# A pentagonal number theorem for tribone tilings

Jesse Kim
Department of Mathematics
University of California, San Diego
California, U.S.A.
<EMAIL_ADDRESS>
James Propp
Department of Mathematical Sciences
University of Massachusetts, Lowell
Massachusetts, U.S.A.
<EMAIL_ADDRESS>Supported by a Simons Collaboration Grant.

###### Abstract

Conway and Lagarias showed that certain roughly triangular regions in the
hexagonal grid cannot be tiled by shapes Thurston later dubbed tribones. Here
we study a two-parameter family of roughly hexagonal regions in the hexagonal
grid and show that a tiling by tribones exists if and only if the two
parameters associated with the region are the paired pentagonal numbers
$k(3k\pm 1)/2$.

##  1 Introduction

Figure 1: A benzel tiled by (tri)bones.

This article treats tilings such as the one shown in Figure 1 in which the
tiles are tribones consisting of three successive unit hexagons with collinear
centers. (Here we use “tribone” in the sense introduced by Thurston [6]; we
will often call such tiles simply bones.) These tilings were studied by Conway
and Lagarias who showed that for all integers $N\geqslant 1$, the roughly
triangular region $T_{N}$ exemplified for $N=6$ in Figure 2 cannot be tiled by
tribones (which they called $L_{3}$’s). In Figure 2 the region $T_{6}$ is
tiled by a mixture of tribones and additional tiles Conway and Lagarias called
$T_{2}$’s (which we will call stones), where three of the stones point to the
right and one points to the left. Conway and Lagarias showed that the number
of right-pointing stones minus the number of left-pointing stones is the same
for all tilings of a simply-connected region. Since this quantity (a rescaled
version of the Conway-Lagarias invariant) is nonzero for the particular tiling
of $T_{6}$ shown in Figure 2, it must be nonzero for all tilings of $T_{6}$,
and in particular, there can be no pure tribone tilings. Similar arguments
show that $T_{N}$ cannot be tiled by tribones for any positive value of $N$.

Figure 2: The honeycomb triangle $T_{6}$ tiled by stones and bones.

(In the preceding paragraph and throughout this paper we conflate the subtly
different perspectives of Conway and Lagarias [2] and Thurston [6] while
introducing some notions of our own; readers who want a fine-grained
understanding of the lineage of the ideas in this paper should read those two
articles, as well as the earlier article [4].)

Here we study a two-parameter family of regions introduced in [5] called
benzels; for instance, Figure 1 shows the (5,7)-benzel. A variant of Gauss’
shoelace formula allows one to compute the signed area (aka algebraic area)
enclosed by a closed polygonal path, and by “twisting” the formula in the
manner prescribed by the work of Conway, Lagarias, and Thurston one can
compute the values of the Conway-Lagarias invariant for all benzels. It is
then simple to determine the values of the parameters that cause the Conway-
Lagarias invariant to vanish. It emerges that the $(a,b)$-benzel can be tiled
by tribones if and only if $a$ and $b$ equal $k(3k\pm 1)/2$, that is, when $a$
and $b$ are paired pentagonal numbers. (In referring to this result as a
“pentagonal number theorem” we are hearkening back to Euler’s famed theorem
about the series expansion of $\prod_{n=1}^{\infty}(1-q^{n})$, but we know of
no connection between the two theorems.) It then remains to show that when $a$
and $b$ are of this form, at least one tribone tiling exists; several students
at the Open Problems in Algebraic Combinatorics 2022 meeting found solutions
to this challenge, including the first author.

Section 2 defines benzels, which are parametrized by integer pairs $a,b$
satisfying $2\leqslant a\leqslant 2b$ and $2\leqslant b\leqslant 2a$. In
section 3 we express the boundary of the $(a,b)$-benzel via a concatenation of
unit vectors and compute the signed area enclosed by the boundary path
(Theorem 1). Section 4 presents a reformulation of the Conway-Lagarias
invariant as the signed area enclosed by the “shadow path” associated with the
boundary of the region being tiled. Section 5 combines the results of the
three preceding sections to obtain a formula for the value of the Conway-
Lagarias invariant (Theorem 2).

Section 6 analyzes the polynomials computed in section 5 to show that the
Conway-Lagarias invariant vanishes only when $a=k(3k-1)/2$ and $b=k(3k+1)/2$
or vice versa. Section 7 shows, conversely, that at least one tribone tiling
exists when $a$ and $b$ are of this form. This proves that the algebraic
condition on $a$ and $b$ is both necessary and sufficient for the existence of
tilings of the $(a,b)$-benzel by bones (Theorem 3).

The companion article [5] discusses other classes of stones-and-bones tilings
of benzels, offering some (mostly conjectural) exact formulas for special
cases in which certain prototiles are permitted/prohibited and $a$ and $b$
take on certain kinds of values. The two articles adopt slightly different
conventions. In that other article, the focus is on the hexagonal cells and it
makes sense to locate the three middle cells of a benzel so that their centers
are at 1, $\omega$, and $\omega^{2}$ in the complex plane. In this article the
vertices and edges along the boundaries of the cells play a leading role so it
makes more sense to have edges joining the vertex 0 to the vertices 1,
$\omega$, and $\omega^{2}$ in the benzel graph dual to the benzel 2-complex.
The reader should have little difficulty translating between the two pictures
using appropriate reflections and sign-changes where needed.

##  2 Stones, bones, and benzels

Conway and Lagarias [2] refer to the first two tiles shown in Figure 3 as
$T_{2}$ and $L_{3}$. Thurston [6] keeps the name $T_{2}$ but calls the $L_{3}$
a tribone. Ardila and Stanley [1] call the $T_{2}$ a tribone and the $L_{3}$ a
trihex. Meanwhile, in the recreational mathematics literature, a trihex
denotes any of the three tiles shown in Figure 3. Curiously, although the
literature on polyforms has given names to the tetrahexes, it has not given
names to the trihexes. We therefore take the liberty of dubbing them stones,
bones, and phones, respectively. (“Bone” is sometimes used as a synonym for
domino but we think this ambiguity is unlikely to cause confusion since the
contexts are very different.) Phones will not play a major role in this
article but it seems sensible to assign coordinated names to all three tiles
for possible future purposes.

Figure 3: The three kinds of trihex: stone, bone, and phone.

