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\title{Tilings of an equilateral triangle by at most five\\
lattice trapezoids with $60^\circ$ base angles:\\
a complete structural classification}
\author{Gonzalo Barr\'{\i}a\\[2pt]
\small Instituto Profesional AIEP \quad ORCID 0009-0003-8878-737X}
\date{1 October 2026}

\begin{document}
\maketitle

\begin{abstract}
We study tilings of an equilateral triangle of side $n$ in the triangular grid by
$k$ lattice trapezoids with base angles $60^\circ$, the objects behind the OEIS
sequences A389392 ($k=4$) and A391498 ($k=5$). We prove two angle identities valid
for every such tiling and a lemma relating the number of boundary vertices to the
number of pieces having a side on the boundary. With these tools we show that there
are exactly $1$, $2$ and $13$ combinatorial types of tilings for $k=3,4,5$. For $k=3$
every tiling is a pinwheel. For $k=4$ every tiling belongs to one of the two
categories used in A389392, a fact that had previously been taken for granted. For
$k=5$ the thirteen types refine the eight categories of A391498; in three of them a
piece has no side on the boundary. As a consistency check, the volumes of the
parameter polytopes together with the generic multiplicities reproduce the leading
coefficient $7/36$ of the conjectured quasi-polynomial for A391498. All lemmas were
checked against an exhaustive enumeration of the tilings with pairwise distinct
pieces for $n\le 15$. Finally, the classification turns the thirteen types into
eight explicit families of sets of shapes, and an inclusion--exclusion over them
reduces the conjectured generating function of A391498 to twelve elementary
counting statements, and we settle all twelve: the resulting closed formula
reproduces the sequence for every $n\le127$.
\end{abstract}

\section{Setting and main result}\label{sec:setting}

Throughout, $n\ge 1$ is an integer and we use oblique coordinates: $(x,y)$ denotes
the point $x\,\mathbf u+y\,\mathbf v$, where $\mathbf u=(1,0)$ and
$\mathbf v=(\tfrac12,\tfrac{\sqrt3}{2})$. Let
\[
  T=T_n=\{(x,y):\ x\ge 0,\ y\ge 0,\ x+y\le n\},
\]
with corners $A=(0,0)$, $B=(n,0)$ and $C=(0,n)$. \emph{Lattice lines} are the lines
$x=c$, $y=c$ and $x+y=c$, i.e.\ the lines parallel to a side of $T$; two of them meet
at angles of $60^\circ$ and $120^\circ$. A \emph{lattice segment} is a segment lying
on a lattice line.

For integers $1\le s\le b-1$ we write $\trap{b}{s}$ for the trapezoid with sides on
lattice lines, long base $b$, short base $b-s$ and legs of length $s$. Its angles are
$60^\circ$ at the ends of the long base and $120^\circ$ at the ends of the short
base. A convex polygon with sides on lattice lines has all its angles in
$\{60^\circ,120^\circ\}$, and if it has $m_{60}$ corners of the first kind and
$m_{120}$ of the second, then $2m_{60}+m_{120}=6$. Hence it is an equilateral
triangle, a quadrilateral with two angles of each kind (a trapezoid if its two
$60^\circ$ corners are adjacent, a parallelogram otherwise), a pentagon or a hexagon.

A \emph{$k$-tiling} of $T$ is a decomposition $T=Q_1\cup\dots\cup Q_k$ into such
trapezoids (the \emph{pieces}) with pairwise disjoint interiors. The sequences
A389392 and A391498~\cite{oeis,rodrigo} count, for $k=4$ and $k=5$ respectively, the
sets of $k$ pairwise different shapes $\trap{b}{s}$ that occur as the shapes of the
pieces of some $k$-tiling of $T_n$. Distinctness plays no role in the structural
results below; it only matters for the counting.

\paragraph{The graph of a tiling.} Let $V$ consist of $A$, $B$, $C$ and all corners of
pieces. Every side of a piece is subdivided by the points of $V$ that it contains; the
resulting segments are the \emph{edges} of a plane graph $G$ with vertex set $V$,
whose bounded faces are the pieces. A vertex is a \emph{boundary vertex} if it lies on
$\partial T$ and is not a corner of $T$, and an \emph{interior vertex} if it does not
lie on $\partial T$. An edge is a \emph{boundary edge} if it is contained in
$\partial T$, and an \emph{interior edge} otherwise.

\paragraph{Combinatorial types.} The \emph{combinatorial type} of a tiling is the
plane graph $G$ together with the lattice direction of each of its edges, up to the
six symmetries of $T$. Thus two tilings have the same type if and only if, after a
symmetry of $T$ is applied to one of them, they differ only in the lengths of their
edges. The angle of a piece at one of its corners is the angle between the two edges
there; hence $G$, together with the angles of all pieces at all their corners,
determines the directions of all edges as soon as the direction of one boundary edge
is fixed, and therefore determines the combinatorial type.

\begin{theorem}\label{thm:main}
For $k=3$, $4$ and $5$ there are exactly $1$, $2$ and $13$ combinatorial types of
$k$-tilings of equilateral triangles. They are described in
Theorems~\ref{thm:k3}, \ref{thm:k4}, \ref{thm:k5I} and~\ref{thm:k5II} and in
Corollary~\ref{cor:13}.
\end{theorem}

Sections~\ref{sec:local}--\ref{sec:k34} develop the tools and settle $k=3$ and $k=4$;
they close the gap in the treatment of A389392, where the exhaustiveness of the two
categories was assumed (Remark~\ref{rem:A389392}). Section~\ref{sec:k5} deals with
$k=5$, Section~\ref{sec:types} compares the thirteen types with the eight categories
of A391498, Section~\ref{sec:verif} describes the computer verification,
Section~\ref{sec:outlook} discusses the counting problem and
Section~\ref{sec:counting} turns the classification into a closed formula for
A391498.

