# Tilings of Benzels via the Abacus Bijection

Colin Defant  Department of Mathematics, Massachusetts Institute of
Technology, Cambridge, MA 02139, USA<EMAIL_ADDRESS>,  Rupert Li
Massachusetts Institute of Technology, Cambridge, MA 02139, USA
<EMAIL_ADDRESS>,  James Propp  Department of Mathematical Sciences, UMass
Lowell, Lowell, MA 01854, USA<EMAIL_ADDRESS>and  Benjamin Young
Department of Mathematics, University of Oregon, Eugene OR 97403 USA
<EMAIL_ADDRESS>
###### Abstract.

Propp recently introduced regions in the hexagonal grid called _benzels_ and
stated several enumerative conjectures about the tilings of benzels using two
types of prototiles called _stones_ and _bones_. We resolve two of his
conjectures and prove some additional results that he left tacit. In order to
solve these problems, we first transfer benzels into the square grid. One of
our primary tools, which we combine with several new ideas, is a bijection
(rediscovered by Stanton and White and often attributed to them although it is
considerably older) between $k$-ribbon tableaux of certain skew shapes and
certain $k$-tuples of Young tableaux.

##  1\. Introduction

In [1], Conway and Lagarias studied some tiling problems in which the regions
that are to be tiled are, like the tiles themselves, composed of regular
hexagons in a hexagonal grid; such regions are sometimes called _polyhexes_.
Using combinatorial group theory in a clever and novel fashion, they were able
to prove necessary and sufficient conditions for tileability. Later, Thurston
[11] provided alternative perspectives and further results, and (more
relevantly to our work here) Lagarias and Romano [4] determined the exact
number of tilings for a particular one-parameter family of polyhex tiling
problems. Meanwhile, physicists had worked in an essentially equivalent (more
precisely, dual) setting, studying _trimer covers_ [12] of the regular
6-valent planar graph, but the physicists’ interests were in asymptotic
enumeration, not exact formulas for finite subgraphs of the 6-valent grid;
furthermore, their proofs relied on the controversial Bethe ansatz, which
tends to give correct answers but has not been rigorously established in all
the contexts in which it has been applied.

In [8], inspired by the large and still-growing literature on exact
enumeration of tilings (see [6] for an overview), Propp proposed using the
tiles studied by Conway and Lagarias to tile different sorts of regions in the
hexagonal grid not considered in earlier literature, and he made numerous
conjectures regarding the exact number of tilings of those regions. We prove
some of Propp’s conjectures here.

It is convenient to view the hexagonal grid as a tiling of the complex plane
with regular hexagons of side length $1$. We can uniquely specify the exact
positions of the unit hexagons, which we call the _cells_ of the grid, by
declaring that one of the hexagons has vertices $1\pm
1,1\pm\omega,1\pm\omega^{2}$, where $\omega=e^{2\pi i/3}$. Suppose $a$ and $b$
are positive integers satisfying $2\leq a\leq 2b$ and $2\leq b\leq 2a$.
Following [8], we define the _$(a,b)$ -benzel_ to be the union of the cells
lying entirely within the hexagon with vertices
$a\omega+b,-a\omega^{2}-b,a\omega^{2}+b\omega,-a-b\omega,a+b\omega^{2},-a\omega-b\omega^{2}$.
(This is the polyhex that we will attempt to tile using polyhexes consisting
of three hexagonal cells, aka _trihexes_.) Note that this hexagon is centered
at $0$, is invariant under rotation by $120^{\circ}$, and has sides whose
lengths alternate between $2a-b$ and $2b-a$. Figure 1 shows the
$(7,8)$-benzel. It can be shown (see [7]) that the number of cells in the
$(a,b)$-benzel is

(1) $\begin{cases}\frac{-a^{2}+4ab-b^{2}-a-b}{2}&\text{if }a+b\equiv 0\text{
or }2\pmod{3},\\\ \frac{-a^{2}+4ab-b^{2}-a-b+2}{2}&\text{if }a+b\equiv
1\pmod{3}.\end{cases}$

Figure 1. The $(7,8)$-benzel. Figure 2. The right stone, left stone, vertical
bone, rising bone, and falling bone, from left to right, respectively.

The prototiles in [1] and [8] consist of three cells and are of two forms: a
_stone_ consists of three hexagonal cells arranged in a triangle while a
_bone_ consists of three hexagonal cells arranged in a line. (Conway and
Lagarias called them $T_{2}$’s and $L_{3}$’s and Thurston [11] called them
$T_{2}$’s and _tribones_.) We will consider the various translationally-
inequivalent rotations of these shapes to be distinct prototiles. Thus, there
are really five different prototiles, which are shown in Figure 2. We call
them the _right stone_ , _left stone_ , _vertical bone_ , _rising bone_ , and
_falling bone_. Tilings of regions in the hexagonal grid using these
prototiles will be referred to as _stones-and-bones tilings_.

Conway and Lagarias showed that for any simply-connected region in the
hexagonal grid that admits a stones-and-bones tiling, the number of right
stones minus the number of left stones in such a tiling is invariant (i.e.,
depends only on the region and not the specific tiling). This is the
_Conway–Lagarias invariant_ of the region. Propp [7] showed that the
Conway–Lagarias invariant of the $(a,b)$-benzel is

$\displaystyle\begin{cases}\frac{3a^{2}-6ab+3b^{2}-a-b}{6}&\text{if }a+b\equiv
0\pmod{3},\\\ \frac{-a^{2}+4ab-b^{2}-a-b+2}{6}&\text{if }a+b\equiv
1\pmod{3},\\\ \frac{3a^{2}-6ab+3b^{2}+a+b-2}{6}&\text{if }a+b\equiv
2\pmod{3}.\end{cases}$

Note that when $a+b\equiv 1\pmod{3}$, the number of cells in the
$(a,b)$-benzel is three times its Conway–Lagarias invariant, which implies
(because each tile uses $3$ cells) that a stones-and-bones tiling must consist
entirely of right stones. We will prove that such a tiling exists and is
unique.

######  Theorem 1.1.

Let $a$ and $b$ be integers such that $2\leq a\leq 2b$, $2\leq b\leq 2a$, and
$a+b\equiv 1\pmod{3}$. The $(a,b)$-benzel has a unique stones-and-bones
tiling, and this tiling consists entirely of right stones.

Frequently, tiling problems will restrict the set of possible prototiles.
There are $2^{5}-1=31$ nonempty subsets of our set of 5 prototiles. However,
as benzels have threefold rotational symmetry—which preserves the two
stones—the types of bones allowed does not affect the answer to enumerative
questions; all that matters is the number of different bones that are allowed.
Hence, there are really only $2\cdot 2\cdot 4-1=15$ inequivalent tiling
problems to consider. Moreover, the $(a,b)$-benzel and $(b,a)$-benzel are
reflections of each other across the real axis; as this reflection preserves
the two stones as well as the number of types of allowed bones, each tiling
problem is the same for the $(a,b)$-benzel as the $(b,a)$-benzel. We will
often make use of this symmetry in order to assume without loss of generality
that $a\leq b$.

Much of our work centers around the enumeration of tilings of benzels in which
we allow just two types of bones and one type of stone. Since we only care
about the number of allowed bones, we may assume that we are using rising
bones and falling bones. Propp conjectured (see [8, Problems 2 and 3]) that if
$(a,b)$ is of the form $(3n,3n)$ or $(3n+1,3n+2)$, then the number of tilings
of the $(a,b)$-benzel using left stones, rising bones, and falling bones is
$(2n)!!=2^{n}n!$. One of our main results resolves this conjecture; in fact,
we will determine the number of such tilings of the $(a,b)$-benzel for all
choices of $a$ and $b$.

######  Theorem 1.2 (cf. [8, Problems 2 and 3]).

Let $a$ and $b$ be integers with $2\leq a\leq b\leq 2a$. If there is an
integer $n$ such that $(a,b)=(3n,3n)$ or $(a,b)=(3n+1,3n+2)$, then there are
exactly $(2n)!!$ tilings of the $(a,b)$-benzel by left stones, rising bones,
and falling bones. Otherwise, there are no such tilings of the $(a,b)$-benzel.

On the other hand, we have the following theorem regarding tilings with right
stones, rising bones, and falling bones.

######  Theorem 1.3.

Let $a$ and $b$ be integers with $2\leq a\leq b\leq 2a$. If $b=2a$ or
$a+b\equiv 1\pmod{3}$, then there is a unique tiling of the $(a,b)$-benzel by
right stones, rising bones, and falling bones. If $b<2a$ and $a+b\equiv
0\pmod{3}$, then there are no such tilings.

Theorem 1.3 says nothing about what happens when $a+b\equiv 2\pmod{3}$.; Propp
[8] gave the following conjecture in this case.

######  Conjecture 1.4 ([8, Problem 5]).

Let $a$ and $b$ be integers with $2\leq a\leq b\leq 2a$ and $a+b\equiv
2\pmod{3}$. Let $n$ and $k$ be the integers such that $(a,b)=(n+3k,2n+3k-1)$.
The $(a,b)$-benzel has

$\prod_{i=1}^{k}\frac{(2i)!(2i+2n-2)!}{(i+n-1)!(i+n+k-1)!}$

tilings by right stones, rising bones, and falling bones.

Our approach to proving Theorems 1.2 and 1.3 is to first transfer benzels from
the hexagonal grid into the square grid. This allows us to use tools from the
theory of ribbon tilings of Young diagrams. In particular, if $\lambda$ is a
partition with $k$-core $\kappa$ and $k$-quotient
$(\lambda^{(0)},\ldots,\lambda^{(k-1)})$, then there is a bijection (dubbed
the “abacus bijection” by Gordon James [3] and rediscovered by Dennis Stanton
and Dennis White (see [10] and [2]) that maps $k$-ribbon tableaux of shape
$\lambda/\kappa$ to $k$-tuple Young tableaux of shape
$(\lambda^{(0)},\ldots,\lambda^{(k-1)})$. For other authors’ descriptions of
the bijection, see [3] or [5]. We felt the need to provide our own discussion
and proof; we hope that our treatment of this bijection will be useful to
researchers. More relevant to our purpose is a related bijection that we call
$\operatorname{\mathsf{Compress}}$, which is apparently new. This bijection is
between $k$-ribbon tilings of a Young diagram of a particular type and
$(k-1)$-tilings of a related Young diagram; it is essentially a version of the
abacus bijection that “forgets” the ordering of the tiles in a ribbon tableau.
Our proofs of Theorems 1.2 and 1.3 employ these bijections as tools and
combine them with several new ideas. Although we have made no attempt toward
proving 1.4, we expect that it could be fruitful to transfer the problem into
the square grid and employ techniques similar to those that we use.

