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\begin{center}
\vskip 1cm{\LARGE\bf Partial Complements and Transposable Dispersions}
\vskip 1cm
\large
Clark Kimberling and John E. Brown\\
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 $A=\{a(i,j)\},$ for $i\geq 0$ and $j\geq 0,$ is the dispersion of a
strictly increasing sequence $r=(r(0),r(1),r(2),\ldots )$ of integers$,$
where $r(0)=1$ and infinitely many postive integers are not terms of $r.$ \
Let $R$ be the set of such sequences, and define $t$ on $R$ by $tr(n)=a(0,n)$
for $n\geq 0.$ Let $F$ be the subset of $R$ consisting of sequences $r$
satisfying $ttr=r.$ \ The set $F$ is characterized in terms of ordered
arrangements of numbers $i+j\theta $. \ For fixed $i\geq 0,$ the sequence $%
a(i,j),$ for $j\geq 1,$ is the $(i+1)$st partial complement of $r.$ \
Central to the characterization of $F$ is the role of the families of
figurate (or polygonal) number sequences and the centered polygonal number
sequences. \ Finally, it is conjectured that for every $r$ in $R,$ the
iterates $t^{(2m)}r$ converge to a sequence in $F$. \ \ 
\end{abstract}

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



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\section{Introduction}

Let $N$ denote the set of positive integers. \ Suppose $r=(r(0),r(1),r(2),%
\ldots )$ is a strictly increasing sequence of integers with $r(0)=1$ and
infinite complement in $N.$ \ Let $r^{\ast }$ be the sequence obtained by
arranging in increasing order the complement of $r$. \ Let%
\begin{equation*}
a(0,0)=1,\text{ }a(0,1)=r^{\ast }(1),
\end{equation*}%
and inductively,%
\begin{equation*}
a(0,j)=r^{\ast }(a(0,j-1))
\end{equation*}%
for $j\geq 1.$ \ The sequence $a(0,j),$ for $j\geq 1,$ is the \textit{1st
partial complement} of $r.$ \ For arbitrary $i\geq 0,$ suppose that the
sequence $a(i,j)$ for $j\geq 0$ is defined, and let $a(i+1,0)$ be the least
number in $N$ that is not $a(h,j)$ for any $(h,j)$ satisfying $0\leq h\leq i$
and $j\geq 0.$ \ Define%
\begin{equation}
a(i+1,j)=r^{\ast }(a(i+1,j-1))
\end{equation}%
for $j\geq 1.$ \ In this inductive manner, an array $A=\{a(i,j)\},$ for $%
i\geq 0,$ $j\geq 0,$ is defined. \ It is called the \textit{dispersion} of $%
r^{\ast }.$ \ (In [3], where the terms dispersion and interspersion are
introduced, the indexing is by $i\geq 1,j\geq 1$ instead of $i\geq 0,$ $%
j\geq 0.$) \ For fixed $i\geq 0$ and variable $j\geq 1,$ the sequence $%
a(i,j) $ is the $(i+1)$\textit{st partial complement} of $r.$

To summarize, the dispersion $A$ of the complement of $r,$ henceforth
denoted by $A(r),$ consists of first column $r$ together with the terms of $%
r^{\ast }$ dispersed into partial complements; row $i$ of $A(r)$ consists of
first term $r(i)$ followed by the $(i+1)$st partial complement of $r$. \ It
is sometimes desirable to use a recurrence for $a(i+1,j)$ that refers
directly to $r.$ \ To develop such a recurrence, for any strictly increasing
sequence $\rho $ in $N,$ let $\#(\rho (n)\leq m)$ denote the number of $%
n\geq 1$ satisfying $\rho (n)\leq m.$ \ Clearly, $\rho (h)$ is the \textit{%
least} $m$ satisfying $\#(\rho (n)\leq m)=h.$ \ Now take $\rho =r^{\ast }$
and $h=a(i,j)$ to see that%
\begin{equation*}
a(i,j+1)=\text{ least }m\text{ satisfying {\LARGE (}}\#(r^{\ast }(n)\leq
m)=a(i,j)\text{{\LARGE )}}.
\end{equation*}%
As $\#(r(n)\leq m)+\#(r^{\ast }(n)\leq m)=m,$ we obtain%
\begin{eqnarray}
a(i,j+1) &=&\text{ least }m\text{ satisfying {\LARGE (}}m-\#(r(n)\leq
m)=a(i,j)\text{{\LARGE )}}  \notag \\
&=&\min \text{{\Large \{}}m:\max \{n:r(n)\leq m\}=a(i,j)-m+1\text{{\large \}}%
}.
\end{eqnarray}%
This recurrence is especially useful in coding computer programs that
generate dispersions.

Let $R$ be the set of sequences $r$ described in the first sentence. \ Let $%
tr$ denote the first partial complement of $r,$ and let $F$ be the family of
sequences $r$ for which $ttr=t.$ \ A main objective of this paper is to
characterize $F$ in terms of sequences associated with multisets of the form%
\begin{equation}
S_{\theta }=\{i+j\theta :i\geq 0,\text{ }j\geq 0\},
\end{equation}%
where $\theta $ is a positive real number. \ The numbers in $S_{\theta }$
are distinct if $\theta $ is irrational; otherwise, write $\theta =c/d,$
where $c$ and $d$ are relatively prime positive integers. \ Then a number $%
i+j\theta $ occurs more than once in $S_{\theta }$ if and only if $i\geq c,$
and in this case the representations of $i+j\theta $ are these:%
\begin{equation*}
i,\text{ }i-c+d\theta ,\text{ }i-2c+2d\theta ,\cdots ,\text{ }i-\left\lfloor
i/c\right\rfloor c+\left\lfloor i/c\right\rfloor d\theta .
\end{equation*}%
In order to treat these as distinct objects, we represent each $h+k\theta $
as an ordered pair $(h,k),$ so that the representations of $i+j\theta $
become%
\begin{equation*}
(i,0),\text{ (}i-c,d),\text{ (}i-2c,2d),\cdots ,\text{ (}i-\left\lfloor
i/c\right\rfloor c,\left\lfloor i/c\right\rfloor d).
\end{equation*}%
Define an order relation $\prec $ on the set $\{(i,j):$ \ $i\geq 0,$ $j\geq
0\}$ by%
\begin{eqnarray*}
(i_{1},j_{1}) &\prec &(i_{2},j_{2})\text{ if }i_{1}+j_{1}\theta
<i_{2}+j_{2}\theta \\
(i_{1},j_{1}) &\prec &(i_{2},j_{2})\text{ if }i_{1}+j_{1}\theta
=i_{2}+j_{2}\theta \text{ and }j_{1}<j_{2}.
\end{eqnarray*}%
We extend this definition to the case that $\theta $ is irrational, noting
that the condition $i_{1}+j_{1}\theta =i_{2}+j_{2}\theta $ does not occur. \
Now for any real $\theta >0,$ the \textit{rank array of }$\theta $ is
defined by%
\begin{equation*}
A_{\theta }=\{a_{\theta }(i,j):i\geq 0,\text{ }j\geq 0\},
\end{equation*}%
where%
\begin{equation*}
a_{\theta }(i,j)=\text{ rank of }(j,i)\text{ under }\prec .
\end{equation*}%
(Defining $a_{\theta }(i,j)$ as the rank of $(j,i)$ rather than that of $%
(i,j)$ facilitates later developments, such as equation (13) and connections
with Farey sequences. For irrational $\theta ,$ the condition ``rank of $%
(j,i)$ under $\prec $'' can be replaced by ``rank of $j+i\theta $ under
[ordinary] $<$''; in this case, $r_{\theta }(n)$ is simply the rank of $%
n\theta $ among all the numbers $h+k\theta .$)

As an example, let $c=4$ and $d=3.$ \ Then%
\begin{eqnarray*}
(0,0) &\prec &(1,0)\prec (0,1)\prec (2,0)\prec (1,1)\prec (0,2)\prec
(3,0)\prec (2,1)\prec (1,2)\prec \\
(4,0) &\prec &(0,3)\prec (3,1)\prec (2,2)\prec (5,0)\prec (1,3)\prec
(4,1)\prec (0,4)\prec (3,2)\prec \\
(6,0) &\prec &(2,3)\prec (5,1)\prec (1,4)\prec (4,2)\prec (0,5)\prec
(7,0)\prec (3,3)\prec (6,1)\prec \\
(2,4) &\prec &(5,2)\prec (1,5)\prec (8,0)\prec (4,3)\prec (0,6)\prec \cdots
\end{eqnarray*}%
Numbering these from $1$ to $33$ shows that the rank array $A_{4/3}$ starts
like this:%
\begin{equation*}
\begin{tabular}{lllllllll}
$1$ & $2$ & $4$ & $7$ & $10$ & $14$ & $19$ & $25$ & $31$ \\ 
$3$ & $5$ & $8$ & $12$ & $16$ & $21$ & $27$ &  &  \\ 
$6$ & $9$ & $13$ & $18$ & $23$ & $29$ &  &  &  \\ 
$11$ & $15$ & $20$ & $26$ & $32$ &  &  &  &  \\ 
$17$ & $22$ & $28$ &  &  &  &  &  &  \\ 
$24$ &  &  &  &  &  &  &  &  \\ 
$33$ &  &  &  &  &  &  &  & 
\end{tabular}%
\end{equation*}

\begin{theorem}
\textit{Suppose }$\theta >0.$\textit{\ \ Then }$A_{\theta }$\textit{\ is an
interspersion.}
\end{theorem}

\textbf{Proof:} \ We shall show that each of the propositions (I1)-(I4) that
define an interspersion in [3] is satisfied:\smallskip

(I1)\ \ The rows of $A_{\theta }$ comprise a partition of $N,$ as $N$ is the
set of ranks of numbers in $S_{\theta },$ and these ranks are partitioned by
the rows of $A_{\theta }.$\smallskip

(I2) \ Every row of $A_{\theta }$ is an increasing sequence, as $(j,i)\prec
(j+1,i)$ for all $i\geq 0,$ $j\geq 0.$\smallskip

(I3) \ Every column of $A_{\theta }$ is an increasing sequence, as $%
(j,i)\prec (j,i+1)$ for all $i\geq 0,$ $j\geq 0.$\smallskip

(I4) \ If $(u_{j})$ and $(v_{j})$ are distinct rows of $A_{\theta }$ and if $%
p$ and $q$ are indices for which $u_{p}<v_{q}<u_{p+1},$ then $%
u_{p+1}<v_{q+1}<u_{p+2}.$ \ That is, in the present context, if $i\neq
i^{\prime }$ and $a(i,p)<a(i^{\prime },q)<a(i,p+1),$ or equivalently,%
\begin{equation*}
(p,i)\prec (q,i^{\prime })\prec (p+1,i),
\end{equation*}%
then%
\begin{equation*}
(p+1,i)\prec (q+1,i^{\prime })\prec (p+2,i),
\end{equation*}%
or equivalently, $a(i,p+1)<a(i^{\prime },q+1)<a(i,p+2).$ \ \ \ \ \ \ \ \ $%
\square $\bigskip

\begin{corollary}
\textit{For }$i\geq 0,$ \textit{let s be the sequence obtained by deleting
the initial term of row }$i$\textit{\ of }$A_{\theta }.$\textit{\ \ Then }$s$
\textit{is the }$(i+1)$st\textit{\ partial complement of column }$0$ of $%
A_{\theta }.$
\end{corollary}

