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\begin{center}
\vskip 1cm{\LARGE\bf 
Lucas Representations of Positive Integers
}
\vskip 1cm
\large
Clark Kimberling\\
Department of Mathematics\\
University of Evansville \\
Evansville, Indiana 47722 \\
USA \\
\href{mailto:ck6@evansville.edu}{\tt ck6@evansville.edu}
\end{center}

\vskip .2 in

\begin{abstract}
Various kinds of representations of positive integers using nonconsecutive
Lucas numbers are used to define arrays related to the Wythoff array.
The columns of these arrays, or their order arrays, partition the
positive integers.  Limiting densities are found for numbers whose Lucas
representations all have the same least term.
\end{abstract}

\section{Introduction}\label{section:Introduction}
Throughout, the set of positive integers is denoted by $\mathbb{N}$, and $i, k, m, n, u$ represent elements of $\mathbb{N}$.  The golden ratio, $(1+\sqrt 5)/2$, is denoted by $\tau$. The sequence $(L_n)$ of Lucas numbers is given by $L_0=2, L_1=1,$ and $L_n=L_{n-1}+L_{n-2}$ for $n \geq 2$. In 1972, Carlitz, Scoville, and Hoggatt \cite{Carlitz} proved the following uniqueness theorem for representations of positive integers as sums of nonconsecutive Lucas numbers.
\begin{theorem}\label{CSH}
Every $n$ has a unique representation in exactly one of these two forms:
\begin{align}
n &= L_{k_1}+\cdots+L_{k_u}+L_0 \text{ (or } n=2), \label{form1a} \\
& \text { where } k_i-k_{i+1} \geq 2 \text{ for }  1 \leq i \leq u-1, k_u \geq 3  \label{form1b}
 \end{align}
or
\begin{align}
n &= L_{k_1}+\cdots+L_{k_u} \label{form2a} \\
& \text { where } k_i-k_{i+1} \geq 2 \text{ for } 1 \leq i \leq u-1, k_u \geq 1. \label{form2b}
\end{align}
\end{theorem}

Following the notation in \cite{Carlitz}, let $B_0$ be the sequence of numbers $n$ given by (\ref{form1a}) and (\ref{form1b}), and let $B_k$ be the sequence given by (\ref{form2a}) and (\ref{form2b}); e.g.,
\begin{align*}
B_0 &= (2,6,9,13,17,20,24,27,31,35,38,42,46,49, \cdots) \\
B_1 &= (1,5,8,12,16,19,23,26,30,34,37,41,45,48, \cdots) \\
B_2 &= (3,10,14,21,28,32,39,43,50,57,61,68,75,79, \cdots).
\end{align*}
In general, $B_k$ consists of all $n$ such that the least term in (\ref{form1a}), or (\ref{form2a}), is $L_k$.  Clearly the sequences $B_k$ partition $\mathbb{N}$.

The representations in Theorem \ref{CSH} are patterned after Zeckendorf representations, which we review as follows.  The sequence $(F_n)$ of Fibonacci numbers is given by $F_0=0,F_1=1$, and $F_n=F_{n-1}+F_{n-2}$ for $n \geq 2$.  The Zeckendorf representation of $n$ is the unique sum
\begin{align}
n &= F_{k_1}+\cdots+F_{k_u} \label{form3a}\\
& \text { where } k_i-k_{i+1} \geq 2 \text{ for } i=1\cdots u-1, k_u \geq 2.  \label{form3b}
 \end{align}

For every $n$, the greedy algorithm can be used to find the successive terms in all three representations (\ref{form1a}) and (\ref{form1b}), (\ref{form2a}) and (\ref{form2b}), and (\ref{form3a}) and (\ref{form3b}).

Historically, the Zeckendorf representation dates back to Zeckendorf's work as early as 1939, but he did not submit the result for publication until April, 1972; remarkably, his reference section includes the 1972 papers \cite{CSH} and \cite{Carlitz}. Zeckendorf's theorems \cite{Zeck} are quoted here:
\begin{quote}
\small{
TH\'EOR\`EME I.a. \textit{Tout nombre naturel $N$ peut \^etre repr\'esent\'e par une somme de nombres de Fibonacci distincts non cons\'ecutifs.} \\
TH\'EOR\`EME I.b. \textit{Pour tout nombre naturel, cette somme est unique.} \\
TH\'EOR\`EME II.a. \textit{Tout nombre naturel $N$ peut \^etre repr\'esent\'e par une somme de nomberes de Lucas distincts non cons\'ecutifs.} \\
TH\'EOR\`EME II.b. \textit{La repr\'esent\'e des nombres naturels par une somme de nombres de Lucas non cons\'ecutifs est unique, sauf pour les nombres $L_{2v+1}+1$}.}
\end{quote}
Aside from the Lucas representations in Theorem \ref{CSH}, another kind of Lucas representation is given by Luo \cite{Luo}, in which some but not all $n$ have a unique representation; indeed those $n$ having more than one representation have exactly two representations.

The main purpose of this article can now be stated: to partition $\mathbb{N}$, or some subset of $\mathbb{N}$, as the set of columns of certain arrays (as in Tables \ref{tab:Table2}-\ref{tab:Table5}) obtained from various kinds of Lucas representations, and to consider corresponding order arrays, densities, and limiting densities. These results can be compared to similar results already well known for the Wythoff array.