We mention in passing that, like us, Conway and Lagarias transferred their
tiling problem to the square grid. However, there are two key differences
between their use of the tactic and ours. The first difference is that, as
Thurston later demonstrated, the use of the square grid is not an essential
feature of the translation of tiling problems into combinatorial group theory
problems; in contrast, our adaptation of the theory of cores and quotients
requires working in the square grid (where Ferrers graphs and Young diagrams
live). The second difference is that in [1], all three bones are transferred
to the square grid, at the cost of making one of them disconnected; in
contrast, such disconnected tiles are forbidden in the theory of ribbon
tilings.

The outline for the rest of the paper is as follows. Section 2 is brief; its
purpose is to explain how to translate certain tiling problems in the
hexagonal grid into tiling problems in the square grid. In Section 3, we
provide a thorough discussion of $k$-cores, $k$-quotients, and the abacus
bijection. Section 4 defines and develops the
$\operatorname{\mathsf{Compress}}$ bijection. Section 5 is devoted to proving
Theorem 1.1. We use the properties of $\operatorname{\mathsf{Compress}}$
discussed in Section 4 to prove Theorem 1.2 in Section 6 and to prove Theorem
1.3 in Section 7.

##  2\. Transferring Tilings to the Square Grid

In this article, we draw the square grid in the complex plane so that the unit
square cells are oriented like diamonds. More precisely, each square grid cell
has vertices of the form $(a\pm 1)+bi$ and $a+(b\pm 1)i$, where $a$ and $b$
are integers such that $a+b$ is odd. We call each such square cell a _box_. A
great deal of theory has already been developed for tilings of regions in the
square grid; in order to make use of it, we will need a way of converting
tilings in the hexagonal grid into tilings in the square grid. Notice that
there are unit-width vertical strips of the complex plane that are traversed
only by horizontal edges of the hexagonal grid, not by edges of the other two
orientations. Removing these vertical strips and compressing the plane
appropriately yields a bijective mapping from the hexagons in the hexagonal
grid to rhombuses in the resulting rhombic grid. Rescaling the axes suitably
transforms these rhombuses into squares of side length $\sqrt{2}$. Let us then
rotate the resulting grid by $90^{\circ}$ counterclockwise.

$\xrightarrow{\text{delete
strips}}$$\xrightarrow{\text{compress}}$$\xrightarrow{\text{scale and
rotate}}$

Figure 3. The steps that transform the hexagonal grid into the square grid.
The bold rising bone in the hexagonal grid becomes a negative bone in the
square grid.

The prototiles of Figure 2, transferred to this square grid, are shown in
Figure 4. Notice that the vertical bone becomes a union of three disconnected
boxes. When at most two types of bones are allowed, we can assume without loss
of generality that the vertical bone is forbidden, which allows the resulting
square grid tiling problem to exclusively use prototiles of connected boxes.
In fact, when the vertical bone is forbidden, we can specifically leverage the
theory of _ribbon tilings_. A _ribbon_ is a connected union of boxes in the
square grid in which no two boxes have the same $x$-coordinate; a _$k$
-ribbon_ is a ribbon that contains exactly $k$ unit squares. Our four
prototiles (excluding the vertical bone) in the square grid are precisely the
four $3$-ribbons. With these conventions, we will rename the right stone, left
stone, rising bone, and falling bone the _mountain stone_ , _valley stone_ ,
_negative bone_ , and _positive bone_.

Figure 4. Transforming the prototiles in Figure 2 into the square grid yields
these five prototiles. From left to right, the first, second, fourth, and
fifth are the mountain stone, valley stone, negative bone, and positive bone,
respectively.

##  3\. The Abacus Bijection

The version of the bijection that we use is due to Gordon James (see [3]) but
different forms of it seem to have been discovered independently by various
people working in the field of modular representation theory around 1950; this
community includes H. Farahat, J. S. Frame, D. E. Littlewood, T. Nakayama, M.
Osima, G. de B. Robinson, R. A. Staal, and R. M. Thrall. The interested reader
may find more details in the book [9] and the references it contains.

We identify integer partitions with their Young diagrams. We will draw the
Young diagram of a partition $\lambda$ in Russian notation so that it lives
within the square grid as we have chosen to draw it. Let us position the Young
diagram so that its bottom-most point is $0$ and so that the boxes
representing the first part of the partition have their lower-left edges lying
along the ray $\mathbb{R}_{\geq 0}\,e^{3\pi i/4}$, as in Figure 5.

A _coordinatized abacus word_ is a function $w$ from the set
$\mathbb{Z}+1/2=\\{n+1/2:n\in\mathbb{Z}\\}$ to the alphabet
$\\{\circ,\bullet\\}$ with the property that $w(-n-1/2)=\bullet$ and
$w(n+1/2)=\circ$ for all sufficiently large $n$. Such words also arise in the
study of the exclusion process; in that literature, $\bullet$ is called a
_particle_ and $\circ$ is called a _vacancy_. We can represent such a word as
a bi-infinite sequence of the symbols $\circ$ and $\bullet$, where we place a
period between the symbols $w(-1/2)$ and $w(1/2)$ to indicated the position
$0$. For example, $\cdots\bullet\bullet\bullet\circ\bullet\\!\\!\hskip
1.084pt.\hskip 1.084pt\\!\\!\bullet\circ\circ\circ\cdots$ represents the
coordinatized abacus word $w$ given by $w(-3/2)=w(n+1/2)=\circ$ for all
integers $n\geq 1$ and $w(1/2)=w(-1/2)=w(m-1/2)$ for all integers $m\leq-2$.
An _abacus word_ is an orbit of a coordinatized abacus word under the natural
$\mathbb{Z}$-action given by the shift map. As with coordinatized abacus
words, we can represent abacus words as bi-infinite sequences using the
symbols $\circ$ and $\bullet$; however, we no longer include the period in
this representation.

Given a partition $\lambda$, we obtain an abacus word $\overline{w}_{\lambda}$
by traveling along its northern border from left to right and recording
whether we go up or down at each step by writing the symbol $\circ$ to record
an up step and the symbol $\bullet$ to record a down step. From this abacus
word $\overline{w}_{\lambda}$, we obtain a coordinatized abacus word
$w_{\lambda}$ by insisting that $w_{\lambda}(\ell)$ records whether the step
whose midpoint has real part $\ell$ is up or down. For example, if
$\lambda=(5,5,3,3,2)$ is the partition shown in Figure 5, then
$w_{\lambda}=\cdots\bullet\bullet\bullet\circ\circ\bullet\bullet\circ.\\!{}{}\circ\bullet\circ\bullet\bullet\circ\circ\circ\cdots$.

Each abacus word $\overline{w}$ (which is, by definition, an orbit under the
shift map) has a unique representative of the form $w_{\lambda}$ for some
partition $\lambda$. Indeed, in the partition $\lambda$, there must be the
same number of up steps to the left of position $0$ as down steps to the right
of position $0$, so

$|\\{\ell<0:w_{\lambda}(\ell)=\circ\\}|=|\\{\ell\geq
0:w_{\lambda}(\ell)=\bullet\\}|;$

this value is also the number of boxes of $\lambda$ that are directly above
the origin. We call $w_{\lambda}$ the _canonical coordinatization_ of $w$.

Figure 5. The Young diagram of the partition $\lambda=(5,5,3,3,2)$. We have
$w_{\lambda}=\cdots\bullet\bullet\bullet\circ\circ\bullet\bullet\circ.\\!\circ\bullet\circ\bullet\bullet\circ\circ\circ\cdots$.

Let us fix a positive integer $k$ and split the word $w_{\lambda}$ into $k$
words $w^{(0)}_{\lambda},\ldots,w^{(k-1)}_{\lambda}$ so that each
$w^{(j)}_{\lambda}$ is obtained by reading every $k$-th symbol in
$w_{\lambda}$. In other words, $w_{\lambda}^{(j)}$ is the word obtained by
reading $w_{\lambda}(k\ell+j+1/2)$ for all $\ell\in\mathbb{Z}$. As above, we
can coordinatize each of the words $w_{\lambda}^{(j)}$ so that it corresponds
to an integer partition $\lambda^{(j)}$. For each $0\leq j\leq k-1$, there is
a unique integer $c_{j}$ such that if we set $w^{(j)}_{\lambda}(\ell-
c_{j}+1/2)=w_{\lambda}(k\ell+j+1/2)$ for all $\ell$, then
$w_{\lambda}^{(j)}=w_{\lambda^{(j)}}$, the abacus word associated with the
partition $\lambda^{(j)}$. The integers $c_{0},\ldots,c_{k-1}$ are called the
_$k$ -charges_ of $\lambda$; they satisfy $c_{0}+\cdots+c_{k-1}=0$. The tuple
$(\lambda^{(0)},\ldots,\lambda^{(k-1)})$ is called the _$k$ -quotient_ of
$\lambda$. For an example of this construction, let $k=3$, and let
$\lambda=(5,5,3,3,2)$ as shown in Figure 5. We have
$w_{\lambda}^{(0)}=\cdots\bullet\bullet\bullet\circ\bullet\circ\circ\circ\cdots$,
$w_{\lambda}^{(1)}=\cdots\bullet\bullet\bullet\circ\bullet\bullet\bullet\circ\circ\circ\cdots$,
and $w_{\lambda}^{(2)}=\cdots\bullet\bullet\bullet\circ\circ\circ\cdots$. Then
$\lambda^{(0)}=(1)$, $\lambda^{(1)}=(3)$, and $\lambda^{(2)}=\emptyset$ are
shown in Figure 6. The $3$-charges are $c_{0}=c_{1}=1$ and $c_{2}=-2$.

Figure 6. The $3$-quotient of the partition $\lambda=(5,5,3,3,2)$ is the
triple $(\lambda^{(0)},\lambda^{(1)},\lambda^{(2)})=((1),(3),\emptyset)$. The
$3$-charges are $c_{0}=c_{1}=1$ and $c_{2}=-2$.