\textbf{Proof:} \ By Theorem 1 of [3], the array $A_{\theta }$ is a
dispersion, and as shown in the proof in [3],%
\begin{equation*}
a(i,j+1)=t(a(i,j)),
\end{equation*}%
where $(t(k))$ denotes the sequence of numbers, arranged in increasing
order, in the set $N\backslash \{a(0,j):$ $j\geq 0\},$ this set being the
complement of column $0$ of $A_{\theta }$. \ \ \ \ \ \ \ \ $\square $%
\bigskip 

Let $r_{\theta }$ denote column $0$ of $A_{\theta }.$ \ By Corollary 1, row $%
0$ of $A_{\theta }$ is the sequence $tr_{\theta },$ a property to be used in
the next proof.\bigskip 

\begin{corollary}
\textit{Suppose }$\theta $ is a positive irrational number$.$\textit{\ \
Then }$tr_{\theta }=r_{1/\theta }$ and $ttr_{\theta }=r_{\theta }.$
\end{corollary}

\textbf{Proof:} \ Referring to the array $A_{1/\theta },$ we have, for $%
m\geq 0$ and $n\geq 0,$%
\begin{eqnarray*}
a_{1/\theta }(m,n) &=&(\text{rank of }m+n/\theta \text{ in }S_{1/\theta }) \\
&=&(\text{rank of }m\theta +n\text{ in }S_{\theta }) \\
&=&a_{\theta }(n,m).
\end{eqnarray*}%
That is, $A_{1/\theta }$ is the transpose of $A_{\theta }.$ \ Consequently, $%
A_{1/(1/\theta )}=A_{\theta },$ so that $ttr_{\theta }=r_{\theta }.$ \ \ \ \
\ \ \ \ \ $\square $\bigskip

A dispersion $\{a(i,j)\}$ is \textit{transposable} if its transpose, $%
\{a(j,i)\},$ is a dispersion. \ According to Corollary 3, the dispersions of 
$A_{\sqrt{2}}$ and $A_{1/\sqrt{2}}$ are transposable; northwest corners of
these arrays are shown in Example 1 \ We shall see in section 3 that there
are transposable dispersions other than those given by the proof of
Corollary 3.\bigskip 

\begin{center}
\textbf{Example 1:} \ The arrays $A_{\theta }$ and $A_{1/\theta }$ for $\
\theta =\sqrt{2}$

$%
\begin{tabular}{ccccccccccccccc}
$1$ & $2$ & $4$ & $7$ & $10$ & $\ldots $ &  &  &  & $1$ & $3$ & $6$ & $11$ & 
$17$ & $\ldots $ \\ 
$3$ & $5$ & $8$ & $12$ & 16 & $\ldots $ &  &  &  & $2$ & $5$ & $9$ & $15$ & $%
22$ & $\ldots $ \\ 
$6$ & $9$ & $13$ & $18$ & $23$ & $\ldots $ &  &  &  & $4$ & $8$ & $13$ & $20$
& $28$ & $\ldots $ \\ 
$11$ & $15$ & $20$ & $26$ & $32$ & $\ldots $ &  &  &  & $7$ & $12$ & $18$ & $%
26$ & $35$ & $\ldots $ \\ 
$17$ & $22$ & $28$ & $35$ & $42$ & $\ldots $ &  &  &  & $10$ & $16$ & $23$ & 
$32$ & $42$ & $\ldots $ \\ 
$\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\ddots $ &  & 
&  & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\ddots $%
\end{tabular}%
$\bigskip
\end{center}

Initial terms of $tr$ can be written out from initial terms of $r,$ for any $%
r$ in $R,$ by means of a handly little algorithm. \ We use the sequence $%
r=(1,2,4,7,10,\ldots )$ of column $0$ of $A_{\sqrt{2}}$ to exemplify:\bigskip

\begin{quote}
\textbf{Step 1.} \ Write initial terms of the complement, $r^{\ast },$ with
counting numbers above:%
\begin{equation*}
\begin{tabular}{|c|c|c|cc|c|cccc|c|ccc}
\cline{1-1}\cline{3-3}\cline{6-6}\cline{11-11}
$1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12$ & $%
13$ & $14$ \\ \cline{1-1}\cline{3-3}\cline{6-6}\cline{11-11}
$3$ & $5$ & $6$ & $8$ & $9$ & $11$ & $12$ & $13$ & $15$ & $16$ & $17$ & $18$
& $20$ & $21$ \\ \cline{1-1}\cline{3-3}\cline{6-6}\cline{11-11}
\end{tabular}%
\end{equation*}

\textbf{Step 2.} \ Determine the chain $1\rightarrow 3\rightarrow
6\rightarrow 11\rightarrow 17\rightarrow \cdots ,$ as indicated by the
paired rectangles. \ These numbers form the sequence $tr,$ alias row $0$ of $%
A_{\sqrt{2}},$ alias (after the initial $1$) the 1st partial complement of $%
r.$ \ (Note that repeating the procedure starting with the pair $(2,5)$
yields the 2nd partial complement of $r,$ and so on.)\bigskip
\end{quote}

If $\theta \in N,$ then the sequences $r_{\theta }$ and $r_{1/\theta },$ we
shall prove in section 2, are closely related to certain well-known
sequences. \ We continue this introduction with an overview of those
sequences, with reference to indexing systems in [5] and [6]. \ Sequences of
polygonal (or figurate) numbers, typified by%
\begin{eqnarray*}
P_{1} &=&(1,3,6,10,15,\ldots )=\text{triangular numbers (A000217 in [6], 253
in [5])} \\
P_{2} &=&(1,4,9,16,25,\ldots )=\text{square numbers (A000290 in [6], 338 in
[5])} \\
P_{3} &=&(1,5,12,22,35,\ldots )=\text{pentagonal numbers (A000326 in [6],
339 in [5]),}
\end{eqnarray*}%
are given for $k\geq 1$ by%
\begin{equation}
P_{k}(n)=k\left( 
\begin{array}{c}
n+1 \\ 
2%
\end{array}%
\right) +n+1
\end{equation}%
for $n\geq 0.$ \ Sequences of central polygonal numbers,%
\begin{eqnarray*}
Q_{1} &=&(1,3,7,13,21,\ldots )=\text{central polygonal numbers (A002061 in
[6])} \\
Q_{2} &=&(1,4,10,19,31,\ldots )=\text{central triangular numbers (A005448 in
[6])} \\
Q_{3} &=&(1,5,13,25,41,\ldots )=\text{central square numbers (A001844 in
[6]),}
\end{eqnarray*}%
are given for $k\geq 1$ by%
\begin{equation}
Q_{k}(n)=(k+1)\left( 
\begin{array}{c}
n+1 \\ 
2%
\end{array}%
\right) +1
\end{equation}%
for $n\geq 0,$ so that%
\begin{equation}
Q_{k}(n)=P_{k+1}(n)-n.
\end{equation}

The sequences $P_{k}$ and $Q_{k}$ are easily expressed in terms of $P_{1}$
and $Q_{1}$:%
\begin{eqnarray*}
P_{k}(n) &=&P_{1}(n)+(k-1)P_{1}(n-1), \\
Q_{k}(n) &=&Q_{1}(n)+(k-1)P_{1}(n-1),
\end{eqnarray*}%
for $k\geq 1,$ $n\geq 1.$ \ For a colorful introduction to the geometry
associated with the numbers $P_{k}$ and $Q_{k},$ see [1, pp. 38-42]. \
Consider next the sequences%
\begin{eqnarray*}
\widehat{P}_{1} &=&(1,2,4,7,11,\ldots )=\text{lazy caterer sequence (A000124
in [6], 386 in [5])} \\
\widehat{P}_{2} &=&(1,2,3,5,7,\ldots )=\text{(A001401 in [6], 354 in [5])} \\
\widehat{P}_{3} &=&(1,2,3,4,6,\ldots )=\text{(A008748 in [6])}
\end{eqnarray*}%
and%
\begin{eqnarray*}
\widehat{Q}_{1} &=&(1,2,4,6,9,\ldots )=\text{quarter-squares sequence
(A002620 in [6], 105 in [5])} \\
\widehat{Q}_{2} &=&(1,2,3,5,7,\ldots )=\text{(A001840 in [6], 207 in [5])} \\
\widehat{Q}_{3} &=&(1,2,3,4,6,\ldots )=\text{(A001972 in [6], 208 on [5]).}
\end{eqnarray*}%
These typify families defined for $k\geq 1$ by%
\begin{eqnarray}
\widehat{P}_{k}(n) &=&\widehat{P}_{k}(n-1)+\QOVERD\lfloor \rfloor {n+k-1}{k},
\\
\widehat{Q}_{k}(n) &=&\widehat{Q}_{k}(n-1)+\QOVERD\lfloor \rfloor
{n+k+1}{k+1},
\end{eqnarray}%
for $n\geq 1,$ where $\widehat{P}_{k}(0)=\widehat{Q}_{k}(0)=1.$ \ For $k=1,$
the recurrence (7) gives $\widehat{P}_{1}(n)=\widehat{P}_{1}(n-1)+n,$ whence
by induction,%
\begin{equation}
\widehat{P}_{1}(n)=P_{1}(n)-n.
\end{equation}

Also, equation (7) implies that $\widehat{P}_{k}$ is the sequence obtained
by arranging in increasing order the numbers in the set%
\begin{equation}
\{1\}\cup \{P_{k}(n)+i(n+1)+1:0\leq i\leq k-1,n\geq 1\}.
\end{equation}%
Likewise, equation (8), or alternatively the identity%
\begin{equation}
\widehat{Q}_{k}(n)=\widehat{P}_{k+1}(n+1)-1,
\end{equation}%
can be used to prove that the sequence $\widehat{Q}_{k}$ results by ordering
the numbers in the set%
\begin{equation}
\{1\}\cup \{Q_{k}(n)+in+i-1:0\leq i\leq k,n\geq 1\}.
\end{equation}

The sequences $\widehat{Q}_{k},$ for $k\geq 2,$ count certain restricted
partitions called \textit{denumerants }in [2, pp. 108-124] and [5]. \
References listed in [6] and [5] lead to extensive literature on the
sequences $P_{k}$ and $Q_{k}.$ \ However, observations of relationships
(e.g., Theorem 4) between $P_{k}$ and $\widehat{P}_{k}$ (and between $Q_{k}$
and $\widehat{Q}_{k})$ to be proved in section 2 may be new.