\input{completeness_k3k4}

\section{Five pieces}\label{sec:k5}

\begin{theorem}[five pieces with a full strip]\label{thm:k5I}
If a $5$-tiling of $T$ contains a full strip $\trap{n}{s}$, then this is its only
full strip, and the other four pieces form a tiling of the triangle of side $n-s$ of
form \textup{(A)} or \textup{(B)}.
\end{theorem}

\begin{proof}
This follows from Lemma~\ref{lem:strip} and Theorem~\ref{thm:k4}.
\end{proof}

\begin{theorem}[five pieces without full strip]\label{thm:k5II}
Up to a symmetry of $T$, every $5$-tiling of $T$ without full strip is of exactly one
of the following seven forms. In the first five forms the tiling has a pinwheel
skeleton $R_1,R_2,R_3$ (Lemma~\ref{lem:skeleton}), and the regions not mentioned are
single pieces.
\begin{itemize}[leftmargin=3.4em,labelwidth=2.8em,labelsep=0.6em,align=left]
\item[\textup{(D)}] Two regions are each divided into two pieces by a parallel cut.
\item[\textup{(S)}] One region is divided into three pieces by two parallel cuts.
\item[\textup{(V)}] One region $R$ is divided into three pieces by two disjoint
  segments which start on its base-arm and end on its long base, each of them
  parallel to the leg of $R$ farther from it; the middle piece is a trapezoid whose
  long base lies on the base-arm.
\item[\textup{($\Cb$)}] One region $R$ is divided by a segment $\gamma$, which starts
  on the base-arm of $R$ and is parallel to its boundary leg $\lambda$, into a
  trapezoid, which is a piece, and a parallelogram $\Pi$ adjacent to $\lambda$; and
  $\Pi$ is divided into two pieces by a segment in the third lattice direction
  joining $\gamma$ to $\lambda$.
\item[\textup{($\Co$)}] As in \textup{($\Cb$)}, but with $\lambda$ the leg-arm of $R$
  instead of its boundary leg.
\item[\textup{(F)}] Some piece $F$ does not meet $\partial T$. Its corners are a
  $\mathrm{Y}$ vertex $P$ and three $\mathrm{T}$ vertices $V_1,V_2,V_3$
  (counterclockwise), with angles at $P,V_1,V_2,V_3$ equal to
  $120^\circ,60^\circ,60^\circ,120^\circ$. The sides $PV_1$, $V_1V_2$ and $V_2V_3$ of
  $F$ are prolonged beyond $V_1$, $V_2$ and $V_3$, respectively, up to $\partial T$,
  and the third edge at $P$ ends on $\partial T$.
\item[\textup{($\Delta$)}] Some piece $M$ meets $\partial T$ in a single point
  $D^*$, a vertex of type $B_3$. The other corners of $M$ are a $\mathrm{T}$ vertex
  $Q$ and two $\mathrm{Y}$ vertices $P_j,P_i$; the long base of $M$ is $D^*Q$, which
  is the stem of $Q$, and its short base is $P_jP_i$. The side $QP_j$ lies on the line
  of $Q$, whose other half ends on $\partial T$, and the third edges at $P_i$ and
  $P_j$ end on $\partial T$.
\end{itemize}
Each of these seven forms is a single combinatorial type.
\end{theorem}

\begin{proof}
\emph{Step 0: profiles.} By Lemma~\ref{lem:nochord} there is no chord, and by
Corollary~\ref{cor:mge1} and \eqref{eq:m} we have $m\in\{1,2\}$, $B=3+m$ and
$f=2-m$. Substituting $B_1=3+m-B_3$ into \eqref{eq:I2} gives
\begin{equation}\label{eq:reduced}
  2B_3+t+3t'+2x+4w+6z=4-m .
\end{equation}
Together with \eqref{eq:I1} this leaves exactly the following seven possibilities
(all counts not mentioned are $0$):
\begin{center}
\begin{tabular}{@{}lll@{\qquad\qquad}lll@{}}
\toprule
\multicolumn{3}{@{}l}{$m=2$} & \multicolumn{3}{l@{}}{$m=1$}\\
\midrule
(i)   & $B_1=5$, $t=2$   & $y=1$ & (iv)  & $B_1=4$, $t=3$            & $y=1$\\
(ii)  & $B_1=5$, $x=1$   & $y=1$ & (v)   & $B_1=4$, $t'=1$           & $y=2$\\
(iii) & $B_1=4$, $B_3=1$ & $y=2$ & (vi)  & $B_1=4$, $t=x=1$          & $y=1$\\
      &                  &       & (vii) & $B_1=3$, $B_3=1$, $t=1$   & $y=2$\\
\bottomrule
\end{tabular}
\end{center}
In the left-hand side of \eqref{eq:slots}, interior vertices of types $\mathrm{T}$,
$\mathrm{Y}$, $\mathrm{T}'$, $\mathrm{X}$ contribute $3,3,4,4$.

\medskip\noindent
\emph{Step 1: cases (v) and (vi) are impossible.} In both cases there are three
interior vertices, one of them of type $\mathrm{Y}$, and \eqref{eq:slots} reads
$10-2e_H=4$. Thus $e_H=3$, and the three interior vertices form a triangle through a
$\mathrm{Y}$ vertex, contrary to Lemma~\ref{lem:slots}(a).

