{Lemma.} \ \textit{Suppose that }$p$\textit{\ and }$q$\textit{\ are
integers, that }$p>0,$\textit{\ and that }$A=\{a(g,h)\}$\textit{\ is an
array consisting of all the positive integers, each exactly once. \ Suppose
further that the first column of }$A$\textit{\ is increasing and that every
row sequence satisfies the recurrence}%
\begin{equation*}
x_{n}=px_{n-1}+qx_{n-2}
\end{equation*}%
\textit{for }$n\geq 3.$\textit{\ \ Finally, suppose for arbitrary indices }$%
g,g_{1},g_{2},$\textit{\ and }$h$\textit{\ that}%
\begin{equation}
\text{if }a(g_{1},h)<a(g,1)<a(g_{1},h+1),\text{ then }%
a(g_{1},h+1)<a(g,2)<a(g_{1},h+2).  \tag{26}
\end{equation}%
\textit{Then }$A$\textit{\ is a dispersion.}\bigskip 

\textbf{Proof:} \ An array is a dispersion if and only if it is an
interspersion. \ In \cite{Kim1}, the four defining properties of an
interspersion are given, and the first three are assumed in our present
hypothesis. \ It remains to prove the fourth, which is as follows:%
\begin{equation}
\text{if }(\sigma _{i})\text{ and }(\tau _{i})\text{ are rows of }A\text{
and }\sigma _{g}<\tau _{h}<\sigma _{g+1},\text{ then }\sigma _{g+1}<\tau
_{h+1}<\sigma _{g+2}.  \tag{27}
\end{equation}%
(Property (27) essentially means that, beginning at the first term of any
row having greater initial term than that of another row, all the following
terms individually separate the individual terms of the other row; in this
sense, \textit{every} pair of rows are mutually interspersed, as in Examples
1-4.)

Now suppose $(\sigma _{i})$ and $(\tau _{i})$ are distinct rows, that $g\geq
1$ and $h\geq 1,$ and that%
\begin{eqnarray}
\sigma _{g} &<&\tau _{h}<\sigma _{g+1},  \TCItag{28} \\
\sigma _{g+1} &<&\tau _{h+1}<\sigma _{g+2}.  \TCItag{29}
\end{eqnarray}%
\ Then since%
\begin{equation*}
\sigma _{g+2}=p\sigma _{g+1}+q\sigma _{g}\text{ \ \ and \ \ }\tau
_{g+2}=p\tau _{g+1}+q\tau _{g},
\end{equation*}%
we have $\sigma _{g+2}<\tau _{h+2}<\sigma _{g+3},$ and, inductively,%
\begin{equation*}
\sigma _{g^{\prime }}<\tau _{h^{\prime }}<\sigma _{g^{\prime }+1}
\end{equation*}%
for all $g^{\prime }\geq g$ and $h^{\prime }\geq h.$ $\ $This is to say, if
(28) and (29) hold, then the rows $(\sigma _{i})$ and $(\tau _{i})$ are
interspersed beginning at term $\tau _{h}.$ \ The hypothesis (26) implies
(28) and (29) for $h=1,$ so that (27) holds. \ \ \ \ \ \ \ \ \ $\blacksquare 
$\bigskip 

\textbf{Theorem 6.} \ \textit{Suppose that }$Q\equiv 0$\textit{\ mod }$4.$%
\textit{\ \ Then the fixed-}$j$\textit{\ array for }$Q$\textit{\ is a
dispersion.}\bigskip