\section{Wythoff array and Lucas-Wythoff arrays}\label{section:section2}

In 1980, Morrison defined the Wythoff array $(w(n,k))$ by the formulas
$$w(n,1)=\lfloor\lfloor n\tau\rfloor\tau\rfloor \text{ and } w(n,2)=\lfloor\lfloor n\tau\rfloor\tau^2\rfloor$$
together with the Fibonacci recurrence
$$w(n,k)=w(n,k-1)+w(n,k-2) \text{ for } k \geq 3.$$
The Wythoff array, which shows all the winning pairs for the Wythoff game, has been widely studied; see, for example, the Comments and Links at \seqnum{A035513} in \cite{OEIS} and \enquote{Wythoff visions} \cite{DFGHKL} In Theorem \ref{Theorem1}, we state and verify a formula for $w(n,k)$ that appears elsewhere (e.g., \cite{Kim1}) without proof.  We begin with lemmas that account for the first two columns of the Wythoff array. The notation $\{x\}$ is used for the fractional part of a real number $x$, defined by $\{x\}=x-\lfloor x \rfloor$.
\begin{lemma}\label{Lemma1}
If $n \geq 1$, then
$\lfloor\lfloor n\tau\rfloor\tau\rfloor = \lfloor n\tau \rfloor+n-1$.
\end{lemma}
\begin{proof}
Since the fractional part $\{n\tau\}$ of $n\tau$ is in $(0,1)$, we have
$$\lfloor n-\{n\tau\}(\tau-1)\rfloor = n-1.$$
The identity $\tau^2 = \tau+1$ then gives
\begin{align*}
n-1 &= \lfloor n \tau^2 - n \tau - \{n\tau\}(\tau-1) \rfloor \\
&= \lfloor (n \tau - \{n\tau\})(\tau-1) \rfloor \\
&= \lfloor \lfloor n \tau \rfloor(\tau-1) \rfloor \\
&= \lfloor \lfloor n \tau \rfloor \tau \rfloor -\lfloor n\tau \rfloor.
\end{align*}
\end{proof}
\begin{lemma}\label{Lemma2}
$\lfloor\lfloor n\tau\rfloor \tau^2\rfloor = 2\lfloor n\tau \rfloor+n-1$.
\end{lemma}
\begin{proof}
Using Lemma \ref{Lemma1}, we have
\begin{align*}
\lfloor\lfloor n\tau\rfloor \tau^2\rfloor &= \lfloor \lfloor n \tau \rfloor (\tau+1) \rfloor \\
&= \lfloor\lfloor n\tau\rfloor \tau\rfloor+\lfloor n \tau \rfloor \\
&= (\lfloor n \tau \rfloor + n - 1) + \lfloor n \tau \rfloor.
\end{align*}
\end{proof}


\begin{theorem}\label{Theorem1}
The Wythoff array is given by
\begin{equation}\label{eq:WythoffEquation}
w(n,k)=\lfloor n\tau \rfloor F_{k+1}+(n-1)F_k
\end{equation}
for $n \geq 1, k \geq 1$.
\end{theorem}

\begin{proof}
For every $n$, equation (\ref{eq:WythoffEquation}) holds for $k=1$ and $k=2$, by the two lemmas. Assume (\ref{eq:WythoffEquation}) for all $n$ and arbitrary $k \geq 2$.  Then
\begin{align*}
w(n,k+1) &= w(n,k)+ w(n,k-1) \\
&= \lfloor n \tau \rfloor F_{k+1}+(n-1)F_k + \lfloor n \tau \rfloor F_k+(n-1)F_{k-1} \\
&= \lfloor n \tau \rfloor (F_{k+1}+F_k)+(n-1)(F_k+F_{k-1}) \\
&= \lfloor n \tau \rfloor F_{k+2}+(n-1)
F_{k+1};
\end{align*}
so that inductively, (\ref{eq:WythoffEquation}) holds for all $k$.
\end{proof}

\begin{table} \centering
\begin{tabular}{| c c c c c c c c c c c |}
\hline
1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & \ldots \\
4 & 7 & 11 & 18 & 29 & 47 & 76 & 123 & 199 & 322 & \\
6 & 10 & 16 & 26 & 42 & 68 & 110 & 178 & 288 & 466 & \\
9 & 15 & 24 & 39 & 63 & 102 & 165 & 267 & 432 & 699 &\\
12 & 20 & 32 & 52 & 84 & 136 & 220 & 356 & 576 & 932 & \\
14 & 23 & 37 & 60 & 97 & 157 & 254 & 411 & 665 & 1076 & \\
\vdots & & & & & & & & & & \\
\hline
\end{tabular}
\caption{Wythoff array}
\label{tab:Table1}
\end{table}

Referring to Table \ref{tab:Table1}, the Fibonacci numbers and Lucas numbers occupy rows 1 and 2, respectively, and every row satisfies the recurrence $r_n=r_{n-1}+r_{n-2}$, and every $n$ occurs exactly once.  Furthermore, for all $k$, column $k$ consists of those numbers $m$ having $F_{k+1}$ as least term in the Zeckendorf representation ((\ref{form3a}) and (\ref{form3b})), as proved in \cite{Kim1}. Indeed, the Zeckendorf array as defined in \cite{Kim1} is identical to the Wythoff array.