We imbed our tilings in the complex plane. Let $\omega=e^{2\pi i/3}$; all
points of interest will belong to the lattice $L=\mathbb{Z}+\mathbb{Z}\omega$.
More specifically, consider the index-three sublattice
$L_{0}=\mathbb{Z}+\mathbb{Z}(1-\omega)$ which together with its translates
$L_{1}=L_{0}+1$ and $L_{-1}=L_{0}-1$ partition $L$; we form a graph whose
vertices are the elements of $L_{0}\cup L_{1}$ with edges connecting two
vertices whenever the distance between them is 1. A point $\alpha$ in $L_{-1}$
is the center of a hexagon with corner points $\alpha\pm 1$,
$\alpha\pm\omega$, and $\alpha\pm\omega^{2}$; a point $\alpha$ in $L_{0}$
borders three hexagons, one of which is to its left; and a point $\alpha$ in
$L_{1}$ borders three hexagons, one of which is to its right. See Figure 4, in
which points are marked 0, $1$, or $-1$ according to whether they belong to
$L_{0}$, $L_{1}$, or $L_{-1}$.

Figure 4: The lattice $L=\mathbb{Z}+\mathbb{Z}\omega$. Figure 5: The
(5,7)-benzel and its bounding hexagon.

For positive integers $a,b$ satisfying $2\leqslant a\leqslant 2b$ and
$2\leqslant b\leqslant 2a$ we define the $(a,b)$-benzel as the union of all
the unit hexagons that lie completely within the large hexagon with vertices
$a\omega+b$, $-a\omega^{2}-b$, $a\omega^{2}+b\omega$, $-a-b\omega$,
$a+b\omega^{2}$, and $-a\omega-b\omega^{2}$, as shown in Figure 5 for $a=5$,
$b=7$. We refer to this hexagon as the bounding hexagon of the benzel. The
bounding hexagon has three sides of length $2a-b$ alternating with three sides
of length $2b-a$. Note that the figure includes three stray edges, or “spurs”,
at $5\omega+7$, $5\omega^{2}+7\omega$, and $5+7\omega^{2}$; although such
edges are not part of the proper boundary of the benzel, it is sometimes
convenient to include them.

We say that a benzel is of class 0, 1, or $-1$ according to the residue of
$a+b$ mod 3. If a benzel is of class $c$, then three of the vertices of the
bounding hexagon ($a\omega+b$, $a\omega^{2}+b\omega$, and $a+b\omega^{2}$)
belong to $L_{c}$ and the other three ($-a\omega^{2}-b$, $-a-b\omega$, and
$-a\omega-b\omega^{2}$) belong to $L_{-c}$. Note that $2a-b\equiv
2b-a\equiv-c$ mod 3.

The $(a,b)$-benzel has 120-degree rotational symmetry. The boundary of a
benzel consists of six stretches that follow the successive sides of the
bounding hexagon. The three panels of Figure 6 show the upper and lower
stretches of the boundary of the $(a,b)$-benzel in the respective cases $c=0$,
$c=1$, and $c=-1$. The upper and lower boundaries of the benzel contain
$(2b-a+c)/3$ and $(2a-b+c)/3$ horizontal edges respectively. The other four
stretches are just rotated copies of those two; the six stretches alternate
between ones that look like the upper stretch and ones that look like the
lower stretch.

Figure 6: The upper and lower stretches of the boundary of a class 0 benzel
with $2b-a=6$ (left), a class 1 benzel with $2b-a=5$ (middle), and a class
$-1$ benzel with $2b-a=7$ (right).

##  3 A Gauss formula for the hexagonal grid

For regions described by boundary-paths made of short segments, it’s helpful
to express the signed area as a double-sum of scalar cross-products of those
segments (where the scalar cross-product
${\rm\bf{v}}_{1}\times{\rm\bf{v}}_{2}$ of ${\rm\bf{v}}_{1}=(x_{1},y_{1})$ and
${\rm\bf{v}}_{2}=(x_{2},y_{2})$ in $\mathbb{R}^{2}$ is defined as
$x_{1}y_{2}-x_{2}y_{1}$). To be specific, suppose we have vectors
${\rm\bf{v}}_{i}$ ($1\leqslant i\leqslant n$) satisfying
$\sum_{i=1}^{n}{\rm\bf{v}}_{i}=\rm\bf{0}$. Then, with
${\rm\bf{w}}_{i}={\rm\bf{v}}_{1}+\dots+{\rm\bf{v}}_{i}$, the signed area
enclosed by the $n$-gon with vertices ${\rm\bf{w}}_{1}$, ${\rm\bf{w}}_{2}$, …,
${\rm\bf{w}}_{n}=\rm\bf{0}$ is
$\frac{1}{2}\sum_{i}{\rm\bf{w}}_{i}\times{\rm\bf{v}}_{i+1}$, which can be
rewritten as $\frac{1}{2}\sum_{1\leqslant i<j\leqslant
n}{\rm\bf{v}}_{i}\times{\rm\bf{v}}_{j}$. One may call this (with slight
historical distortion) a Gauss area formula for the signed area enclosed by
the closed curve.

As an example, consider the square grid with unit vectors ${\rm\bf{i}}=(1,0)$
and ${\rm\bf{j}}=(0,1)$, and consider the $k$-by-$k$ square with vertices
$(0,0)$, $(k,0)$, $(k,k)$, $(0,k)$, and $(0,0)$. The counterclockwise boundary
of the square consists of $k$ occurrences of ${\rm\bf{i}}$, $k$ occurrences of
${\rm\bf{j}}$, $k$ occurrences of $-{\rm\bf{i}}$, and $k$ occurrences of
$-{\rm\bf{j}}$, in that order. If we ignore the terms of the double sum
$\sum_{1\leqslant i<j\leqslant n}{\rm\bf{v}}_{i}\times{\rm\bf{v}}_{j}$ that
vanish because they are of the form ${\rm\bf{v}}\times{\rm\bf{v}}$ or
${\rm\bf{v}}\times-{\rm\bf{v}}$, we obtain $k^{2}$ occurrences of
${\rm\bf{i}}\times{\rm\bf{j}}$, $k^{2}$ occurrences of
${\rm\bf{i}}\times-{\rm\bf{j}}$, $k^{2}$ occurrences of
${\rm\bf{j}}\times-{\rm\bf{i}}$, and $k^{2}$ occurrences of
$-{\rm\bf{i}}\times-{\rm\bf{j}}$. This amounts to $k^{2}$ times $1-1+1+1=2$,
so we find that the signed area is $\frac{1}{2}2k^{2}=k^{2}$, as it should be.
For brevity, we will write the boundary as
$({\rm\bf{i}})^{k}({\rm\bf{j}})^{k}(-{\rm\bf{i}})^{k}(-{\rm\bf{j}})^{k}$,
where superscripts denote string repetition. A more complicated example is the
Aztec diamond of order $k$; the reader may wish to check that applying the
Gauss formula to the string
$({\rm\bf{i}},{\rm\bf{j}})^{k}({\rm\bf{j}},-{\rm\bf{i}})^{k}(-{\rm\bf{i}},-{\rm\bf{j}})^{k}(-{\rm\bf{j}},{\rm\bf{i}})^{k}$
gives the predicted signed area $2k^{2}+2k$. For a general path, the signed
area $\frac{1}{2}\sum_{1\leqslant i<j\leqslant
n}{\rm\bf{v}}_{i}\times{\rm\bf{v}}_{j}$ is equal to the sum, over all square
cells $C$, of the winding number of the path around $C$; although there are
infinitely many cells $C$ in the infinite square grid, the winding number of
any particular finite path around $C$ vanishes for all but finitely many $C$,
so the sum is well-defined.