\textbf{Proof: \ }Let $\{d(g,h\}$ be the fixed-$j$ array for $Q,$ and
suppose that $g,g_{1},$ and $h$ are indices such that%
\begin{equation}
d(g_{1},h)<d(g,1)<d(g_{1},h+1).  \tag{30}
\end{equation}%
With reference to the lemma, we wish to prove that%
\begin{equation}
d(g_{1},h+1)<d(g,2)<d(g_{1},h+2).  \tag{31}
\end{equation}%
Abbreviate (30) and (31) as%
\begin{equation}
n<w<n^{\prime }  \tag{32}
\end{equation}%
and%
\begin{equation}
n^{\prime }<w^{\prime }<n^{\prime \prime },  \tag{33}
\end{equation}%
where, by Theorem 5,%
\begin{eqnarray*}
n^{\prime } &=&un+y\left\lfloor Q_{1}n\right\rfloor , \\
w^{\prime } &=&uw+y\left\lfloor Q_{1}w\right\rfloor , \\
n^{\prime \prime } &=&2un^{\prime }-n.
\end{eqnarray*}%
The obvious inequality%
\begin{equation*}
0<u(w-n)+y(\left\lfloor Q_{1}w\right\rfloor -\left\lfloor
Q_{1}n\right\rfloor )
\end{equation*}%
easily implies $n^{\prime }<w^{\prime }.$ \ To prove that $w^{\prime
}<n^{\prime \prime },$ we appeal to (19) to see that%
\begin{equation}
(u-yQ_{1})(Q_{1}n-\left\lfloor Q_{1}n\right\rfloor )<1,  \tag{34}
\end{equation}%
which readily yields%
\begin{equation*}
unQ_{1}+yQ_{1}\left\lfloor Q_{1}n\right\rfloor <yQ_{1}^{2}+u\left\lfloor
Q_{1}n\right\rfloor +1.
\end{equation*}%
The hypothesis that%
\begin{equation}
w<un+y\left\lfloor Q_{1}n\right\rfloor   \tag{35}
\end{equation}%
implies%
\begin{equation*}
Q_{1}w<unQ_{1}+yQ_{1}\left\lfloor Q_{1}n\right\rfloor ,
\end{equation*}%
so that by (34),%
\begin{equation*}
Q_{1}w<ynQ_{1}^{2}+u\left\lfloor Q_{1}n\right\rfloor +1,
\end{equation*}%
which implies%
\begin{equation*}
\left\lfloor Q_{1}w\right\rfloor -u\left\lfloor Q_{1}n\right\rfloor
<ynQ_{1}^{2}.
\end{equation*}%
Multiply through by $y$ and apply (12) to obtain%
\begin{equation*}
n+y\left\lfloor Q_{1}w\right\rfloor <u^{2}n+uy\left\lfloor
Q_{1}n\right\rfloor ,
\end{equation*}%
equivalent to%
\begin{equation}
u^{2}n+uy\left\lfloor Q_{1}n\right\rfloor +y\left\lfloor Q_{1}w\right\rfloor
<2u^{2}n+2uy\left\lfloor Q_{1}n\right\rfloor -n.  \tag{36}
\end{equation}%
Returning to the hypothesis, we have%
\begin{equation*}
uw<u^{2}n+uy\left\lfloor Q_{1}n\right\rfloor ,
\end{equation*}%
so that%
\begin{equation*}
uw+y\left\lfloor Q_{1}w\right\rfloor <u^{2}n+uy\left\lfloor
Q_{1}n\right\rfloor +y\left\lfloor Q_{1}w\right\rfloor .
\end{equation*}%
This and (36) give%
\begin{equation*}
uw+y\left\lfloor Q_{1}w\right\rfloor <2u(un+y\left\lfloor
Q_{1}n\right\rfloor )-n,
\end{equation*}%
which is $w^{\prime }<n^{\prime \prime }.$ \ To summarize, (30) implies
(31). \ Therefore, the lemma applies, and $\{d(g,h)\}$ is a dispersion. \ \
\ \ \ \ \ \ \ $\blacksquare $\bigskip 

\section{Fixed-$k$ Dispersions for $Q\equiv 0$ mod $4$}

Loosely speaking, if we hold $k$ fixed instead of $j,$ then the methods of
section 4 yield another kind of dispersion. \ The purpose of this section is
to define these fixed-$k$ arrays and to prove that they are dispersions. \
Theorems 7 and 8 are similar to Theorems 4 and 5, and the fixed-$k$ array is
then defined with reference to Theorems 7 and 8. \ Thereafter, Example 4
using $Q=8$ is presented, and then lemma of section 4 is used to prove
Theorem 9, similar to Theorem 6.\bigskip 

\textbf{Theorem 7.} \ \textit{Suppose that }$Q\equiv 0$\textit{\ mod }$4$%
\textit{\ and that }$n,j,k$\textit{\ satisfy (1). \ Let }$u,$ $y,$\textit{\
and }$r$ \textit{be as in Theorem 4, and let}%
\begin{eqnarray}
n_{1} &=&un+y\left\lfloor Q_{1}n\right\rfloor +y,  \TCItag{37} \\
j_{1} &=&(Q_{1}n_{1})^{2}-\left\lfloor Q_{1}n_{1}\right\rfloor ^{2}. 
\TCItag{38}
\end{eqnarray}%
\textit{Then}%
\begin{equation}
(j_{1}+k+1)^{2}-4k=Qn_{1}^{2}.  \tag{1A}
\end{equation}%
\bigskip 