\medskip\noindent
\emph{Step 2: case (ii) is impossible.} Let $P$ and $X$ be the interior vertices.
Now $7-2e_H=5$, so $PX$ is an edge, and the three other edges at $X$ end at boundary
vertices. At a vertex of type $\mathrm{X}$ the four angles are, in cyclic order,
$60^\circ,120^\circ,60^\circ,120^\circ$ or $60^\circ,60^\circ,120^\circ,120^\circ$. In
the first case the four edges lie on two lines through $X$; in the second they lie on
one line together with two rays on opposite sides of it. A line through $X$ not
containing $PX$ would be a chord. Hence we are in the second case, and $PX$ lies on
the line, which is therefore an arm of $P$ ending on $\partial T$. The other two arms
of $P$ are single edges ending on $\partial T$, so we have a pinwheel skeleton, and
the two rays at $X$ divide the two regions adjacent to this arm into two pieces each.
By Corollary~\ref{cor:convex}(b) both rays are parallel cuts. Since a parallel cut
joins the two legs of its region, the arm is a leg of both adjacent regions, contrary
to Lemma~\ref{lem:skeleton}.

\medskip\noindent
\emph{Step 3: case (iii) is impossible.} The interior vertices are two $\mathrm{Y}$
vertices $P_1,P_2$, and there is one vertex $D^*$ of type $B_3$. Now $6-2e_H=4+2$, so
$e_H=0$, and the two interior edges at $D^*$ end at interior vertices, one at $P_1$
and one at $P_2$ (two non-collinear segments issuing from $D^*$ cannot end at the
same point). Let $M$ be the piece having a $60^\circ$ corner at $D^*$ between these
two edges. The corners of $M$ adjacent to $D^*$ are $P_1$ and $P_2$, where $M$ has
angles of $120^\circ$. But in a trapezoid each $60^\circ$ corner is adjacent to the
other $60^\circ$ corner.

\medskip\noindent
\emph{Step 4: case (i) leads to \textup{(D)}, \textup{(S)}, \textup{(V)} or
\textup{($\Cb$)}.} Let $P$ be the $\mathrm{Y}$ vertex and $Q_1,Q_2$ the $\mathrm{T}$
vertices. Now $9-2e_H=5$, so $e_H=2$ and $H$ is a path.

(4a) \emph{$P$ is the middle vertex of $H$.} Each $Q_i$ has two edges ending at
boundary vertices, so by Lemma~\ref{lem:slots}(b) the edge $Q_iP$ lies on the line of
$Q_i$, which continues up to $\partial T$, and the stem $\sigma_i$ of $Q_i$ ends on
$\partial T$. Thus all arms of $P$ end on $\partial T$ and we have a pinwheel
skeleton, with $Q_1$ and $Q_2$ on different arms. The only interior edges not
contained in arms are $\sigma_1$ and $\sigma_2$. If they lie in different regions,
each of these two regions is divided into two pieces, the third region is a piece,
and Corollary~\ref{cor:convex}(b) gives (D). Suppose that they lie in the same region
$R$; then they start on the two different arms of $R$. Each $\sigma_i$ is a lattice
segment joining two points of $\partial R$, so it is of one of the types (i)--(iii)
of Lemma~\ref{lem:chords}(b). Type (iii) is excluded, because the triangle cut off
would be a union of at most two pieces (Corollary~\ref{cor:convex}(a)). A segment
starting at an interior point of the leg-arm cannot be of type (ii), and a segment
starting at an interior point of the base-arm cannot be of type (i). Hence one of
$\sigma_1,\sigma_2$ joins the two legs of $R$ and the other joins its two bases, so
they cross; this is impossible, since there are no further interior vertices.

(4b) \emph{A $\mathrm{T}$ vertex, say $Q_1$, is the middle vertex of $H$.} Since
$Q_2$ has two edges ending at boundary vertices, $Q_1Q_2$ lies on the line of $Q_2$
(Lemma~\ref{lem:slots}(b)). Let $\gamma$ be the segment from $Q_1$ through $Q_2$ to
$\partial T$, and let $\sigma_2$ be the stem of $Q_2$. The vertex $Q_1$ has one more
edge, which ends at a boundary vertex $D_1$. If the line of $Q_1$ consisted of
$Q_1D_1$ and $Q_1Q_2$, then $D_1Q_1$ together with $\gamma$ would be a chord. Hence
the line of $Q_1$ contains $Q_1P$, and there are two possibilities.

\emph{The line of $Q_1$ is $PQ_1\cup Q_1D_1$.} Then $Q_1Q_2$ is the stem of $Q_1$.
All arms of $P$ end on $\partial T$; let $R$ be the region containing $\gamma$. All
interior edges not contained in arms lie in $R$, so $R$ consists of three pieces: the
part of $R$ on one side of $\gamma$, and the two pieces into which $\sigma_2$ divides
the part $R'$ on the other side. As in (4a), $\gamma$ is of type (i) or (ii). If it
were of type (i), $R'$ would be a trapezoid with bases parallel to $\gamma$, and by
Corollary~\ref{cor:convex}(b) $\sigma_2$ would be parallel to $\gamma$, which is
absurd because $\sigma_2$ starts on $\gamma$. So $\gamma$ is of type (ii): $Q_1$ lies
on the base-arm of $R$, and $\gamma$ is parallel to a leg $\lambda$ of $R$ and
divides $R$ into a parallelogram $\Pi$ adjacent to $\lambda$ and a trapezoid $Z$. A
parallelogram is not a piece, so $R'=\Pi$ and $Z$ is a piece. By
Corollary~\ref{cor:convex}(c), $\sigma_2$ lies in the third lattice direction and
joins $\gamma$ to the opposite side $\lambda$ of $\Pi$. Since $\sigma_2$ ends on
$\partial T$, $\lambda$ is the boundary leg of $R$: this is ($\Cb$).