Figure 7: The six unit vectors.

For the hexagonal grid we will need three unit vectors ${\rm\bf{a}}$,
${\rm\bf{b}}$, and ${\rm\bf{c}}$, pointing from $0$ to $1$, $\omega$ and
$\omega^{2}$, respectively, under our identification of $\mathbb{C}$ with the
plane containing the tiling; see Figure 7. To remove pesky factors of
$\sqrt{3}$, we redefine $\times$ so that
${\rm\bf{a}}\times{\rm\bf{b}}={\rm\bf{b}}\times{\rm\bf{c}}={\rm\bf{c}}\times{\rm\bf{a}}=+1$,
${\rm\bf{b}}\times{\rm\bf{a}}={\rm\bf{c}}\times{\rm\bf{b}}={\rm\bf{a}}\times{\rm\bf{c}}=-1$,
and
${\rm\bf{a}}\times{\rm\bf{a}}={\rm\bf{b}}\times{\rm\bf{b}}={\rm\bf{c}}\times{\rm\bf{c}}=0$
(with other cases being determined by the assumption that $\times$ is
bilinear); this bilinear form is antisymmetric. If each hexagonal cell is
assigned area 1, then the signed area enclosed by a path given by the vectors
${\rm\bf{v}}_{1},\dots,{\rm\bf{v}}_{n}$ (with
${\rm\bf{v}}_{i}\in\\{\pm{\rm\bf{a}},\pm{\rm\bf{b}},\pm{\rm\bf{c}}\\}$ for
$1\leqslant i\leqslant n$) is $\frac{1}{6}\sum_{1\leqslant i<j\leqslant
n}{\rm\bf{v}}_{i}\times{\rm\bf{v}}_{j}$. This is equal to the sum, over all
hexagonal cells $C$, of the winding number of the path around $C$. For
instance, consider the unit hexagon centered at $-1$ as in Figure 8. The
counterclockwise path around the hexagon that starts and ends at the origin
(the rightmost corner of the hexagon) is associated with the sequence of
vectors
${\rm\bf{b}},-{\rm\bf{a}},{\rm\bf{c}},-{\rm\bf{b}},{\rm\bf{a}},-{\rm\bf{c}}$.
The nonvanishing terms of the sum $\sum_{1\leqslant i<j\leqslant
6}{\rm\bf{v}}_{i}\times{\rm\bf{v}}_{j}$ are ${\rm\bf{b}}\times-{\rm\bf{a}}$,
${\rm\bf{b}}\times{\rm\bf{c}}$, ${\rm\bf{b}}\times{\rm\bf{a}}$,
${\rm\bf{b}}\times-{\rm\bf{c}}$, $-{\rm\bf{a}}\times{\rm\bf{c}}$,
$-{\rm\bf{a}}\times-{\rm\bf{b}}$, $-{\rm\bf{a}}\times-{\rm\bf{c}}$,
${\rm\bf{c}}\times-{\rm\bf{b}}$, ${\rm\bf{c}}\times{\rm\bf{a}}$,
$-{\rm\bf{b}}\times{\rm\bf{a}}$, $-{\rm\bf{b}}\times-{\rm\bf{c}}$, and
${\rm\bf{a}}\times-{\rm\bf{c}}$. Using the fact that $\times$ is
antisymmetric, we can simplify the sum, obtaining
$2({\rm\bf{a}}\times{\rm\bf{b}}+{\rm\bf{b}}\times{\rm\bf{c}}+{\rm\bf{c}}\times{\rm\bf{a}})=6$,
so that $\frac{1}{6}\sum_{1\leqslant i<j\leqslant
n}{\rm\bf{v}}_{i}\times{\rm\bf{v}}_{j}=1$, which is indeed the sum of the
winding numbers of the paths around all the hexagonal cells, which is 1 for
the cell centered at $-1$ and 0 for all other cells.

Figure 8: The counterclockwise path enclosing a hexagon.

To save space on the page, we will sometimes write $-{\rm\bf{a}}$,
$-{\rm\bf{b}}$, and $-{\rm\bf{c}}$ as ${\rm\bf{a}}^{\prime}$,
${\rm\bf{b}}^{\prime}$, and ${\rm\bf{c}}^{\prime}$, respectively.

Take $a,b$ with $a+b\equiv 0$ mod 3. The upper stretch of the $(a,b)$-benzel,
starting at the upper right corner of the bounding hexagon and ending at the
upper left corner of the bounding hexagon (see the first panel of Figure 6) is
$({\rm\bf{c}},{\rm\bf{a}}^{\prime},{\rm\bf{b}},{\rm\bf{a}}^{\prime})^{t}$ with
$t=(2b-a)/3$. The lower stretch of the $(a,b)$-benzel, starting at the lower
left corner of the bounding hexagon and ending at the lower right corner of
the bounding hexagon is
$({\rm\bf{a}},{\rm\bf{c}}^{\prime},{\rm\bf{a}},{\rm\bf{b}}^{\prime})^{s}$ with
$s=(2a-b)/3$. Thus the counterclockwise boundary of the benzel, starting and
ending at the rightmost corner of the bounding hexagon, is the concatenation