\textbf{Proof:} \ In view of (1) and (1A), it suffices to prove that%
\begin{equation}
(j+k+1)^{2}-Qn^{2}=(j+k_{1}-1)^{2}-Qn_{1}^{2}.  \tag{39}
\end{equation}%
Following the method of proof of Theorem 4, we find (39) equivalent to%
\begin{eqnarray*}
(j_{1}+(1+\left\lfloor Q_{1}n\right\rfloor )^{2}-Q_{1}^{2}n^{2}+1)^{2}
&=&(2+2\left\lfloor Q_{1}n\right\rfloor )^{2}+Q_{1}n_{1}^{2}-Q_{1}n^{2} \\
&=&4(u+u\left\lfloor Q_{1}n\right\rfloor +yQ_{1}^{2}n)^{2}.
\end{eqnarray*}%
Taking square roots leads to%
\begin{equation}
j_{1}=-\left\lfloor Q_{1}n\right\rfloor
^{2}+Q_{1}^{2}n^{2}+2(u-1)\left\lfloor Q_{1}n\right\rfloor
+2ynQ_{1}^{2}n+2(u-1)  \tag{40}
\end{equation}%
as an equivalent of (39). \ Thus, using (36), what we must prove is that%
\begin{eqnarray}
\left\lfloor Q_{1}n_{1}\right\rfloor ^{2} &=&Q_{1}^{2}(un+y\left\lfloor
Q_{1}n\right\rfloor +y)^{2}+\left\lfloor Q_{1}n\right\rfloor
^{2}-Q_{1}^{2}n^{2}  \notag \\
&&-2(u-1)\left\lfloor Q_{1}n\right\rfloor -2ynQ_{1}^{2}n-2(u-1).  \TCItag{41}
\end{eqnarray}%
Expanding the right side of (41) and using (12) leads to factoring the
right-hand side as a square, and then taking square roots leaves the
following equivalent of (39):%
\begin{equation*}
\left\lfloor Q_{1}n_{1}\right\rfloor ^{2}=u\left\lfloor Q_{1}n\right\rfloor
+ynQ_{1}^{2}+u-1.
\end{equation*}%
Thus, in view of (37), we need only prove that%
\begin{equation*}
\left\lfloor Q_{1}(un+y\left\lfloor Q_{1}n\right\rfloor +y)\right\rfloor
=u\left\lfloor Q_{1}n\right\rfloor +ynQ_{1}^{2}+u-1;
\end{equation*}%
but this now follows easily from%
\begin{equation*}
(u-yQ_{1})(1-\epsilon )>0,
\end{equation*}%
where $\epsilon =Q_{1}n-\left\lfloor Q_{1}n\right\rfloor .$ \ \ \ \ \ \ \ \
\ $\blacksquare $\bigskip 

\textbf{Theorem 8.} \ \textit{Suppose that }$Q,j,k,u,y,n,$\textit{\ and }$%
n_{1}$\textit{\ are as in Theorem 5, and let}%
\begin{equation}
n_{2}=un_{1}+y\left\lfloor Q_{1}n_{1}\right\rfloor +y.  \tag{42}
\end{equation}%
\textit{\ Then}%
\begin{equation}
n_{2}=2un_{1}-n.  \tag{21}
\end{equation}

A proof similar to that of Theorem 7 is omitted. \ We do note, however, that
(42) and (21) yield%
\begin{equation*}
n=un_{1}-y\left\lfloor Q_{1}n_{1}\right\rfloor -y,
\end{equation*}%
an inversion formula for (37), as (22) is for (8).\bigskip 

We are now in a position to define, by construction, the fixed-$k$ array for 
$Q.$ \ \ Each $n$ is uniquely represented as in (1) by a pair $(j,k)$, which
we write as $(j_{n},k_{n}).$ \ By Theorem 7, for each $k_{n},$ there are
infinitely many pairs $(n^{\prime },j^{\prime })$ such that (1) and (2)
hold; that is,%
\begin{equation}
(j_{n^{\prime }}+k_{n}+1)^{2}-4k_{n}=Q(n^{\prime })^{2}.  \tag{46}
\end{equation}