\emph{The line of $Q_1$ is $PQ_1\cup Q_1Q_2$.} Then the segment from $P$ through $Q_1$
and $Q_2$ to $\partial T$ is an arm of $P$, from which the stems $\sigma_1=Q_1D_1$
and $\sigma_2$ branch off. The other two arms of $P$ are single edges ending on
$\partial T$, so we have a pinwheel skeleton. If $\sigma_1$ and $\sigma_2$ lay in the
two different regions adjacent to this arm, each of them would divide its region into
two pieces, and by Corollary~\ref{cor:convex}(b) the arm would be a leg of both
regions, contrary to Lemma~\ref{lem:skeleton}. So they lie in one region $R$, which
they divide into three pieces, and they are disjoint. If the arm is the leg-arm of
$R$, both are parallel cuts (types (ii) and (iii) being excluded as in (4a)): this is
(S). If the arm is the base-arm of $R$, both are of type (ii). After a symmetry of the
lattice we may write $R=\{0\le y\le s,\ x\ge 0,\ x+y\le b\}$ with the base-arm on the
line $y=s$; then each of the two segments lies on a line $x=p$ or on a line $x+y=q$.
If the segment nearer to the leg $\{x=0\}$ lay on a line $x=p$, the piece between them
would be the parallelogram $R\cap\{x\le p\}$; similarly for the leg $\{x+y=b\}$.
Hence the two segments lie on lines $x+y=q$ and $x=p$ with $q<p$, and the middle piece
$R\cap\{x\le p,\ x+y\ge q\}$ is a trapezoid whose long base lies on the line $y=s$:
this is (V).

\medskip\noindent
\emph{Step 5: case (iv) leads to \textup{(F)} or \textup{($\Co$)}.} Let $P$ be the
$\mathrm{Y}$ vertex and $Q_1,Q_2,Q_3$ the $\mathrm{T}$ vertices. Now $12-2e_H=4$, so
$e_H=4$, and the graph $H$, which has four vertices and four edges, contains exactly
one cycle. A triangle through $P$ is excluded by Lemma~\ref{lem:slots}(a). If the
cycle were the triangle $Q_1Q_2Q_3$, then $P$, which would have two edges ending on
$\partial T$, would lie outside it; the open triangle would contain no vertex and no
edge, and it would be a triangular piece. Hence the cycle is a $4$-cycle through all
interior vertices. It bounds a region containing no vertex and no edge (such an edge
would be a fifth edge of $H$), i.e.\ a piece $F$ with corners $P,Q_1,Q_2,Q_3$; as $F$
has four corners, none of these points lies in the relative interior of a side of
$F$. The piece $F$ does not meet $\partial T$, and each $Q_i$ has exactly one edge
ending on $\partial T$. Since $Q_i$ is a corner of $F$, the two sides of $F$ at $Q_i$
are the stem of $Q_i$ and one half of its line, and the edge ending on $\partial T$
is the other half of the line. So at each $Q_i$ exactly one of the two sides of $F$
through $Q_i$ is prolonged beyond $Q_i$ up to $\partial T$.

Label the corners of $F$ counterclockwise $V_0=P,V_1,V_2,V_3$, put $e_j=V_jV_{j+1}$
(indices modulo $4$), and let $\theta_j$ be the angle of $F$ at $V_j$, so that
$\theta_0=120^\circ$. No side of $F$ is prolonged at both ends, since this would give a
chord, and no side is prolonged at $P$. Hence the sides prolonged at $V_1,V_2,V_3$
are, respectively,
\[
  \text{(a) } e_0,e_1,e_2;\qquad \text{(b) } e_0,e_1,e_3;\qquad
  \text{(c) } e_0,e_2,e_3;\qquad \text{(d) } e_1,e_2,e_3 .
\]
A reflection of $T$ reverses the cyclic order and exchanges (a) with (d) and (b) with
(c), so it suffices to treat (a) and (c). Let $D$ be the endpoint on $\partial T$ of
the third edge at $P$.

\emph{Pattern (c).} The sides $e_0$ and $e_3$ are prolonged at $V_1$ and $V_3$, so
all arms of $P$ end on $\partial T$, and $F$ lies in the region $R$ bounded by the
arms containing $e_0$ and $e_3$. The prolongation of $e_2$ at $V_2$ yields a segment
$\gamma$ from $V_3$ through $V_2$ to $\partial T$, and the stem $V_2V_1$ of $V_2$
joins $\gamma$ to the other arm of $R$. All interior edges not contained in arms lie
in $R$, so $R$ consists of three pieces: the part of $R$ on the side of $\gamma$ not
containing $P$, and the two pieces into which $V_1V_2$ divides the part $R'$
containing $P$. Exactly as in Step~4, $\gamma$ is of type (ii), $V_3$ lies on the
base-arm, $\gamma$ is parallel to a leg $\lambda$ of $R$, and $R'$ is the
parallelogram $\Pi$ adjacent to $\lambda$. Since $P\in R'$ is the corner of $R$ where
the base-arm meets the leg-arm, $\lambda$ is the leg-arm. By
Corollary~\ref{cor:convex}(c), $V_1V_2$ lies in the third lattice direction and joins
$\gamma$ to $\lambda$: this is ($\Co$), and $F$ is the piece of $\Pi$ containing $P$.

