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\theoremstyle{plain}
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\begin{center}
\vskip 1cm{\LARGE\bf 
Fault-Free Tilings of the $3 \times n$ Rectangle\\
\vskip .1in
With Squares and Dominos
}
\vskip 1cm
\large
Oluwatobi Jemima Alabi\\
Government Science Secondary School\\
Pyakasa, Abuja\\
Nigeria\\
\href{mailto:tobialabi609@gmail.com}{\tt tobialabi609@gmail.com}  \\
\ \\
Greg Dresden\\ 
Washington \& Lee University \\
Lexington, VA 24450\\
USA\\ 
\href{mailto:dresdeng@wlu.edu}{\tt dresdeng@wlu.edu} \\
\end{center}

\vskip 0.2in

\begin{abstract} 
A (completely) fault-free tiling of a board is a tiling with no vertical
or horizontal faults. We find a fourth-order recurrence relation for
the number of ways to tile such a board using squares and dominos,
and we do the same for a vertical-fault-free board (which could have
horizontal faults).
\end{abstract}

\section{Introduction}\label{intro}
Tiling problems appear frequently in the mathematical literature. 
McQuistan and Lichtman \cite{ML} found the number of tilings of
a $2 \times n$ rectangle with squares and dominos (given by \seqnum{A030186}), and 
Read \cite{Read} did the same for $3 \times n$ (see \seqnum{A033506}). 
And of course, the tilings of a $1 \times n$ rectangle are counted by the Fibonacci numbers.

One variation on these kinds of tiling problems is to ask for {\em fault-free}  tilings, in which there are no ``clean breaks" of the tiling along any horizontal or vertical line. 
\begin{figure}[H]
\centering
\includegraphics[height=0.6in]{3x6.eps}
\caption{A $3 \times 6$ tiling with a horizontal fault but no vertical fault.}
\label{3x6}
\end{figure}


Figure \ref{3x6} gives a tiling of a $3 \times 6$ rectangle (using only dominos) which has a horizontal fault but no vertical fault (we would call this a {\em vertical-fault-free} tiling).


Recent articles on fault-free tilings with dominos include Graham's paper \cite{Graham} on rectangles, 
a preprint by Montelius \cite{Montelius} which expands on Graham's work to consider cylinders, 
tori, and M\"obius strips, and a paper by Atkinson and Lunnon \cite{AL} which 
enumerates fault-free rectangles in which each fault crosses a fixed number of dominos. 
Graham points out that there does not exist a  fault-free tiling of a 
$3 \times n$ rectangle using only dominos. However, if we use dominos and squares,
there are quite a few fault-free tilings, and that is the subject of our paper. Here is one fault-free tiling:
\begin{figure}[H]
\centering
\includegraphics[height=0.6in]{3x6ff11.eps}
\caption{A $3 \times 6$ tiling which has neither vertical nor horizontal faults.}
\label{3x6ff}
\end{figure}

In what follows, we will count the number of 
such tilings (with dominos and squares) of the $3 \times n$ rectangle. 
We let $U_n$  represent the number of completely fault-free tilings, and 
$V_n$ represent the number of vertical-fault-free tilings. 
(We are using the expression {\em completely} fault-free, instead of just fault-free, to avoid confusion with the expression {\em vertical}-fault-free. Keep in mind that most authors use the expression fault-free instead of 
completely fault-free.)

\section{Initial work on counting the tilings}

Let us look at some initial values for $U_n$ and $V_n$. 
Fortunately, the tilings at $n=1$ are easy. Figure \ref{V1} shows all possible tilings, and so we see that
$U_1 = 0$ and $V_1 = 3$.
\begin{figure}[H]
\centering
\includegraphics[height=0.6in]{3x1.eps}
\caption{All possible tilings of a $3 \times 1$ rectangle with squares and dominos.}\label{V1}
\end{figure}

For $n=2$, all the vertical-fault-free tilings are given in Figure \ref{V2}. Since none of them are also horizontally-fault-free, we have $U_2 = 0$ and $V_2 = 13$.
\begin{figure}[H]
\centering
\includegraphics[height=1.4in]{3x2attempt.eps}
\caption{All possible vertical-fault-free tilings of a $3 \times 2$ rectangle.}\label{V2}
\end{figure}