We now define the \textit{1st Lucas-Wythoff array}, $(r(n,k))$, by columns: (column 1) $= B_1$, (column 2) $= B_0$, and, for $k \geq 2$, (column $k$) $= B_k$.  See Table 2.

\begin{table} \centering
\begin{tabular}{| c c c c c c c c c c c |}
\hline
1 & 2 & 3 & 4 & 7 & 11 & 18 & 29 & 47 & 76 & \ldots \\
5 & 6 & 10 & 15 & 25 & 40 & 65 & 105 & 170 & 275 & \\
8 & 9 & 14 & 22 & 36 & 58 & 94 & 152 & 246 & 398 & \\
12 & 13 & 21 & 33 & 54 & 87 & 141 & 228 & 369 & 597 &\\
16 & 17 & 28 & 44 & 72 & 116 & 188 & 304 & 492 & 796 & \\
\vdots & & & & & & & & & &\\
\hline
\end{tabular}
\caption{1st Lucas-Wythoff array}
\label{tab:Table2}
\end{table}

The inclusion of column 2 ensures that every $n$ in $\mathbb{N}$ occurs (exactly once) in the array and that all rows and columns are strictly increasing.  However, column 2 interrupts the Fibonacci row recurrence seen in the Zeckendorf array.  Here, instead, we have $r(n,4) = r(n,1)+r(n,3)$ for all $n$, and $r(n,k)=r(n,k-1)+r(n,k-2)$ for $n \geq 1$ and $k \geq 5$.  Deleting column 2 results in the \textit{2nd Lucas-Wythoff array}, $(r^*(n,k))$, for which we have a formula much like (\ref{eq:WythoffEquation}), shown in Theorem \ref{Theorem2}. See Table \ref{tab:Table3}.

\begin{table} \centering
\begin{tabular}{| c c c c c c c c c c c |}
\hline
1 & 3 & 4 & 7 & 11 & 18 & 29 & 47 & 76 & 123 & \ldots \\
5 & 10 & 15 & 25 & 40 & 65 & 105 & 170 & 275 & 445 & \\
8 & 14 & 22 & 36 & 58 & 94 & 152 & 246 & 398 & 644 & \\
12 & 21 & 33 & 54 & 87 & 141 & 228 & 369 & 597 & 966 & \\
16 & 28 & 44 & 72 & 116 & 188 & 304 & 492 & 796 & 1288 & \\
19 & 32 & 51 & 83 & 134 & 217 & 351 & 568 & 919 & 1487 & \\
23 & 39 & 62 & 101 & 163 & 264 & 427 & 691 & 1118 & 1809 & \\
\vdots & & & & & & & & & & \\
\hline
\end{tabular}
\caption{2nd Lucas-Wythoff array}
\label{tab:Table3}
\end{table}


\begin{theorem}\label{Theorem2}
The 2nd Lucas-Wythoff array is given by
\begin{equation}\label{eq:LucasWythoffEquation}
r^*(n,k)=\lfloor n\tau \rfloor L_k+(n-1)L_{k-1},
\end{equation}
for $n \geq 1, k \geq 1$.
\end{theorem}

\begin{proof}
Let $b(n,k)$ denote the $n$th term of the sequence $B_k$. First we prove that $b(n,k)=w(n,k-2)+w(n,k)$ for $k \geq 1$.  Following \cite{Carlitz}, let
$$a(n)=\lfloor \tau n \rfloor, b(n)=\lfloor \tau^2 n \rfloor, \text{ and let}$$
\begin{align*}
A_{2t} &= (ab^{t-1}a(n): n \geq 1, t \geq 1) \\
A_{2t+1} &= (b^{t-1}a(n): n \geq 1, t \geq 1),
\end{align*}
\noindent where concatenation of functions abbreviates composition.
As proved in \cite{CSH}, we have $w(n,k)=A_k$.  Likewise \cite{Carlitz},
\begin{align*}
B_0 &= (a^2(n)+n : n \geq 1) \\
B_1 &= (a^2(n)+n-1 : n \geq 1) \\
B_{2t} &= (b^{t-1}a(n)+b^ta(n): n \geq 1, t \geq 1) \\
B_{2t+1} &= (ab^{t-1}a(n)+ab^ta(n): n \geq 1, t \geq 1).
\end{align*}
Thus,
\begin{align*}
B_{2t} &= A_{2t-1}+A_{2t+1} = (w(n,2t-1)+(w(n,2t+1)) \\
B_{2t+1} &= A_{2t}+A_{2t+1} = (w(n,2t)+w(n,2t+1)),
\end{align*}
so that $b(n,k)=w(n,k-2)+w(n,k)$ for all $n$ and $k \geq 3$. By Theorem \ref{Theorem1},
\begin{align*}
b(n,k) &= \lfloor n \tau \rfloor F_{k-1}+(n-1)F_{k-2}+ \lfloor n \tau \rfloor F_{k+1}+(n-1)F_k \\
&= \lfloor n \tau \rfloor(F_{k-1}+F_{k+1})+(n-1)(F_{k-2}+F_k) \\
&= \lfloor n \tau \rfloor L_k+(n-1)L_{k-1}.
\end{align*}
\end{proof}

The first rows of Table \ref{tab:Table3} provide some interesting examples involving the Wythoff array ($W$, as in Table \ref{tab:Table1}):
\begin{center}
Row 1 $\sim$ row 2 of $W$, as they have in common $(4,7,\ldots)$\\
Row 2 $\sim$ row 16 of $W$, with common tail $(40,65,\ldots)$\\
Row 3 $\sim$ row 9 of $W$, with common tail $(22,36,\ldots)$\\
Row 4 $\sim$ row 13 of $W$, with common tail $(33,54,\ldots)$ .
\end{center}

Because \textit{every} positive Fibonacci sequence is represented in $W$, every row of both Lucas-Wythoff arrays must, in the sense indicated by the examples, be tail-equivalent to a row of $W$.  We leave open further investigation of this equivalence, which can be cast as follows: in each row of either Lucas-Wythoff array, where does a Wythoff pair first occur? (After the first pair, all subsequent pairs in a row are Wythoff pairs.)