As a final introductory note, interspersions are closely related to fractal
sequences; see [7] for a list of references. \ Thus, results proved below
for interspersions (hence dispersions) could to be stated in terms of
fractal sequences.

\section{Rank Sequences}

In this section, the word \textit{rank} refers to the ordinary
less-than-or-equal relation, $\leq $ (not the $\theta $-dependent relation $%
\prec $ defined in section 1, which will be considered further in section
4.) \ Recall from section 1 that for rational $\theta ,$ the multiset $%
S_{\theta }$ in (3) contains repeated elements, so that an element may have
more than one rank.

Suppose $\theta >0$ and $n\geq 0$. \ The \textit{minrank sequence of }$%
\theta ,$ denoted by $m_{\theta },$ is defined by%
\begin{equation*}
m_{\theta }(n)=\text{ least }h\text{ such that }n\text{ has rank }h\text{ in 
}S_{\theta },
\end{equation*}%
and the \textit{maxrank sequence of }$\theta ,$ denoted by $M_{\theta },$ is
defined by%
\begin{equation*}
M_{\theta }(n)=\text{ greatest }h\text{ such that }n\text{ has rank }h\text{
in }S_{\theta }.
\end{equation*}%
If $\theta $ is a positive integer, then the \textit{lower self-rank
sequence of }$\theta $ is defined by%
\begin{equation*}
\ell _{\theta }(n)=m_{\theta }(\theta n),
\end{equation*}%
and the \textit{upper self-rank sequence of }$\theta $, by%
\begin{equation*}
\mathcal{L}_{\theta }(n)=M_{\theta }(\theta n).
\end{equation*}%
As $\mathcal{L}_{\theta }$ is clearly the sequence $r_{\theta }$ of section
1, we shall henceforth write it as $r_{\theta }.$ \ Of course, $m_{\theta
}=M_{\theta }$ (and $\ell _{\theta }=r_{\theta }$) if and only if $\theta $
is irrational.

For any real $\theta >0,$%
\begin{equation*}
r_{\theta }(n)=\#\{(i,j):\text{ }i+j\theta \leq n\theta \}.
\end{equation*}%
Consider the set of $(i,j)$ satisfying $(n-1)\theta <i+j\theta \leq n\theta
, $ or equivalently, $(n-j-1)\theta <i\leq (n-j)\theta .$ \ There is one
such $i$ for $j=n,$ and for $j=0,1,\ldots ,n-1,$ the number of such $i$ is $%
\left\lfloor (n-j)\theta \right\rfloor -\left\lfloor (n-j-1)\theta
\right\rfloor .$\ \ Summing gives a simple recurrence%
\begin{equation}
r_{\theta }(n)=r_{\theta }(n-1)+\left\lfloor n\theta \right\rfloor +1
\end{equation}%
for $n\geq 1,$ from which follows%
\begin{equation}
r_{\theta }(n)=n+1+\dsum\limits_{i=1}^{n}\left\lfloor i\theta \right\rfloor .
\end{equation}%
Alternatively, we may write%
\begin{eqnarray*}
r_{\theta }(n) &=&\#\{(i,j):\text{ }i+j\theta \leq n\theta \} \\
&=&\#\{(i,j):\text{ }j\leq n-i/\theta \},
\end{eqnarray*}%
so that the numbers $j$ to be counted are $0,1,2,\ldots ,\left\lfloor
n-i/\theta \right\rfloor ,$ for $i=0,1,\ldots ,\left\lfloor n\theta
\right\rfloor ,$ and%
\begin{eqnarray}
r_{\theta }(n) &=&\dsum\limits_{i=0}^{\left\lfloor n\theta \right\rfloor
}(1+\left\lfloor n-i/\theta \right\rfloor )  \notag \\
&=&\left\lfloor n\theta \right\rfloor +1+\dsum\limits_{i=0}^{\left\lfloor
n\theta \right\rfloor }\left\lfloor n-i/\theta \right\rfloor .
\end{eqnarray}%
(Equations much like (14) and (15) appear in [4]; the sequence $r_{3/2}$ is
A077043 in [6].)

Clearly,%
\begin{equation}
M_{\theta }(n)=M_{1/\theta }(n/\theta )=r_{1/\theta }(n).
\end{equation}%
If $\theta $ is rational, write $\theta =c/d,$ where $c$ and $d$ are
relatively prime numbers in $N.$ \ Then there are a total of $\left\lfloor
n/c\right\rfloor +1$ pairs $(i,j)$ satisfying $i+j\theta =n,$ so that there
are $M_{\theta }(n)-(\left\lfloor n/c\right\rfloor +1)$ pairs $(i,j)$
satisfying $i+j\theta <n.$ \ Consequently,%
\begin{equation}
m_{c/d}(n)=M_{c/d}(n)-\left\lfloor n/c\right\rfloor =r_{d/c}(n)-\left\lfloor
n/c\right\rfloor .
\end{equation}%
\bigskip

\begin{center}
\textbf{Table 1. \ Sequences associated with }$\theta =2\smallskip $

\begin{tabular}{|l||c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$n$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$
\\ \hline
$S_{\theta }(n)$ & $0$ & $1$ & $2$ & $2$ & $3$ & $3$ & $4$ & $4$ & $4$ & $5$
& $5$ & $5$ \\ \hline
$m_{\theta }(n)$ & $1$ & $2$ & $3$ & $5$ & $7$ & $10$ & $13$ & $17$ & $21$ & 
$26$ & $31$ & $37$ \\ \hline
$M_{\theta }(n)$ & $1$ & $2$ & $4$ & $6$ & $9$ & $12$ & $16$ & $20$ & $25$ & 
$30$ & $36$ & $42$ \\ \hline
$\ell _{\theta }(n)$ & $1$ & $3$ & $7$ & $13$ & $21$ & $31$ & $43$ & $57$ & $%
73$ & $91$ & $111$ & $133$ \\ \hline
$r_{\theta }(n)$ & $1$ & $4$ & $9$ & $16$ & $25$ & $36$ & $49$ & $64$ & $81$
& $100$ & $121$ & $144$ \\ \hline
\end{tabular}
\end{center}

As suggested by Table 1, the self-rank sequences of a positive integer, $k,$
are, loosely speaking, the sequences $P_{k}$ and $Q_{k}$ in section 1. \ A
more precise statement of this connection takes the form of Theorem
2.\bigskip

\begin{theorem}
\textit{Suppose }$\theta \in N$\textit{. \ The upper self-rank sequence of }$%
\theta $\textit{\ is given by}%
\begin{equation*}
r_{\theta }(n)=P_{\theta }(n),\text{ }n\geq 0.
\end{equation*}%
\textit{The lower self-rank sequence of }$\theta $\textit{\ is given by}%
\begin{eqnarray*}
\ell _{\theta }(n) &=&Q_{\theta -1}(n),\text{ }n\geq 0,\text{ if }\theta
\geq 2, \\
\ell _{1}(n) &=&\widehat{P}_{1}(n),\text{ }n\geq 0.
\end{eqnarray*}
\end{theorem}

\textbf{Proof: \ }Suppose $n\geq 0.$ \ Putting $\left\lfloor i\theta
\right\rfloor =i\theta $ in (14) gives%
\begin{eqnarray*}
r_{\theta }(n) &=&n+1+\theta \dsum\limits_{i=0}^{n}i. \\
&=&P_{\theta }(n),\text{ by (4).}
\end{eqnarray*}

As the rank of the first occurrence of $\theta n$ in $S_{\theta },$ the
number $\ell _{\theta }(n)$ is easily obtained from $r_{\theta }(n),$ the
total number of occurrences of $\theta n$ being $n+1;$ thus%
\begin{equation*}
\ell _{\theta }(n)=r_{\theta }(n)-n=P_{\theta }(n)-n.
\end{equation*}%
In case $\theta \geq 2,$ we therefore have $\ell _{\theta }(n)=Q_{\theta
-1}(n),$ by (6), and if $\theta =1,$ then $\ell _{\theta }(n)=\widehat{P}%
_{1}(n),$ by (9). \ \ \ \ \ \ \ \ \ $\square $\bigskip

\begin{theorem}
\textit{Suppose }$\theta $\textit{\ is an integer }$\geq 2.$\textit{\ \ If }$%
n\geq 0,$ \textit{then}%
\begin{eqnarray}
M_{\theta }(n) &=&r_{1/\theta }(n)=\widehat{Q}_{\theta -1}(n), \\
m_{1/\theta }(n) &=&Q_{\theta -1}(n), \\
M_{1/\theta }(n) &=&P_{\theta }(n), \\
m_{\theta }(n) &=&\widehat{P}_{\theta }(n).
\end{eqnarray}
\end{theorem}

\textbf{Proof: \ }That $M_{\theta }(n)=r_{1/\theta }(n)$ has already been
established. \ Equations (14) and (8) give 
\begin{equation*}
r_{1/\theta }(n)=r_{1/\theta }(n-1)+\QOVERD\lfloor \rfloor {n}{\theta }+1%
\text{ \ \ and \ \ }\widehat{Q}_{\theta -1}(n)=\widehat{Q}_{\theta
-1}(n-1)+\QOVERD\lfloor \rfloor {n}{\theta }+1,
\end{equation*}%
showing that the sequences $r_{1/\theta }$ and $\widehat{Q}_{\theta +1}$
have a common recurrence relation. \ As they also have identical initial
terms, (18) follows.