Suppose there is a $k$-ribbon tile $T$ at the top of the Young diagram of
$\lambda$ such that removing $T$ results in the Young diagram of a smaller
partition. Let us remove it. Iterating this process, we can continue removing
$k$-ribbon tiles until it is no longer possible to do so. The resulting
partition $\kappa$ is called the _$k$ -core_ of $\lambda$; it is known that
$\kappa$ does not depend on the order in which the $k$-ribbons were removed.
We will consider the skew shape $\lambda/\kappa$. It will be convenient to
think of $\kappa$ as a tile inside of $\lambda$.

A _$k$ -ribbon tableau_ of shape $\lambda/\kappa$ is a tuple
$(T_{1},\ldots,T_{m})$ of $k$-ribbon tiles such that the set
$\\{\kappa,T_{1},\ldots,T_{m}\\}$ forms a tiling of $\lambda$ and such that
for every $1\leq i\leq m$, the set $\\{\kappa,T_{1},\ldots,T_{i}\\}$ forms a
tiling of some Young diagram. One can imagine adding the tiles
$\kappa,T_{1},\ldots,T_{m}$ one by one to build $\lambda$ from the bottom up.
Let $\operatorname{RTab}_{k}(\lambda/\kappa)$ denote the set of $k$-ribbon
tableaux of shape $\lambda/\kappa$.

A _strict Young tableau_ of shape $\lambda$ is a filling of the square boxes
of $\lambda$ (which is again drawn in Russian notation) with distinct positive
integers so that numbers strictly increase as we move up (northeast or
northwest). We write $\operatorname{cont}(\mathcal{T})$ for the _content_ of a
strict Young tableau $\mathcal{T}$, which is just the set of integers
appearing in $\mathcal{T}$.

Suppose $(\alpha_{0},\ldots,\alpha_{k-1})$ is a $k$-tuple of partitions. We
define a _$k$ -tuple Young tableau_ of shape
$(\alpha_{0},\ldots,\alpha_{k-1})$ to be a $k$-tuple
$(\mathcal{T}_{0},\ldots,\mathcal{T}_{k-1})$ of strict Young tableaux such
that each $\mathcal{T}_{j}$ has shape $\alpha_{j}$ and
$\operatorname{cont}(\mathcal{T}_{0})\cup\cdots\cup\operatorname{cont}(\mathcal{T}_{k-1})=[|\alpha_{0}|+\cdots+|\alpha_{k-1}|]$.
(Here we use $[n]$ to denote $\\{1,2,\dots,n\\}$.) Note that these conditions
force the sets
$\operatorname{cont}(\mathcal{T}_{0}),\ldots,\operatorname{cont}(\mathcal{T}_{k-1})$
to be pairwise disjoint. Let
$\operatorname{YT}_{k}(\alpha_{0},\ldots,\alpha_{k-1})$ be the set of
$k$-tuple Young tableaux of shape $(\alpha_{0},\ldots,\alpha_{k-1})$.

As above, let $\lambda$ be a partition with $k$-quotient
$(\lambda^{(0)},\ldots,\lambda^{(k-1)})$ and $k$-core $\kappa$. The _abacus
bijection_ is a bijection
$\operatorname{\mathsf{Ab}}\colon\operatorname{RTab}_{k}(\lambda/\kappa)\to\operatorname{YT}_{k}(\lambda^{(0)},\ldots,\lambda^{(k-1)})$.
To describe it, let us start with a $k$-ribbon tableau
$(T_{1},\ldots,T_{m})\in\operatorname{RTab}_{k}(\lambda/\kappa)$. The
intuitive idea is to remove the tiles $T_{m},\ldots,T_{1}$ in this order and
insert the number $r$ into one of the boxes of one of
$\lambda^{(0)},\ldots,\lambda^{(k-1)}$ at the point in time when we remove
$T_{r}$. In order to be rigorous, it is convenient to define the bijection
recursively. Thus, let us assume that we have already defined the abacus
bijection on $\operatorname{RTab}_{k}(\mu/\kappa)$ whenever
$|\mu/\kappa|<|\lambda/\kappa|$.

Let $\ell\in\mathbb{Z}$ and $j\in\\{0,\ldots,k-1\\}$ be such that $k\ell+j$ is
the real part of the leftmost point in the $k$-ribbon tile $T_{m}$; the real
part of the rightmost point in $T_{m}$ must be $k(\ell+1)+j+1$. There are
$k+1$ steps in the northern boundary of the Young diagram of $\lambda$ that
are part of the $k$-ribbon tile $T_{m}$; they are encoded by the symbols
$w_{\lambda}(k\ell+j+1/2),\ldots,w_{\lambda}(k(\ell+1)+j+1/2)$. It is
straightforward to see that $w_{\lambda}(k\ell+j+1/2)=\circ$ and
$w_{\lambda}(k(\ell+1)+j+1/2)=\bullet$. Let $\mu$ be the partition whose Young
diagram is obtained from that of $\lambda$ by removing $T_{m}$. Then $w_{\mu}$
agrees with $w_{\lambda}$ except that $w_{\mu}(k\ell+j+1/2)=\bullet$ and
$w_{\mu}(k(\ell+1)+j+1/2)=\circ$. The $k$-charges of $\mu$ are the same as the
$k$-charges $c_{0},\ldots,c_{k-1}$ of $\lambda$. Indeed, if we replace
$\lambda$ by $\mu$, the quantities $|\\{\ell<0:w_{\lambda}^{(j)}(\ell-
c_{j}+1/2)=\circ\\}|$ and $|\\{\ell\geq 0:w_{\lambda}^{(j)}(\ell-
c_{j}+1/2)=\bullet\\}|$ will not change if $\ell\neq-1$ and will both decrease
by $1$ if $\ell=-1$; more generally, adding or removing $k$-ribbons from the
Young diagram of a partition $\nu$ to yield a new Young diagram of a partition
$\nu^{\prime}$ does not change the $k$-charges. This implies $\lambda$, $\mu$,
and $\kappa$ have the same $k$-charges. It follows that
$\lambda^{(r)}=\mu^{(r)}$ for all $r\neq j$ and that the Young diagram of
$\mu^{(j)}$ is obtained from that of $\lambda^{(j)}$ by removing a single box
$B$. Note that the real parts of the leftmost and rightmost points in $B$ are
$\ell-c_{j}$ and $\ell-c_{j}+2$, respectively. Since
$(T_{1},\ldots,T_{m-1})\in\operatorname{RTab}_{k}(\mu/\kappa)$ and
$|\mu/\kappa|<|\lambda/\kappa|$, we can apply the abacus bijection inductively
to obtain a $k$-tuple Young tableau
$\operatorname{\mathsf{Ab}}(T_{1},\ldots,T_{m-1})\in\operatorname{YT}_{k}(\mu^{(0)},\ldots,\mu^{(k-1)})$.
The $k$-tuple Young tableau
$\operatorname{\mathsf{Ab}}(T_{1},\ldots,T_{m})\in\operatorname{YT}_{k}(\lambda^{(0)},\ldots,\lambda^{(k-1)})$
is now obtained from $\operatorname{\mathsf{Ab}}(T_{1},\ldots,T_{m-1})$ by
adding the box $B$ and filling it with the number $m$.

######  Example 3.1.

Let $k=3$ and $\lambda=(5,5,3,3,2)$, as in Figure 5. We saw above that
$\lambda^{(0)}=(1)$, $\lambda^{(1)}=(3)$, and $\lambda^{(2)}=\emptyset$. The
$3$-core of $\lambda$ is $\kappa=(4,2)$. One standard $3$-ribbon tableau
$(T_{1},T_{2},T_{3},T_{4})$ of shape $\lambda/\kappa$ is shown at the top of
Figure 7; the corresponding $3$-tuple Young tableau
$\operatorname{\mathsf{Ab}}(T_{1},T_{2},T_{3},T_{4})$ appears at the bottom of
the same figure. To illustrate the recursive description of the abacus
bijection, let $\mu$ be the partition obtained by removing $T_{4}$ from
$\lambda$. Then
$w_{\lambda}=\cdots\bullet\bullet\bullet\circ\circ\bullet\bullet\circ\circ\bullet\circ\bullet\bullet\circ\circ\circ\cdots$
and
$w_{\mu}=\cdots\bullet\bullet\bullet\bullet\circ\bullet\circ\circ\circ\bullet\circ\bullet\bullet\circ\circ\circ\cdots$;
the latter is obtained from the former by changing one $\circ$ to a $\bullet$
and changing one $\bullet$ to a $\circ$. Preserving notation from above, we
have $\ell=-2$ and $j=1$.

Figure 7. An example of the abacus bijection for $\lambda=(5,5,3,3,2)$. At top
is a $3$-ribbon tableau of shape $\lambda$; at bottom is the corresponding
$3$-tuple Young tableau of shape $((1),(3),\emptyset)$.