$({\rm\bf{b}},{\rm\bf{a}}^{\prime},{\rm\bf{b}},{\rm\bf{c}}^{\prime})^{s}({\rm\bf{c}},{\rm\bf{a}}^{\prime},{\rm\bf{b}},{\rm\bf{a}}^{\prime})^{t}({\rm\bf{c}},{\rm\bf{b}}^{\prime},{\rm\bf{c}},{\rm\bf{a}}^{\prime})^{s}({\rm\bf{a}},{\rm\bf{b}}^{\prime},{\rm\bf{c}},{\rm\bf{b}}^{\prime})^{t}({\rm\bf{a}},{\rm\bf{c}}^{\prime},{\rm\bf{a}},{\rm\bf{b}}^{\prime})^{s}({\rm\bf{b}},{\rm\bf{c}}^{\prime},{\rm\bf{a}},{\rm\bf{c}}^{\prime})^{t}$

involving $12s+12t$ unit vectors. One could in principle work out the value of
the double sum $\frac{1}{6}\sum_{1\leqslant i<j\leqslant
12s+12t}{\rm\bf{v}}_{i}\times{\rm\bf{v}}_{j}$, but it is also possible to
determine the general form of the answer and then fit the undetermined
coefficients. Specifically, one can lump together the ${12s+12t\choose 2}$
products ${\rm\bf{v}}_{i}\times{\rm\bf{v}}_{j}$ into ${6\choose 2}\cdot 4\cdot
4+{6\choose 1}\cdot 4\cdot 4=256$ types (according to which of the 24 symbols
in the above expression give rise to ${\rm\bf{v}}_{i}$ and ${\rm\bf{v}}_{j}$
respectively) such that the number of terms
${\rm\bf{v}}_{i}\times{\rm\bf{v}}_{j}$ of each type is a quadratic function of
$s$ and $t$. This implies that the area of the benzel is itself a quadratic
function of $s$ and $t$, and hence a quadratic function of $a$ and $b$. This
quadratic function has six undetermined coefficients that can be determined by
finding the area of half a dozen specific benzels and solving the resulting
system of linear equations. One finds in this way that the area of a class 0
benzel is equal to $(-a^{2}+4ab-b^{2}-a-b)/2$ (a positive integer divisible by
3).

One can analyze benzels of class 1 and class $-1$ in a similar way. For both
of these cases, one can augment the boundary path by spurs that join the path
to one of the corners of the bounding hexagon; these spurs are traversed in
one direction and then immediately traversed in the opposite direction, so
they enclose no cells and make no contribution to the area.

Theorem 1: Let $R$ be a benzel of class $c$. Then the area of $R$ is

$\left\\{\begin{array}[]{ll}(-a^{2}+4ab-b^{2}-a-b)/2&\mbox{if $c$ is 0 or
$-1$},\\\ (-a^{2}+4ab-b^{2}-a-b+2)/2&\mbox{if $c$ is 1.}\end{array}\right.$

##  4 Shadow words and the Conway-Lagarias invariant

Conway and Lagarias show that in any simply-connected region $R$ that can be
tiled by stones and bones, the number of right-pointing stones minus the
number of left-pointing stones depends only on the region being tiled, not the
particular tiling. We call this the rescaled Conway-Thurston invariant and
denote it by $i(R)$. Thurston shows how to compute this invariant using the
Cayley graphs of the two groups

$A=\langle a,b,c\>|\>a^{2}=b^{2}=c^{2}=(abc)^{2}=1\rangle$

and

$T_{0}=\langle
a,b,c\>|\>a^{2}=b^{2}=c^{2}=(ab)^{3}=(bc)^{3}=(ca)^{3}=1\rangle$

(where the formal involutions $a$, $b$, and $c$ are not to be confused with
the integers $a,b,c$ introduced in section 2 but are closely related to the
lax vectors $\pm{\rm\bf{a}}$, $\pm{\rm\bf{b}}$, and $\pm{\rm\bf{c}}$ with
${\rm\bf{a}}$, ${\rm\bf{b}}$, and ${\rm\bf{c}}$ as in section 3).
Specifically, one draws $R$ in $\Gamma(A)$, the Cayley graph of $A$, and picks
a basepoint $p$ on the boundary of $R$; one expresses the boundary path that
goes around $R$ once counterclockwise from $p$ to itself as a product of
$a$’s, $b$’s, and $c$’s; one reinterprets the letters of this word as the
generators of the group $T_{0}$; one draws a path in $\Gamma(T_{0})$, the
Cayley graph of $T_{0}$, associated with that word (the new path must in fact
be a closed path, for non-obvious reasons); and lastly one computes the signed
area enclosed by the new path; this signed area, which we call the
(unrescaled) Conway-Lagarias invariant and denote by $I(R)$, is equal to
$3i(R)$.

Figure 9: Weaving versus winding.

Thurston remarks “It is curious that even though the groups $A$ and $T_{0}$
and the labeled graphs $\Gamma(A)$ and $\Gamma(T_{0})$ are different, when the
labels are stripped they become isomorphic.” Specifically, both are isomorphic
to the hexagon graph, and there is a direct pictorial way to convert the old
path into the new without explicit recourse to the groups $A$ and $T_{0}$.
Given a basepoint $p\in L_{0}$ on the boundary of $R$, let $e_{1},\dots,e_{n}$
be the sequence of edges traversing the boundary of $R$ counterclockwise from
$p$ back to itself, and let $e_{0}=e_{n}$ and $e_{n+1}=e_{1}$. Call this path
$\pi$. For $1\leqslant i\leqslant n$, say that $\pi$ weaves at step $i$ if
$e_{i-1}$ and $e_{i+1}$ are parallel and say that $\pi$ winds otherwise, i.e.,
if $e_{i-1}$, $e_{i}$, and $e_{i+1}$ are consecutive edges of some hexagon;
see Figure 9, in which the edges are given orientations and marked with the
associated vectors. Suppose we have another path
$e^{\prime}_{0},\dots,e^{\prime}_{n}$ (call it $\pi^{\prime}$ for short) in
the hexagon graph that starts from a vertex $p^{\prime}\in L_{0}$, and suppose
that $\pi^{\prime}$ shadows $\pi$ for all $1\leqslant i\leqslant n$ in the
sense that $\pi^{\prime}$ weaves at step $i$ if $\pi$ winds at step $i$ and
$\pi^{\prime}$ winds at step $i$ if $\pi$ weaves at step $i$. (This is the
“twist” referred to in section 1.) Then the unrescaled Conway-Lagarias
invariant of $R$ equals the signed area enclosed by $\pi^{\prime}$.

Figure 10: A path and its shadow.