\emph{Pattern (a).} Now $e_3$ is the stem of $V_3$, so the arm of $P$ along $e_3$ ends
at $V_3$. Let $D_1,D_2,D_3$ be the endpoints on $\partial T$ of the prolongations of
$e_0,e_1,e_2$. The interior edges are the four sides of $F$ and the four segments
$PD$, $V_1D_1$, $V_2D_2$, $V_3D_3$, which join $F$ to $\partial T$ and divide
$T\setminus F$ into four pieces:
\begin{itemize}[leftmargin=2em]
\item $O_b$, bounded by $PD$, an arc of $\partial T$ from $D$ to $D_1$, and the
  straight segment $D_1V_1P$ (which contains $V_1$ in its interior);
\item $O_c$, bounded by $V_1D_1$, an arc from $D_1$ to $D_2$, and the segment
  $D_2V_2V_1$;
\item $O_d$, bounded by $V_2D_2$, an arc from $D_2$ to $D_3$, and the segment
  $D_3V_3V_2$;
\item $O_a$, bounded by $V_3D_3$, an arc from $D_3$ to $D$, and the segments $DP$ and
  $PV_3$.
\end{itemize}
Each of them is a quadrilateral whose corners, apart from corners of $T$, are
$P,D,D_1$ for $O_b$; $V_1,D_1,D_2$ for $O_c$; $V_2,D_2,D_3$ for $O_d$; and
$V_3,D_3,D,P$ for $O_a$. Hence the four arcs contain $1,1,1,0$ corners of $T$,
respectively. The angle of $O_c$ at $V_1$ is $180^\circ-\theta_1$. In $O_c$ the corner
$K$ of $T$ is a $60^\circ$ corner whose neighbours are $D_1$ and $D_2$, so the corner
opposite to $K$, which is $V_1$, has angle $120^\circ$; hence $\theta_1=60^\circ$. In
the same way $\theta_2=60^\circ$, and therefore $\theta_3=120^\circ$. In $O_a$ the
corner $V_3$ has angle $60^\circ$ and its neighbours are $D_3$ and $P$ (angle
$120^\circ$), so $O_a$ has angle $60^\circ$ at $D_3$ and $120^\circ$ at $D$. All
boundary vertices are of type $B_1$, so the two pieces meeting at each of
$D,D_1,D_2,D_3$ have angles adding up to $180^\circ$ there. Consequently $O_b$ has
angle $60^\circ$ at $D$ and $120^\circ$ at $D_1$; $O_c$ has $60^\circ$ at $D_1$ and
$120^\circ$ at $D_2$; and $O_d$ has $60^\circ$ at $D_2$ and $120^\circ$ at $D_3$, in
accordance with $O_a$. All angles are determined, and this is (F).

\medskip\noindent
\emph{Step 6: case (vii) leads to \textup{($\Delta$)}.} Let $P_1,P_2$ be the
$\mathrm{Y}$ vertices, $Q$ the $\mathrm{T}$ vertex and $D^*$ the vertex of type
$B_3$. Now $9-2e_H=3+2$, so $e_H=2$. The two interior edges at $D^*$ end at two
different interior vertices. Let $M$ be the piece with a $60^\circ$ corner at $D^*$.
By the argument of Step~3, these two vertices are not both $\mathrm{Y}$ vertices, so
they are $Q$ and some $P_i$, and $M$ has angle $120^\circ$ at its corner $P_i$.
Consequently, if $X$ denotes the corner of $M$ following $D^*$ along $D^*Q$, then $M$
has angle $60^\circ$ at $X$, and the fourth corner $Y$ of $M$, adjacent to $X$ and to
$P_i$, has angle $120^\circ$.

Suppose first that $D^*Q$ lies on the line of $Q$. If $M$ contains $Q$ in the relative
interior of a side, then $X$ is the vertex following $Q$ on this line. It is not on
$\partial T$, since the line would then be a chord, so it is a $\mathrm{Y}$ vertex,
where $M$ has angle $120^\circ$, a contradiction. Otherwise $X=Q$, and the side $QY$
of $M$ is the stem of $Q$. The vertex $Y$ is not on $\partial T$: it is not a corner of
$T$, only one interior edge arrives at a boundary vertex other than $D^*$, and the
sides $QY$ and $YP_i$ of $M$ are not contained in $\partial T$. So $Y$ is an interior
vertex, with angle $120^\circ$ in $M$ and different from $Q$ and from $P_i$ (otherwise
$M$ would be a triangle); thus $Y$ is the other $\mathrm{Y}$ vertex $P_j$, and $H$
consists of the edges $QP_j$ and $P_jP_i$. Then the third edge at $Q$ ends on
$\partial T$, and together with $D^*Q$ it forms the line of $Q$, which is a chord.
This is impossible.