For $n=3$ and $n=4$ there are exactly two completely fault-free tilings as shown in 
Figure \ref{U3U4}, so 
$U_3 = U_4 = 2$. (It turns out that $V_3 = 26$ and $V_4 = 66$ but we will show that later.)
\begin{figure}[H]
\centering
\includegraphics[height=0.6in]{U3U42.eps}   
\caption{The completely fault-free tilings of a $3 \times 3$ and $3 \times 4$ rectangle.}
\label{U3U4}
\end{figure}



Now, to tile a fault-free $3 \times n$ board, we must take three very important issues into consideration.
\begin{enumerate}
\item First, for a $3 \times n$ tiling to be vertical-fault-free, there must not be any vertical dominoes in the middle of the board;  they can only be at the beginning or end (if at all). As seen in Figure \ref{Vbad}, using vertical dominos in the middle of a tiling causes vertical faults  (represented by dashed lines). 
\begin{figure}[H]
\centering
\includegraphics[height=0.8in]{badvert.eps}
\caption{Interior vertical dominos create interior vertical faults.}\label{Vbad}
\end{figure}


\item  Second, for a tiling to be horizontal-fault-free,
then since there can not be vertical dominos in the middle we must instead have
(unaligned) vertical dominos on the ends, as shown in the two diagrams
in Figure \ref{3xnHV}.
\begin{figure}[H]
\centering
\includegraphics[height=0.6in]{3xnHV.eps}
\caption{Vertical dominos on each end are required for completely fault-free tilings.}\label{3xnHV}
\end{figure}




\item Finally, there are six recurring profiles  which we will use to count tilings 
for $n \geq 3$, based on what happens when one tries to break the tiling, vertically, at position $m$ for $1 \leq m \leq n-1$. Figure \ref{6sit3} illustrates the six profiles (similar to those used by Read in \cite{Read}), which we arbitrarily label $A$ through $F$.
\begin{figure}[H]
\centering
\includegraphics[height=0.8in]{6sit3.eps}
\caption{Six profiles in which a tiling fails to have a vertical fault.}
\label{6sit3}
\end{figure}
\end{enumerate} 


\section{Vertical-fault-free tilings}\label{section3}
Let us now determine the number of vertical-fault-free tilings $V_n$ for a $3 \times n$ rectangle. From our drawings earlier, we know that $V_1 = 3$ and $V_2 = 13$. 

First, we define $A_n, \, B_n, \dots, F_n$ to be the number of 
ways to tile a vertical-fault-free $3\times n$ rectangle ending in profiles $A, B, \dots, F$ from Figure \ref{6sit3}, respectively, where the one or two rectangles in each profile extend one unit beyond the end on the right. 
 For example, 
$A_1 = 2$ and $A_2 = 4$, thanks to the tilings in Figure \ref{S11S12},
and $B_1 = 1$ and $B_2 = 5$ as seen in Figure \ref{S21S22}. Since profile $C$ is just a reflection (around a horizontal axis) of profile $A$, then we also
have $C_1 = 2$ and $C_2  = 4$. 
\begin{figure}[H]
\centering
\includegraphics[height=0.8in]{S11S12.eps}
\caption{$A_{1} = 2$ and $A_{2} = 4$; flip the picture to see that $C_1=2$ and $C_2=4$.}
\label{S11S12}
\end{figure}

\begin{figure}[H]
\centering
\includegraphics[height=0.8in]{S21S22.eps}
\caption{$B_{1} = 1$ and $B_{2} = 5$.}
\label{S21S22}
\end{figure}

But what about profile D? If we study Figure \ref{D2D3}, we see
that $D_2 =2$ and $D_3 = 4$, and furthermore we can see that in general,
$D_n = A_{n-1}$ because any profile-$D$ tiling must have a 
horizontal domino (in orange) above the two horizontal dominos (in yellow) as
shown, to avoid a vertical fault to the left of those two horizontal dominos. 
\begin{figure}[H]
\centering
\includegraphics[height=0.8in]{D2D3.eps}
\caption{$D_{2} =  A_1 = 2$ and $D_{3} = A_2  = 4$.}
\label{D2D3}
\end{figure}