For example, suppose $R$ is the (3,3)-benzel (aka $T_{3}$), as shown in the
left panel of Figure 10, let the basepoint $p$ be its rightmost point, and let
$\pi$ be the path that encircles $R$ counterclockwise from $p$ to itself, with
edges numbered 1 through 18. Let $\pi^{\prime}$ be the path shown in the right
panel of Figure 10, starting and ending at the marked point. It is easy to
check that $\pi^{\prime}$ shadows $\pi$ and that $\pi^{\prime}$ encircles
three hexagons in the counterclockwise direction, enclosing signed area 3.
This is also the total area of all the right-pointing stones minus the total
area of all the left-pointing stones in each of the (three) tilings of the
(3,3)-benzel by stones and bones. Note that in terms of unit vectors, the path
$\pi$ is given by

${\rm\bf{b}},-{\rm\bf{a}},{\rm\bf{b}},-{\rm\bf{a}},{\rm\bf{b}},-{\rm\bf{a}},{\rm\bf{c}},-{\rm\bf{b}},{\rm\bf{c}},-{\rm\bf{b}},{\rm\bf{c}},-{\rm\bf{b}},{\rm\bf{a}},-{\rm\bf{c}},{\rm\bf{a}},-{\rm\bf{c}},{\rm\bf{a}},-{\rm\bf{c}}\>$

while the path $\pi^{\prime}$ is given by

${\rm\bf{b}},-{\rm\bf{a}},{\rm\bf{c}},-{\rm\bf{b}},{\rm\bf{a}},-{\rm\bf{c}},{\rm\bf{a}},-{\rm\bf{c}},{\rm\bf{b}},-{\rm\bf{a}},{\rm\bf{c}},-{\rm\bf{b}},{\rm\bf{c}},-{\rm\bf{b}},{\rm\bf{a}},-{\rm\bf{c}},{\rm\bf{b}},-{\rm\bf{a}};$

such representations of paths as strings of unit vectors will be helpful
shortly.

It should be stressed that $p$ can be any point in $\partial(R)\cap L_{0}$ and
$p^{\prime}$ can be any point in $L_{0}$. Moreover, for fixed $p$ and
$p^{\prime}$, there are six paths $\pi^{\prime}$ that shadow a given $\pi$: if
$e^{\prime}_{1}$ is any of the three edges emanating from $p^{\prime}$
(joining $p^{\prime}$ to some other vertex $q^{\prime}$, say), and
$e^{\prime}_{2}$ is either of the two remaining edges emanating from
$q^{\prime}$, then there is a unique way to choose $e^{\prime}_{3},\dots$ so
that the resulting path shadows $\pi$. Magically, the signed area enclosed by
$\pi^{\prime}$ is independent of all those choices.

It should also be mentioned that if we choose basepoints $p$ and $p^{\prime}$
belonging to $L_{1}$ rather than $L_{0}$ then the signed area enclosed by
$\pi^{\prime}$ will be $-I(R)$ rather than $I(R)$.

We will not go into the details of the proof, except to mention three facts:
the shadow of the counterclockwise boundary of a right-pointing stone
(starting and ending at a boundary point in $L_{0}$) encloses signed area 3;
the shadow of the counterclockwise boundary of a left-pointing stone (starting
and ending at a boundary point in $L_{0}$) encloses signed area $-3$; and the
shadow of the counterclockwise boundary of a bone encloses signed area 0. It
would be possible to re-prove the main facts of [2] and [6] that pertain to
tribone tilings in an elementary fashion (without use of combinatorial group
theory or combinatorial topology) along lines laid out in [3] and [4].

##  5 Computing the shadow-path area

When we apply the ideas of the previous section to compute the Conway-Lagarias
invariant of the $(a,b)$-benzel, an extra complication arises from the spurs
seen in Figures 5and 6. A spur in a path consists of two consecutive
occurrences of the same edge. Our definition of shadow-paths assumed that
$\pi$ was the boundary of a region $R$ and hence did not involve spurs; we
want to allow there to be some $i$’s for which $e_{i}=e_{i+1}$. Say that such
a spur is isolated if the shortened path $e_{i-2},e_{i-1},e_{i+2},e_{i+3}$
obtained by removing $e_{i}$ and $e_{i+1}$ satisfies $e_{i-2}\neq e_{i-1}\neq
e_{i+2}\neq e_{i+3}$. Let us suppose that all spurs in $\pi$ are isolated. We
will say that $\pi^{\prime}$ shadows $\pi$ if $\pi^{\prime}$ has isolated
spurs exactly where $\pi$ does (and no other spurs at all), and if for all $i$
such that $e_{i}=e_{i+1}$ the shortened path
$e^{\prime}_{i-2},e^{\prime}_{i-1},e^{\prime}_{i+2},e^{\prime}_{i+3}$ shadows
$e_{i-2},e_{i-1},e_{i+2},e_{i+3}$.

We can now describe the boundary words of the benzels and their shadows,
expressed as sequences of unit vectors whose spurs are all isolated. The
reader may find it helpful to consult Figures 11, 12, and 13 which show the
(6,6)-benzel, (5,5)-benzel, and (4,4)-benzel and are representative of benzels
of class 0, 1, and $-1$, respectively. In the interest of legibility, we omit
commas in sequences of unit vectors.

For class 0 benzels we choose our basepoint $p\in L_{0}$ to be the rightmost
corner of the bounding hexagon. Let $s=(2a-b)/3$ and $t=(2b-a)/3$. The
boundary word of the $(a,b)$-benzel is

$({\rm\bf{b}}{\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime})^{s}({\rm\bf{c}}{\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{a}}^{\prime})^{t}({\rm\bf{c}}{\rm\bf{b}}^{\prime}{\rm\bf{c}}{\rm\bf{a}}^{\prime})^{s}({\rm\bf{a}}{\rm\bf{b}}^{\prime}{\rm\bf{c}}{\rm\bf{b}}^{\prime})^{t}({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{s}({\rm\bf{b}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{c}}^{\prime})^{t}$

which is shadowed by

$({\rm\bf{a}}{\rm\bf{b}}^{\prime}{\rm\bf{c}}{\rm\bf{b}}^{\prime})^{s}({\rm\bf{b}}{\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime})^{t}({\rm\bf{b}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{c}}^{\prime})^{s}({\rm\bf{c}}{\rm\bf{b}}^{\prime}{\rm\bf{c}}{\rm\bf{a}}^{\prime})^{t}({\rm\bf{c}}{\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{a}}^{\prime})^{s}({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{t}.$

For class 1 benzels we choose our basepoint $p\in L_{1}$ so that $p+1$ is the
rightmost corner of the bounding hexagon. Let $s=(2a-b-2)/3$ and
$t=(2b-a-2)/3$. The boundary word of the $(a,b)$-benzel is