Hence $D^*Q$ is the stem of $Q$. Then $M$ lies on the side of the stem, $X=Q$, and the
side $QY$ of $M$ lies on the line of $Q$. As before, $Y=P_j$ with $j\ne i$, and $H$
consists of $QP_j$ and $P_jP_i$. So $M=D^*QP_jP_i$ has long base $D^*Q$ and short base
$P_jP_i$. The other half of the line of $Q$ and the third edges at $P_j$ and at $P_i$
end at three boundary vertices $D'$, $D'''$, $D''$, all of type $B_1$ (the two
interior edges at $D^*$ are already known). These three segments and the sides of
$M$ are all the interior edges, and they divide $T\setminus M$ into four pieces:
$X_1$, bounded by $D^*Q$, $QD'$ and an arc from $D'$ to $D^*$; $X_H$, bounded by the
straight segment $D'QP_j$ (which contains $Q$ in its interior), by $P_jD'''$ and by an
arc from $D'''$ to $D'$; $X_2$, bounded by $P_jD'''$, an arc from $D'''$ to $D''$,
$D''P_i$ and $P_iP_j$; and $X_3$, bounded by $P_iD''$, an arc from $D''$ to $D^*$, and
$D^*P_i$. Counting corners as in Step~5, these four arcs contain $1,1,0,1$ corners of
$T$. By Lemma~\ref{lem:local}(b), $X_1$ and $X_3$ have angle $60^\circ$ at $D^*$. The
corners of $X_1$ are $Q,D',K,D^*$ in cyclic order, with $K$ a corner of $T$; its two
$60^\circ$ corners are $K$ and $D^*$, so $X_1$ has angle $120^\circ$ at $D'$. Hence
$X_H$ has angle $60^\circ$ at $D'$; its corners are $D',P_j,D''',K'$, its $60^\circ$
corners are $D'$ and $K'$, and so it has angle $120^\circ$ at $D'''$. Then $X_2$, with
corners $D''',P_j,P_i,D''$, has angle $60^\circ$ at $D'''$ and hence also at $D''$; and
$X_3$ has angle $120^\circ$ at $D''$, in accordance with its corners $D'',P_i,D^*,K''$
having angles $120^\circ,120^\circ,60^\circ,60^\circ$. All angles are determined, and
this is ($\Delta$).

\medskip\noindent
\emph{Conclusion.} The forms (D), (S), (V) and ($\Cb$) have profile (i), the forms (F)
and ($\Co$) have profile (iv), and ($\Delta$) has profile (vii). Within profile (i),
$P$ is the middle vertex of $H$ only in (D); the three interior vertices are collinear
in (S) and (V) but not in ($\Cb$); and (S) differs from (V) in whether the arm
containing them is the leg-arm or the base-arm of the divided region. Within profile
(iv), all arms of $P$ end on $\partial T$ in ($\Co$) but not in (F). So the seven
forms are pairwise different. In each case the arguments above determine $G$ and all
angles up to a symmetry of $T$: for the first five forms the pinwheel skeleton is
unique up to symmetry (Theorem~\ref{thm:k3}), and the admissible choices of the
divided regions are permuted transitively by the rotations of $T$. Hence each form is
a single combinatorial type. Finally, each form occurs (Figure~\ref{fig:types}).
\end{proof}

\begin{corollary}\label{cor:13}
Up to symmetry there are exactly thirteen combinatorial types of $5$-tilings. Six of
them contain a full strip: the strip together with a $4$-tiling of form \textup{(A)}
whose strip is parallel to it (one type) or lies on one of the other two sides of the
remaining triangle (two types, distinguished by the chirality of the pinwheel), and
the strip together with a $4$-tiling of form \textup{(B)} in which any one of the
three pinwheel pieces is cut (three types). The other seven are the types of
Theorem~\ref{thm:k5II}.
\end{corollary}

\begin{proof}
By Theorem~\ref{thm:k5I}, a tiling with a full strip on $AB$ is determined, up to edge
lengths, by the form of the $4$-tiling of the triangle $T'$ at $C$ and by its
position; the reflection of $T$ fixing $AB$ acts on these data, and different orbits
give different types because the full strip is unique. For (A), the second strip lies
on the side of $T'$ parallel to $AB$ or on one of its other two sides, and the
pinwheel has one of two chiralities; the reflection exchanges the two chiralities and
the two non-parallel sides, which leaves $1+2=3$ orbits. For (B), the pinwheel has
two chiralities and the cut piece is one of three; the reflection acts freely on these
six configurations, which leaves three orbits.
\end{proof}

\section{The thirteen types and the categories of A391498}\label{sec:types}

Table~\ref{tab:types} lists the thirteen types with the labels T1--T13 used in our
programs (numbered in order of first appearance in the enumeration), their profiles,
the smallest $n$ for which each type occurs with pairwise distinct pieces, the
generic multiplicity $\mu$ discussed in Section~\ref{sec:outlook}, and the category of
A391498 containing the type. The last column was obtained for $n=12$ by comparing,
set by set, the output of Rodrigo's program for the eight categories with our
classification: each category consists exactly of the sets realized by the types
listed (a set belongs to several categories when it is realized by tilings of
different types).

\begin{table}[ht]
\centering
\begin{tabular}{@{}llcccl@{}}
\toprule
Type & Form & $(B_1,B_3,t,y)$ & first $n$ & $\mu$ & Category\\
\midrule
T1      & full strip $+$ (A), parallel strips      & $(6,0,1,1)$ & 6  & 9 & AA1\\
T4, T5  & full strip $+$ (A), non-parallel strips & $(4,0,3,1)$ & 6  & 9 & AA1\\
T6, T9  & full strip $+$ (B)                       & $(5,0,2,1)$ & 7  & 3 & AA2\\
T11     & full strip $+$ (B)                       & $(4,0,3,1)$ & 7  & 3 & AA2\\
T2      & (D)                                      & $(5,0,2,1)$ & 6  & 1 & DD1\\
T8      & (S)                                      & $(5,0,2,1)$ & 7  & 1 & CC3\\
T13     & (V)                                      & $(5,0,2,1)$ & 10 & 6 & CC1\\
T7      & ($\Cb$)                                  & $(5,0,2,1)$ & 7  & 4 & CC2\\
T12     & ($\Co$)                                  & $(4,0,3,1)$ & 7  & 4 & CC2\\
T3      & (F)                                      & $(4,0,3,1)$ & 6  & 1 & BB2\\
T10     & ($\Delta$)                               & $(3,1,1,2)$ & 7  & 1 & BB1\\
\bottomrule
\end{tabular}
\caption{The thirteen types of $5$-tilings. In all of them $t'=x=w=z=0$. The column
$\mu$ gives the number of tilings, up to symmetry, realizing a generic set of the
type: for T1, T4, T5 the nine tilings are three of each type; for T6, T9, T11 one of
each; for T7, T12 two of each; for T13 all six are of type T13.}
\label{tab:types}
\end{table}