To prove that the map
$\operatorname{\mathsf{Ab}}\colon\operatorname{RTab}_{k}(\lambda/\kappa)\to\operatorname{YT}_{k}(\lambda^{(0)},\ldots,\lambda^{(k-1)})$
is bijective, we construct its inverse by reversing the above procedure.
Suppose we start with a $k$-tuple Young tableau
$(\mathcal{T}^{(0)},\ldots,\mathcal{T}^{(k-1)})$ of shape
$(\lambda^{(0)},\ldots,\lambda^{(k-1)})$. Let
$m=|\lambda^{(0)}|+\cdots+|\lambda^{(k-1)}|$, and let $j\in\\{0,\ldots,k-1\\}$
be such that $m\in\operatorname{cont}(\mathcal{T}^{(j)})$. Let $B$ be the box
of $\lambda^{(j)}$ containing $m$ in $\mathcal{T}^{(j)}$. Let
$c_{0},\ldots,c_{k-1}$ be the $k$-charges of $\lambda$, and define $\ell$ so
that $\ell-c_{j}$ is the real part of the leftmost point in $B$. We can
reconstruct the partition $\mu^{(j)}$ by removing $B$ from $\lambda^{(j)}$. We
can then reconstruct $\mu$, which is determined by its $k$-quotient
$(\mu^{(0)},\ldots,\mu^{(k-1)})$ and $k$-charges $c_{0},\ldots,c_{k-1}$ (it
has the same $k$-charges as $\lambda$), by letting $\mu^{(r)}=\lambda^{(r)}$
for all $r\neq j$. Let $\widetilde{\mathcal{T}}^{(j)}$ be the strict Young
tableau of shape $\mu^{(j)}$ obtained by deleting $B$ from
$\mathcal{T}^{(j)}$, and let $\widetilde{\mathcal{T}}^{(r)}=\mathcal{T}^{(r)}$
for all $r\neq j$. Then
$(\widetilde{\mathcal{T}}^{(0)},\ldots,\widetilde{\mathcal{T}}^{(k-1)})$ is a
$k$-tuple Young tableau of shape $(\mu^{(0)},\ldots,\mu^{(k-1)})$. As before,
the words $w_{\lambda}$ and $w_{\mu}$ agree except that
$(w_{\lambda}(k\ell+j+1/2),w_{\mu}(k\ell+j+1/2))=(\circ,\bullet)$ and
$(w_{\lambda}(k(\ell+1)+j+1/2),w_{\mu}(k(\ell+1)+j+1/2))=(\bullet,\circ)$.
This implies that $\lambda$ is obtained from $\mu$ by adding a $k$-ribbon tile
$T_{m}$. Hence, $\mu$ and $\lambda$ have the same $k$-core $\kappa$. We may
assume inductively that
$\operatorname{\mathsf{Ab}}\colon\operatorname{RTab}_{k}(\mu/\kappa)\to\operatorname{YT}_{k}(\mu^{(0)},\ldots,\mu^{(k-1)})$
is a bijection, so we can re-obtain the $k$-ribbon tableau
$(T_{1},\ldots,T_{m-1})\in\operatorname{RTab}_{k}(\mu/\kappa)$ as
$\operatorname{\mathsf{Ab}}^{-1}(\widetilde{\mathcal{T}}^{(0)},\ldots,\widetilde{\mathcal{T}}^{(k-1)})$.
Then
$\operatorname{\mathsf{Ab}}^{-1}(\mathcal{T}^{(0)},\ldots,\mathcal{T}^{(k-1)})$
is the $k$-ribbon tableau $(T_{1},\ldots,T_{m})$.

Finally, we address the base case $|\lambda/\kappa|=0$, i.e.,
$\lambda=\kappa$. This implies
$\operatorname{RTab}_{k}(\lambda/\kappa)=\\{\emptyset\\}$. If any of
$\kappa^{(0)},\ldots,\kappa^{(k-1)}$ are nonempty, then applying our inverse
map to remove a box implies that a $k$-ribbon can be removed from $\kappa$ to
yield a smaller partition $\kappa^{\prime}$, contradicting the definition of
the $k$-core. Thus $\kappa^{(0)}=\cdots=\kappa^{(k-1)}=\emptyset$, and
$\operatorname{YT}_{k}(\lambda^{(0)},\ldots,\lambda^{(k-1)})=\\{(\emptyset,\ldots,\emptyset)\\}$.
We obtain a trivial bijection
$\operatorname{\mathsf{Ab}}\colon\operatorname{RTab}_{k}(\lambda/\kappa)\to\operatorname{YT}_{k}(\lambda^{(0)},\ldots,\lambda^{(k-1)})$.

This proves the following theorem.

######  Theorem 3.2 ([3]).

Let $\lambda$ be an integer partition with $k$-quotient
$(\lambda^{(0)},\ldots,\lambda^{(k-1)})$ and $k$-core $\kappa$. The map
$\operatorname{\mathsf{Ab}}\colon\operatorname{RTab}_{k}(\lambda/\kappa)\to\operatorname{YT}_{k}(\lambda^{(0)},\ldots,\lambda^{(2)})$
defined above is a bijection.

A $k$-ribbon tableau of shape $\lambda/\mu$ is a $k$-ribbon tiling of
$\lambda/\mu$ together with a certain ordering of the tiles. If we are given a
$k$-ribbon tiling, then there is at least one way to order the tiles to obtain
a $k$-ribbon tableau. Indeed, we can define a partial order $\prec$ on the set
of tiles by declaring that $T\prec T^{\prime}$ whenever $T$ has a box that
lies below one of the boxes in $T^{\prime}$. Then the orderings of the tiles
that yield $k$-ribbon tableaux are precisely the linear extensions of this
partially ordered set. Our interest in this article is in unordered tilings,
so our main motivation for discussing the abacus bijection comes from the
following corollary.

######  Corollary 3.3.

Let $\lambda$ be an integer partition with $k$-quotient
$(\lambda^{(0)},\ldots,\lambda^{(k-1)})$ and $k$-core $\kappa$. There exists a
$k$-ribbon tiling of the Young diagram of $\lambda$ if and only if $\kappa$ is
empty, and this occurs if and only if
$|\lambda^{(0)}|+\cdots+|\lambda^{(k-1)}|=\frac{1}{k}|\lambda|$.

###### Proof.

As discussed above, any given skew shape has a $k$-ribbon tableau if and only
if it can be tiled by $k$-ribbons. The definition of the $k$-core $\kappa$
implies that it is the smallest partition contained in $\lambda$ such that
there is a $k$-ribbon tableau of shape $\lambda/\kappa$. This shows that
$\lambda$ can be tiled by $k$-ribbons if and only if $\kappa$ is empty. It
follows from the above discussion of the abacus bijection that
$|\lambda^{(0)}|+\cdots+|\lambda^{(k-1)}|=\frac{1}{k}(|\lambda|-|\kappa|)$, so
$\kappa$ is empty if and only if
$|\lambda^{(0)}|+\cdots+|\lambda^{(k-1)}|=\frac{1}{k}|\lambda|$. ∎

######  Remark 3.4.

The above discussion of the abacus bijection noted that the $k$-quotient of
any $k$-core $\kappa$ is empty, so $\kappa$ is completely determined by the
$k$-charges of $\lambda$, which are the same as the $k$-charges of $\kappa$.
In other words, given the $k$-charges of $\lambda$, we can construct its
$k$-core $\kappa$ to be the unique partition with an empty $k$-quotient and
the same charges as $\lambda$. In particular, $\kappa$ is empty if and only if
the $k$-charges are all zero.

##  4\. The Compress Bijection

Suppose $\lambda$ is a partition with empty $k$-core whose $k$-quotient
$(\lambda^{(0)},\dots,\lambda^{(k-1)})$ satisfies $\lambda^{(j)}=\emptyset$.
Let $\rho$ be the partition with empty $(k-1)$-core whose $(k-1)$-quotient is

$(\lambda^{(0)},\dots,\lambda^{(j-1)},\lambda^{(j+1)},\dots,\lambda^{(k-1)}).$

We will define a bijection $\operatorname{\mathsf{Compress}}$ from the set of
$k$-ribbon tilings of $\lambda$ to the set of $(k-1)$-ribbon tilings of
$\rho$. This bijection is essentially the unordered tiling equivalent of using
the abacus bijection to map a $k$-ribbon tableau to a $k$-tuple Young tableau,
removing the empty $\lambda^{(j)}$ to obtain a $(k-1)$-tuple Young tableau,
and then lifting via the inverse abacus bijection on $(k-1)$-ribbons back to a
$(k-1)$-ribbon tableau. However, without an ordering on our ribbons, a one-
step description of our bijection exists, and it is simpler than this two-step
process involving the abacus bijection.

Similar to how we removed vertical strips of the hexagonal grid that were
traversed by horizontal edges and then compressed the plane to obtain the
square grid in Section 2, our bijection is obtained by removing the vertical
strips of the square grid that correspond to $\lambda^{(k-1)}=\emptyset$ and
then compressing the plane to remove the empty space.

Identifying $\lambda$ with its Young diagram, let us write
$P_{r}=\lambda\cap\\{z\in\mathbb{C}:r\leq\operatorname{Re}(z)<r+1\\}$, where
$\operatorname{Re}(z)$ denotes the real part of $z$. The assumptions that
$\lambda^{(j)}=\emptyset$ and that the $k$-charges of $\lambda$ are all zero
(since its $k$-core is empty) guarantee that $P_{r}$ is a parallelogram
whenever $r\equiv j\pmod{k}$; let us color these parallelograms green. Imagine
deleting these green parallelograms, dividing what remains into vertical
bands, sliding the remaining pieces of $\lambda$ lying left of the imaginary
axis to the southeast, and sliding the remaining pieces of $\lambda$ lying
right of the imaginary axis to the southwest. (Specifically, tile-pieces in
the $i$-th band to the left of the imaginary axis slide $i$ steps southeast,
while tile-pieces in the $i$-th band to the right of the imaginary axis slide
$i$ steps southeast.) See Figure 8 for an example when $k=3$ and $j=1$. It is
straightforward to check that the resulting shape will be the Young diagram of
$\rho$. The bijection $\operatorname{\mathsf{Compress}}$ simply transfers the
tiles in a $k$-ribbon tiling of $\lambda$ through this process so that they
become $(k-1)$-ribbons that tile $\rho$. It is not difficult to check that if
we start with a $(k-1)$-ribbon tiling $\mathcal{D}$ of $\rho$, then there is a
unique way to lift it to a $k$-ribbon tiling $\mathcal{T}$ of $\lambda$ such
that $\operatorname{\mathsf{Compress}}(\mathcal{T})=\mathcal{D}$; every
$(k-1)$-ribbon has a unique square that will be extended into two squares,
where the direction of extension depends on which side of the imaginary axis
the square is on, yielding a $k$-ribbon. Hence, we only need to check that
$\operatorname{\mathsf{Compress}}$ is well-defined; more precisely, we need to
show that the remove-green compression process actually sends all of the tiles
in a $k$-ribbon tiling of $\lambda$ to $(k-1)$-ribbons in $\rho$ (instead of
disconnected shapes).

Figure 8. Applying the bijection $\operatorname{\mathsf{Compress}}$ to the
$3$-ribbon tiling of the partition $\lambda$ on the left yields the $2$-ribbon
tiling of the partition $\rho$ on the right. We have also drawn the green
parallelograms in $\lambda$.