Let $S_{1}$ be the set of all $n^{\prime }$ for which (1A) holds for $n=1$
and some $j_{n^{\prime }}.$ \ By Theorem 7, $n_{1}=u+y\left\lfloor
Q_{1}\right\rfloor \in S_{1}$. \ (Note that in (46), 
\begin{equation*}
k_{1}=\left\lfloor 1+\sqrt{Q}/2\right\rfloor ^{2}-Q/4,
\end{equation*}%
this being the ``fixed $k$" used to define $S_{1}).$ \ By Theorem 8, the
numbers given by the recurrence relation%
\begin{equation*}
x_{i}=2ux_{i-1}-x_{i-1},
\end{equation*}%
with initial values $x_{1}=1$ and $x_{2}=n_{1},$ all lie in $S_{1}.$ \
Possibly they comprise all of $S_{1},$ but if not, let $w$ be the least
number in $S_{1}$ that is not an $x_{i}.$ \ By Theorems 7 and 8, we obtain
another recurrence sequence $\{w_{i}\}$ lying in $S_{1}.$ \ If $S_{1}$
contains a number not in $\{x_{i}\}$ and not in $\{w_{i}\},$ the production
of recurrence sequences with all terms in $S_{1}$ can be continued
inductively, so that $S_{1}$ is partitioned into a set of such sequences. \
Next, let $\upsilon $ be the least positive integer not in $S_{1},$ and let $%
S_{2}$ be the set of all $\upsilon ^{\prime }$ for which (46) holds for $%
n=\upsilon $ and some $k_{\upsilon ^{\prime }}.$ \ Again Theorems 7 and 8
apply, so that $S_{2}$ is partitioned into a set of recurrence sequences.

Continuing in this manner, the set of all positive integers is partitioned
into sets $S_{i},$ which are themselves partitioned into sets of recurrence
sequences. \ Next, arrange these sequences so that their first terms form an
increasing sequence, thus producing an array of all the positive integers,
with increasing first column, and whose rows all satisfying the recurrence
relation $x_{i}=2ux_{i-1}-x_{i-1}.$ \ This array is the fixed-$k$ array$.$%
\bigskip

\textbf{Example 4.} \ The fixed-$k$ array for $Q=8$. 
(This is \seqnum{A120861} in 
\cite{Slo1}.)%
\begin{equation*}
\begin{tabular}{cccccc}
$1$ & $7$ & $41$ & $239$ & $1393$ & $\cdots $ \\ 
$2$ & $12$ & $70$ & $408$ & $2378$ &  \\ 
$3$ & $19$ & $111$ & $647$ & $3771$ &  \\ 
$4$ & $24$ & $140$ & $816$ & $4756$ &  \\ 
$6$ & $31$ & $181$ & $1055$ & $6149$ &  \\ 
$7$ & $36$ & $210$ & $1224$ & $7134$ &  \\ 
$8$ & $48$ & $280$ & $1632$ & $9512$ &  \\ 
$9$ & $53$ & $309$ & $1801$ & $10497$ &  \\ 
$11$ & $60$ & $350$ & $2040$ & $11890$ &  \\ 
$12$ & $65$ & $379$ & $2209$ & $12875$ &  \\ 
$13$ & $77$ & $449$ & $2617$ & $15253$ &  \\ 
$\vdots $ &  &  &  &  & 
\end{tabular}%
\end{equation*}

\textbf{Theorem 9.} \ \textit{Suppose that }$Q\equiv 0$\textit{\ mod }$4.$%
\textit{\ \ Then the fixed-}$k$\textit{\ array for }$Q$\textit{\ is a
dispersion.}\bigskip

A proof using%
\begin{eqnarray*}
n^{\prime } &=&un+y\left\lfloor Q_{1}n\right\rfloor +y, \\
w^{\prime } &=&uw+y\left\lfloor Q_{1}w\right\rfloor +y, \\
n^{\prime \prime } &=&2un^{\prime }-n
\end{eqnarray*}%
is very similar to that of Theorem 6 and is omitted.

\section{Dispersions for $Q\equiv 1$ mod $4$}

If $Q$ $\equiv 1$\textit{\ }mod\textit{\ }$4,$ the parity of $n$ leads to
two cases, as in Theorem 2, and to more subtle results of the sort given in
Theorems 4 and 5 (and Theorems 7 and 8). \ We offer the following
conjectures:\bigskip