\begin{figure}[ht]
\centering
\includegraphics[width=\textwidth]{tipos_13.png}
\caption{Representatives of the thirteen types for $n=13$; the pieces are labelled
$b\times s$.}
\label{fig:types}
\end{figure}

\begin{remark}\label{rem:ops}
(1) The forms (V), ($\Cb$) and ($\Co$) arise in the same way: a region $R$ of the
skeleton is divided by a segment of type (ii), parallel to a leg $\lambda$, into a
trapezoid and a parallelogram $\Pi$, and $\Pi$ is divided in the third lattice
direction. In (V) the second cut joins the two sides of $\Pi$ lying on the bases of
$R$ (for either choice of $\lambda$); in ($\Cb$) and ($\Co$) it joins the other two
sides of $\Pi$, and $\lambda$ is the boundary leg or the leg-arm, respectively. This
agrees with CC1 consisting of (V) and CC2 of ($\Cb$) and ($\Co$).

(2) A $5$-tiling arises from a $4$-tiling by one of the two operations underlying
A389392 (adding a full strip, or dividing a piece by a parallel cut) if and only if its
type is T1, T2, T4, T5, T6, T8, T9 or T11. Indeed, adding a full strip gives the six
types with a full strip; and if two pieces form a trapezoid, merging them gives a
tiling of form (A) or (B) by Theorem~\ref{thm:k4}, and dividing one piece of such a
tiling by a parallel cut gives T1 (cutting the strip of (A)), a full strip together
with (B) (cutting a pinwheel piece of (A)), (D) (cutting an uncut piece of (B)) or (S)
(cutting one half of the cut piece of (B)). The remaining five types T3, T7, T10, T12,
T13 arise in neither way. In agreement with this, our program found no pair of pieces
forming a trapezoid in any of the $2556+936+390+936+432$ tilings of these types for
$n=13$, whereas every tiling of types T1, T2, T6, T8, T9, T11 contains such a pair.
This is why exhaustiveness for $k=5$ cannot be proved by a simple induction on $k$ and
requires the global argument of Theorem~\ref{thm:k5II}.

(3) In (F) and ($\Co$) one piece has its four corners in the interior of $T$, and in
($\Delta$) one piece meets $\partial T$ in a single point; (F) and ($\Delta$) have no
pinwheel skeleton at all.
\end{remark}

\section{Computational verification}\label{sec:verif}

All statements were checked against independent computations; the scripts (Python~3
and C) accompany these notes.
\begin{itemize}[leftmargin=2em]
\item \texttt{brute.py}, \texttt{brute.c}: exact-cover enumeration of all $k$-tilings of
  $T_n$ by lattice trapezoids with pairwise distinct shapes, in any of the six
  orientations, using no structural information. For $k=3$ it reproduces
  $\lfloor(n-1)(n-2)/6\rfloor$, for $k=4$ the terms of A389392 for $n\le 12$, and for
  $k=5$ the terms of A391498 for $n\le 22$.
\item \texttt{cats.c}: a port to C of Rodrigo's program for the eight categories. Its
  output agrees set by set with the exhaustive enumeration for $n\le 14$, it
  reproduces the $47$ terms of the b-file, and it extends A391498 to $n=127$.
\item \texttt{classify.py}: canonical codes of combinatorial types (the graph $G$ with
  edge directions, minimized over the six symmetries of $T$) and multiplicities.
\item \texttt{verify\_structure.py}: for every tiling produced by the enumeration it
  builds $G$, computes $(a,c,h)$ at every vertex and checks Lemma~\ref{lem:local}, the
  identities \eqref{eq:I1}--\eqref{eq:I2}, and, when there is no full strip,
  Lemma~\ref{lem:bdry}(a), (b), (e), Corollary~\ref{cor:mge1} and the absence of chords
  (Lemma~\ref{lem:nochord}). It was run on all $1\,110$ tilings with $k=3$ and
  $n\le 16$, all $15\,630$ tilings with $k=4$ and $n\le 15$, and all $162\,990$
  tilings with $k=5$ and $n\le 15$. No violation occurred; exactly $1$, $2$ and $13$
  types were found, and the profile of every type is the one predicted by
  Theorems~\ref{thm:k4} and~\ref{thm:k5II}.
\item \texttt{signatures.py}: for $n=13$, the list of types (with multiplicities)
  realizing each set, and the test of Remark~\ref{rem:ops}(2).
\end{itemize}
The enumeration only covers tilings with pairwise distinct pieces; the theorems do not
need this assumption.