\section{Admissible representations}\label{section:section3}
The requirement $k_u \geq 1$ in connection with the representation (\ref{form2b}) shows that the number $2$ is disallowed as a term.  If $2$ is allowed, then uniqueness is lost; e.g., $5$ can be represented by both $4+1$ and $3+2$. Luo \cite{Luo} proved that if $2$ is allowed, then each $n$ has at most two representations, so that any $n$ having at least two representations must have exactly two.  We shall identify them explicitly.

\begin{definition}\label{admissible}
A representation
\begin{equation}\label{form4a}
n = L_{k_1}+\cdots+L_{k_u}
\end{equation}
is an admissible representation of $n$ if
\begin{equation}\label{form4b}
k_i-k_{i+1} \geq 2 \text{ for } 1 \leq i \leq u-1, k_1 \geq 0.
\end{equation}
\end{definition}
Note that, unlike (\ref{form2b}), in (\ref{form4b}), the index $k_1$ can be $0$. Clearly, both of the representations ((\ref{form1a}) and (\ref{form1b})) and ((\ref{form2a}) and (\ref{form2b})) are admissible.

\begin{theorem}\label{Theorem7}
If $n=L_{k_1}+\cdots+L_{k_u}$, where $u \geq 2$, $k_u=1$, and $k_{u-1}$ is odd, then $n$ has exactly two admissible representations.
\end{theorem}

\begin{proof}
Suppose that $k_1=1$ and $k_2$ is odd. We consider two cases.

\bigskip\noindent
Case 1: $u=2$. Here,
$$n=1+L_{2i+1} \text{ for some } i \ge 1.$$
As a first induction step, if $i=1$, then $n=5=1+4=2+3,$ two representations.  Assume for arbitrary $i \geq 1$ that $n=1+L_{2i+1}$ has a second admissible representation, $n=s$, where $1$ is not a term of $s$, and the greatest term of $s$ is less than $L_{2i+1}$. Then
\begin{align}
1+L_{2i+3} &= 1+L_{2i+1}+L_{2i+2}\\ \label{eq1}
&= s+L_{2i+2}.
\end{align}
This shows that $1+L_{2i+3}$ has two admissible representations.

\bigskip\noindent
Case 2:  $u \geq 3$.  Here, suppose that
\begin{equation}\label{eq: eq2}
n'=1+L_{2i+1}+s'
\end{equation}
\noindent where $i \geq 1$ and the Lucas representation ((\ref{form2a}) and (\ref{form2b})) of $s'$ has least term $\geq L_{2i+3}$.  Then by (\ref{eq1}), a second admissible representation of $n'$ is $s+L_{2i+2}$.

For both cases, if $n$ or $n'$ has a third admissible representation, then one of the representations ((\ref{form1a}) and (\ref{form1b})) or ((\ref{form2a}) and (\ref{form2b})) is not unique, contrary to Theorem \ref{CSH}. Therefore, numbers of the forms in cases 1 and 2 have exactly two admissible representations.
\end{proof}

\begin{theorem}
(Converse of Theorem \ref{Theorem7}) If $n=L_{k_1}+\cdots+L_{k_u}$ has two admissible representations, then one of them has $u \geq 2$, $k_u=1$, and $k_{u-1}$ odd.
\end{theorem}

\begin{proof}
Suppose that $n$ has two admissible representations,
$$n=L_{k_1}+\cdots+L_{k_u} = L_{i_1}+\cdots+L_{i_v}.$$
By Theorem \ref{CSH}, either $k_u=0$ or $i_v=0$; assume the latter, so that
$$n=L_{i_1}+\cdots+L_{i_v-1}+2, \text{ where } i_v-1 \geq 3; i.e., L_{i_{v-1}} \geq 4.$$
By the uniqueness of ((\ref{form1b}) and (\ref{form2b})), we must have $i_v-1=2$ because of (\ref{form1b}), leading to two cases.

\bigskip\noindent
Case 1: $n=3+2$, so that $n=4+1$, as asserted.

\bigskip\noindent
Case 2: $n = w+3+2$, where, by (\ref{form2a}) and (\ref{form2b}), the number $w$ has a representation (\ref{form2a}) with least term $L_m$ for some $m \geq 4.$ If $m \geq 5$, then
$$n = \cdots + L_m+3+2 = \cdots + L_m+4+1, \text{ as asserted}.$$
This leaves the possibility that $m=4$, so that $n = w' + 7 + 3 + 2$, where, again by (\ref{form2a}) and (\ref{form2b}), the number $w'$ has a representation (\ref{form2a}) with least term $L_m$ for some $m \geq 6.$ This procedure can be continued, leading to the asserted form of representation in fewer than $2i_1$ steps.
\end{proof}

Examples:
\begin{align*}
12 &= 11+1 = 7+3+2 \\
16 &= 11+4+1 = 11+3+2 \\
19 &= 18+1 \text { (unique)} \\
30 &= 29+1 = 18+7+3+2 \\
34 &= 29+4+1 = 29+3+2
\end{align*}

\begin{definition}\label{LuoArray}
The Luo-Lucas array, $(\ell(n,k))$, consists of the numbers that have exactly two admissible representations: column $k$ of the array is the increasing sequence of numbers $n$ whose representation (\ref{form2a}) has least term $L_{2k+1}+1.$
\end{definition}