Let us fix a $k$-ribbon tiling $\mathcal{T}$ of $\lambda$. Suppose we order
the tiles in $\mathcal{T}$ to form a $k$-ribbon tableau. When we apply the
abacus bijection to this tableau, each tile will correspond to one of the
boxes in the resulting $k$-tuple Young tableau. More precisely, a tile $T$
will correspond to one of the boxes in the Young tableau of shape
$\lambda^{(r)}$, where $r\in\\{0,1,\dots,k-1\\}$ is congruent modulo $k$ to
the real part of the leftmost point in $T$. Since $\lambda^{(j)}=\emptyset$,
we cannot have $r=j$. Equivalently, $T$ cannot intersect two of the green
parallelograms. It must intersect at least one of the green parallelograms
because the real parts of the leftmost and rightmost points of $T$ are $k$
apart and because the horizontal distance between two consecutive green
parallelograms is only $k-1$. This proves the following lemma.

######  Lemma 4.1.

Let $\lambda$ be a partition with empty $k$-core and with $k$-quotient
$(\lambda^{(0)},\dots,\lambda^{(k-1)})$ satisfying $\lambda^{(j)}=\emptyset$.
If $\mathcal{T}$ is a $k$-ribbon tiling of $\lambda$, then each tile in
$\mathcal{T}$ intersects a unique green parallelogram.

Now choose a fixed green parallelogram $P=P_{r}$ in $\lambda$. The side
lengths of $P$ are $\sqrt{2}$ and $2\alpha$ for some nonnegative integer
$\alpha$. We can break $P$ into $\alpha$ smaller parallelograms whose side
lengths are $\sqrt{2}$ and $2$; say these smaller parallelograms are
$P^{1},\ldots,P^{\alpha}$, listed from bottom to top. Let $L^{s}$ and $R^{s}$
be the left and right vertical sides of $P^{s}$, respectively. Each tile in
$\mathcal{T}$ that intersects $P$ must contain exactly one of
$L^{1},\ldots,L^{\alpha}$ and exactly one of $R^{1},\ldots,R^{\alpha}$. It
follows that for each $1\leq s\leq\alpha$, there is a tile $T^{s}$ that
contains both $L^{s}$ and $R^{s}$. When we apply the compression process
removing the green parallelograms, the piece of $T^{s}$ to the left of $P$
will connect with the piece of $T^{s}$ to the right of $P$ to form a
$(k-1)$-ribbon, as desired. See Figure 9 for an illustration when $k=3$.

Figure 9. The remove-green compression process turns $k$-ribbons that
intersect the green parallelogram into $(k-1)$-ribbons.

######  Remark 4.2.

Suppose $T$ is a tile in a $k$-ribbon tiling $\mathcal{T}$ of $\lambda$. By
Lemma 4.1, $T$ intersects a unique green parallelogram $P$. In $P$ appears to
the left (respectively, right) of the imaginary axis, then it follows from the
preceding discussion that the line segments in the the boundary of $T$ that
intersect $P$ must have negative (respectively, positive) slope. Lemma 4.1 and
the present remark encompass the main properties of the $\mathsf{Compress}$
bijection that we will use in Sections 6 and 7.

######  Remark 4.3.

The preceding discussion implies that the Young diagrams with $k$-quotients

$(\mu^{(0)},\mu^{(1)},\dots,\mu^{(k-2)},\emptyset),(\mu^{(0)},\dots,\mu^{(k-3)},\emptyset,\mu^{(k-2)}),\ldots,(\emptyset,\mu^{(0)},\dots,\mu^{(k-2)})$

with vanishing $k$-charges all have the same number of $k$-ribbon tilings, as
each such set of tilings is in bijection with the set of $(k-1)$-ribbon
tilings of the Young diagram with $(k-1)$-quotient
$(\mu^{(0)},\dots,\mu^{(k-2)})$ and vanishing $(k-1)$-charges.

######  Remark 4.4.

The $\operatorname{\mathsf{Compress}}$ bijection can be generalized to remove
multiple empty $k$-quotients simultaneously. Removing $k$-quotients
$\lambda^{(j_{1})}=\cdots=\lambda^{(j_{n})}=\emptyset$, we simply remove all
$P_{r}$ with $r\equiv j_{\ell}\pmod{k}$ for some $\ell$ and then slide the
remaining pieces of $\lambda$ together accordingly. This is equivalent to
composing $n$ $\operatorname{\mathsf{Compress}}$ bijections together, each
time removing one of the empty parts of the quotient.

The relationship between the abacus bijection and the compress bijection is
summarized in the following commutative diagram:

The map from $\operatorname{RTab}_{k}(\lambda)$ (respectively,
$\operatorname{RTab}_{k-1}(\rho)$) to the set of $k$-ribbon tilings of
$\lambda$ (respectively, $(k-1)$-ribbon tilings of $\rho$) simply forgets the
ordering of the tiles. The map $\widetilde{\mathsf{Compress}}$ acts in the
same way as the map $\operatorname{\mathsf{Compress}}$, except it preserves
the ordering of the tiles as it compresses $k$-ribbon tiles into
$(k-1)$-ribbon tiles.

##  5\. Tilings of Benzels Using Only Right Stones

When $a+b\equiv 1\pmod{3}$, we can use the Conway–Lagarias invariant to see
that in any stones-and-bones tiling of the $(a,b)$-benzel (allowing all five
types of stones and bones), the benzel must be tiled entirely by right stones.
In this section, we prove Theorem 1.1, which states that such a tiling exists
and is unique.

###### Proof of Theorem 1.1.

The $(2,2)$-benzel consists of three hexagonal cells in the shape of a right
stone, so the theorem is obvious when $a=b=2$. Thus, we may assume $a+b>4$ and
proceed by induction on $a+b$. By the discussion above, we simply need to show
that there is only one way to tile the $(a,b)$-benzel with right stones. The
$(a,b)$-benzel and the $(b,a)$-benzel are related by a reflection across the
real axis, and this reflection preserves right stones. Therefore, it suffices
to consider the case when $a\leq b$.

We illustrate the argument in Figure 10. The bottom side of the bounding
hexagon of the $(a,b)$-benzel touches $\frac{2a-b+1}{3}$ of the hexagonal
cells of the benzel. Any tiling of the $(a,b)$-benzel with right stones must
cover these $\frac{2a-b+1}{3}$ cells with distinct right stones, which are
aligned in a row. Since benzels and stones have $120^{\circ}$-rotational
symmetry, the same argument shows that the upper left side and the upper right
side of the benzel must also each be tiled with a line of $\frac{2a-b+1}{3}$
right stones. Removing these three lines of right stones results in the
$(a,b-3)$-benzel, which we know by induction has a unique tiling by right
stones. ∎

Figure 10. The $(8,11)$-benzel. We have indicated the three lines of right
stones appearing on the bottom and in the upper left and upper right corners
of the benzel; each line contains $\frac{2\cdot 8-11+1}{3}=2$ right stones.
Removing these lines of right stones results in the $(8,8)$-benzel, as
indicated by the blue hexagon.

This immediately implies that for any tiling problem where only certain
specified prototiles are allowed, there is a unique tiling of the
$(a,b)$-benzel for $a+b\equiv 1\pmod{3}$ if and only if the right stone is an
allowed prototile, and no tilings otherwise.

######  Remark 5.1.

Recall that the parameters $a,b$ of an $(a,b)$-benzel must satisfy $2\leq
a\leq 2b$ and $2\leq b\leq 2a$. For the three residues of $a+b$ modulo 3, the
boundary cases are the $(n,2n-2)$-benzel, the $(n,2n-1)$-benzel, and the
$(n,2n)$-benzel for $n\geq 2$. For each $n$, these three benzels coincide; as
Theorem 1.1 applies to the $(n,2n-2)$-benzel, it also implies that the
$(n,2n-1)$-benzel and $(n,2n)$-benzel each have a unique stones-and-bones
tiling, which consists entirely of right stones.

######  Remark 5.2.

It should be noted that all the stones in the unique tiling constructed above
are in phase with each other. That is, if we 3-color the cells in the
hexagonal grid so that no two adjoining hexagons have the same color (which
can be done in an essentially unique way), then all the stones in the tiling
constructed above exhibit the same color-pattern. Putting it differently, the
tiling of the benzel extends to a periodic tiling of the plane by right stones
whose fundamental domain can be chosen to be a single tile.

##  6\. Two Bones and the Left Stone

The goal of this section is to enumerate tilings of the $(a,b)$-benzel using
two types of bones along with the left stone. Without loss of generality, we
may assume the vertical bone is forbidden. Thus, our goal is to prove Theorem
1.2, which provides the enumeration of tilings using rising bones, falling
bones, and left stones. It follows immediately from Theorem 1.1 that there are
no such tilings when $a+b\equiv 1\pmod{3}$ (since the only stones-and-bones
tiling of the $(a,b)$-benzel uses only right stones). If $a+b\equiv
2\pmod{3}$, the Conway–Lagarias invariant of the $(a,b)$-benzel is

$\frac{1}{6}(3a^{2}-6ab+3b^{2}+a+b-2)=\frac{1}{6}\left(3(a-b)^{2}+a+b-2\right)\geq\frac{a+b-2}{6}\geq\frac{1}{3}>0.$

Therefore, in this case, a stones-and-bones tiling of the $(a,b)$-benzel
requires a strictly positive number of right stones, so again no such tiling
exists. This completes the proof of Theorem 1.2 when $a+b\not\equiv
0\pmod{3}$, so we may assume throughout the rest of this section that
$a+b\equiv 0\pmod{3}$.

After transferring the problem to the square grid as in Section 2, we are left
to consider tilings using valley stones, negative bones, and positive bones,
i.e., the three prototiles other than the mountain stone. Henceforth, we will
refer to such tilings with these three allowed prototiles as _mountainless
tilings_.