Next,%
\begin{eqnarray*}
m_{1/\theta }(n) &=&r_{\theta }(n)-n,\text{ by (17)} \\
&=&P_{\theta }(n)-n,\text{ by Theorem 2} \\
&=&Q_{\theta -1}(n)\text{ by (6), and (18) is proved.}
\end{eqnarray*}%
Further,%
\begin{eqnarray*}
M_{1/\theta }(n) &=&r_{\theta }(n),\text{ by (12)} \\
&=&P_{\theta }(n),\text{ by Theorem 2, so that (19) holds,}
\end{eqnarray*}%
and%
\begin{eqnarray*}
m_{\theta }(n) &=&r_{1/\theta }(n)-\left\lfloor n/\theta \right\rfloor ,%
\text{ by (17)} \\
&=&\widehat{Q}_{\theta -1}(n)-\left\lfloor n/\theta \right\rfloor ,\text{ by
(18)} \\
&=&\widehat{P}_{\theta }(n+1)-1-\left\lfloor n/\theta \right\rfloor \text{,
by (11)} \\
&=&\widehat{P}_{\theta }(n),\text{ by (7).}\ \ \ \ \ \ \ \ \ \square
\end{eqnarray*}

Theorems 4 and 5 show the manner in which the sequences $P_{k},Q_{k},%
\widehat{P}_{k},\widehat{Q}_{k}$ are related to minrank and maxrank
sequences. \ We turn next toward Theorem 6, which establishes that two of
the sequences are partial complements of the other two. \ A preliminary
example may be helpful. \ We begin by writing the complement $P_{2}^{\ast }$
of $P_{2}$ in labeled rows of consecutive integers:

\begin{equation*}
\begin{tabular}{lllllllllll}
\cline{3-4}
$\mathcal{S}_{0}\text{:}$ &  & \multicolumn{1}{|l}{2} & \multicolumn{1}{|l}{3
} & \multicolumn{1}{|l}{} &  &  &  &  &  &  \\ \cline{3-3}\cline{3-5}
$\mathcal{S}_{1}\text{:}$ &  & \multicolumn{1}{|l}{5} & \multicolumn{1}{|l}{6
} & \multicolumn{1}{|l}{7} & \multicolumn{1}{|l}{8} &  &  &  &  &  \\ 
\cline{3-3}\cline{5-6}
$\mathcal{S}_{2}\text{:}$ &  & \multicolumn{1}{|l}{10} & \multicolumn{1}{|l}{
11} & 12 & \multicolumn{1}{|l}{13} & \multicolumn{1}{|l}{14} & 15 &  &  & 
\\ \cline{3-3}\cline{6-7}
$\mathcal{S}_{3}\text{:}$ &  & \multicolumn{1}{|l}{17} & \multicolumn{1}{|l}{
18} & 19 & 20 & \multicolumn{1}{|l}{21} & \multicolumn{1}{|l}{22} & 23 & 24
&  \\ \cline{3-3}\cline{7-7}
\multicolumn{1}{c}{$\mathcal{\vdots }$} &  &  &  &  &  &  &  &  &  & 
\end{tabular}%
\end{equation*}%
Two equally spaced numbers from $\mathcal{S}_{n}$ are boxed. \ These
numbers, one can easily check, taken together and preceded by $1,$ form the
first partial complement of $P_{2}.$ \ By equation (7), they also form the
sequence $\widehat{P}_{2}.$ \ We generalize this method in the next
proof.\bigskip

\begin{theorem}
\textit{Suppose }$k\geq 1.$\textit{\ \ The 1st partial complement of }$P_{k}$%
\textit{\ is }$\widehat{P}_{k},$\textit{\ and the 1st partial complement of }%
$Q_{k}$\textit{\ is }$\widehat{Q}_{k}.$
\end{theorem}

\textbf{Proof: \ }The complement $P_{k}^{\ast }$ of $P_{k}$ consists of
segments $\mathcal{S}_{n}$ of $k(n+1)$ consecutive integers, given by%
\begin{equation*}
\mathcal{S}_{n}=\{P_{k}(n)+1,P_{k}(n)+2,\ldots ,P_{k}(n+1)-1\}
\end{equation*}%
for $n\geq 0.$ $\ $Thus, the initial segment $\mathcal{S}_{0}$ consists of
the $k$ numbers $2,3,\ldots ,k+1,$ for which we have%
\begin{equation*}
tP_{k}(h)=h+1=\widehat{P}_{k}(h)\text{ for }1\leq h\leq k.
\end{equation*}%
(Recall that $tr$ denotes the first partial complement of a sequence $r$;
the notation $tr(h)$ abbreviates $(tr)(h).$) \ As the number $k+2$ is not in 
$P_{k}^{\ast },$ we have from $\mathcal{S}_{1}$ the $k+1$ numbers%
\begin{equation*}
k+3,\text{ }k+5,\text{ \ldots , }3k+1
\end{equation*}%
satisfying%
\begin{eqnarray*}
tP_{k}(k+1) &=&k+3, \\
tP_{k}(k+3) &=&k+5, \\
&\vdots & \\
tP_{k}(3k-1) &=&3k+1=P_{k}(2)-2.
\end{eqnarray*}%
In order to extend these patterns to represent the appropriate numbers in
each segment $\mathcal{S}_{n},$ let%
\begin{equation*}
s_{n,j}=P_{k}(n)+1+(j-1)(n+1),\text{ \ for }1\leq j\leq k,\text{ }n\geq 0,
\end{equation*}%
and arrange these in segments of length $k+1:\medskip $

\begin{center}
\textbf{Table 2. \ The numbers }$s_{n,j}$%
\begin{equation*}
\begin{tabular}{|l|l|l|l|}
\hline
$s_{0,1}=2$ & $s_{0,2}=3$ & $\ldots $ & $s_{0,k}=k+1$ \\ \hline
$s_{1,1}=k+3$ & $s_{1,2}=k+5$ & $\ldots $ & $%
\begin{array}{l}
s_{1,k}=3k+1 \\ 
\text{ \ \ \ \ \ \ \ \ }=P_{k}(2)-2%
\end{array}%
$ \\ \hline
$s_{2,1}=P_{k}(2)+1$ & $s_{2,2}=P_{k}(2)+4$ & $\ldots $ & $%
\begin{array}{l}
s_{2,k}=P_{k}(2)+3k-2 \\ 
\text{ \ \ \ \ \ \ \ \ }=P_{k}(3)-3%
\end{array}%
$ \\ \hline
\multicolumn{1}{|c|}{$\vdots $} & \multicolumn{1}{|c|}{$\vdots $} & 
\multicolumn{1}{|c|}{$\vdots $} & \multicolumn{1}{|c|}{$\vdots $} \\ \hline
$s_{n,1}=P_{k}(n)+1$ & $s_{n,2}=P_{k}(n)+n+2$ & $\ldots $ & $%
\begin{array}{l}
s_{n,k}=P_{k}(n)+(k-1)n+k+1 \\ 
\text{ \ \ \ \ \ \ \ \ }=P_{k}(n+1)-n-1%
\end{array}%
$ \\ \hline
\multicolumn{1}{|c|}{$\vdots $} & \multicolumn{1}{|c|}{$\vdots $} & 
\multicolumn{1}{|c|}{$\vdots $} & \multicolumn{1}{|c|}{$\vdots $} \\ \hline
\end{tabular}%
\end{equation*}
\end{center}

The numbers in the following tableau satisfy the chain of equations defining
first partial complement:%
\begin{equation}
\begin{tabular}{|l|l|l|l|}
\hline
$tP_{k}(1)=s_{0,1}$ & $tP_{k}(s_{0,1})=s_{0,2}$ & $\ldots $ & $%
tP_{k}(s_{0,k-1})=s_{0,k}$ \\ \hline
$tP_{k}(s_{0,k})=s_{1,1}$ & $tP_{k}(s_{1,1})=s_{1,2}$ & $\ldots $ & $%
tP_{k}(s_{1,k-1})=s_{1,k}$ \\ \hline
$tP_{k}(s_{1,k})=s_{2,1}$ & \multicolumn{1}{|c|}{$\ldots $} & 
\multicolumn{1}{|c|}{$\ldots $} & \multicolumn{1}{|c|}{$\ldots $} \\ \hline
\end{tabular}%
\end{equation}%
In order to confirm that the sequence $tP_{k}$ has the values indicated by
(22), these equations must be proved:%
\begin{equation*}
\begin{tabular}{ll}
(i) & $tP_{k}(s_{n,k})=s_{n+1,1}$ for $n\geq 0,$ as in column 1; \\ 
(ii) & $tP_{k}(s_{n,j})=s_{n,j+1}$ for $1\leq j\leq k-1,$ $n\geq 0,$ as in
columns $2$ to $k.$%
\end{tabular}%
\end{equation*}%
We begin with (ii). \ Within $\mathcal{S}_{n}$ lie the consecutive integers%
\begin{equation*}
s_{n,j},s_{n,j}+1,\ldots ,s_{n,j+1}.
\end{equation*}%
Inductively, the first of these is $tP_{k}(s_{n,j-1}),$ so that in $%
P_{k}^{\ast },$ these integers are indexed as indicated by the top row of
the next array:%
\begin{equation*}
\begin{tabular}{|l|l|l|l|l|}
\hline
$s_{n,j-1}$ & $s_{n,j-1}+1$ & $s_{n,j-1}+2$ & $\ldots $ & $%
s_{n,j-1}+n+1=s_{n,j}$ \\ \hline
$s_{n,j}$ & $s_{n,j}+1$ & $s_{n,j}+2$ & $\ldots $ & $s_{n,j}+n+1=s_{n,j+1}$
\\ \hline
\end{tabular}%
\end{equation*}%
In other words, because the integers in both rows are consecutive, the
inductively assumed relation $tP_{k}(s_{n,j-1})=s_{n,j}$ implies $%
tP_{k}(s_{n,j})=s_{n,j+1}.$

Now regarding (i), we modify (22) to obtain%
\begin{equation*}
\begin{tabular}{|l|l|l|l|}
\hline
$s_{n,k-1}$ & $\ldots $ & $s_{n,k-1}+n$ & $s_{n,k-1}+n+1=s_{n,k}$ \\ \hline
$s_{n,k}$ & $\ldots $ & $s_{n,k}+n$ & $s_{n,k}+n+2=s_{n+1,1}$ \\ \hline
\end{tabular}%
\end{equation*}%
The point here is that the integers in row 1 are consecutive, but that those
in row 2, consecutive up to the penultimate term, skip over the number $%
s_{n,k}+n+1=P_{k}(n).$

As indicated by Table 2, $tP_{k}$ is the sequence having initial term $1$
and difference sequence consisting of $k$ $1$s followed by $k$ $2$s followed
by $k$ $3$s, and so on. \ The same characterization holds, by (7), for the
sequence $\widehat{P}_{k}.$ \ Therefore, $\widehat{P}_{k}=tP_{k}$.