Glancing back at Figure \ref{6sit3}, we see that we can find other recurrence relations for these six sequences $A_n, B_n, \dots, F_n$. If we look at Figure \ref{BackStep3}, we can see that a tiling that ends in profile $A$ must, in order to prevent a vertical fault before that yellow domino, have either one or two (orange) horizontal dominos below, and slightly to the left, of
that original (yellow) horizontal domino, which means that 
$A_n$ must equal  $B_{n-1} + C_{n-1} + D_{n-1}$.
\begin{figure}[H]
\centering
\includegraphics[height=2.8in]{BackStep3.eps}
\caption{Illustrating that $A_{n} =  B_{n-1} + C_{n-1} + D_{n-1}$.}
\label{BackStep3}
\end{figure}

We have shown that $D_n = A_{n-1}$ and that 
$A_n = B_{n-1} + C_{n-1} + D_{n-1}$. Using similar drawing arguments,
we arrive at the six recurrence relations shown in
(\ref{SixR}), which all hold for $n \geq 2$.

\hfill \parbox{3in}{
	\begin{align*}
	A_{n} &=   B_{n-1} + C_{n-1} + D_{n-1}\\
	B_{n} &=   A_{n-1} + C_{n-1} + E_{n-1}\\
	C_{n} &=   A_{n-1} + B_{n-1} + F_{n-1}\\
\end{align*}
	}
\hfill
\parbox{3in}
{	
\begin{equation} \label{SixR}
	\begin{aligned}
		D_{n} &=   A_{n-1}\\
		E_{n} &=   B_{n-1}\\
		F_{n} &=   C_{n-1}
	\end{aligned}
\end{equation}
}

Note that for $n \geq 3$ we can substitute the last three equations from
(\ref{SixR}) into the first three, to 
obtain
\begin{align}
	A_{n} &=   B_{n-1} + C_{n-1} + A_{n-2}, \label{AAn} \\
	B_{n} &=   A_{n-1} + C_{n-1} + B_{n-2}, \label{BBn} \\
	C_{n} &=   A_{n-1} + B_{n-1} + C_{n-2}.   \label{CCn}
\end{align}


At this point, we would like to ``unlink" these linked recurrence formulas in 
equations 
(\ref{AAn}), (\ref{BBn}), and (\ref{CCn}).
After a few minutes of calculations (and with initial values that we drew out by hand), we noticed that $A_n$, $B_n$, and $C_n$ all seem to satisfy the fourth-order 
recurrence relation with signature $(1,4,-1,-1)$, which is to say that
$A_n$ seems to equal $1A_{n-1} + 4A_{n-2} -1A_{n-3} -1A_{n-4}$ and likewise for $B_n$ and $C_n$. 
We now prove that this observation is correct.
\begin{lemma}\label{lemAB}
With $A_n$, $B_n$, and $C_n$ defined in equations 
(\ref{AAn}), (\ref{BBn}), and (\ref{CCn}), then for $n \geq 5$, we have 
\begin{align*}
A_n &= A_{n-1} + 4A_{n-2} - A_{n-3} - A_{n-4},\\
B_n &= B_{n-1} + 4B_{n-2} - B_{n-3} - B_{n-4},\\
C_n &= C_{n-1} + 4C_{n-2} - C_{n-3} - C_{n-4}.
\end{align*}
\end{lemma} 
\begin{proof}
We give a direct proof that follows immediately from
(\ref{AAn}), (\ref{BBn}), and (\ref{CCn}).
Starting with
equation (\ref{AAn}), we have 
\begin{align*}
A_n &=   B_{n-1} + C_{n-1} +  A_{n-2}, \\
\intertext{and if we substitute $B_{n-1} = A_{n-2} + C_{n-2} + B_{n-3}$ from  (\ref{BBn}) and a similar expression for $C_{n-1}$ from 
(\ref{CCn}), we obtain }
A_n &=  (A_{n-2} + C_{n-2} + B_{n-3}) + (A_{n-2} + B_{n-2} + C_{n-3}) + A_{n-2}.
\end{align*}
We now substitute
$B_{n-3} = A_{n-2} - C_{n-3} - A_{n-4}$ and 
$B_{n-2} = A_{n-1} - C_{n-2} - A_{n-3}$
from  (\ref{AAn})
to  obtain
\begin{multline*}
A_n =  (A_{n-2} + C_{n-2} +(A_{n-2} - C_{n-3} - A_{n-4})) \\
+ (A_{n-2} + (A_{n-1} - C_{n-2} - A_{n-3})) + C_{n-3})  \ + \ A_{n-2},
\end{multline*}
which simplifies nicely to become our desired recurrence relation. A similar method would give us the same
recurrence relations for $B_n$ and $C_n$. 
\end{proof}