\textbf{Conjecture 1.} \ \textit{Suppose that }$Q\equiv 1$\textit{\ mod }$4$%
\textit{\ and that }$n,j,k$\textit{\ satisfy (2). \ Let }$u$\textit{\ be the
least positive integer }$x$\textit{\ such that}%
\begin{equation*}
x^{2}-4=Qy^{2}
\end{equation*}%
\textit{\ for some positive integer }$y,$\textit{\ let}%
\begin{eqnarray*}
r &=&(u+y\sqrt{Q})/2, \\
n_{1} &=&\left\lfloor rn\right\rfloor -\left\lfloor yf(n)\right\rfloor ,
\end{eqnarray*}%
\textit{where }$f(n)$\textit{\ denotes the fractional part of }$(1+\sqrt{Q}%
)n/2,$\textit{\ and let}%
\begin{equation*}
k_{1}=\left\{ 
\begin{tabular}{ll}
$(\left\lfloor 1/2+Q_{1}n_{1}\right\rfloor +1/2)^{2}-(Q_{1}n_{1})^{2}$ & if $%
n_{1}$ is odd, \\ 
$(1+\left\lfloor Q_{1}n_{1}\right\rfloor )^{2}-(Q_{1}n_{1})^{2}$ & if $n_{1}$
is even.%
\end{tabular}%
\right. 
\end{equation*}%
\textit{Then}%
\begin{equation*}
(j+k_{1}-1)^{2}+4j=Qn_{1}^{2}.
\end{equation*}%
\textit{Moreover, if}%
\begin{equation*}
n_{2}=\left\lfloor rn_{1}\right\rfloor -\left\lfloor yf(n_{1})\right\rfloor ,
\end{equation*}%
\textit{then the recurrence (21) holds.}\bigskip 

Conjecture 1 depends on the existence of the pairs $(u,y)$. \ Regarding $u$
and $y$ as sequences with terms $u(n)$ and $y(n),$ where $n=(Q-1)/4,$ we
note that $u$ and $y$ are registered in [3] as \seqnum{A077428} and
\seqnum{A078355}. \
Several initial terms are shown here:%
\begin{equation*}
\begin{tabular}{||c||c|c|c|c|c|c|c|c|c|}
\hline
$Q$ & $5$ & $13$ & $17$ & $21$ & $29$ & $33$ & $37$ & $41$ & $45$ \\ \hline
$u$ & $3$ & $11$ & $66$ & $5$ & $27$ & $46$ & $146$ & $4098$ & $7$ \\ \hline
$y$ & $1$ & $3$ & $16$ & $1$ & $5$ & $8$ & $24$ & $640$ & $1$ \\ \hline
\end{tabular}%
\end{equation*}%
\bigskip

\textbf{Conjecture 2.}\textit{\ \ Suppose that }$Q\equiv 1$\textit{\ mod }$4$%
\textit{\ and that }$n,j,k$\textit{\ satisfy (1). \ Let }$u,y,r$\textit{\ be
as in Conjecture 1, let}%
\begin{equation*}
n_{1}=\left\lfloor rn\right\rfloor -\left\lfloor yf(n)\right\rfloor +y,
\end{equation*}%
\textit{where }$f(n)$\textit{\ denotes the fractional part of }$(1+\sqrt{Q}%
)n/2,$\textit{\ and let}%
\begin{equation*}
j_{1}=\left\{ 
\begin{tabular}{ll}
$(Q_{1}n_{1})^{2}-(\left\lfloor 1/2+Q_{1}n_{1}\right\rfloor -1/2)^{2}$ & if $%
n_{1}$ is odd, \\ 
$(Q_{1}n_{1})^{2}-\left\lfloor Q_{1}n_{1}\right\rfloor ^{2}$ & if $n_{1}$ is
even.%
\end{tabular}%
\right. 
\end{equation*}%
\textit{Then}%
\begin{equation*}
(j_{1}+k+1)^{2}-4k=Qn_{1}^{2}.
\end{equation*}%
\textit{Moreover, if}%
\begin{equation*}
n_{2}=\left\lfloor rn_{1}\right\rfloor -\left\lfloor yf(n_{1})\right\rfloor
+y,
\end{equation*}%
\textit{then the recurrence (21) holds.}\bigskip 

Assuming the two conjectures valid, we construct fixed-$j$ and fixed-$k$
arrays exactly as in sections 4 and 5, and we conjecture that they are
dispersions.\bigskip

\textbf{Example 5.} \ The fixed-$j$ array for $Q=13.$%
\begin{equation*}
\begin{tabular}{ccccc}
$1$ & $10$ & $109$ & $1189$ & $\cdots $ \\ 
$2$ & $20$ & $218$ & $2378$ &  \\ 
$3$ & $30$ & $327$ & $3567$ &  \\ 
$4$ & $43$ & $469$ & $5116$ &  \\ 
$5$ & $53$ & $578$ & $6305$ &  \\ 
$6$ & $63$ & $687$ & $7494$ &  \\ 
$7$ & $76$ & $829$ & $9043$ &  \\ 
$8$ & $86$ & $938$ & $10232$ &  \\ 
$9$ & $96$ & $1047$ & $11421$ &  \\ 
$11$ & $119$ & $1298$ & $14159$ &  \\ 
$\vdots $ &  &  &  & 
\end{tabular}%
\end{equation*}%
\bigskip