We turn now to the proposition that the 1st partial complement of $Q_{k}$\
is $\widehat{Q}_{k}.$ \ The complement $Q_{k}^{\ast }$ of $Q_{k}$ consists
of segments $\mathcal{T}_{n}$ of $(k+1)(n+1)-2$ consecutive integers, given
by%
\begin{equation*}
\mathcal{T}_{n}=\{Q_{k}(n)+1,Q_{k}(n)+2,\ldots ,Q_{k}(n+1)-1\}
\end{equation*}%
for $n\geq 0.$ \ The method of proof used above for $P_{k}$\ is $\widehat{P}%
_{k}$ applies to these segments, leading to the conclusion, via (8), that $%
\widehat{Q}_{k}=tQ_{k}$. \ \ \ \ \ \ \ \ \ $\square $

\section{Farey trees}

In section 1, it is proved that if $\theta $ is a positive irrational
number, then $ttr_{\theta }=r_{\theta },$ or equivalently, that the
dispersion $A_{\theta }$ is transposable$.$ \ In case $\theta \in N,$ the
sequences $r_{\theta }$ and $r_{1/\theta }$ are those discussed in section
2. \ Here in section 3, we consider $r_{\theta }$ when $\theta $ is a
rational number and introduce certain limiting sequences, to be denoted by $%
r_{\theta ^{-}}.$ \ In section 4, we outline a possible proof that the only
sequences $r$ satisfying the equation $ttr=r$ are of the forms $r_{\theta }$
and $r_{\theta ^{-}}.$

Suppose $n\geq 1.$ \ The set $\mathcal{F}_{n}$ of Farey fractions of order $%
n $ consists of the rational numbers $c/d$ for which $0\leq c\leq d,$ $1\leq
d\leq n,$ and $c$ and $d$ are relatively prime. \ For example, the Farey
fractions of order $5$ are%
\begin{equation*}
0,\text{ }\frac{1}{5},\text{ }\frac{1}{4},\text{ }\frac{1}{3},\text{ }\frac{2%
}{5},\text{ }\frac{1}{2},\text{ }\frac{3}{5},\text{ }\frac{2}{3},\text{ }%
\frac{3}{4},\text{ }\frac{4}{5},\text{ }1.
\end{equation*}%
The numbers in $\mathcal{F}_{n},$ taken consecutively as endpoints,
determine a partition of the interval $[0,1)$; for example, for $n=5,$ the
subintervals are%
\begin{equation}
\lbrack 0,\frac{1}{5}),[\frac{1}{5},\frac{1}{4}),[\frac{1}{4},\frac{1}{3}),[%
\frac{1}{3},\frac{2}{5}),[\frac{2}{5},\frac{1}{2}),[\frac{1}{2},\frac{3}{5}%
),[\frac{3}{5},\frac{2}{3}),[\frac{2}{3},\frac{3}{4})\text{,}[\frac{3}{4},%
\frac{4}{5}),[\frac{4}{5},1).
\end{equation}%
The connection between Farey fractions and sequences $r_{\theta }$ is
indicated by the following example: \ for $0\leq \theta <1,$ the number $%
r_{\theta }(5)$ has one of the ten values%
\begin{equation*}
6,7,8,9,10,12,13,14,15,16
\end{equation*}%
according to which of the intervals in (23) contains $\theta .$

Corresponding to successive sets $\mathcal{F}_{n},$ we show six levels of
the $0$\textit{-Farey tree}, which represents all the sequences $r_{\theta }$
for which $\theta \in \lbrack 0,1):$\bigskip

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\bigskip

Level $n$ consists of $|\mathcal{F}_{n}|$ numbers, ranging from $n+1$ to $%
\widehat{P}_{1}(n).$ \ When progressing from level $n$ to level $n+1,$ a
branching occurs at a number if and only if the corresponding $n$-level
interval contains a fraction (in reduced form) having denominator $n+1.$ \
For example, if $\theta \in \lbrack 1/5,1/4)$ then $r_{\theta }(8)=13,$ in
level $8$ of the Farey tree. \ The fraction $2/9$ lies in $[1/5,1/4),$ so
that a branching occurs at $13$:%
\begin{equation*}
r_{\theta }(9)=\left\{ 
\begin{tabular}{ll}
$15$ & if $\theta \in \lbrack 1/5,2/9)$ \\ 
$16$ & if $\theta \in \lbrack 2/9,1/4).$%
\end{tabular}%
.\right.
\end{equation*}

It is easy to adapt the Farey tree for numbers $\theta ^{\prime }$ in an
interval $\lfloor h,h+1),$ where $h$ is a positive integer$;$ viz.,\
equation (14) yields%
\begin{equation}
r_{\theta ^{\prime }}(n)=r_{\theta }(n)+h\left( 
\begin{array}{c}
n+1 \\ 
2%
\end{array}%
\right) ,
\end{equation}%
where $\theta =\theta ^{\prime }-h,$ so that the desired tree, which we call
the $h$-Farey tree, results by adding $n(n+1)/2$ to the $n$-level nodes of
the Farey tree, for all $n\geq 0.$ \ It follows from (24) that in the $h$%
\textit{-Farey tree} at level $n$ the numbers range from $P_{h}(n)$ up to $%
Q_{h}(n).$

The set of infinite paths from $1$ down through a Farey tree are of three
types:%
\begin{equation*}
\begin{tabular}{rl}
(i) & paths that eventually stay left \\ 
(ii) & paths that eventually stay right \\ 
(iii) & all other paths.%
\end{tabular}%
\end{equation*}

The description, ``paths that eventually stay left,'' applies to any path
such that, after \textit{some} level in a Farey tree, \textit{every} time a
branching occurs, the path takes the left branch, and likewise for ``paths
that eventually stay right''.

In order to interpret these paths as representations of three kinds of
sequences $r_{\theta },$ recall that each node $x$ of a Farey tree
corresponds to an interval $\lfloor u,w).$ \ If $x$ connects to only one
successor, $y,$ in the next level, then $y$ corresponds to the same
interval, $\lfloor u,w).$ \ The only other possibility is that $x$ is a
branching node, meaning that $x$ has two successors$,$ $y$ and $z,$ in the
next level, and that there is a rational number $v$ such that $y\in \lfloor
u,v)$ and $z\in \lfloor v,w).$

Clearly then, paths of type (i) represent $r_{\theta }$ when $\theta $ is a
rational number, and the property, ``eventually stay left'' corresponds to $%
\theta $ lying in the interval $\lfloor \theta ,w)$ for every $w.$

Next, consider an eventually-stay-right path. From some node on, the
corresponding intervals are of the form $\lfloor v_{m},w)$ where $(v_{m})$
is a nondecreasing sequence of rationals with rational limit $w.$ \ The
sequence corresponding to these intervals, which we denote by $r_{(w-)},$ is
therefore given by%
\begin{equation}
r_{(w-)}=\underset{m\rightarrow \infty }{\lim }r_{v_{m}}.
\end{equation}%
For example, $r_{(2-)}=Q_{1}$ (whereas $r_{2}=P_{2}$)$.$

Paths of type (iii) represent $r_{\theta }$ when $\theta $ is an irrational
number, the limit point of a nest of intervals $\lfloor u_{m},w_{m})$ having
rational endpoints.\bigskip

\begin{theorem}
\textit{Suppose }$\theta =c/d,$\textit{\ where }$c$\textit{\ and }$d$\textit{%
\ in }$N$\textit{\ are relatively prime. \ Then}%
\begin{equation}
r_{(c/d-)}(n)=r_{d/c}(n)-\left\lfloor n/c\right\rfloor .
\end{equation}
\end{theorem}

\textbf{Proof: \ }Clearly (26) holds for $n=0.$ \ Suppose $n\geq 1,$ and let%
\begin{equation}
\mu =\min \{kd/c-\left\lfloor kd/c\right\rfloor :\text{ \ }1\leq k\leq n,%
\text{ }c\nmid k\}.
\end{equation}%
Let $p,q$ in $N$ be relatively prime satisfying $d/c-p/q<\mu .$ \ (That is,
the interval $[p/q,c/d)$ is of the form $[v_{m},w)$ used for (25) to define $%
r_{(c/d-)}.$) \ Then%
\begin{eqnarray}
r_{(c/d-)}(n) &=&r_{p/q}(n)  \notag \\
&=&n+1+\left\lfloor p/q\right\rfloor +\left\lfloor 2p/q\right\rfloor +\cdots
+\left\lfloor np/q\right\rfloor .
\end{eqnarray}%
By (27),%
\begin{equation*}
\left\lfloor kp/q\right\rfloor =\left\{ 
\begin{tabular}{ll}
$\left\lfloor kd/c\right\rfloor $ & if $c\nmid kd;$ \\ 
$\left\lfloor kd/c\right\rfloor -1$ & if $c\mid kd$%
\end{tabular}%
\right. 
\end{equation*}%
for $1\leq k\leq n.$ \ As $c$ and $d$ are relatively prime, the values of $k$
for which $c\mid kd$ are $c,2c,\ldots ,\left\lfloor n/c\right\rfloor c,$ so
that (28) gives%
\begin{eqnarray*}
r_{(c/d-)}(n) &=&n+1+\left\lfloor d/c\right\rfloor +\left\lfloor
2d/c\right\rfloor +\cdots +\left\lfloor nd/c\right\rfloor -\left\lfloor
n/c\right\rfloor  \\
&=&r_{d/c}(n)-\left\lfloor n/c\right\rfloor ,\text{ by (14). \ \ \ \ \ \ \ \
\ }\square 
\end{eqnarray*}

\section{Rank Arrays and the Family $F$}

Throughout section 4, the word \textit{rank} refers to the order relation $%
\prec $ except where otherwise indicated. $\ $(The relation $\prec ,$
dependent on rational $\theta $, is defined in section 1 on the set of
ordered pairs $(i,j),$ $i\geq 0,$ $j\geq 0.$)\bigskip

\begin{theorem}
\textit{Suppose }$\theta =c/d,$\textit{\ where }$c$\textit{\ and }$d$\textit{%
\ in }$N$\textit{\ are relatively prime. \ Then the 1st partial complement
of }$r_{\theta }$\textit{\ is given by}%
\begin{equation}
tr_{\theta }(n)=m_{\theta }(n)=r_{1/\theta }(n)-\left\lfloor
n/c\right\rfloor .
\end{equation}
\end{theorem}

\textbf{Proof: \ }The sequence $tr_{\theta }$ is row $0$ of $A_{\theta },$
so that $tr_{\theta }(n)$ is the rank of $(n,0),$ for all $n\geq 0.$ \ This
means that $tr_{\theta }(n)$ is the rank of the first occurrence of the
number $n$ when the numbers in $S_{\theta }$ are ranked under $\leq .$ \ In
other words, $tr_{\theta }(n)=m_{\theta }(n),$ where $m_{\theta }$ is the
minrank sequence of section 2. \ Thus, (29) follows from (17). \ \ \ \ \ \ \
\ \ $\square $\bigskip

\begin{corollary}
\textit{Suppose }$\theta =c/d,$\textit{\ where }$c$\textit{\ and }$d$\textit{%
\ in }$N$\textit{\ are relatively prime. \ Then the 1st partial complement
of }$r_{\theta }$\textit{\ is given by}%
\begin{equation}
tr_{\theta }(n)=r_{(\theta -)}(n)
\end{equation}%
for $n\geq 0.$
\end{corollary}

\textbf{Proof: \ }This is an obvious consequence of Theorems 5 and 6. \ \ \
\ \ \ \ \ \ $\square $\bigskip

\begin{lemma}
\textit{Suppose }$r\in R$\textit{\ and }$n_{1}\in N.$\textit{\ \ Then there
exists }$n_{0}$\textit{\ in }$N$\textit{\ such that if }$s$\textit{\ in }$R$%
\textit{\ satisfies }$s(n)=r(n)$\textit{\ for }$n=0,1,\ldots ,n_{0},$\textit{%
\ then }$s^{\ast }(n)=r^{\ast }(n)$\textit{\ for }$n=1,2,\ldots ,n_{1}.$
\end{lemma}