We can now begin to calculate $V_n$. From earlier, we know that $V_1 = 3$ and $V_2 = 13$. With some clever drawings, we can establish the following 
formula that relates $V_n$ to the $A_n$, $B_n$, and $C_n$'s discussed earlier. 
\begin{lemma}\label{lemABV} For $n \geq 3$ we have 
\[
V_n = 2A_{n-1} + B_{n-1} +  2C_{n-1} +  A_{n-2} + B_{n-2} + C_{n-2}.
\]
\end{lemma} 
\begin{proof}
We will first establish that for $n \geq 3$,
\begin{equation}\label{BigV}
V_n = 2A_{n-1} + B_{n-1} + 2C_{n-1} + D_{n-1} + E_{n-1} + F_{n-1}.
\end{equation}
To see this, we imagine a  vertical-fault-free tiling of the entire length
of a $3 \times n$ rectangle (with $n \geq 3$), and we separate it into cases
 based on how many horizontal dominos are at the
right-hand end of the rectangle. We can not have three horizontal dominos (as then 
the tiling would have a vertical fault at $n-2$), nor can we have zero 
horizontal dominos (as then the tiling would have a vertical fault at $n-1$),
and so there must be either one or two horizontal dominos. 

Suppose there is one horizontal domino at the far right-hand side,
and suppose it is located in the upper right corner. Figure \ref{2Sn1} shows that the number of such tilings 
is $2A_{n-1}$, since there could be either two stacked squares, or a single vertical domino, below the horizontal domino and so we have to double the number
of tilings in profile $A_{n-1}$. 
\begin{figure}[H]
\centering
\includegraphics[height=1.8in]{2Sn1.eps}
\caption{Vertical-fault-free tilings with an upper-right horizontal domino.}
\label{2Sn1}
\end{figure}

Using similar diagrams, it is not hard to show that a horizontal domino
at the far-right side in the middle gives $B_{n-1}$ ways to tile the rectangle, 
a horizontal domino in the bottom right corner has $2C_{n-1}$ tilings, and so on. This gives us equation (\ref{BigV}).

We can now use the last three equations in (\ref{SixR}) to replace the 
$D_{n-1}$, $E_{n-1}$, and $F_{n-1}$ with $A_{n-2}$, $B_{n-2}$, and $C_{n-2}$ respectively, and so equation (\ref{BigV}) becomes
\begin{equation}\label{SmallV}
V_n = 2A_{n-1} + B_{n-1} +  2C_{n-1} +  A_{n-2} + B_{n-2} + C_{n-2},
\end{equation}
as desired.
\end{proof}

We are now ready to establish the following properties for our sequence $V_n$.

\begin{theorem} With $V_n$ the number of ways to tile a vertical-fault-free $3 \times n$ rectangle with squares and dominos, 
\begin{enumerate} 
\item  The first six values of $V_n$, starting at $n=1$, are 
$3, 13, 26, 66, 154, 380$.
\item  For $n \geq 7$, we have $V_n = V_{n-1} + 4V_{n-2} - V_{n-3} - V_{n-4}$.
\end{enumerate} 
\end{theorem} 

We note that this theorem verifies the recursion formula (and subsequent values) in \seqnum{A335747}.

\begin{proof} We consider each part separately.
\begin{enumerate} 
\item  The values of $V_1$ and $V_2$ (namely, 3 and 13) have already been calculated  in Figures \ref{V1} and \ref{V2}.
As for $V_3, V_4, V_5, V_6$, and $V_7$, we can use Lemma \ref{lemABV} to calculate
them from $A_n, B_n$, and $C_n$. 
Fortunately, $A_n, B_n$, and $C_n$ satisfy the recurrences in equations
(\ref{AAn}), (\ref{BBn}), and (\ref{CCn}),
and so the only hand-calculations we need are for $n=1$ and 2.
Thanks to Figures \ref{S11S12} and \ref{S21S22}, we know that 
$A_1 = 2$, $A_2 = 4$, and $B_1 = 1$, $B_2 = 5$. Recall also  
from the discussion at the beginning of Section \ref{section3} and
in the caption for Figure \ref{S11S12} that $C_n = A_n$. 
So, the recurrence
relations in  (\ref{AAn}) and (\ref{BBn}) give us 
the sequences $\{2,4,11,24, 62 \dots\}$ for $A_n$ and $C_n$, and 
$\{1, 5, 9, 27, 57, \dots\}$ for $B_n$. Applying these values to 
Lemma \ref{lemABV} gives us $V_3 = 26$, $V_4 = 66$, $V_5 = 154$, and $V_6 = 380$.