$({\rm\bf{c}}^{\prime}{\rm\bf{b}}{\rm\bf{a}}^{\prime}{\rm\bf{b}})^{s}{\rm\bf{c}}^{\prime}{\rm\bf{b}}({\rm\bf{a}}^{\prime}{\rm\bf{c}}{\rm\bf{a}}^{\prime}{\rm\bf{b}})^{t}{\rm\bf{a}}^{\prime}{\rm\bf{c}}({\rm\bf{a}}^{\prime}{\rm\bf{c}}{\rm\bf{b}}^{\prime}{\rm\bf{c}})^{s}{\rm\bf{a}}^{\prime}{\rm\bf{c}}({\rm\bf{b}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime}{\rm\bf{c}})^{t}{\rm\bf{b}}^{\prime}{\rm\bf{a}}({\rm\bf{b}}^{\prime}{\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}})^{s}{\rm\bf{b}}^{\prime}{\rm\bf{a}}({\rm\bf{c}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime}{\rm\bf{a}})^{t}{\rm\bf{c}}^{\prime}{\rm\bf{b}}$

which is shadowed by

$({\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{a}}^{\prime}{\rm\bf{c}})^{s}{\rm\bf{a}}^{\prime}{\rm\bf{b}}({\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime}{\rm\bf{b}})^{t}{\rm\bf{a}}^{\prime}{\rm\bf{b}}({\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{b}})^{s}{\rm\bf{c}}^{\prime}{\rm\bf{a}}({\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime}{\rm\bf{a}})^{t}{\rm\bf{c}}^{\prime}{\rm\bf{a}}({\rm\bf{b}}^{\prime}{\rm\bf{c}}{\rm\bf{b}}^{\prime}{\rm\bf{a}})^{s}{\rm\bf{b}}^{\prime}{\rm\bf{c}}({\rm\bf{b}}^{\prime}{\rm\bf{c}}{\rm\bf{a}}^{\prime}{\rm\bf{c}})^{t}{\rm\bf{b}}^{\prime}{\rm\bf{c}}.$

For class $-1$ benzels we choose our basepoint $p\in L_{1}$ to be the
rightmost corner of the bounding hexagon. Let $s=(2a-b-1)/3$ and
$t=(2b-a-1)/3$. The boundary word of the $(a,b)$-benzel is

$({\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime}{\rm\bf{b}})^{s}{\rm\bf{a}}^{\prime}({\rm\bf{b}}{\rm\bf{a}}^{\prime}{\rm\bf{c}}{\rm\bf{a}}^{\prime})^{t}{\rm\bf{b}}({\rm\bf{b}}^{\prime}{\rm\bf{c}}{\rm\bf{a}}^{\prime}{\rm\bf{c}})^{s}{\rm\bf{b}}^{\prime}({\rm\bf{c}}{\rm\bf{b}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{t}{\rm\bf{c}}({\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime}{\rm\bf{a}})^{s}{\rm\bf{c}}^{\prime}({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime})^{t}{\rm\bf{a}}$

which is shadowed by

$({\rm\bf{b}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime}{\rm\bf{c}})^{s}{\rm\bf{b}}^{\prime}({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{t}{\rm\bf{a}}({\rm\bf{a}}^{\prime}{\rm\bf{c}}{\rm\bf{a}}^{\prime}{\rm\bf{b}})^{s}{\rm\bf{a}}^{\prime}({\rm\bf{c}}{\rm\bf{b}}^{\prime}{\rm\bf{c}}{\rm\bf{a}}^{\prime})^{t}{\rm\bf{c}}({\rm\bf{c}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime}{\rm\bf{a}})^{s}{\rm\bf{c}}^{\prime}({\rm\bf{b}}{\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime})^{t}{\rm\bf{b}}.$

Figure 11: The boundary of the (6,6)-benzel. Figure 12: The boundary of the
(5,5)-benzel. Figure 13: The boundary of the (4,4)-benzel.

We do not need to use these shadow words in any explicit way; all that matters
is that, like the boundary words they shadow, each intermixes fixed strings,
strings raised to the power of $s$, and strings raised to the power of $t$; as
in section 3, we can conclude that the signed area must be a quadratic
function of $s$ and $t$, which in each of the three cases amounts to a
quadratic function of $a$ and $b$. (If one wishes to do this explicitly, one
must remember that when $c$ is 1 or $-1$, the basepoint $p$ is in $L_{1}$
rather than $L_{0}$, which introduces a sign-change.)

For each $c$ in $\\{0,1,-1\\}$, one can compute $I(R)$ for half a dozen small
benzels of class $c$. (The most pleasant way to do this is to tile $R$ with
bones and stones; then $i(R)$ is the number of right-pointing stones minus the
number of left-pointing stones, and $I(R)$ is $3i(R)$.) Equating these values
of $I(R)$ with the values taken on by a quadratic with undetermined
coefficients, one obtains a linear system with 6 equations and 6 unknowns,
which one can solve, obtaining the values of the coefficients and proving the
following theorem.

Theorem 2: Let $R$ be a benzel of class $c$. Then $I(R)$, the (unrescaled)
Conway-Lagarias invariant of $R$, is

$\left\\{\begin{array}[]{ll}(-3a^{2}+6ab-3b^{2}+a+b)/2&\mbox{if $c$ is 0,}\\\
(a^{2}-4ab+b^{2}+a+b-2)/2&\mbox{if $c$ is 1,}\\\
(-3a^{2}+6ab-3b^{2}-a-b+2)/2&\mbox{if $c$ is $-1$.}\end{array}\right.$

##  6 When the invariant vanishes

We now ask: For what values of $a$ and $b$ does the Conway-Lagarias invariant
of the $(a,b)$-benzel vanish?

The easiest case to dispose of is $a+b\equiv 1$ mod 3. Recall from Theorem 2
that in this case we have $I(R)=(a^{2}-4ab+b^{2}+a+b-2)/2$, so we are trying
to find positive integer solutions to $a^{2}-4ab+b^{2}+a+b-2=0$ satisfying the
contraints $2\leqslant a\leqslant 2b$, $2\leqslant b\leqslant 2a$. We cannot
have $b=2a$ since that would give $a+b=3a\not\equiv 1$, so $b\leqslant 2a-1$,
and likewise $a\leqslant 2b-1$; but this implies that
$a^{2}-4ab+b^{2}+a+b-2=a(a-2b+1)+b(b-2a+1)-2<0$.

Just as easy to dispose of is the case $a+b\equiv-1$ mod 3. Theorem 2 tells us
that in this case we have $I(R)=(-3a^{2}+6ab-3b^{2}-a-b+2)/2$, so we seek
positive integers $a,b$ satisfying $-3a^{2}+6ab-3b^{2}-a-b+2=0$ with
$2\leqslant a\leqslant 2b$ and $2\leqslant b\leqslant 2a$. If we solve for $b$
in terms of $a$ we get a quadratic equation with discriminant
$(6a-1)^{2}-12(3a^{2}+a-2)=25-24a$, which is negative since $a\geqslant 2$.