\textbf{Example 6.} \ The fixed-$k$ array for $Q=13.$%
\begin{equation*}
\begin{tabular}{ccccc}
$1$ & $13$ & $142$ & $1549$ & $\cdots $ \\ 
$2$ & $23$ & $251$ & $2738$ &  \\ 
$3$ & $33$ & $360$ & $3927$ &  \\ 
$4$ & $46$ & $502$ & $5476$ &  \\ 
$5$ & $56$ & $611$ & $6665$ &  \\ 
$6$ & $66$ & $720$ & $7854$ &  \\ 
$7$ & $79$ & $862$ & $9403$ &  \\ 
$8$ & $89$ & $971$ & $10592$ &  \\ 
$9$ & $99$ & $1080$ & $11781$ &  \\ 
$10$ & $112$ & $1222$ & $13330$ &  \\ 
$11$ & $122$ & $1331$ & $14519$ &  \\ 
$\vdots $ &  &  &  & 
\end{tabular}%
\end{equation*}

\section{Examples}

This section consists of Examples 7-14 illustrating Theorems 1 and 2 for
small values of $Q,$ followed by fractal sequences associated with the fixed-%
$j$ and fixed-$k$ arrays for $Q=5.$\bigskip

\textbf{Example 7.} \ Taking $Q=8$ in Theorem 1, for each $n$, there is a
unique pair $(j,k)=(j(n),k(n))$ such that $(j+k+1)-4k=8n^{2}.$ \ The
sequences $j$ and $k$ are given by%
\begin{equation*}
\begin{tabular}{||c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$n$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12
$ & $13$ & $\cdots $ \\ \hline
$j$ & $1$ & $4$ & $2$ & $7$ & $1$ & $8$ & $17$ & $7$ & $18$ & $4$ & $17$ & $%
32$ & $14$ & $\cdots $ \\ \hline
$k$ & $2$ & $1$ & $7$ & $4$ & $14$ & $9$ & $2$ & $16$ & $7$ & $25$ & $14$ & $%
1$ & $23$ & $\cdots $ \\ \hline
\end{tabular}%
\end{equation*}%
The sequences $j$ and $k$ appear in \cite{Slo1} as \seqnum{A087056} and
\seqnum{A087059},
respectively.\bigskip 

\textbf{Example 8.} \ Taking $Q=12$ in Theorem 1, for each $n$, there is a
unique pair $(j,k)=(j(n),k(n))$ such that $(j+k+1)-4k=12n^{2}.$ \ The
sequences $j$ and $k$ are given by%
\begin{equation*}
\begin{tabular}{||c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$n$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12
$ & $13$ & $\cdots $ \\ \hline
$j$ & $2$ & $3$ & $2$ & $2$ & $11$ & $8$ & $3$ & $23$ & $18$ & $11$ & $2$ & $%
32$ & $23$ & $\cdots $ \\ \hline
$k$ & $1$ & $4$ & $9$ & $1$ & $6$ & $13$ & $22$ & $4$ & $13$ & $24$ & $37$ & 
$9$ & $22$ & $\cdots $ \\ \hline
\end{tabular}%
\end{equation*}%
The sequences $j$ and $k$ appear in \cite{Slo1} as \seqnum{A120864}
and \seqnum{A120865},
respectively.\bigskip 

\textbf{Example 9.} \ Taking $Q=20$ in Theorem 1, for each $n$, there is a
unique pair $(j,k)=(j(n),k(n))$ such that $(j+k+1)-4k=20n^{2}.$ \ The
sequences $j$ and $k$ are given by%
\begin{equation*}
\begin{tabular}{||c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$n$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12
$ & $13$ & $\cdots $ \\ \hline
$j$ & $1$ & $4$ & $9$ & $16$ & $4$ & $11$ & $20$ & $31$ & $5$ & $16$ & $29$
& $44$ & $4$ & $\cdots $ \\ \hline
$k$ & $4$ & $5$ & $4$ & $1$ & $19$ & $16$ & $11$ & $4$ & $36$ & $29$ & $20$
& $9$ & $55$ & $\cdots $ \\ \hline
\end{tabular}%
\end{equation*}%
The sequences $j$ and $k$ appear in \cite{Slo1} as \seqnum{A120866}
and \seqnum{A120867},
respectively.\bigskip 