\begin{table} \centering
\begin{tabular}{| c c c c c c c c c |}
\hline
5 & 12 & 30 & 77 & 200 & 522 & 1365 & 3572 & \ldots \\
16 & 41 & 106 & 276 & 721 & 1886 & 4936 & 12921 & \\
23 & 59 & 153 & 399 & 1043 & 2729 & 7143 & 18699 & \\
34 & 88 & 229 & 598 & 1564 & 4093 & 10714 & 28048 & \\
45 & 117 & 305 & 797 & 2085 & 5457 & 14285 & 37397 & \\
52 & 135 & 352 & 920 & 2407 & 6300 & 16492 & 43175 & \\
\vdots & & & & & & & & \\
\hline
\end{tabular}
\caption{Luo-Lucas array}
\label{tab:Table4}
\end{table}

\begin{theorem}\label{LuoTheorem}
The Luo-Lucas array is given by
\begin{equation}\label{eq:Luo}
\ell(n,k)=1+\lfloor n \tau \rfloor L_{2k+1}+(n-1)L_{2k}, \hspace{2mm} n \geq 1, k \geq 1.
\end{equation}
\end{theorem}
\begin{proof}
Column $k$ of $(\ell(n,k))$ has first term $m(1,k)=L_{2k+1}+1$, and all the terms thereafter are, in order, of the form $m(1,k)+U$, where $U$ ranges through the positive integers $u$ having least term $L \geq L_{2k+3}$ in the Lucas-Wythoff representation of $U$. These numbers are, in the same order, the numbers in column $2k+1$ of the 2nd Lucas-Wythoff array.  Therefore, by Theorem \ref{Theorem2}, equation (\ref{eq:Luo}) holds.
\end{proof}

Every number in the Luo-Lucas array is in column $1$ of the Lucas-Wythoff array \ref{tab:Table3}.  The remaining numbers in column $1$, excluding the initial $1$, form a sequence $(d(n,k))$ given by
$$d(n,k)=(8,19,26,37,48,55,66,\ldots),$$
whose terms can be naturally arranged to form a dual, $(\ell^*(n,k))$  of the Luo-Lucas array, given by
$$\ell^*(n,k) = 1 + \lfloor n \tau \rfloor L_{2k+2}+(n-1)L_{2k+1}, n \geq 1, k \geq 1.$$

\begin{table} \centering
\begin{tabular}{| c c c c c c c c |}
\hline
8 & 19 & 48 & 124 & 323 & 844 & 2208 & \ldots \\
26 & 66 & 171 & 446 & 1166 & 3051 & 7986 & \\
37 & 95 & 247 & 645 & 1687 & 4415 & 11557 & \\
55 & 142 & 370 & 967 & 2530 & 6622 & 17335 & \\
73 & 189 & 493 & 1289 & 3373 & 8829 & 23113 & \\
84 & 218 & 569 & 1488 & 3894 & 10193 & 26684 & \\
\vdots & & & & & & \\
\hline
\end{tabular}
\caption{Dual of Luo-Lucas array}
\label{tab:Table5}
\end{table}


\section{Order arrays}\label{section:section4}
Following the definition at \seqnum{A333029}, suppose that $(a(n,k))$, for $n \geq 1, k \geq 1$, is an array of distinct numbers. If each $a(n,k)$ is replaced by its position when all the numbers $a(n,k)$ are ordered by $<$, the resulting array is the \textit{order array of $(a(n,k))$}.

In this section, we shall see that the order array of the 2nd Lucas-Wythoff array is the Wythoff array. A proof depends on several lemmas. The first three involve properties of the Fibonacci numbers that follow readily from the fact that the fractions $F_{k+1}/F_k$ are the convergents to $\tau$.  Proofs of these three are omitted.

\begin{lemma}\label{odd1}
If $k \geq 3$ is odd, then $0 < \tau F_{k-2}-F_{k-1}+1$.
\end{lemma}

\begin{lemma}\label{odd2}
If $k \geq 3$ is odd, then $F_{k-2}-1 < \tau (F_{k-3}+1)$.
\end{lemma}

\begin{lemma}\label{even}
If $k \geq 3$ is even, then $-1 \leq \tau F_{k-3}-F_{k-2} < 1$.
\end{lemma}

Let $\beta(n,k)$ be the number of numbers $i$ in $B_0$ such that $i \leq r^*(n,k)$. Several lemmas will be used to prove the following equation, to be used in proving Theorem \ref{order1}:
\begin{equation}\label{eq:basic}
\beta(n,k)=\lfloor n \tau \rfloor F_{k-1}+(n-1)F_{k-2},
\end{equation}
for $n \geq 1$ and $k \geq 1$, where $F_0=0$ and $F_{-1}=1$.

\begin{lemma}\label{A1}
If $n \geq 1$ and $k \geq 3$, then $\beta(n,k) = \lfloor \frac{\lfloor n \tau \rfloor L_k+(n-1)L_{k-1}+1}{\tau + 2} \rfloor$.
\end{lemma}
\begin{proof}
The numbers in $B_0=(B(m))$, for $m \geq 1$, are given by
\begin{align*}
B(m) &= \lfloor \tau \lfloor \tau m \rfloor \rfloor + m \\
&= \lfloor m \tau \rfloor + m - 1 + m \text{ by Lemma } \ref{Lemma1},
\end{align*}
so that we seek the number of numbers $m$, hence the greatest such $m$, satisfying
$$\lfloor m \tau \rfloor + 2m - 1 \leq \lfloor n \tau \rfloor L_k + (n-1)L_{k-1}+1+\{m \tau \}.$$
This inequality can be recast as
$$m(\tau + 2) \leq \lfloor n \tau \rfloor L_k + (n-1)L_{k-1} + 1 - \{m \tau \},$$
and dividing by $\tau + 2$ finishes the proof.
\end{proof}