\textbf{Proof: \ }Let $n_{0}$ be the least $n$ for which $r(n)>r^{\ast
}(n_{1}).$ \ The numbers $r^{\ast }(1),$ $r^{\ast }(2),\ldots ,r^{\ast
}(n_{1})$ are then uniquely determined by the numbers $r(0),$ $r(1),\ldots
,r(n_{0}),$ as the former simply occupy in increasing order the positions
not occupied by the latter in the list $1,2,3,\ldots ,r(n_{0})$ of
consecutive integers. \ \ \ \ \ \ \ \ \ $\square $\bigskip

\begin{lemma}
\textit{Suppose }$r\in R$\textit{\ and }$n^{\prime }\in N.$\textit{\ \ Then
there exists }$n_{0}$\textit{\ in }$N$\textit{\ such that if }$s$\textit{\
in }$R$\textit{\ satisfies }$s(n)=r(n)$\textit{\ for }$n=0,1,\ldots ,n_{0},$%
\textit{\ then }$ts(n)=tr(n)$\textit{\ for }$n=0,1,\ldots ,n^{\prime }.$
\end{lemma}

\textbf{Proof: \ }Let $n_{1}$ be the number such that $tr(n^{\prime
})=r^{\ast }(n_{1}).$ \ The numbers $tr(n)$ for $n=0,1,\ldots ,n^{\prime }$
are among, and are uniquely determined by, the numbers $r^{\ast }(n)$ for $%
n=1,2,\ldots ,n_{1}.$ \ Let $n_{0}$ be as in Lemma 10, so that the numbers $%
tr(n)$ for $n=0,1,\ldots ,n^{\prime }$ are uniquely determined by the
numbers $r(n)$ for $n=0,1\ldots ,n_{0}.$ \ \ \ \ \ \ \ \ \ $\square $%
\bigskip 

\begin{corollary}
\textit{Suppose }$\theta =c/d,$\textit{\ where }$c$\textit{\ and }$d$\textit{%
\ in }$N$\textit{\ are relatively prime. \ Then}%
\begin{equation}
ttr_{\theta }=r_{\theta }.
\end{equation}
\end{corollary}

\textbf{Proof: \ }Suppose $n_{0}$ and $n_{1}$ are positive integers$.$ \ By
Corollary 9, there exist $c^{\prime }$ and $d^{\prime },$ relatively prime
in $N,$ such that $d^{\prime }/c^{\prime }<d/c$ and%
\begin{equation}
tr_{c/d}(n)=r_{d^{\prime }/c^{\prime }}(n)\text{ for }n=0,1,\ldots ,n_{0}.
\end{equation}%
It might be tempting to say that we apply $t$ to (32) to get%
\begin{equation*}
ttr_{c/d}(n)=tr_{d^{\prime }/c^{\prime }}(n)\text{ for }n=0,1,\ldots ,n_{0},
\end{equation*}%
but this mistake overlooks the meaning of the notation $tr(n)$ as $(tr)(n),$
not $t(r(n)).$ \ Instead, we can, in accord with Lemma 11, and do, take $%
n_{0}$ large enough that (32) implies%
\begin{equation}
ttr_{c/d}(n)=tr_{d^{\prime }/c^{\prime }}(n)\text{ for }n=0,1,\ldots ,n_{1}.
\end{equation}%
Let $\gamma $ and $\delta ,$ relatively prime in $N,$ satisfy $\gamma
/\delta >c/d$ and%
\begin{equation}
r_{\gamma /\delta }(n)=r_{c/d}(n)\text{ for }n=0,1,\ldots ,n_{1}.
\end{equation}%
By Corollary 9, there exist $c^{\prime \prime }$ and $d^{\prime \prime },$
relatively prime in $N,$ such that $c^{\prime \prime }/d^{\prime \prime
}<c^{\prime }/d^{\prime }$ and%
\begin{equation}
tr_{d^{\prime }/c^{\prime }}(n)=r_{c^{\prime \prime }/d^{\prime \prime }}(n)%
\text{ for }n=0,1,\ldots ,n_{1}.
\end{equation}%
As $c^{\prime }/d^{\prime }>c/d,$ we can and do choose $c^{\prime \prime }$
and $d^{\prime \prime }$ so that $c^{\prime \prime }/d^{\prime \prime }>c/d$
and $c^{\prime \prime }/d^{\prime \prime }>\gamma /\delta .$ \ Then by
(33)-(35),%
\begin{eqnarray*}
ttr_{c/d}(n) &=&r_{c^{\prime \prime }/d^{\prime \prime }}(n) \\
&=&r_{c/d}(n)\text{ for }n=0,1,\ldots ,n_{1}.
\end{eqnarray*}%
As $n_{1}$ is arbitrary, (31) follows. \ \ \ \ \ \ \ \ \ $\square $\bigskip 

Corollaries 9 and 12 extend Corollary 3 from the case that $\theta $ is
irrational to the case that $\theta $ is any positive real number. \ In
other words, $r_{\theta }$ is in the family $F$ of fixed points under the
operator $tt,$ or, in yet other words, the $1$st partial complement of the $1
$st partial complement of any $r_{\theta }$ is $r_{\theta }$ itself. \ The
next corollary identifies additional members of $F,$ and in section 5, we
conjecture that there are no others.\bigskip 

\begin{corollary}
\textit{Suppose }$\theta =c/d,$\textit{\ where }$c$\textit{\ and }$d$\textit{%
\ in }$N$\textit{\ are relatively prime. \ Then}%
\begin{equation}
ttr_{(\theta -)}=r_{(\theta -)}.
\end{equation}
\end{corollary}

\textbf{Proof: \ }Suppose $n_{0\text{,}}$ $n_{1},$ and $n_{2}$ are positive
integers. \ Let $c^{\prime }$ and $d^{\prime },$ relatively prime in $N,$
satisfy%
\begin{eqnarray*}
r_{c^{\prime }/d^{\prime }}(n) &=&r_{(\theta -)}(n)\text{ for }n=0,1,\ldots
,n_{0}\text{ and} \\
ttr_{c^{\prime }/d^{\prime }}(n) &=&ttr_{(\theta -)}(n)\text{ for }%
n=0,1,\ldots ,n_{1}.
\end{eqnarray*}%
By Corollary 12, we can, and do, choose $n_{0}$ and $n_{1}$ large enough that%
\begin{equation*}
ttr_{(\theta -)}(n)=r_{c^{\prime }/d^{\prime }}(n)\text{ for }n=0,1,\ldots
,\max \{n_{0},n_{1},n_{2}\}.
\end{equation*}%
As $n_{0},n_{1},n_{2}$ are arbitrarily large, (36) follows. \ \ \ \ \ \ \ \
\ $\square $\bigskip 

For any array $A=\{a(i,j)\},$ let $TA$ denote the transpose, $\{a(j,i)\},$
of $A.$ \ Let $A_{(\theta -)}$ denote the interspersion whose column $0$ is
the sequence $r_{(\theta -)}.$ \ Membership in the family $F$ can now be
stated in terms of transposable interspersions:\bigskip

\begin{corollary}
\textit{Suppose }$\theta $ is a positive real number. $\ $\textit{Then }$%
TA_{(\theta -)}=A,$ $TTA_{\theta }=A_{\theta },$ and $TTA_{(\theta
-)}=A_{(\theta -)}.$
\end{corollary}

\textbf{Proof: \ }If $\theta $ is a positive integer, the three asserted
equations follow from Theorem 4, as in this case, $r_{\theta }$ and $%
r_{(\theta -)}$ are $P_{\theta }$ and $\widehat{P}_{\theta },$ (and $%
r_{1/\theta }$ and $r_{(1/\theta -)}$ are $Q_{\theta }$ and $\widehat{Q}%
_{\theta }).$ \ For irrational, $\theta ,$ we have $A_{(\theta
-)}=A_{1/\theta }$, and the proposition follows from Corollary 12$.$ \ For
all other positive $\theta ,$ the asserted equations follow immediately from
Corollaries 9, 12, and 13.\bigskip 

\begin{center}
\textbf{Example 2:} \ The interspersions $A_{3/2}$ and $A_{(3/2-)}$

$%
\begin{tabular}{ccccccccccccccc}
$1$ & $2$ & $4$ & $6$ & $9$ & $\ldots $ &  &  &  & $1$ & $3$ & $7$ & $12$ & $%
19$ & $\ldots $ \\ 
$3$ & $5$ & $8$ & $11$ & $15$ & $\ldots $ &  &  &  & $2$ & $5$ & $10$ & $16$
& $24$ & $\ldots $ \\ 
$7$ & $10$ & $14$ & $18$ & $23$ & $\ldots $ &  &  &  & $4$ & $8$ & $14$ & $%
21 $ & $30$ & $\ldots $ \\ 
$12$ & $16$ & $21$ & $26$ & $32$ & $\ldots $ &  &  &  & $6$ & $11$ & $18$ & $%
26$ & $36$ & $\ldots $ \\ 
$19$ & $24$ & $30$ & $36$ & $43$ & $\ldots $ &  &  &  & $9$ & $15$ & $23$ & $%
32$ & $43$ & $\ldots $ \\ 
$\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\ddots $ &  & 
&  & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\ddots $%
\end{tabular}%
$\bigskip

\textbf{Example 3:} \ The interspersions $A_{2/3}$ and $A_{(2/3-)}$

$%
\begin{tabular}{ccccccccccccccc}
$1$ & $3$ & $6$ & $11$ & $17$ & $\ldots $ &  &  &  & $1$ & $2$ & $4$ & $7$ & 
$10$ & $\ldots $ \\ 
$2$ & $5$ & $9$ & $15$ & $22$ & $\ldots $ &  &  &  & $3$ & $5$ & $8$ & $12$
& $16$ & $\ldots $ \\ 
$4$ & $8$ & $13$ & $20$ & $28$ & $\ldots $ &  &  &  & $6$ & $9$ & $13$ & $18$
& $23$ & $\ldots $ \\ 
$7$ & $12$ & $18$ & $26$ & $35$ & $\ldots $ &  &  &  & $11$ & $15$ & $20$ & $%
26$ & $32$ & $\ldots $ \\ 
$10$ & $16$ & $23$ & $32$ & $42$ & $\ldots $ &  &  &  & $17$ & $22$ & $28$ & 
$35$ & $42$ & $\ldots $ \\ 
$\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\ddots $ &  & 
&  & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\ddots $%
\end{tabular}%
$\bigskip
\end{center}

\begin{theorem}
\textit{Suppose }$\theta $\textit{\ is a positive real number. \ Neighboring
terms in column }$j$ \textit{of the rank array }$A_{\theta }$\textit{\
satisfy the equation}%
\begin{equation}
a(i,j)-a(i-1,j)=\left\lfloor i\theta \right\rfloor +j+1
\end{equation}%
for $i\geq 1,$ $j\geq 0.$
\end{theorem}