\item  To establish the recurrence relation for $V_n$, we look once more at Lemma \ref{lemABV}. Since $V_n$ is defined in terms of $A_{n-1}$ and $A_{n-2}$,
$B_{n-1}$ and $B_{n-2}$, and $C_{n-1}$ and $C_{n-2}$, and since these
all satisfy
our fourth-order recurrence formula thanks to Lemma 
\ref{lemAB}, then so long as $n-2 \geq 5$ (that is to say, so long as 
$n \geq 7$) we have that $V_n$ also satisfies the same fourth-order
recurrence formula. 
\end{enumerate} 
\end{proof}

\section{Completely fault-free tilings} 

We now turn our attention to finding $U_n$,  the number of completely fault-free tilings of a $3 \times n$ rectangle. We have already shown in Section \ref{intro} that
$U_1 = U_2 = 0$, and (thanks to Figure \ref{U3U4}) that  $U_3 = U_4 = 2$.
As seen in Figure \ref{3xnHV}, 
we must have vertical dominos on each end for a completely fault-free tiling, and those two diagrams suggest that we can count $U_n$ by instead counting the
number of completely fault-free tilings that have a vertical domino in the lower left corner (as seen in the left-hand diagram in Figure \ref{3xnHV}) and 
then doubling that number.

With this in mind, we let $T_n$ equal the number of completely fault-free tilings that have a vertical domino in the lower left corner. From 
the previous paragraph, we have that $T_1 = T_2 = 0$ and $T_3 = T_4 = 1$. 
Figure \ref{3x5ff} shows that $T_5 = 5$, and Figure \ref{3x6allff} at the 
end of the paper demonstrates that $T_6 = 8$.
\begin{figure}[H]
\centering
\includegraphics[height=1.4in]{3x5ff.eps}
\caption{Completely fault-free $3 \times 5$ tilings with a lower-left vertical domino.}
\label{3x5ff}
\end{figure}

As we now show, it is quite easy to adapt our methods from Section \ref{section3} to the current problem. 
Just as we defined $A_n, B_n, \dots, F_n$ to count the number of vertical-fault-free 
tilings ending in profiles $A, B, \dots, F$ from Figure \ref{6sit3}, 
we now define 
$A_n', B_n', \dots, F_n'$ to give us  the number of vertical-fault-free 
tilings {\em beginning with a vertical domino in the lower left corner} and 
 ending in those same profiles $A, B, \dots, F$.
 
 We know (thanks to Figure \ref{3xnHV}) that any {\em completely} fault-free 
 tiling that begins with a vertical domino in the lower left corner, 
 must end with a vertical domino in the upper right corner. Hence,
 $T_n = C_{n-1}'$ as seen in Figure \ref{TnCnm1}.
\begin{figure}[H]
\centering
\includegraphics[height=0.8in]{TnCnm1.eps}
\caption{$T_n = C_{n-1}'$.}
\label{TnCnm1}
\end{figure}
 
At this point, we recognize that the three sequences 
$A_n', B_n'$, and $C_n'$  satisfy exactly the same linked recurrence relations
(\ref{AAn}), (\ref{BBn}), and (\ref{CCn}) as do 
$A_n$, $B_n$, and $C_n$ from Section \ref{section3}; recall that we established those by studying diagrans such as Figure \ref{BackStep3} and of course the scenario from that figure is exactly the same for our ``new" sequences $A_n', B_n', \dots, F_n'$. And just as we showed that $A_n, B_n$ and $C_n$ satisfy
our particular fourth-order recurrence relation in 
Lemma \ref{lemAB}, we can assume that the same applies for $A_n', B_n'$, and 
$C_n'$. 

Now, since $T_n = C_{n-1}'$, we know that $T_n$ also satisfies our 
particular fourth-order recurrence relation. To find the initial values for $T_n$ we simply need to find the initial values for $C_n'$, which means
we also need those value for $A_n'$ and $B_n'$.