\begin{lemma}\label{A2}
The equation (\ref{eq:basic}) holds for $k=1$ and all $n$.
\end{lemma}
\begin{proof}
\begin{align*}
0 &< 1 - \{n \tau \} < \tau + 2 \\
(n-1)(\tau + 2) &< n \tau + \{n \tau \} + 2n - 1 < n(\tau + 2)\\
n-1 &< \frac{n \tau + \{n \tau \} +2n-1}{\tau+2} < n,\\
\end{align*}
so that (\ref{eq:basic}), the right-hand side of which is $n-1$ when $k=1$, holds.
\end{proof}

\begin{lemma}\label{A3}
The equation (\ref{eq:basic}) holds for $k=2$ and all $n$.
\end{lemma}
\begin{proof}
First, $\tau \lfloor n \tau \rfloor < \tau^2 n = (\tau + 1)n $, and also
\begin{align*}
n &= n \tau^2 - n \tau < (\tau-1)(n \tau+ \{n \tau \}) + \tau + 2, \text{ so that} \\
\tau \lfloor n \tau \rfloor &< \lfloor n \tau \rfloor +n < \tau \lfloor n \tau \rfloor + \tau + 2.
\end{align*}
Adding $2\lfloor n \tau \rfloor$  and dividing by $\tau + 2$ give
$$\lfloor n \tau \rfloor < \frac{3\lfloor n \tau \rfloor + n}{\tau+2} < \lfloor n \tau \rfloor + 1,$$
so that (\ref{eq:basic}), the right-hand side of which is $\lfloor n \tau \rfloor$ when $k=2$, holds.
\end{proof}

\begin{lemma}\label{A4}
If $k \geq 3$, then
$\beta(n,k) = \lfloor \frac{(n-1 +3 \lfloor n \tau \rfloor) F_{k-1}+(2n-2 +  \lfloor n \tau \rfloor)F_{k-1}+1}{\tau + 2} \rfloor$.
\end{lemma}
\begin{proof}
This results readily by substituting $L_k=3F_{k-1}+F_{k-2}$ and $L_{k-1}=2F_{k-2}+F_{k-1}$ into the formula in Lemma \ref{A1}.
\end{proof}

\begin{lemma}\label{A5}
The equation (\ref{eq:basic}), for $k \geq 3$ and all $n$, is equivalent to
\begin{equation}\label{eq:maininequality}
0 < \tau F_{k-2} - F_{k-1} + 1 + (F_k - \tau F_{k-1})\{n \tau\} < \tau + 2.
\end{equation}
\end{lemma}
\begin{proof}
Equation (\ref{eq:basic}) and Lemma \ref{A4} give the following inequalities equivalent to (\ref{eq:basic}):
\begin{align*}
\lfloor n \tau \rfloor F_{k-1}+(n-1)F_{k-2} &< \frac{(n-1 +3 \lfloor n \tau \rfloor) F_{k-1}+(2n-2 +  \lfloor n \tau \rfloor)F_{k-1}+1}{\tau + 2} \\
&< \lfloor n \tau \rfloor F_{k-1}+(n-1)F_{k-2} + 1 \\
\end{align*}
Multiplying by $\tau + 2$ and then expanding the products and canceling like terms leave
\begin{align*}
\tau \lfloor n \tau \rfloor F_{k-1} + \tau (n-1)F_{k-2} &< (\lfloor n \tau \rfloor + n - 1)F_{k-1} + \lfloor n \tau \rfloor F_{k-2} + 1 \\
&< \tau \lfloor n \tau \rfloor F_{k-1} + \tau (n-1)F_{k-2} + \tau + 2,
\end{align*}
which is successively equivalent to each of these:
\begin{align*}
0 &< ((1 - \tau)\{n \tau\}-1)F_{k-1} + (\tau + \{n \tau\})F_{k-2} + 1 < \tau + 2 \\
0 &< \tau F_{k-2} - F_{k-1} + 1 + (F_{k-2} + (1 - \tau)F_{k-1})\{n \tau\} < \tau + 2  \\
0 &< \tau F_{k-2} - F_{k-1} + 1 + (F_k - \tau F_{k-1})\{n \tau\} < \tau + 2.
\end{align*}
\end{proof}

\begin{theorem}\label{order1}
The order array of the 2nd Lucas-Wythoff array (Table \ref{tab:Table3} is the Wythoff array.
\end{theorem}

\begin{proof}
By Lemma \ref{A5}, we must prove that (\ref{eq:maininequality}) holds for $k \geq 3$ and arbitrary $n$. (For $k=1$ and $k=2$, the proof is already established by Lemmas \ref{A2} and \ref{A3}.)

\bigskip\noindent
Case 1: $k$ odd. Here, $F_k- \tau F_{k-1} > 0$, so that by Lemma \ref{odd1},
$$0 < \tau F_{k-2} - F_{k-1} + 1 <  \tau F_{k-2} - F_{k-1} + 1 + (F_k - \tau F_{k-1})\{n \tau\}.$$
Also,
$$\tau F_{k-2} - F_{k-1} + 1 + (F_k - \tau F_{k-1})\{n \tau\} < \tau F_{k-2} - F_{k-1} + 1 + (F_k - \tau F_{k-1}),$$
and this last expression simplifies to $F_{k-2} - \tau F_{k-3} + 1$, which by Lemma \ref{odd2} is $< \tau + 2$.