Finally we come to the case $a+b\equiv 0$ mod 3. Theorem 2 tells us that in
this case we have $I(R)=(-3a^{2}+6ab-3b^{2}+a+b)/2$. Suppose $a$ and $b$ are
positive integers satisfying $-3a^{2}+6ab-3b^{2}+a+b=0$ with $2\leqslant
a\leqslant 2b$ and $2\leqslant b\leqslant 2a$. If we solve for $b$ in terms of
$a$ we get a quadratic equation with discriminant
$(6a+1)^{2}-12(3a^{2}-a)=24a+1$; this can give us a rational value of $b$ only
if $24a+1$ is a perfect square. Since $24a+1$ is an odd number not divisible
by 3, its integer square roots must be of the form $6k\pm 1$. Equating $24a+1$
with $(6k+1)^{2}$ gives $a=(3k^{2}+k)/2$. Plugging this into the quadratic
gives us an equation with two solutions, one of which is $b=(9k^{2}+9k+2)/6$
which is not an integer since the numerator is not a multiple of 3 and hence
not divisible by 6; the other solution is $b=(3k^{2}-k)/2$. Hence we must have
$(a,b)=((3k^{2}+k)/2,(3k^{2}-k)/2)$ for some integer $k$. Likewise, equating
$24a+1$ with $(6k-1)^{2}$ gives $(a,b)=((3k^{2}-k)/2,(3k^{2}+k)/2)$. (Of
course both families of solutions are the same, as can be seen by replacing
$k$ by $-k$.) These solution-pairs are plotted in Figure 14; they lie along a
parabola.

Figure 14: Integer solutions to $-3a^{2}+6ab-3b^{2}+a+b=0$. The solutions
(0,0), (1,2), and (2,1) are spurious for our purposes.

##  7 Constructing a tiling

Figure 15: Tiling a (12,15)-benzel with bones. The marked points lie on three
parabolas.

Figure 15 shows a (12,15)-benzel that has been symmetrically divided into
three sectors, each of which can be tiled by bones in a single orientation.
The path joining the origin to the rightmost point of the benzel is given by
the sequence

$({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{0}({\rm\bf{a}}{\rm\bf{c}}^{\prime})({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{1}({\rm\bf{a}}{\rm\bf{c}}^{\prime})({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{2}({\rm\bf{a}}{\rm\bf{c}}^{\prime}).$

More generally, consider the path

$({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{0}({\rm\bf{a}}{\rm\bf{c}}^{\prime})({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{1}({\rm\bf{a}}{\rm\bf{c}}^{\prime})\cdots({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{k-1}({\rm\bf{a}}{\rm\bf{c}}^{\prime})$

We claim that this path and the two paths obtained by rotating it 120 and 240
degrees around the origin divide the $(a,b)$-benzel with $a=k(3k-1)/2$ and
$b=k(3k+1)/2$ into three sectors each of which can be tiled by bones in a
single orientation.

First, note that the sum of the vectors in the aforementioned path is

$\displaystyle\ \ \,(1+3+5+\dots+(2k-1)){\rm\bf{a}}$
$\displaystyle+(0+1+2+\dots+(k-1)){\rm\bf{b}}^{\prime}$
$\displaystyle+(1+2+3+\dots+k){\rm\bf{c}}^{\prime}$ $\displaystyle=$
$\displaystyle k^{2}{\rm\bf{a}}-(k(k-1)/2){\rm\bf{b}}-(k(k+1)/2){\rm\bf{c}}$
$\displaystyle=$ $\displaystyle k^{2}-(k(k-1)/2)\omega-(k(k+1)/2)\omega^{2}$
$\displaystyle=$
$\displaystyle-(k^{2}+k(k-1)/2)\omega-(k^{2}+k(k+1)/2)\omega^{2}$
$\displaystyle=$ $\displaystyle-((3k^{2}-k)/2)\omega-((3k^{2}+k)/2)\omega^{2}$
$\displaystyle=$ $\displaystyle-a\omega-b\omega^{2}$

so that the path does indeed lead from the center of the benzel to the
rightmost point of the bounding hexagon.

Figure 16: One-third of a (12,15)-benzel.

Put $s=(2a-b)/3=k(k-1)/2$ and $t=(2b-a)/3=k(k+1)/2$. Consider the path

$P_{1}=({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{0}({\rm\bf{a}}{\rm\bf{c}}^{\prime})({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{1}({\rm\bf{a}}{\rm\bf{c}}^{\prime})\cdots({\rm\bf{a}}{\rm\bf{c}}^{\prime}{\rm\bf{a}}{\rm\bf{b}}^{\prime})^{k-1}({\rm\bf{a}}{\rm\bf{c}}^{\prime})({\rm\bf{b}}{\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime})^{s}$

joining the origin to the upper right corner of the bounding hexagon by way of
the right corner of the bounding hexagon, and the path

$P_{2}=({\rm\bf{b}}{\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime})^{0}({\rm\bf{b}}{\rm\bf{a}}^{\prime})({\rm\bf{b}}{\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime})^{1}({\rm\bf{b}}{\rm\bf{a}}^{\prime})\cdots({\rm\bf{b}}{\rm\bf{a}}^{\prime}{\rm\bf{b}}{\rm\bf{c}}^{\prime})^{k-1}({\rm\bf{b}}{\rm\bf{a}}^{\prime})({\rm\bf{a}}{\rm\bf{b}}^{\prime}{\rm\bf{a}}{\rm\bf{c}}^{\prime})^{t}$

joining the origin to the upper right corner of the bounding hexagon by way of
the upper left corner of the bounding hexagon, as shown in Figure 16. Let
$R^{\prime}$ be the sector of the benzel contained between the paths $P_{1}$
and $P_{2}$. We need to show that $R^{\prime}$ can be tiled by translates of
the bone whose axis points in the ${\rm\bf{a}}\\!-\\!{\rm\bf{b}}$ direction.
Assign each point $i{\rm\bf{a}}+j{\rm\bf{b}}+k{\rm\bf{c}}$ “height” $i-j$
(readily checked to be well-defined), so that two points that differ by
$\pm{\rm\bf{c}}$ have the same height; a bone whose axis points in the
${\rm\bf{a}}\\!-\\!{\rm\bf{b}}$ direction has two $\pm{\rm\bf{c}}$-edges on
which the height is constant, and the heights of the two edges differ by 6.
Mark each $\pm{\rm\bf{c}}$-edge on the boundary of $R^{\prime}$ with the
common height of its two endpoints. We need to show that the successive
heights of the $\pm{\rm\bf{c}}$-edges that lie along the path $P_{1}$
($1,2,5,6,9,12,9,6,3$ in the example) and the successive heights of the
$\pm{\rm\bf{c}}$-edges that lie along the path $P_{2}$
($-5,-10,-13,-12,-9,-6,-3,0,3$ in the example) agree mod 6; that is, we need
to check that subtracting the latter from the former gives a sequence of
multiples of 6. (In the example, subtracting the two sequences and dividing by
6 gives $1,2,3,3,3,3,2,1,0$; the terms count the number of
${\rm\bf{a}}\\!-\\!{\rm\bf{b}}$ bones in each diagonal band of $R^{\prime}$.)
To prove the divisibility assertion, it suffices to show that the same
assertion holds if we insert a 0 at the beginning of each of the two original
sequences and then replace each resulting sequence by its sequence of first
differences; in the example, the two difference sequences are
$1,1,3,1,3,3,-3,-3,-3$ and $-5,-5,-3,1,3,3,3,3,3$, and subtracting the second
from the first gives $6,6,6,0,0,0,-6,-6,-6$.