\textbf{Proof: \ }Suppose $i\geq 1.$ \ By (13),%
\begin{equation*}
a(i,0)-a(i-1,0)=r_{\theta }(i)-r_{\theta }(i-1)=\left\lfloor i\theta
\right\rfloor +1.
\end{equation*}%
Thus, equation (37) is equivalent to%
\begin{equation}
\Re (j,i)-\Re (j,i-1)=\Re (0,i)-\Re (0,i-1)+j,
\end{equation}%
where $\Re (h,k)$ denotes the rank of $(h,k)$ under $\prec $ $.$ \ Let $%
v=\left\lfloor i\theta \right\rfloor .$ \ The $v$ ordered pairs $(h,k)$
counted by $\Re (0,i)-\Re (0,i-1)$ we represent as%
\begin{equation}
(0,i-1)\prec (h_{1},k_{1})\prec (h_{2},k_{2})\prec \cdots \prec
(h_{v},k_{v})=(0,i).
\end{equation}%
Clearly, (39) implies%
\begin{equation}
(1,i-1)\prec (h_{1}+1,k_{1})\prec (h_{2}+1,k_{2})\prec \cdots \prec
(h_{v}+1,k_{v})=(1,i).
\end{equation}%
It is easy to check that there is exactly one integer $q$ satisfying%
\begin{equation}
(1,i-1)\prec (0,i+q)\prec (1,i),
\end{equation}%
namely%
\begin{equation*}
q=\left\{ 
\begin{tabular}{cl}
$0$ & if $\theta =1;$ \\ 
$\lfloor 1/\theta \rfloor $ & otherwise.%
\end{tabular}%
\right. 
\end{equation*}%
\ 

Thus, by (40) and (41), there are $v+1$ ordered pairs $(h,k)$, satisfying%
\begin{equation}
(1,i-1)\prec (h,k)\prec (1,i).
\end{equation}%
Now if $(h,k)$ is an ordered pair satisfying (42), then either $h=0,$ so
that $k=q,$ or else $(h-1,k)$ is one of the ordered pairs in (39); thus
every $(h,k)$ satisfying (40) is one of the $v+1,$ so that (37) holds for $%
j=1.$

The method used to get from (39) to (42), namely, adding $1$ to all first
coordinates and then inserting the sole ordered pair having first coordinate 
$0$, applies inductively, so that (38) and (37) hold for all $j\geq 0.$ \ \
\ \ \ \ \ \ \ $\square $\bigskip

\begin{corollary}
\textit{Suppose }$\theta $\textit{\ is a positive real number. \ Neighboring
terms in column }$j$ \textit{of the array }$A_{(\theta -)}=\{b(i,j)\}$%
\textit{\ satisfy the equation}%
\begin{equation}
b(i,j)-b(i-1,j)=\left\lceil i/\theta \right\rceil +j,
\end{equation}%
where $\left\lceil \;\right\rceil $ denotes the ceiling function, for $i\geq
1,$ $j\geq 0.$
\end{corollary}

\textbf{Proof: \ }Suppose $n_{0}\in N.$ \ Let $c^{\prime }$ and $d^{\prime }$%
, relatively prime in $N,$ satisfy%
\begin{equation*}
r_{c^{\prime }/d^{\prime }}(n)=r_{(\theta -)}(n)
\end{equation*}%
for $n=0,1,\ldots ,n_{0}.$ \ In the rank array $A_{c^{\prime }/d^{\prime
}}=\{a^{\prime }(i,j)\},$ each $a^{\prime }(i,j)$ that is $\leq r_{c^{\prime
}/d^{\prime }}(n_{0})$ has the same position $(i,j)$ that the same number
has in $A_{(\theta -)}$; that is, $b(i,j)=a^{\prime }(i,j)$ for $a^{\prime
}(i,j)$ ranging from $1$ up to $r_{c^{\prime }/d^{\prime }}(n_{0}).$ \ By
(37),%
\begin{equation*}
a^{\prime }(i,j)=a^{\prime }(i-1,j)=\lfloor ic^{\prime }/d^{\prime }\rfloor
+j+1,
\end{equation*}%
so that $b(i,j)-b(i-1,j)-j=\lfloor ic^{\prime }/d^{\prime }\rfloor +1,$
showing that%
\begin{equation*}
b(i,j)-b(i-1,j)-j
\end{equation*}%
is invariant of $j.$ \ Now $b(i,0)=r_{d/c}-\lfloor i/c\rfloor ,$ by (26), so
that%
\begin{eqnarray*}
b(i,0)-b(i-1,0) &=&r_{d/c}(i)-r_{d/c}(i-1)-(\lfloor i/c\rfloor -\lfloor
(i-1)/c\rfloor \\
&=&\lfloor id/c\rfloor +1-\lfloor i/c\rfloor +\lfloor (i-1)/c\rfloor .
\end{eqnarray*}%
Consequently,%
\begin{equation*}
b(i,j)-b(i-1,j)=\lfloor id/c\rfloor -\lfloor i/c\rfloor +\lfloor
(i-1)/c\rfloor +j+1,
\end{equation*}%
and it is easy to verify that the right-hand side simplifies as in (43). \ \
\ \ \ \ \ \ \ $\square $\bigskip

\begin{corollary}
\textit{Suppose }$\theta $\textit{\ is a positive real number. }$\ $\textit{%
The terms of the dispersion }$A_{\theta }$ \textit{satisfy the equation}%
\begin{equation*}
a(i,j)=a(i,0)+a(0,j)+ij-1
\end{equation*}%
for $i\geq 0,$ $j\geq 0.$
\end{corollary}

\textbf{Proof: \ }By (37),%
\begin{equation*}
a(h,j)-a(h-1,j)=\lfloor h\theta \rfloor +j+1
\end{equation*}%
for $h=1,2,\ldots ,i.$ \ Summing over those values of $h$ gives%
\begin{eqnarray*}
a(i,j) &=&a(0,j)+\lfloor \theta \rfloor +\lfloor 2\theta \rfloor +\cdots
+\lfloor i\theta \rfloor +i(j+1) \\
&=&a(0,j)+r_{\theta }(i)-i-1+i(j+1),\text{ by (13)} \\
&=&a(i,0)+a(0,j)+ij-1.\ \ \ \ \ \ \ \ \ \square 
\end{eqnarray*}

\begin{corollary}
\textit{Suppose }$\theta =c/d,$ where $c$ and $c$ are relatively prime in $N$%
\textit{. }$\ $\textit{The terms of the dispersion }$A_{(\theta
-)}=\{b(i,j)\}$ \textit{satisfy the equation}%
\begin{equation*}
b(i,j)=b(i,0)+b(0,j)+ij+\lfloor i/c\rfloor 
\end{equation*}%
for $i\geq 0,$ $j\geq 0.$
\end{corollary}

\textbf{Proof: \ }By (43),%
\begin{equation*}
b(h,j)-b(h-1,j)=\left\lceil h/\theta \right\rceil +j
\end{equation*}%
for $h=1,2,\ldots ,i.$ \ Summing over those values of $h$ as in the proof of
Corollary 7.2 and simplfying via the identify%
\begin{equation*}
\left\lceil x\right\rceil =\left\{ 
\begin{tabular}{cl}
$x$ & if $x$ is an integer; \\ 
$\lfloor x\rfloor +1$ & otherwise%
\end{tabular}%
\right. 
\end{equation*}%
yield the asserted equation.$\ \ \ \ \ \ \ \ \ \square $\bigskip 

\section{Convergence}

It appears likely that if $r\in R,$ then the sequence of sequences,%
\begin{equation*}
r,\text{ }ttr,\text{ }tt(ttr),\text{ }\ldots ,
\end{equation*}%
whose general term we abbreviate as $t^{(2n)}r,$ converges, and that the
limiting sequence is one of the sequences $r_{\theta }$ or $r_{(\theta -)}$
for some positive real number $\theta .$ \ If so, then the family $F$ of
sequences satisfying $ttr=r$ clearly contains no sequences other than those
already accounted for.

Proof that $(t^{(2n)}r)$ converges seems elusive. \ This section offers
lemmas which may someday be found useful in a proof but are of independent
interest in any case. \ The items for which proof is sought are then
presented as Conjectures 22-26.

Suppose that $u$ and $v$ are sequences in $R$ and that their initial
segments of some length are identical \ Then some initial segments of the
complements, $u^{\ast }$ and $v^{\ast },$ must be identical. \ The following
lemma provides some insight.\bigskip

\begin{lemma}
\textit{Suppose }$u$\textit{\ and }$v$\textit{\ are sequences in }$R$ 
\textit{such that }$u(i)=v(i)$\textit{\ for }$i=0,1,\ldots ,n,$\textit{\ and 
}$u(n+1)\geq v(n+1).$\textit{\ \ Then}%
\begin{equation*}
u^{\ast }(j)=v^{\ast }(j)\text{\textit{\ \ \ \ }for}\mathit{\ }j=1,2,\ldots
,v(n+1)-n-2.
\end{equation*}
\end{lemma}

\textbf{Proof: \ }For $h=0,1,\ldots ,n,$ the $v(h+1)-v(h)-1$ numbers $v(h)+1$
to $v(h+1)-1$, taken in order, comprise an initial segment of both $v^{\ast
} $ and $u^{\ast }.$ \ The length of this common segment is%
\begin{equation}
\dsum\limits_{h=0}^{n}[v(h+1)-v(h)-1]=v(n+1)-n-2.\text{ \ \ \ \ \ \ \ \ \ }%
\square
\end{equation}

\begin{lemma}
\textit{Suppose }$c$\textit{\ and }$d$\textit{\ in }$N$\textit{\ are
relatively prime. \ Suppose }$r\in R$\textit{\ and that there exists }$n\geq
2$\textit{\ such that }$r(i)=r_{c/d}(i)$\textit{\ and }$r(n+1)\geq
r_{c/d}(n+1).$\textit{\ \ Then }$r^{\ast }(j)=r_{c/d}^{\ast }(j)$\textit{\
for }$j=1,2,\ldots ,J,$\textit{\ where}%
\begin{equation}
J=r_{c/d}(n)+\lfloor (n+1)c/d\rfloor -n-1.
\end{equation}
\end{lemma}

\textbf{Proof: \ }In Lemma 19, take $u=r$ and $v=r_{c/d}.$ \ Then $%
v(n+1)=r_{c/d}(n+1)$ in (44)$,$ and (13) implies (45). \ \ \ \ \ \ \ \ \ $%
\square $\bigskip 

\begin{lemma}
\textit{Suppose }$c$\textit{\ and }$d$\textit{\ in }$N$\textit{\ are
relatively prime and }$c>d$\textit{. \ Then}%
\begin{eqnarray}
r_{c/d}^{\ast }(i) &=&i+1\text{ for }i=1,2,\ldots ,r_{c/d}(1)-2,  \notag \\
r_{c/d}^{\ast }(r_{c/d}(m)-m+J) &=&r_{c/d}(m)+J+1 \\
\text{for }J &=&0,1,\ldots ,\lfloor (m+1)c/d\rfloor -1,\text{ }%
m=1,2,3,\ldots .
\end{eqnarray}
\end{lemma}