\bigskip\noindent
Case 2: $k$ even. Here, $F_k- \tau F_{k-1} < 0$, so that by Lemma \ref{even},
\begin{align*}
0 &< F_{k-2} - \tau F_{k-3} + 1 < \tau F_{k-2} - F_{k-1}+ 1 + F_k - \tau F_{k-1} \\
&< \tau F_{k-2} - F_{k-1} + 1 + (F_k - \tau F_{k-1})\{n \tau\}.
\end{align*}
Also,
$$\tau F_{k-2} - F_{k-1} + 1 + (F_k - \tau F_{k-1})\{n \tau\} < \tau F_{k-2} - F_{k-1} + 1 + (F_k - \tau F_{k-1}).$$
Abbreviating this last expression as $E$, the desired inequality $E < \tau + 2$ is equivalent to
$$F_k - F_{k-1} - 1 < \tau(F_{k-1} - F_{k-2}),$$
which is equivalent to $F_{k-2} - 1 \leq \tau F_{k-3}$, which holds by Lemma \ref{even}.
\end{proof}

The reader may wish to prove the following proposition.

\begin{theorem}\label{order2}
The Wythoff difference array, \seqnum{A080164}, is the order array of both the Luo-Lucas array (Table \ref{tab:Table4}) and its dual (Table \ref{tab:Table5}).
\end{theorem}


\section{Densities and limiting densities}\label{section:section5}
Suppose that $s = (s_k)$ for $k \geq 1$, is a sequence in $\mathbb{N}$. Define
$$c(s,m) = \text{number of numbers in } s \text{ that are } \leq m,$$
and define the \textit{density of $s$ in $[1,m]$}, by
\begin{equation}\label{dense1}
D(s,m) = \frac{m}{c(s,m)}.
\end{equation}
In order to estimate densities of column sequences of the Wythoff array, $(w(n,k))$, we start with a lemma:
\begin{lemma}\label{B1}
For every $n$ and $k$, $w(n+1,k)-w(n,k) \in \{F_{k+2},F_{k+3}\}$.
\end{lemma}
\begin{proof}
By Theorem \ref{Theorem1},
\begin{align*}
w(n+1,k)-w(n,k) &= \lfloor (n+1)\tau \rfloor F_{k+1} + nF_k - (\lfloor n \tau \rfloor F_{k+1} + (n-1)F_k)\\
&= (\tau + \{(n+1)\tau \} - \{n \tau \})F_{k+1} + F_k\\
&= \delta F_{k+1} + F_k, \text{ where } \delta \in \{1,2\}.
\end{align*}
\end{proof}

\begin{example}
For fixed $n$ and all $k$, let $s_k=w(n,k)$.  The density
$$D(w(n,k),m),$$
which is the proportion of numbers in column $k$ (that is, numbers whose Zeckendorf representation has $F_{k+1}$ as least term) of the Wythoff array that are $\leq m$, is estimated as follows. Let $n$ be the number satisfying
$$w(n,k) \leq m < w(n+1,k).$$
Then by Lemma \ref{B1},
$$\frac{n}{w(n,k)+F_{k+3}} \leq \frac{n}{w(n+1,k)} \leq D(w(n,k),m) = \frac{n}{m} < \frac{n}{w(n,k)}.$$
Applying Theorem \ref{Theorem1} and dividing by $n$ lead to a limiting density:
$$\lim_{n \to \infty} D(w(n,k)) = \frac{1}{\tau F_{k+1}+F_k}.$$
Since the columns of $(w(n,k))$ partition $\mathbb{N}$, we have
$$\sum_{k=1}^\infty \frac{1}{\tau F_{k+1}+F_k} = 1.$$
\label{one}
\end{example}

\begin{example}
We turn now to the second column, $B_0 = (r(2,k)) = \seqnum{A188378},$
of the 1st Lucas-Wythoff array:
$$D(r(2,k),n)=\frac{n}{\lfloor \tau \lfloor n \tau \rfloor \rfloor + n},$$
so that the limiting density is
$$\frac{1}{\tau^2 + 1} = \frac{2}{5+\sqrt{5}} \approx{27.64}\%.$$

Next, consider the 2nd Lucas-Wythoff array, $(r^*(n,k))$.
\begin{lemma}\label{B3}
For every $n$ and $k$, $r^*(n+1,k)-r^*(n,k) \in \{L_{k+1},L_{k+2}\}$.
\end{lemma}
\begin{proof}
By Theorem \ref{Theorem2},
\begin{align*}
r^*(n+1,k)-r^*(n,k) &= \lfloor (n+1)\tau \rfloor L_k + nL_{k-1} - (\lfloor n \tau \rfloor L_k + (n-1)L_{k-1})\\
&= (\tau + \{n \tau \})L_k - L_{k-1}\\
&= \delta L_k + L_{k-1}, \text{ where } \delta \in \{1,2\}.
\end{align*}
\end{proof}
\label{two}
\end{example}

\begin{example}
For fixed $n$ and all $k$, let $s_k=r^*(n,k)$.  The density
$$D(r^*(n,k),m),$$
which is the proportion of numbers in column $k$ (that is, numbers whose Lucas representation has $L_k$ as least term) of the 2nd Lucas-Wythoff array that are $\leq m$, is estimated as follows. Let $n$ be the number satisfying
$$r^*(n,k) \leq m < r^*(n+1,k).$$
Then by Lemma \ref{B3},
$$\frac{n}{r^*(n,k)+L_{k+2}} \leq \frac{n}{r^*(n+1,k)} \leq D(r^*(n,k),m) = \frac{n}{m} < \frac{n}{r^*(n,k)}.$$
Applying Theorem \ref{Theorem1} and dividing by $n$ lead to a limiting density:
$$\lim_{n \to \infty} D(r^*(n,k)) = \frac{1}{\tau L_k+L_{k-1} }.$$
Since the columns of $(r^*(n,k))$ partition $\mathbb{N}-B_0$, we have, as a corollary of Example~\ref{two},
$$\sum_{k=1}^\infty \frac{1}{\tau L_k+L_{k-1}} = 1- \frac{1}{\tau^2+1} \approx{72.36}\%.$$
\label{three}
\end{example}