The first of the two difference sequences can be obtained by computing the
height-changes in the subwords between successive occurrences of
$\pm{\rm\bf{c}}$ in $P_{1}$. Those $k^{2}$ subwords yield height-changes

$1;1,3;1,3,3;1,3,3,3;\cdots;1,3,\dots,3;-3,\dots,-3$

where the “$3,\dots,3$” consists of $k-1$ successive 3’s and the
“$-3,\dots,-3$” consists of $s$ successive $-3$’s.

The second of the two difference sequences can be similarly obtained by
computing the height-changes in the subwords between successive occurrences of
$\pm{\rm\bf{c}}$ in $P_{2}$. Those $k^{2}$ subwords are yield height-changes

$-5;-5,-3;-5,-3,-3;-5,-3,-3,-3;\cdots;-5,-3,\dots,-3;1,3,\dots,3$

where the “$-3,\dots,-3$” consists of $k-2$ successive $-3$’s and the
“$3,\dots,3$” consists of $t-1$ successive 3’s.

The difference between the two sequences consists of $k(k-1)/2$ successive
6’s, followed by $k$ successive 0’s, followed by $k(k-1)/2$ successive $-6$’s.
Since all terms are divisible by 6, we have proved that $R^{\prime}$ can be
tiled by bones (specifically, bones parallel to
${\rm\bf{a}}\\!-\\!{\rm\bf{b}}$), which in turn shows that $R$ can be tiled by
bones.

Combining the preceding result with the result of section 6, we conclude:

Theorem 3: An $(a,b)$-benzel can be tiled by bones if and only if we have
$a=k(3k-1)/2$ and $b=k(3k+1)/2$ or vice versa for some integer $k\geqslant 2$.

We have shown that when the Conway-Lagarias invariant of a benzel is zero the
benzel has a tiling that uses only bones. It appears that when the Conway-
Lagarias invariant of a benzel is positive the benzel has a tiling that uses
only bones and right-pointing stones and that when the Conway-Lagarias
invariant of a benzel is negative the benzel has a tiling that uses only bones
and left-pointing stones, but we do not have a proof. (See Problems 4 and 7 in
[5].) In light of the remarks made at the end of section 4 we know that these
conditions are necessary, but we do not know whether they are sufficient.

##  8 Further questions

David desJardins’ TilingCount program was useful in leading the second author
to conjecture that bone tilings of the $(a,b)$-benzel do not exist except when
$a$ and $b$ are paired pentagonal numbers. However, we were not able to get
much data regarding the number of bone tilings that exist when $a$ and $b$ are
of that form, simply because the area of the
$((3k^{2}-k)/2$,$(3k^{2}+k)/2)$-benzel grows like the fourth power of $k$, and
the number of tilings appears to grow like an exponential function of the
area. For $k=2$, the number of tilings is 2; for $k=3$, the number of tilings
is $42705=(3)^{2}(5)(13)(73)$; for $k=4$, the number of tilings is
$7501790059160666750=(2)(5)^{3}(13)(19)(373)(1559)(208916623)$; and for $k=5$,
the program runs out of memory.

Figure 17: The two tilings of the (5,7)-benzel.

This article began with a tiling of the (5,7)-benzel. In fact, there are two
such tilings; Figure 17 shows both of them. One can see that the three
outermost tiles are actually forced, in the sense that both tilings contain
those tiles. Similarly, for the (12,15)-benzel, the tile at the upper right of
Figure 15 is forced. It would be good to know more generally which tiles are
forced in a bone tiling of a benzel.

Given that both tilings of the (5,7)-benzel contain the same number of bones
in each of the three orientations, one might wonder whether this is true for
larger benzels. In fact this surmise is false; there exist bone tilings of the
(12,15)-benzel with different numbers of bones in the three orientations.

Finally, one finds that some tiles near the boundary of the (12,15)-benzel,
although not actually forced, are highly likely to occur. For instance,
consider the bone in Figure 15 immediately to the left of the aforementioned
forced tile; it occurs in 42587 of the 42705 tilings, which means that one
sees it more than 99.7 percent of the time if one picks a tiling uniformly at
random. This suggests that there is a (possibly large) vicinity of the
boundary that is statistically “frozen” in the sense that a uniformly random
tiling exhibits very little variety there. Is this frozen region (assuming it
exists) a thin layer near the boundary of the benzel, or does it penetrate a
macroscopic distance into the interior?

### Acknowledgements

The authors thanks David desJardins, without whose code this main result would
never have come to light; Timothy Chow, who suggested the term “phone” to
refer to the third trihex tile; Jonathan Boretsky, Son Nguyen, and Ashley
Tharp who along with the first author explored bone tilings of
$(k(3k-1)/2,k(3k+1)/2)$-benzels; Colin Defant and Rupert Li, who gave helpful
comments on earlier versions of this writeup; and Juri Kirillov, who made some
useful observations. Thanks also to the organizers of the Open Problems in
Algebraic Combinatorics 2022 conference for inviting the second author to
speak and for funding his participation; the second author did not know how to
prove the main result of section 7 until students attending the conference,
including the first author, figured it out.

## References

  * [1] Federico Ardila and Richard Stanley, Tilings, Math. Intelligencer 32 (2005).
  * [2] John Conway and Jeffrey Lagarias, Tiling with polyominoes and combinatorial group theory, J. Combin. Theory (Ser. A) 53 (1990), no. 2, 183–208.
  * [3] Christopher Moore and Igor Pak, Ribbon tile invariants from the signed area, J. Combin. Theory (Ser. A) 98 (2002), 1–16.
  * [4] James Propp, A pedestrian approach to a method of Conway, or, a tale of two cities, Mathematics Magazine 70 (1997), 327–340.
  * [5] James Propp, Trimer covers in the triangular grid: twenty mostly open problems, to appear in the proceedings of the 2022 Open Problems in Algebraic Combinatorics conference. http://arxiv.org/abs/2206.06472
  * [6] William Thurston, Conway’s tiling groups, Amer. Math. Monthly 97 (1990), no. 8, 757–773.