We will use the $\operatorname{\mathsf{Compress}}$ bijection from Section 4
with $k=3$, i.e., between certain 3-ribbon tilings and 2-ribbon tilings (also
called _domino tilings_) of smaller shapes. Note that there are two 2-ribbons,
the 2-ribbon analogs of the positive and negative bones. We will set $j=1$, so
our bijection will be from 3-ribbon tilings of an integer partition $\lambda$
with 3-quotient $(\lambda^{(0)},\emptyset,\lambda^{(2)})$ and vanishing
$3$-charges. As we will see later, the integer partition that ends up being
relevant for our purposes is the partition $\lambda_{n}$ with 3-quotient
$(\square_{n},\emptyset,\square_{n})$ and vanishing $3$-charges, where
$\square_{n}=(n,n,\dots,n)$ is the integer partition consisting of $n$ parts
of size $n$. The Young diagram of $\lambda_{n}$ consists of $n$ nested
V-shapes $V_{1},\ldots,V_{n}$, listed from top to bottom. For each $1\leq
i\leq n-1$, we will assign the color red to the boundary between $V_{i}$ and
$V_{i+1}$; we call this boundary a _red border_. See Figure 11 for an example
with $n=3$. Let us label the boxes in $\lambda_{n}$ with the letters A, B, C
so that the label of a box is determined by reading the real part of its
leftmost point modulo $3$, with A, B, C corresponding to $0$, $1$, $2$,
respectively.

Figure 11. The Young diagram of the partition $\lambda_{3}$ breaks up into
three V-shaped regions $V_{1},V_{2},V_{3}$, which are separated by red
borders. We have drawn the green parallelograms and labeled each box with A,
B, or C.

######  Lemma 6.1.

If $\mathcal{T}$ is a $3$-ribbon tiling of $\lambda_{n}$, then no bones in
$\mathcal{T}$ cross any red borders. If $v$ is a valley stone in $\mathcal{T}$
that crosses a red border, then there is exactly one box in $v$ lying above
that red border, and this one box has label $\mathrm{C}$. If $w$ is a mountain
stone in $\mathcal{T}$ that crosses a red border, then there is exactly one
box in $w$ lying below that red border, and this one box has label
$\mathrm{C}$.

###### Proof.

By Lemma 4.1, every tile must intersect a unique green parallelogram.
Furthermore, Remark 4.2 tells us that each tile must intersect its green
parallelogram with the correct slope. This immediately implies that no bone
may cross a red border.

Now suppose there is a valley stone $v$ that crosses one of the red borders.
Because this stone cannot intersect two green parallelograms, its center box
must have label A or B. If the label is A, then the center box must appear to
the right of the imaginary axis since, otherwise, the valley stone would cross
the green parallelogram with the wrong slope (contradicting Remark 4.2).
Similarly, if the center box has label B, then it must appear to the left of
the imaginary axis. In either case, there is exactly one box in $v$ lying
above the red border that $v$ crosses, and this one box must have label C.

A completely analogous argument proves the last statement of the lemma
concerning the mountain stone $w$, except for one new possibility that must be
considered: a mountain stone whose top cell is the bottom cell of one of the
V-shapes. However, this possibility can be eliminated by noting that such a
mountain stone intersects two green parallelograms, one on each side,
violating Lemma 4.1. ∎

######  Proposition 6.2.

The number of mountainless tilings of $\lambda_{n}$ is $(2n)!!$.

###### Proof.

Consider a mountainless tiling of $\lambda_{n}$. We claim that every tile must
be contained in a single V-shaped region $V_{m}$; in other words, no tile can
cross a red border. To see this, suppose by way of contradiction that our
mountainless tiling contains at least one tile that crosses a red border. Let
$t$ be such that the highest red border that is crossed is the one between
$V_{t}$ and $V_{t+1}$, and let $X=V_{1}\cup\cdots\cup V_{t}$. According to
Lemma 6.1, the only stones that cross the red border between $V_{t}$ and
$V_{t+1}$ are valley stones. Furthermore, each such valley stone uses exactly
one box from $X$, and each such box is labeled C. However, each tile lying
entirely within $X$ has exactly one box labeled A, one box labeled B, and one
box labeled C. This is a contradiction because one can readily check that
there are equal numbers of boxes labeled A, B, and C lying in $X$.

We conclude the proof by showing that $V_{m}$ has $2m$ mountainless tilings.
We may tile $V_{m}$ by placing a single valley stone as far left as possible
(e.g., the leftmost blue silhouette in Figure 12), and then tiling the
remaining region with bones. This implies the Conway–Lagarias invariant of the
region $V_{m}$ is $-1$, so any mountainless tiling of $V_{m}$ must use exactly
one valley stone. It is straightforward to see that there are $2m$ positions
for this one valley stone (those centered at a cell labeled B on the left half
or at a cell labeled A on the right half) that admit a tiling of the remaining
region with bones; moreover, each of these positions for the valley stone
gives rise to a unique mountainless tiling. For example, the 6 positions for
the valley stone in $V_{3}$ are shown in Figure 12. Placing a valley stone in
any other position would split $V_{m}$ into two parts such that the number of
cells in each part is not divisible by $3$, so it would not yield a tiling.
Thus, there are $2m$ mountainless tilings of $V_{m}$, which implies (since no
tile can cross a red border) that $\lambda$ has $(2n)!!$ mountainless tilings.
∎

Figure 12. There are $6$ positions where a valley stone could appear in a
mountainless tiling of $V_{3}$ (two of these positions overlap).

Recall that we are assuming $2\leq a\leq b\leq 2a$ and $a+b\equiv 0\pmod{3}$;
say $a+b=3(N+1)$ with $N\geq 1$. Let $B_{a,b}$ be the region obtained by
transferring the $(a,b)$-benzel into the square grid as in Section 2. We can
embed $B_{a,b}$ inside of the Young diagram of $\lambda_{N}$. More precisely,
it is not hard to see that there are exactly three boxes of maximal height in
$B_{a,b}$, corresponding to the three rightmost cells of the $(a,b)$-benzel,
and there are exactly three boxes of maximal height in $\lambda_{N}$. There is
a unique way to embed $B_{a,b}$ into $\lambda_{N}$ so that the three maximal-
height boxes in $B_{a,b}$ coincide with those of $\lambda_{N}$. Figures 13 and
14 illustrate this for $(a,b)=(7,8)$ and $(a,b)=(8,10)$, respectively; in each
of these figures, the embedded image of $B_{a,b}$ is colored in blue in the
diagram on the right.

Figure 13. The $(7,8)$-benzel (left) and its embedded image $B_{7,8}$ in
$\lambda_{4}$ (right), shown in blue. There are $L=2$ columns of orange valley
stones on the left side of $\lambda_{4}$ and $R=1$ column of pink valley
stones on the right side of $\lambda_{4}$. The region obtained by removing
$B_{7,8}$ and the orange and pink valley stones is the Young diagram of
$\theta_{7,8}$, which is shaded gray.

We can divide the region $\lambda_{N}\setminus B_{a,b}$ into three pieces. The
first piece is a region to the left of $B_{a,b}$ that can be tiled by
$\binom{L+1}{2}$ valley stones, where $L=\frac{2b-a}{3}-1$; these valley
stones (shown in orange in Figures 13 and 14) are arranged in columns of sizes
$1,2,\ldots,L$, from left to right. The second piece is a region to the right
of $B_{a,b}$ that can be tiled by $\binom{R+1}{2}$ valley stones, where
$R=\frac{2a-b}{3}-1$; these valley stones (shown in pink in Figures 13 and 14)
are arranged in columns of sizes $R,R-1,\ldots,1$, from left to right. The
third piece is the Young diagram of a partition $\theta_{a,b}$ (shaded in
Figures 13 and 14). It is most convenient to describe the partition
$\theta_{a,b}$ via its abacus word, which is

(2)
$w_{\theta_{a,b}}=\cdots\bullet\bullet\bullet(\circ\bullet\bullet)^{L}\bullet(\circ\bullet\bullet)^{R}(\circ\circ\bullet)^{L}\circ(\circ\circ\bullet)^{R}\circ\circ\circ\cdots.$

Recall that the _Durfee square_ of a partition $\mu$ is the $s\times s$ square
Young diagram inside of $\mu$, where $s$ is the largest integer such that
$\mu$ has at least $s$ parts of size at least $s$. We call $s$ the _size_ of
the Durfee square. The following lemma, which we record for later use, can be
read off immediately from the abacus word in (2) (using the fact that
$L+R=N-1$).

######  Lemma 6.3.

Let $a$ and $b$ be integers with $2\leq a\leq b\leq 2a$ and $a+b\equiv
0\pmod{3}$. The Durfee square of $\theta_{a,b}$ has size $N-1$, where
$N=\frac{a+b}{3}-1$.

Our motivation for embedding $B_{a,b}$ into $\lambda_{N}$ comes from the
following lemma, which states that there exists a $3$-ribbon tiling of
$\lambda_{N}\setminus B_{a,b}$. This will allow us to extend $3$-ribbon
tilings of $B_{a,b}$ to $3$-ribbon tilings of $\lambda_{N}$, which can then be
analyzed via the abacus bijection and the remove-green bijection
$\operatorname{\mathsf{Compress}}$.

######  Lemma 6.4.

Let $a$ and $b$ be integers with $2\leq a\leq b\leq 2a$ and $a+b\equiv
0\pmod{3}$. Let $N=\frac{a+b}{3}-1$. As above, let $B_{a,b}$ be the embedded
image of the $(a,b)$-benzel inside of $\lambda_{N}$. There exists a $3$-ribbon
tiling of the region $\lambda_{N}\setminus B_{a,b}$.

###### Proof.

Let $L=\frac{2b-a}{3}-1$ and $R=\frac{2a-b}{3}-1$, and note that $N=L+R+1$. As
above, we can divide $\lambda_{N}\setminus B_{a,b}$ into three pieces, two of
which we know can be tiled using valley stones. Thus, it suffices to show that
there exists a $3$-ribbon tiling of the remaining piece $\theta_{a,b}$. By
Corollary 3.3, we just need to check that
$|\theta_{a,b}^{(0)}|+|\theta_{a,b}^{(1)}|+|\theta_{a,b}^{(2)}|=\frac{1}{3}|\theta_{a,b}|$,
where $(\theta_{a,b}^{(0)},\theta_{a,b}^{(1)},\theta_{a,b}^{(2)})$ is the
$3$-quotient of $\theta_{a,b}$. Alternatively, we could appeal to Remark 3.4
and show that the $3$-charges of $\theta_{a,b}$ are all zero. From the abacus
word $w_{\theta_{a,b}}$ given in (2), we can compute the abacus words of the
partitions in the $3$-quotient of $\theta_{a,b}$:

$\displaystyle w_{\theta_{a,b}}^{(0)}$
$\displaystyle=\cdots\bullet\bullet\bullet\circ^{L}\bullet^{N}\circ\circ\circ\cdots,$
$\displaystyle w_{\theta_{a,b}}^{(1)}$
$\displaystyle=\cdots\bullet\bullet\bullet\circ^{N}\bullet^{R}\circ\circ\circ\cdots,$
$\displaystyle w_{\theta_{a,b}}^{(2)}$
$\displaystyle=\cdots\bullet\bullet\bullet\bullet^{L}\circ^{N}\circ\circ\circ\cdots=\cdots\bullet\bullet\bullet\circ\circ\circ\cdots.$

It is now straightforward to check that the $3$-charges of $\theta_{a,b}$ are
all $0$, which completes the proof.