\textbf{Proof: \ }Consider the sequence of numbers to which $r_{c/d}^{\ast }$
is applied: \ initially,%
\begin{equation*}
1,2,\ldots ,r_{c/d}(1)-2,\text{ followed by}
\end{equation*}%
\begin{eqnarray*}
r_{c/d}(m)-m+J,\text{ for }m &=&1\text{ and }J=0,1,\ldots ,\lfloor
2c/d\rfloor -1, \\
r_{c/d}(m)-m+J,\text{ for }m &=&2\text{ and }J=0,1,\ldots ,\lfloor
3c/d\rfloor -1,
\end{eqnarray*}%
and so on. \ Thus, the sequence in question is the sequence of positive
integers. \ Their images under the function $r_{c/d}^{\ast }$ are determined
by arranging all the numbers $r_{c/d}^{\ast }(i)$ in increasing order$.$ \
This listing is conveniently broken into segments, first from $1$ to $%
r_{c/d}(1)-1,$ then from $r_{c/d}(1)+1$ to $r_{c/d}(2)-1,$ and so on. \
Thus, counting the first segment as segment $1,$ the $(m+1)$st segment, for $%
m\geq 1,$ is as given by (46) and (47). \ \ \ \ \ \ \ \ \ $\square $\bigskip

\begin{conjecture}
\textit{Suppose }$c$\textit{\ and }$d$\textit{\ in }$N$\textit{\ are
relatively prime. \ Suppose that }$r\in R$\textit{\ and that there exists }$%
n\geq 1$\textit{\ such that }$r(i)=r_{c/d}(i)$\textit{\ for }$i=0,1,\ldots
,n,$ \textit{and}%
\begin{equation}
r(n+1)\geq r_{c/d}(n+1).
\end{equation}%
\textit{Then}%
\begin{equation}
tr(i)=tr_{c/d}(i)\mathit{\ }\text{\ \ \ for}\mathit{\ }i=0,1,2,\ldots
,\lfloor (n+1)c/d\rfloor .
\end{equation}
\end{conjecture}

\begin{conjecture}
\textit{Continuing, suppose, instead of }(48)\textit{, that}%
\begin{equation}
r(n+1)\leq r_{c/d}(n+1)-2.
\end{equation}%
\textit{Then}%
\begin{equation}
tr(n+1)=tr_{c/d}(n+1)-1.
\end{equation}%
\bigskip 
\end{conjecture}

\begin{conjecture}
\textit{Continuing, suppose, instead of }(50)\textit{, that}%
\begin{equation}
r(n+1)=r_{c/d}(n+1)-1.
\end{equation}%
\textit{Then}%
\begin{equation}
tr(n+1)=tr_{c/d}(n+1).
\end{equation}%
\bigskip 
\end{conjecture}

\begin{conjecture}
\textit{Continuing, suppose}%
\begin{equation*}
z(i)=tr_{c/d}(i)\text{ for }i=0,1,\ldots ,\lfloor (n+1)c/d\rfloor 
\end{equation*}%
\textit{and}%
\begin{equation}
z(i+1)-z(i)\geq z(i)-z(i-1)
\end{equation}%
for $i\geq \lfloor (n+1)c/d\rfloor .$ \ \textit{Then}%
\begin{equation}
tz(i)=r_{c/d}(i)\mathit{\ }\text{\ \ \ for}\mathit{\ }i=0,1,2,\ldots ,n+1.
\end{equation}%
\smallskip 
\end{conjecture}

\begin{conjecture}
\textit{Suppose }$r\in R.$ \ \textit{Then there exists a positive real
number }$\theta $\textit{\ such that }$\underset{n\rightarrow \infty }{\lim }%
t^{(2n)}r$\textit{\ is one of the sequences }$r_{\theta }$\textit{\ and }$%
r_{(\theta -)}.\smallskip $
\end{conjecture}

\textbf{Possible method of proof: \ }If $r$ is $r_{\theta }$ or $r_{(\theta
-)}$ for some then $\theta ,$ then $ttr=r$ by Corollaries 3, 12, and 13. \
Suppose then that $r$ is not any such $r_{\theta }$ or $r_{(\theta -)}.$ \
Let%
\begin{equation*}
(\gamma ,\delta )=\left\{ 
\begin{tabular}{cl}
$(1,2)$ & if $r(1)=2;$ \\ 
$(r(1)-2,1)$ & if $r(1)\geq 3$.%
\end{tabular}%
\right. 
\end{equation*}%
Then $r(0)=r_{\gamma /\delta }(0)=1$ and $r(1)=r_{\gamma /\delta }(1),$ so
that the set of rational numbers $\gamma /\delta $ satisfying%
\begin{equation}
r(i)=r_{\gamma /\delta }(i)\text{ \ \ for }i=0,1,\ldots ,m,
\end{equation}%
for some $m\geq 1,$ is not empty. \ The set of numbers $m$ for which (56)
holds for some $\gamma /\delta $ is not unbounded, for if it were, there
would be a sequence $\gamma _{m}/\delta _{m}$ having limit $\theta $ such
that $r\in \{r_{\theta },$ $r_{\theta ^{-}}\},$ a contradiction. \ Let $n$
be the greatest $m$ for which (56) holds, and let $c/d,$ where $c$ and $d$
are relatively prime, be a rational number such that%
\begin{equation*}
r(i)=r_{c/d}(i)\text{ \ \ for }0,1,2,\ldots ,n.
\end{equation*}%
As $r(n+1)\neq r_{c/d}(n+1),$ one of the inequalities (48), (50), and (52)
holds, so that we have cases:\smallskip 

\textit{Case 1:} \ $r(n+1)\geq r_{c/d}(n+1).$ \ If Conjecture 22 is valid,
then (49) holds.\smallskip 

\textit{Case 2:} \ $r(n+1)\leq r_{c/d}(n+1)-2.$ \ If Conjecture 23 is valid,
then by (51), the sequences $tr_{c/d}$ and $tr$ satisfy the hypothesis of
Conjecture 24 (with $tr_{c/d}$ and $tr$ substituted for $r_{c/d}$ and $r,$
respectively).\smallskip 

\textit{Case 3:} \ $r(n+1)=r_{c/d}(n+1)-1.$ \ If Conjecture 24 is valid,
then by (53), the sequences $tr_{c/d}$ and $tr$ satisfy the hypothesis of
Conjecture 22 (with $tr_{c/d}$ and $tr$ substituted for $r_{c/d}$ and $r,$
respectively).\smallskip 

Let%
\begin{equation}
r^{\prime }=\left\{ 
\begin{tabular}{ll}
$tr$ & if (48) holds; \\ 
$tttr$ & if (50) holds; \\ 
$tr$ & if (52) holds.%
\end{tabular}%
\right. 
\end{equation}%
The discussion of the three cases shows that $r^{\prime }$ satisfies (48),
hence (49), if Conjecture 22 is valid. \ Let $z=r^{\prime }.$ \ It is easy
to check that (54) holds (for $z=t\rho ,$ for every $\rho $ in $R$). \ Thus,
if Conjecture 25 is valid, then, with reference to (56), the sequence $ttttr$
satisfies%
\begin{equation*}
ttttr(i)=r_{c/d}(i)\text{ \ \ for }i=0,1,\ldots n,n+1
\end{equation*}%
(whereas $r(n+1)$ may not be equal to $r_{c/d}(n+1)$), and a proof of
Conjecture 26 follows by repeated applications of Conjectures 22-25. \ It is
hoped that someone will prove those conjectures!

The interested reader may wish to use the following website:%
\begin{equation*}
\text{\textbf{http://csserver.evansville.edu/\symbol{126}%
brownie/cgi-bin/transpose}}.
\end{equation*}
\ There, the reader can submit the first ten to thirty terms of a sequence $r
$. \ The dispersion of the complementary sequence will appear, of which the
submitted sequence is the first column. \ The reader can then request
iterations and see, in the successive first columns, initial terms of the
sequences $tr,ttr,tttr,$ \ldots 

\bigskip

\textbf{References\medskip }\bigskip

\noindent [1] \ John H. Conway and Richard K. Guy, \textit{The Book of Numbers},
Springer-Verlag, New York, 1996.\bigskip

\noindent [2] \ Louis Comtet, \textit{Advanced Combinatorics}, revised and enlarged
edition, D. Reidel, Dordrecht, 1974.\bigskip

\noindent [3] \ Clark Kimberling, Interspersions and dispersions, \textit{
Proc.\ Amer.\ Math.\ Soc.} {\bf 117} (1993), 313--321.\bigskip

\noindent [4]\ \ Clark Kimberling, Fractal sequences and interspersions, \textit{
Ars Combinatoria} {\bf 45} (1997), 157--168.\bigskip

\noindent [5]\ \ St\'{e}phanie Petit, editor (2003) \textit{Encyclopedia of
Combinatorial Structures}, \\
{\tt http://pauillac.inria.fr/algo/encyclopedia/index.html} .\bigskip

\noindent [6]\ \ N. J. A. Sloane, editor (2003), \textit{The On-Line Encyclopedia of
Integer Sequences, } \\
\texttt{http://www.research.att.com/\symbol{126}%
njas/sequences/}\textit{\ }.\bigskip

\noindent [7]\ \ N. J. A. Sloane, (2003) Classic Sequences In \textit{The On-Line
Encyclopedia of Integer Sequences}: \ (Part 1:) \ The Wythoff Array and The
Para-Fibonacci Sequence, \\
\texttt{http://www.research.att.com/\symbol{126}njas/sequences/classic.html }%
.\bigskip \bigskip


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11B36; Secondary 11B25, 11B37.

\noindent \textit{Keywords:} \ central polygonal numbers, dispersion, interspersion,
Farey sequence, Farey tree, fractal sequence, partial complement, polygonal
numbers, rank array, rank sequence.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000124},
\seqnum{A000217},
\seqnum{A000290},
\seqnum{A000326},
\seqnum{A001401},
\seqnum{A001840},
\seqnum{A001844},
\seqnum{A001972},
\seqnum{A002061},
\seqnum{A002620},
\seqnum{A005448},
\seqnum{A008748},
\seqnum{A077043},
\seqnum{A082152},
\seqnum{A082153},
\seqnum{A082154},
\seqnum{A082155},
\seqnum{A082156},
\seqnum{A086270},
\seqnum{A086271},
\seqnum{A086272},
\seqnum{A086273},
\seqnum{A087465},
\seqnum{A087466},
\seqnum{A087467},
\seqnum{A087468},
\seqnum{A087469},
\seqnum{A087470}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 14 2003;
revised version received  January 19 2004.
Published in {\it Journal of Integer Sequences}, February 18 2004.

\bigskip
\hrule
\bigskip

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


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