\begin{example}
Recall the Luo-Lucas array, $(\ell(n,k))$. Following the proof of Lemma \ref{B3}, it is easy to find that
$$\ell(n+1,k)-\ell(n,k) \in \{L_{2k+2},L_{2k+3}\}$$
and
$$\sum_{k=1}^\infty \frac{1}{\tau L_{2k+1}+L_{2k}} = \frac{1}{3 \tau + 1} = \frac{3 \sqrt{5}-5}{10} \approx{17.08}\%,$$
which agrees with the limit obtained in a different manner by Luo \cite{Luo}.
\label{four}
\end{example}

\begin{example}
Finally, recall the dual of the Luo-Lucas array, $(\ell^*(n,k))$. As in 
Example~\ref{four},
$$\ell^*(n+1,k)-\ell^*(n,k) \in \{L_{2k+2},L_{2k+3}\}$$
and
$$\sum_{k=1}^\infty \frac{1}{\tau L_{2k+2}+L_{2k+1}} = 1-\frac{2}{\sqrt{5}}  \approx{10.55}\% .$$
\label{five}
\end{example}

\section{Concluding conjectures}\label{section:section6}
In addition to the open question posed at the end of Section \ref{section:section2}, we recall conjectures from \seqnum{A214979} and \seqnum{A214981.}

\begin{enumerate}
\item Let $I(n) = \{1,2,\ldots,n\}$. Let $U(n)$ be the number of terms in the unique greedy Lucas representations of the numbers in $I(n)$, and let $V(n)$ be the number of terms in the Zeckendorf representations of $I(n)$. Then $V(n)\geq U(n)$ for all $n$, and $V(n) = U(n)$ for infinitely many $n$.

\item Let $S=(1,2,3,4,5,7,8,11,13,18,\ldots)$ be the sequence, in increasing order, of all Fibonacci and Lucas numbers. Let $W(n)$ be the number of terms in the greedy $S$-representations of the numbers in $I(n)$ (as in \seqnum{A214981}).  Then the limit of $V(n)/W(n)$ exists and is the interval $(1.2,1.4)$.
\end{enumerate}

\section{Acknowledgment}

The author thanks the referee for helpful suggestions.


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   \bibitem{CSH} Leonard Carlitz, Richard Scoville, and V. E. Hoggatt, Jr., Fibonacci representations, \textit{Fibonacci Quart.} \textbf{10} (1972) 1--28.
   Available at
  \url{https://www.fq.math.ca/Scanned/10-1/carlitz1.pdf}.


   \bibitem{Carlitz} Leonard Carlitz, Richard Scoville, and V. E. Hoggatt, Jr., Lucas representations, \textit{Fibonacci Quart.} \textbf{10} (1972) 29--42, 70, 112.  Available at 
  \url{https://www.fq.math.ca/Scanned/10-1/carlitz2-a.pdf}.

\bibitem{DFGHKL} Eric Duch\^ene, Aviezri S. Fraenkel, Vladimir Gurvich,
Nhan Bao Ho, Clark Kimberling, and Urban Larsson, Wythoff visions, in
\textit{Games of No Chance 5}, Math. Sci. Res. Inst. Publ. \textbf{70},
Cambridge University Press, 2019, pp.~35--87.

\bibitem{Kim1} Clark Kimberling, The Zeckendorf array equals the
Wythoff array, \textit{Fibonacci Quart.} \textbf{33} (1995) 3--8.
Available at
\url{https://www.fq.math.ca/Scanned/33-1/kimberling.pdf}.

\bibitem{Luo} David C. Luo, Proportion of nonunique
Zeckendorf related partitions in the Lucas sequence, preprint, 2020.
Available at
\url{https://arxiv.org/pdf/2004.08316.pdf}; to appear in \textit{Fibonacci
Quart.}.

\bibitem{Morrison} David R. Morrison, A Stolarsky array of
Wythoff pairs, in \textit{A Collection of Manuscripts Related
to the Fibonacci Sequence}, The Fibonacci Association, 1980.
Available at
\url{http://web.math.ucsb.edu/~drm/papers/stolarsky.pdf}.

\bibitem{OEIS} N. J. A. Sloane et al.,
\textit{The On-Line Encyclopedia of Integer Sequences}, 2020.
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\bibitem{Zeck} Edouard Zeckendorf, Repr\'esentation des nombres
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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2010 \textit{Mathematics Subject Classification}:
Primary 11A63; Secondary 11B39.

\noindent \textit{Keywords: }
Lucas sequence, Fibonacci sequence, Lucas representation, Zeckendorf representation, Wythoff array.
\bigskip
\hrule
\bigskip

\noindent 
(Concerned with sequences 
\seqnum{A000027},
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A000201},
\seqnum{A001622},
\seqnum{A001950},
\seqnum{A003622},
\seqnum{A035513},
\seqnum{A080164},
\seqnum{A188378},
\seqnum{A214979},
\seqnum{A214981},
\seqnum{A333029},
\seqnum{A335499}, and
\seqnum{A335500}.)

\noindent For sequence numbers of several rows and columns
of the Wythoff array, see \seqnum{A035513}.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received June 14 2020;
revised version received  September 29 2020; October 13 2020.
Published in {\it Journal of Integer Sequences}, October 15 2020.

\bigskip
\hrule
\bigskip

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


\end{document}

                                                                                