\textbf{Example 10.} \ Taking $Q=5$ in Theorem 2, for each $n$, there is a
unique pair $(j,k)=(j(n),k(n))$ such that $(j+k+1)-4k=5n^{2}.$ \ The
sequences $j$ and $k$ are given by%
\begin{equation*}
\begin{tabular}{||c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$n$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12
$ & $13$ & $\cdots $ \\ \hline
$j$ & $1$ & $1$ & $5$ & $4$ & $1$ & $9$ & $5$ & $16$ & $11$ & $4$ & $19$ & $%
11$ & $1$ & $\cdots $ \\ \hline
$k$ & $1$ & $4$ & $1$ & $5$ & $11$ & $4$ & $11$ & $1$ & $9$ & $19$ & $5$ & $%
16$ & $29$ & $\cdots $ \\ \hline
\end{tabular}%
\end{equation*}%
The sequences $j$ and $k$ appear in \cite{Slo1} as \seqnum{A005752} and
\seqnum{A120868},
respectively.\bigskip 

\textbf{Example 11.} \ Taking $Q=13$ in Theorem 2, for each $n$, there is a
unique pair $(j,k)=(j(n),k(n))$ such that $(j+k+1)-4k=13n^{2}.$ \ The
sequences $j$ and $k$ are given by%
\begin{equation*}
\begin{tabular}{||c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$n$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12
$ & $13$ & $\cdots $ \\ \hline
$j$ & $1$ & $4$ & $9$ & $3$ & $9$ & $17$ & $3$ & $12$ & $23$ & $1$ & $13$ & $%
27$ & $43$ & $\cdots $ \\ \hline
$k$ & $3$ & $3$ & $1$ & $12$ & $9$ & $4$ & $23$ & $17$ & $9$ & $36$ & $27$ & 
$16$ & $3$ & $\cdots $ \\ \hline
\end{tabular}%
\end{equation*}%
The sequences $j$ and $k$ appear in \cite{Slo1} as \seqnum{A120869}
and \seqnum{A120870},
respectively.\bigskip 

\textbf{Example 12.} \ The table shows, in row $g$ and column $h,$ the
unique pair $(j,k)$ corresponding to the fixed-$j$ array $\{d(g,h)\}$ for $%
Q=8;$ i.e., $(j+k+1)^{2}-4k=8n^{2},$ where $n=d(g,h)$ is as in Example 3:%
\begin{equation*}
\begin{tabular}{lllll}
$(1,2)$ & $(1,14)$ & $(1,82)$ & $(1,478)$ & $\cdots $ \\ 
$(4,1)$ & $(4,25)$ & $(4,161)$ & $(4,953)$ &  \\ 
$(2,7)$ & $(2,47)$ & $(2,279)$ & $(2,1631)$ &  \\ 
$(7,4)$ & $(7,56)$ & $(7,356)$ & $(7,2104)$ &  \\ 
$(9,6)$ & $(9,89)$ & $(9,553)$ & $(9,3257)$ &  \\ 
$\vdots $ &  &  &  & 
\end{tabular}%
\end{equation*}%
The values of $j,$ fixed for each row, form the sequence \seqnum{A120871}:

\begin{equation*}
(1,4,2,7,8,17,7,18,17,32,14,31,9,28,23,46,16,41,34,63,25,56,14,47,\ldots ).
\end{equation*}%
These numbers are terms of the $j$-sequence in Example 7, but without
duplicates.\bigskip

\textbf{Example 13.} \ The table shows, in row $g$ and column $h,$ the
unique pair $(j,k)$ corresponding to the fixed-$k$ array $\{d(g,h)\}$ for $%
Q=8;$ i.e., $(j+k+1)^{2}-4k=8n^{2},$ where $n=d(g,h)$ is as in Example 4:%
\begin{equation*}
\begin{tabular}{lllll}
$(1,2)$ & $(17,2)$ & $(43,2)$ & $(673,2)$ & $\cdots $ \\ 
$(4,1)$ & $(32,1)$ & $(196,1)$ & $(1152,1)$ &  \\ 
$(2,7)$ & $(46,7)$ & $(306,7)$ & $(1822,7)$ &  \\ 
$(7,4)$ & $(63,4)$ & $(391,4)$ & $(2303,4)$ &  \\ 
$(1,14)$ & $(72,14)$ & $(497,14)$ & $(2969,14)$ &  \\ 
$\vdots $ &  &  &  & 
\end{tabular}%
\end{equation*}%
The values of $k,$ fixed for each column, form the sequence \seqnum{A120872}:

\begin{equation*}
(2,1,7,4,14,9,16,7,25,14,23,8,34,17,47,28,41,18,\ldots ).
\end{equation*}%
These numbers are terms of the $k$-sequence in Example 7, but without
duplicates.\bigskip