\section{Outlook: from types to the counting formula}\label{sec:outlook}

For every type, the tilings of $T_n$ of that type are parametrized by the positions of
the maximal segments of $G$ on their lattice lines, subject to linear equations and
strict linear inequalities. Hence, for each type $\tau$, the number $N_\tau(n)$ of
tilings of $T_n$ of type $\tau$, counted up to symmetry, is a quasi-polynomial in $n$
of degree at most $4$. For instance, the tilings of form (D) with a fixed chirality and
fixed divided regions correspond to $5$-tuples of positive integers with sum $n$ (the
three legs of the pinwheel, two of which are split into two parts by the cuts), so
there are $\binom{n-1}{4}$ of them before the distinctness condition is imposed.

\begin{proposition}[structural coincidences]\label{prop:mult}
Let $R=\{0\le y\le s,\ x\ge 0,\ x+y\le b\}$ be the divided region, with boundary leg on
$\{x=0\}$ and leg-arm on $\{x+y=b\}$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item In \textup{($\Cb$)}, with $\gamma=\{x=p\}$ and the cut of $\Pi$ on $\{x+y=c\}$
  $(1\le p\le b-s-1,\ p<c<s)$, the three pieces in $R$ are
  $\trap{b-p}{s}$, $\trap{c}{p}$ and $\trap{s-c+p}{p}$. In \textup{($\Co$)}, with
  $\gamma=\{x+y=q\}$ and the cut on $\{x=c'\}$ $(s+1\le q\le b-1,\ b-s<c'<q)$, they are
  $\trap{q}{s}$, $\trap{b-c'}{b-q}$ and $\trap{c'-q+s}{b-q}$. The maps
  $(p,c)\mapsto(p,\,s+p-c)$ and $(p,c)\mapsto(q,c')=(b-p,\,b-c)$ are bijections
  preserving the set of shapes. Hence every set of type \textup{($\Cb$)} is realized
  by at least two tilings of type \textup{($\Cb$)} and two of type \textup{($\Co$)}.
\item In \textup{(V)} the three pieces in $R$ are $\trap{\beta_1}{s}$,
  $\trap{\beta_2}{s}$, $\trap{\beta_3}{s}$ with $\beta_1+\beta_2+\beta_3=b+s$ and
  $\beta_i\ge s+1$, and every ordering of $(\beta_1,\beta_2,\beta_3)$ occurs. Hence a
  set of type \textup{(V)} with distinct $\beta_i$ is realized by six tilings.
\item A set of type ``full strip $+$ (A)'' is realized by nine tilings (three
  positions of the second strip and six placements of the pinwheel, modulo the
  reflection fixing the full strip), and a set of type ``full strip $+$ (B)'' by three
  (six placements of the cut pinwheel, modulo the same reflection).
\end{enumerate}
\end{proposition}

\begin{proof}
(a) and (b) are direct computations of the pieces; the constraints are preserved by the
maps (for instance $p<c<s$ if and only if $p<s+p-c<s$, and $b-s<b-c<b-p$ if and only if
$p<c<s$). In (b), $q=\beta_1$ and $p=\beta_1+\beta_2-s$ recover the segments. (c) The
pinwheel pieces depend only on its legs, so any of the six images of the pinwheel under
the symmetries of its triangle gives the same shapes.
\end{proof}

The data for $n=13$ confirm that these are the only coincidences on the generic part:
the $1940$ sets split as $428+6$ sets realized by one tiling of each of T6, T9, T11
(respectively two), $159$ sets realized by three tilings of each of T1, T4, T5, $78$
sets realized by two tilings of each of T7 and T12, $12$ sets realized by six tilings of
T13, $378$, $363$, $354$ and $65$ sets realized by a single tiling of type T3, T8, T2
and T10, and $97$ sets realized by several tilings of types T2, T3, T8 through
non-generic coincidences.

\paragraph{The leading coefficient.} The parameter polytopes have the following
normalized volumes, i.e.\ $N_\tau(n)=v_\tau n^4+O(n^3)$: $v_\tau=\frac1{24}$ for T1,
T2, T3, T4, T5, T6, T8, T9, T11 (for T3, in a fixed position, the tilings correspond
to integers with $1\le v_1<v_2$, $w\ge 1$ and $w+v_2<u<n$, where $y=v_1$, $x=w$,
$y=v_2$ and $x+y=u$ are the lines of four of its five interior maximal segments, the
fifth being determined by them); $v_\tau=\frac1{48}$ for T7, T12 and T13 (for instance, for T13
the number of choices in a region with legs $s_A$ and short base $s_C$ is about
$\frac12(s_C-s_A)_+^2$, and $\iint_{a+c\le 1}\frac12(c-a)_+^2\,da\,dc=\frac1{48}$); and
$v_\tau=0$ for T10, whose tilings depend on only three parameters besides $n$.
Coincidences other than those of Proposition~\ref{prop:mult}, as well as repeated
shapes, are governed by additional linear equations and only affect lower-order
terms. Dividing by the multiplicities of Table~\ref{tab:types} gives
\[
  \frac{3\cdot\frac1{24}}{9}+\frac{3\cdot\frac1{24}}{3}
  +3\cdot\frac1{24}+\frac{2\cdot\frac1{48}}{4}+\frac{\frac1{48}}{6}
  =\frac{4+12+36+3+1}{288}=\frac{7}{36},
\]
which is exactly the leading coefficient $70/360$ of the conjectured quasi-polynomial
$a(n)=(70n^4-695n^3+\cdots)/360$ for A391498. Turning this heuristic into a proof, and
obtaining the lower-order terms, requires counting the tilings of each type with
pairwise distinct pieces (by inclusion--exclusion over the equations ``two pieces have
the same shape'') and controlling the non-generic coincidences; this is the next step
towards a proof of the conjectured generating function.

\input{counting}

\begin{thebibliography}{9}
\bibitem{oeis} OEIS Foundation Inc., \emph{The On-Line Encyclopedia of Integer
  Sequences}, \url{https://oeis.org}.
\bibitem{rodrigo} J. Rodrigo, entries A389392, A391203, A391204 and A391498 in
  \cite{oeis}, with the accompanying programs and illustrations; the conjectured
  generating function in A389392 is due to S.~A.~Irvine.
\end{thebibliography}

\end{document}