Fortunately, this is easy to do. Thanks to Figure \ref{ABC},
we see that $A_1' = 1, B_1' = 0, C_1'=0$ 
and also $A_2' =0, B_2' = 1, C_2' = 1$. 
\begin{figure}[H]
\centering
\includegraphics[height=0.8in]{ABC.eps}
\caption{Initial values for $A_n', B_n', C_n'$.}
\label{ABC}
\end{figure}



We can now use equations 
(\ref{AAn}), (\ref{BBn}), and (\ref{CCn}) to get that the
initial values of $C_n'$, with $n$ from $1$ to $6$, are
$\{0, 1, 1, 5, 8, 26\}$, and so $T_n$ 
has the same values (starting with $n=2$) and satisfies the 
same fourth-order recurrence relation of 
$T_n = T_{n-1} + 4T_{n-2} - T_{n-3} - T_{n-4}$ which (we find by checking) actually holds for all $n \geq 5$ if we recall from earlier that $T_1 = 0$.
If we double these values, we have now proved the following theorem.

\begin{theorem} With $U_n$ the number of ways to tile a completely fault-free $3 \times n$ rectangle with squares and dominos, 
\begin{enumerate} 
\item  The first seven values of $U_n$, starting at $n=1$, are 
$0,0,2,2,10,16,52$.
\item  For $n \geq 5$, we have $U_n = U_{n-1} + 4U_{n-2} - U_{n-3} - U_{n-4}$.
\end{enumerate} 
\end{theorem} 
This is sequence \seqnum{A334396}, and (surprisingly) these numbers $U_n$ are twice \seqnum{A112577}. 

As a final note, we also find it surprising that both our sequences
($V_n$, the number of ways to tile a vertical-fault-free rectangle, and $U_n$, the number of ways to tile a completely fault-free rectangle) satisfy the same
recurrence relation. Surely there is a geometric or tiling explanation for why $U_n = U_{n-1} + 4 U_{n-2} - U_{n-3} - U_{n-4}$ and likewise $V_n$, and we hope an interested reader will be able to find it. 

We conclude with a visual demonstration that $T_6 = 8$ and hence $U_6 = 16$. 
\begin{figure}[H]
\centering
\includegraphics[height=2.2in]{3x6allff.eps}
\caption{A demonstration that $T_6 = 8$.}
\label{3x6allff}
\end{figure}

\section{Acknowledgments}

The authors would like to thank Pioneer Academics for 
 giving us the opportunity to work together. This paper is 
 one of the results of our collaboration from our spring 2020 research class.




\begin{thebibliography}{99}

\bibitem{AL}
M.~D. Atkinson and W.~F. Lunnon,
\newblock Regular fault-free rectangles,
\newblock {\em Math. Gaz.} {\bf 64} (1980), 99--106.
%\newblock {\em Math. Gaz.}, 64(428):99--106, 1980.


\bibitem{Graham}
R.~L. Graham,
\newblock Fault-free tilings of rectangles,
\newblock in D.~A. Klarner, ed., {\em The Mathematical Gardner: A Collection
  in Honor of Martin Gardner}, Wadsworth, 1981, pp. 120--126.

\bibitem{ML}
R.~B. McQuistan and S.~J. Lichtman,
\newblock Exact recursion relation for {$2\times N$} arrays of dumbbells,
\newblock {\em J. Mathematical Phys.}  {\bf 11} (1970), 3095--3099.

\bibitem{Montelius}
E.~Montelius,
\newblock Fault-free tileability of rectangles, cylinders, tori, and M\"obius
  strips with dominoes, preprint, 2019. Available at 
\newblock \url{https://arxiv.org/abs/1912.04445}.

\bibitem{Read}
R.~C. Read,
\newblock The dimer problem for narrow rectangular arrays: a unified method of
  solution, and some extensions,
\newblock {\em Aequationes Math.} {\bf 24} (1982), 47--65.

\end{thebibliography}






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\noindent 2010 {\it Mathematics Subject Classification}: Primary 05B45; Secondary 52C20.

\noindent \emph{Keywords: } square, domino, tiling, fault-free, rectangle.

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\noindent (Concerned with sequences
\seqnum{A030186},
\seqnum{A033506},
\seqnum{A112577},
\seqnum{A124997},
\seqnum{A334396}, and
\seqnum{A335747}.)

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\vspace*{+.1in}
\noindent
Received July 9 2020;
revised versions received October 20 2020; December 22 2020.
Published in {\it Journal of Integer Sequences}, December 28 2020.

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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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