For completeness, let us also outline an argument that uses Corollary 3.3. We
can see directly from the three abacus words listed above that the partitions
in the $3$-quotient of $\theta_{a,b}$ are rectangles and that the number of
boxes in each of these partitions is given by $|\theta_{a,b}^{(0)}|=LN$,
$|\theta_{a,b}^{(1)}|=RN$, and $|\theta_{a,b}^{(2)}|=0$. Because $\lambda_{N}$
has vanishing $3$-charges and $3$-quotient
$(\square_{N},\emptyset,\square_{N})$, it follows from Corollary 3.3 that
$|\lambda_{N}|=3(|\square_{N}|+|\emptyset|+|\square_{N}|)=6N^{2}$. We know by
(1) that $|B_{a,b}|=\frac{-a^{2}+4ab-b^{2}-a-b}{2}$. There are
$\binom{L+1}{2}$ orange valley stones in the left piece of $\lambda_{N}$, and
there are $\binom{R+1}{2}$ pink valley stones in the right piece. Thus,

$|\theta_{a,b}|=6N^{2}-\frac{-a^{2}+4ab-b^{2}-a-b}{2}-3\binom{L+1}{2}-3\binom{R+1}{2};$

one can verify that this equals $3LN+3RN$. Thus,
$|\theta_{a,b}^{(0)}|+|\theta_{a,b}^{(1)}|+|\theta_{a,b}^{(2)}|=LN+RN+0=\frac{1}{3}|\theta_{a,b}|$.
∎

Figure 14. The $(8,10)$-benzel (left) and its embedded image $B_{8,10}$ in
$\lambda_{5}$ (right), shown in blue. There are $L=3$ columns of orange valley
stones on the left side of $\lambda_{5}$ and $R=1$ column of pink valley
stones on the right side of $\lambda_{5}$. The region obtained by removing
$B_{8,10}$ and the orange and pink valley stones is the Young diagram of
$\theta_{8,10}$, which is shaded gray. The V-shaped region $V_{N-L+1}=V_{3}$
is shaded in light blue. There are exactly $4$ boxes in $V_{3}$ that lie
outside of $B_{8,10}$, and they are marked with their labels A, B, C, and C.

We can now prove another piece of Theorem 1.2.

######  Proposition 6.5.

Let $a$ and $b$ be integers with $2\leq a\leq b\leq 2a$, $a+b\equiv
0\pmod{3}$, and $b-a\geq 2$. There are no mountainless tilings of $B_{a,b}$.

###### Proof.

As above, let us embed $B_{a,b}$ into $\lambda_{N}$, where
$N=\frac{a+b}{3}-1$. We will need to consider the labeling of the boxes in
$\lambda_{N}$ with the letters A, B, and C as before. Let $L=\frac{2b-a}{3}-1$
and $R=\frac{2a-b}{3}-1$, and consider the $L$ columns of orange valley stones
and the $R$ columns of pink valley stones as before.

The smallest positive integer $m$ such that $V_{m}$ intersects an orange
valley stone is $N-L+1$, and the smallest positive integer $m$ such that
$V_{m}$ intersects a pink valley stone is $N-R+1$. The condition $a<b$ implies
that $L>R$. Therefore, $V_{N-L+1}$ intersects some of the orange valley
stones, but it does not intersect any of the pink valley stones. More
precisely, there are exactly $4$ boxes that belong to $V_{N-L+1}$ and orange
valley stones; their labels are A, B, C, and C. See Figure 14 for an example
with $a=8$ and $b=10$. We claim that all of the boxes in $V_{N-L+1}$ other
than these $4$ are contained in $B_{a,b}$. This is equivalent to the claim
that the lowest box in $V_{N-L+1}$ is not in the Durfee square of
$\theta_{a,b}$. We know by Lemma 6.3 that the Durfee square of $\theta_{a,b}$
has size $N-1$, so the number of boxes in $\lambda_{N}$ that are centered on
the imaginary axis and lie above the Durfee square of $\theta_{a,b}$ is
$2N-(N-1)=N+1$. Hence, our claim is equivalent to the inequality $2(N-L+1)\leq
N+1$, which is equivalent to our hypothesis that $b-a\geq 2$.

Suppose by way of contradiction that $\mathcal{T}$ is a mountainless tiling of
$B_{a,b}$. It follows from the claim established in the previous paragraph
that $B_{a,b}\cap(V_{1}\cup\cdots\cup V_{N-L+1})$ has strictly fewer boxes
labeled C than boxes labeled A. Since each $3$-ribbon tile uses exactly one
box of each label, this implies that there must be a tile $v$ in $\mathcal{T}$
that has a box labeled A in $B_{a,b}\cap(V_{1}\cup\cdots\cup V_{N-L+1})$ and
has a box labeled C in $B_{a,b}\setminus(V_{1}\cup\cdots\cup V_{N-L+1})$.
Lemma 6.4 tells us that there exists a $3$-ribbon tiling of
$\lambda_{N}\setminus B_{a,b}$, so we can extend $\mathcal{T}$ to a (not
necessarily mountainless) $3$-ribbon tiling $\mathcal{T}^{*}$ of
$\lambda_{N}$. Then $v$ is a tile in $\mathcal{T}^{*}$ that crosses the red
border between $V_{N-L+1}$ and $V_{N-L+2}$, and it is not a mountain stone
because it belongs to the mountainless tiling $\mathcal{T}$. Applying Lemma
6.1 to the tiling $\mathcal{T}^{*}$, we find that $v$ must be a valley stone
that has a box labeled C in $V_{N-L+1}$. This is a contradiction because the
only box labeled C in $v$ is in $B_{a,b}\setminus(V_{1}\cup\cdots\cup
V_{N-L+1})$. ∎

We are now left to consider the cases $(a,b)=(3n,3n)$ and $(a,b)=(3n+1,3n+2)$.
Define $F_{n}$ to be the region in the square grid (considered modulo
translation) obtained by placing $n-1$ columns of negative bones next to each
other, where all bones in the same column have leftmost points with the same
real part and the $m$-th column counted from the left has $n-m$ bones. For
example, Figure 15 shows the regions $F_{2},F_{3},F_{4}$, and $F_{5}$.

Figure 15. The shapes $F_{2},F_{3},F_{4},F_{5}$, listed from left to right.

######  Lemma 6.6.

The region $F_{n}$ has a unique $3$-ribbon tiling, and this tiling only uses
negative bones.

###### Proof.

The lemma is trivial when $n=1$ since $F_{1}$ is empty, and it is also obvious
when $n=2$. Therefore, we may assume $n\geq 3$ and proceed by induction on
$n$. Let $b_{1},\ldots,b_{n-1}$ be the leftmost boxes in $F_{n}$, listed from
top to bottom. Suppose we wish to construct a $3$-ribbon tiling of $F_{n}$. It
is clear that we must use a negative bone whose leftmost box is $b_{1}$. Once
we place that bone in the tiling, we readily see that we must use another
negative bone whose leftmost box is $b_{2}$. Continuing in this fashion, we
see that each box $b_{j}$ must be the leftmost box in a negative bone. Once we
have placed these $n-1$ negative bones into our tiling, the remaining region
has the same shape as $F_{n-1}$, so the proof follows by induction. ∎

In what follows, we slightly abuse notation and write $\lambda_{n}$ for the
Young diagram of the partition $\lambda_{n}$ considered modulo translation
(i.e., we allow the lowest point of $\lambda_{n}$ to be a point other than
$0$).

######  Proposition 6.7.

There are exactly $(2n)!!$ mountainless tilings of $B_{3n,3n}$.

###### Proof.

As above, we can embed $B_{3n,3n}$ inside of $\lambda_{N}$, where $N=2n-1$. We
consider the partition of $\lambda_{N}$ into the V-shapes
$V_{1},\ldots,V_{N}$, which are separated by red borders. The region
$B_{3n,3n}$ breaks into three pieces as follows. The first piece is
$\lambda_{n}=V_{1}\cup\cdots\cup V_{n}$. The second piece is the part of
$B_{3n,3n}\setminus\lambda_{n}$ lying to the left of the imaginary axis; it
has the same shape as the region $F_{n}$ described above, so we refer to it as
$F_{n}$. The third piece is $F_{n}^{\prime}$, which is the part of
$B_{3n,3n}\setminus\lambda_{n}$ lying to the right of the imaginary axis; it
is obtained by reflecting $F_{n}$ across the imaginary axis. Note that $F_{n}$
and $F_{n}^{\prime}$ are not connected to each other. See Figure 16 for an
example with $n=3$.

Suppose $\mathcal{T}$ is a mountainless tiling of $B_{3n,3n}$. Lemma 6.4 tells
us that there exists a $3$-ribbon tiling of $\lambda_{N}\setminus B_{3n,3n}$,
so we can extend $\mathcal{T}$ to a (not necessarily mountainless) $3$-ribbon
tiling $\mathcal{T}^{*}$ of $\lambda_{N}$. Suppose there is a tile $v$ in
$\mathcal{T}^{*}$ that crosses the red border between $V_{n}$ and $V_{n+1}$.
It follows from the construction of $\mathcal{T}^{*}$ that $v$ must be a tile
in $\mathcal{T}$. Hence, $v$ is not a mountain stone. Applying Lemma 6.1 to
the whole tiling $\mathcal{T}^{*}$ of $\lambda_{N}$, we find that $v$ must be
a valley stone that has exactly one box in $\lambda_{n}$; moreover, this one
box has label C.