\textbf{Example 14.} \ The unique pair $(j,k)=(j(n),k(n))$ such that%
\begin{equation*}
(j+k+1)^{2}-4k=5n^{2}
\end{equation*}%
is given in Example 10. \ If the duplicates in $\{j(n)\}$ are expelled, the
remaining sequence gives the $j$ values for the fixed-$j$ array for $Q=5$
(cf., Example 12); this remaining sequence appears to be essentially
\seqnum{A022344}, described as \textquotedblleft Allan Wechsler's `$J$ determinant'\
sequence\textquotedblright . \ To compare the two sequences, check
successive terms of \seqnum{A022344} and \seqnum{A005742}.\bigskip 

We conclude with a few notes about the case $Q=5,$ the least $Q$ covered by
Theorem 2. \ The fixed-$j$ array is the Wythoff difference array
(\seqnum{A080164} in 
\cite{Slo1}) and the fixed-$k$ array is the Fraenkel array (\seqnum{A038150}).
These and their relationship to the equation $(j+k+1)^{2}-4k=5n^{2}$ are
discussed in \cite{Kim2}.

As noted in \cite{Slo2}, there is a fractal sequence associated with every
dispersion. \ Specifically, if $d(g,h)$ is the general term of the
dispersion, and we define $f(n)$ to be the value of $g$ for which $n=d(g,h),$
then $f$ is the associated fractal sequence. \ (A fractal sequence contains
itself as a proper subsequence; e.g., if you delete the first occurrence of
each positive integer in $f,$ the remaining sequence is $f$; iterating this
procedure shows that the sequence properly contains itself infinitely many
times.)\bigskip 

\textbf{Example 15.} \ The fractal sequence associated with the Wythoff
difference array $\{d(g,h)\}$ is \seqnum{A120873}:%
\begin{equation*}
f=(1,1,2,3,1,4,2,5,6,3,7,8,1,9,4,10,11,2,12,5,13,14,6,\ldots ).
\end{equation*}

\textbf{Example 16.} \ The fractal sequence associated with the Fraenkel
array $\{d(g,h)\}$ is \seqnum{A120874}:
\begin{equation*}
f=(1,2,1,3,4,2,5,1,6,7,3,8,9,4,10,2,11,12,5,13,1,14,15,\ldots ).
\end{equation*}

\begin{thebibliography}{9}
\bibitem{Kim1} Clark Kimberling, Interspersions and dispersions,
\ \textit{Proc. Amer. Math. Soc.} {\bf 117} (1993)
313--321.

\bibitem{Kim2} Clark Kimberling, The equation $m^{2}-4k=5n^{2}$ and unique
representations of positive integers, submitted.

\bibitem{Slo1} N. J. A. Sloane, editor, \textit{The On-Line Encyclopedia of
Integer Sequences,} \href{http://www.research.att.com/~njas/sequences}{\tt http://www.research.att.com/\symbol{126}njas/sequences/}

\bibitem{Slo2} N. J. A. Sloane, Classic sequences in the On-Line Encyclopedia
of Integer Sequences, Part 1,
\href{http://www.research.att.com/~njas/sequences/classic.html}{The
Wythoff Array and The Para-Fibonacci Sequence}.
\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11B37; Secondary 11D09, 11D85 .

\noindent \emph{Keywords: } Beatty sequence, complementary equation,
dispersion, interspersion, Pell equation, recurrences.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A005752},
\seqnum{A033313},
\seqnum{A033317},
\seqnum{A038150},
\seqnum{A077428},
\seqnum{A078355},
\seqnum{A080164},
\seqnum{A087076},
\seqnum{A087079},
\seqnum{A098021},
\seqnum{A120858},
\seqnum{A120859},
\seqnum{A120860},
\seqnum{A120861},
\seqnum{A120862},
\seqnum{A120863},
\seqnum{A120864},
\seqnum{A120865},
\seqnum{A120866},
\seqnum{A120867},
\seqnum{A120868},
\seqnum{A120869},
\seqnum{A120870},
\seqnum{A120871},
\seqnum{A120872},
\seqnum{A120873}, and
\seqnum{A120874}.
)

\bigskip
\hrule
\bigskip


\vspace*{+.1in}
\noindent
Received July 10 2006;
revised version received February 20 2007.
Published in {\it Journal of Integer Sequences}, March 13 2007.

\bigskip
\hrule
\bigskip

\noindent
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