We have shown that every tile in $\mathcal{T}^{*}$ that crosses the red border
between $V_{n}$ and $V_{n+1}$ must have exactly one box in $\lambda_{n}$ and
that that box must have label C. However, there are equal numbers of boxes
labeled A, B, and C lying in $\lambda_{n}$, and we know that every tile uses
exactly one box of each label. It follows that there cannot be any tiles in
$\mathcal{T}^{*}$ that cross the red border between $V_{n}$ and $V_{n+1}$.
Thus, the tiling $\mathcal{T}$ can be decomposed into a mountainless tiling of
$\lambda_{n}$, a mountainless tiling of $F_{n}$, and a mountainless tiling of
$F_{n}^{\prime}$. Lemma 6.6 tells us that $F_{n}$ has a unique mountainless
tiling, and, by symmetry, it follows immediately from the same lemma that
$F_{n}^{\prime}$ also has a unique mountainless tiling. This shows that the
number of mountainless tilings of $B_{3n,3n}$ is the same as the number of
mountainless tilings of $\lambda_{n}$, which we know is $(2n)!!$ by
Proposition 6.2. ∎

Figure 16. The region $B_{9,9}$, which we have embedded into $\lambda_{5}$,
breaks into three pieces. The first piece is $\lambda_{3}$, which is shaded
light blue. The other two pieces are $F_{3}$ and $F_{3}^{\prime}$, which are
shaded yellow.

We next want to prove $B_{3n+1,3n+2}$ also has $(2n)!!$ mountainless tilings.
In order to do so, we replace Lemma 6.6 with the following lemma.

######  Lemma 6.8.

The region $B_{3n+1,3n+2}\setminus\lambda_{n}$ has a unique mountainless
tiling, and this tiling only uses bones.

###### Proof.

Rather than describe the shape of $B_{3n+1,3n+2}\setminus\lambda_{n}$ in
words, we simply draw it for $n=1,2,3$ in Figure 17, hoping the pattern is
clear. In each of these three cases, we have tiled the region with bones, and
we have numbered the bones. Imagine trying to construct a mountainless tiling
of $B_{3n+1,3n+2}\setminus\lambda_{n}$. It is straightforward to check that we
must use the bone labeled $1$ in such a tiling. More generally, once we have
added the bones numbered $1,\ldots,\ell$ to the tiling, we are forced to add
in the bone numbered $\ell+1$. This shows that the given tiling is the only
mountainless tiling of $B_{3n+1,3n+2}\setminus\lambda_{n}$; we trust the
reader to see how this argument extends to any choice of a positive integer
$n$. ∎

Figure 17. A tiling of $B_{3n+1,3n+2}\setminus\lambda_{n}$ for $n=1$ (left),
$n=2$ (middle), and $n=3$ (right). In each tiling, the bones are numbered in
such a way that the bone numbered $\ell+1$ is forced to be added to the tiling
once the bones numbered $1,\ldots,\ell$ have already been included.

######  Proposition 6.9.

There are exactly $(2n)!!$ mountainless tilings of $B_{3n+1,3n+2}$.

###### Proof.

Let us transfer the $(3n+1,3n+2)$-benzel into the square grid and embed it as
the region $B_{3n+1,3n+2}$ inside of $\lambda_{N}$, where $N=2n$. We consider
the partition of $\lambda_{N}$ into the $V$-shapes $V_{1},\ldots,V_{N}$, which
are separated by red borders.

Suppose $\mathcal{T}$ is a mountainless tiling of $B_{3n+1,3n+2}$. Lemma 6.4
tells us that there exists a $3$-ribbon tiling of $\lambda_{N}\setminus
B_{3n+1,3n+2}$, so we can extend $\mathcal{T}$ to a (not necessarily
mountainless) $3$-ribbon tiling $\mathcal{T}^{*}$ of $\lambda_{N}$. As in the
proof of Proposition 6.7, one can show that there cannot be any tiles in
$\mathcal{T}^{*}$ that cross the red border between $V_{n}$ and $V_{n+1}$.
Thus, the tiling $\mathcal{T}$ can be decomposed into a mountainless tiling of
$\lambda_{n}$ and a mountainless tiling of
$B_{3n+1,3n+2}\setminus\lambda_{n}$. Lemma 6.8 tells us that
$B_{3n+1,3n+2}\setminus\lambda_{n}$ has a unique mountainless tiling. Hence,
the number of mountainless tilings of $B_{3n+1,3n+2}$ is the same as the
number of mountainless tilings of $\lambda_{n}$, which we know is $(2n)!!$ by
Proposition 6.2. ∎

###### Proof of Theorem 1.2.

When $a+b\not\equiv 0\pmod{3}$, the proof follows from the discussion at the
beginning of this section. When $a+b\equiv 0\pmod{3}$, tilings of the
$(a,b)$-benzel using left stones, rising bones, and falling bones correspond
to mountainless tilings of $B_{a,b}$, so the proof follows from Propositions
6.5, 6.7 and 6.9. ∎

##  7\. Two Bones and the Right Stone

We now wish to enumerate tilings of the $(a,b)$-benzel using two types of
bones along with the right stone. Without loss of generality, we may assume
the vertical bone is forbidden. Thus, our goal is to prove Theorem 1.3, which
provides the enumeration of tilings using rising bones, falling bones, and
right stones whenever $a+b\not\equiv 2\pmod{3}$.

###### Proof of Theorem 1.3.

Suppose $2\leq a\leq b\leq 2a$. If $b=2a$ or $a+b\equiv 1\pmod{3}$, then we
know by Remark 5.1 or Theorem 1.1 that the $(a,b)$-benzel has a unique stones-
and-bones tiling and that this tiling consists entirely of right stones.
Hence, we may assume in what follows that $a+b\equiv 0\pmod{3}$ and $b<2a$.

As in Section 6, let us transfer the $(a,b)$-benzel into the square grid and
embed it as the region $B_{a,b}$ inside of $\lambda_{N}$, where
$N=\frac{a+b}{3}-1$. Recall that we think of $\lambda_{N}$ as the union of $N$
V-shaped regions $V_{1},\ldots,V_{N}$. We say a tiling of a region in the
square grid is _valleyless_ if it consists of only mountain stones, positive
bones, and negative bones. Tilings of the $(a,b)$-benzel using right stones,
rising bones, and falling bones correspond to valleyless tilings of $B_{a,b}$;
our goal is to prove that no such tilings exist.

Suppose by way of contradiction that $\mathcal{T}$ is a valleyless tiling of
$B_{a,b}$. Let $c$ be the number of tiles in $T$ that cross the red border
between $V_{1}$ and $V_{2}$. According to Lemma 6.1, each such tile must be a
mountain stone that uses exactly $2$ boxes from $V_{1}$. Thus, the number of
boxes in $V_{1}$ that belong to tiles that do not cross a red border is
$6-2c$; this must be a multiple of $3$, so $c\in\\{0,3\\}$. Upon inspecting
the shape of $V_{1}$, we immediately see that $c$ cannot be $3$. Hence, there
are no tiles in $T$ that cross the red border between $V_{1}$ and $V_{2}$. It
follows that we can restrict $\mathcal{T}$ to a valleyless tiling of $V_{1}$,
which is our desired contradiction because there are no valleyless tilings of
$V_{1}$. ∎

## Acknowledgements

Colin Defant was supported by the National Science Foundation under Award No.
2201907 and by a Benjamin Peirce Fellowship at Harvard University. James Propp
and Ben Young were supported by Simons Foundation Collaboration Grants. Defant
and Rupert Li conducted this research, in part, while visiting the University
of Minnesota Duluth REU in 2022 with support from Jane Street Capital; we
thank Joe Gallian for providing this wonderful opportunity. We also thank the
organizers of the Open Problems in Algebraic Combinatorics 2022 conference,
where much of this work was launched. Lastly, we thank Igor Pak for his
helpful comments.

## References

  * [1] J. H. Conway and J. C. Lagarias, _Tiling with polyominoes and combinatorial group theory_ , J. Combin. Theory Ser. A 53 (1990), no. 2, 183–208. MR1041445 doi:10.1016/0097-3165(90)90057-4
  * [2] S. Fomin and D. Stanton, _Rim hook lattices_ , Algebra i Analiz 9 (1997), no. 5, 140–150. MR1604377
  * [3] G. James and A. Kerber, _The representation theory of the symmetric group_ , Encyclopedia of Mathematics and its Applications, vol. 16, Cambridge University Press, 1984. MR644144 doi:10.1017/CBO9781107340732
  * [4] J. C. Lagarias and D. S. Romano, _A polyomino tiling problem of Thurston and its configurational entropy_ , J. Combin. Theory Ser. A 63 (1993), no. 2, 338–358. MR1223689 doi:10.1016/0097-3165(93)90065-G
  * [5] I. Pak, _Ribbon tile invariants_ , Trans. Amer. Math. Soc. 352 (2000), no. 12, 5525–5561. MR1781275 doi:10.1090/S0002-9947-00-02666-0
  * [6] J. Propp, _Enumeration of tilings_ , https://faculty.uml.edu/jpropp/eot.pdf, published in Handbook of Enumerative Combinatorics, M. Bona, editor, CRC Press, 2015. MR3408702
  * [7] J. Propp, _A pentagonal number theorem for tribone tilings_ , arXiv:2206.04223 [math.CO] (2022).
  * [8] by same author, _Trimer covers in the triangular grid: twenty mostly open problems_ , arXiv:2206.06472 [math.CO] (2022).
  * [9] G. de B. Robinson, _Representation theory of the symmetric group_ , Mathematical Expositions, No. 12, University of Toronto Press, Toronto, 1961. MR0125885
  * [10] D. W. Stanton and D. E. White, _A Schensted algorithm for rim hook tableaux_ , J. Combin. Theory Ser. A 40 (1985), no. 2, 211--247. MR814412 doi:10.1016/0097-3165(85)90088-3
  * [11] W. P. Thurston, _Conway’s tiling groups_ , Amer. Math. Monthly 97 (1990), no. 8, 757--773. MR1072815 doi:10.2307/2324578
  * [12] A. Verberkmoes and B. Nienhuis, _Bethe ansatz solution of triangular trimers on the triangular lattice_ , Phys. Rev. E 63 (2001), no. 6, 066122\. doi:10.1103/PhysRevE.63.066122