\section{The counting programme}\label{sec:counting}

By Theorems~\ref{thm:k5I} and~\ref{thm:k5II} the shapes of the five pieces of a
$5$-tiling are given by explicit formulas in the parameters of its type. Writing
them out, the thirteen types produce only eight different families of
$5$-element sets of shapes: types T1, T4, T5 share their formula, so do T6, T9,
T11, and so do T7 and T12 by Proposition~\ref{prop:mult}(a). For a pinwheel of a
triangle of side $m$ with legs $(u_1,u_2,u_3)$, $u_1+u_2+u_3=m$, write
\[
  \mathcal P(m;u)=\bigl\{\trap{m-u_2}{u_1},\ \trap{m-u_3}{u_2},\
  \trap{m-u_1}{u_3}\bigr\},
\]
and for a region $\trap{b_i}{u_i}$ of $\mathcal P$ note that $b_i-u_i=u_{i+2}$
(indices modulo $3$).

\begin{center}
\begin{tabular}{@{}llp{7.6cm}@{}}
\toprule
Family & Types & Parameters and shapes\\
\midrule
$\mathrm A$ & T1, T4, T5 &
$a,c\ge1$, $m=n-a-c$, $u$ a composition of $m$:\newline
$\trap{n}{a}$, $\trap{n-a}{c}$, $\mathcal P(m;u)$\\[2pt]
$\mathrm B$ & T6, T9, T11 &
$a\ge1$, $u$ a composition of $m=n-a$, $i$, $1\le h<u_i$:\newline
$\trap{n}{a}$, $\mathcal P\setminus\{P_i\}$, $\trap{b_i}{h}$,
$\trap{b_i-h}{u_i-h}$\\[2pt]
$\mathrm D$ & T2 &
$u$ a composition of $n$, $i<j$, $1\le h_i<u_i$, $1\le h_j<u_j$:\newline
$P_k$ and the two splits of $P_i$, $P_j$\\[2pt]
$\mathrm S$ & T8 &
$u$, $i$, $h_1+h_2+h_3=u_i$:\newline
$P_j$, $P_k$, $\trap{b_i}{h_1}$, $\trap{b_i-h_1}{h_2}$,
$\trap{b_i-h_1-h_2}{h_3}$\\[2pt]
$\mathrm C$ & T7, T12 &
$u$, $i$, $1\le p\le u_{i+2}-1$, $p<c<u_i$:\newline
$P_j$, $P_k$, $\trap{b_i-p}{u_i}$, $\trap{c}{p}$, $\trap{u_i-c+p}{p}$\\[2pt]
$\mathrm V$ & T13 &
$u$, $i$, $\beta_1\le\beta_2\le\beta_3$ with $\beta_r\ge u_i+1$ and
$\sum\beta_r=b_i+u_i$:\newline $P_j$, $P_k$, $\trap{\beta_r}{u_i}$\\[2pt]
$\mathrm F$ & T3 &
$1\le v_1<v_2$, $w\ge1$, $w+v_2<u\le n-1$, $p=u-v_2$:\newline
$\trap{u}{v_1}$, $\trap{n-v_1}{w}$, $\trap{n-p}{n-u}$, $\trap{n-w-v_2}{p-w}$,
$\trap{u-v_1-w}{v_2-v_1}$\\[2pt]
$\Delta$ & T10 &
$1\le\mu<m<\beta\le n-\mu-1$:\newline
$\trap{m}{\mu}$, $\trap{\beta}{m}$, $\trap{n-\beta}{\mu}$,
$\trap{n-m}{\beta-m+\mu}$, $\trap{n-\beta+m-2\mu}{n-\beta-\mu}$\\
\bottomrule
\end{tabular}
\end{center}

Thus $a(n)$, the $n$-th term of A391498, is the number of elements of
$\mathrm A\cup \mathrm B\cup \mathrm D\cup \mathrm S\cup \mathrm C\cup
\mathrm V\cup \mathrm F\cup\Delta$ whose five shapes are pairwise different.
Generating all these sets for $n\le 24$ reproduces the terms of A391498 exactly;
this is a verification of the classification that uses no geometry at all.

Two experimental facts organize the count (both checked for all $n\le 63$, resp.\
$n\le 40$):
\begin{enumerate}[label=\textup{(\roman*)}]
\item \emph{Disjointness.} The only families that meet are $\mathrm D$,
  $\mathrm S$ and $\mathrm F$. Writing $E_{\mathrm{DS}}$, $E_{\mathrm{DF}}$,
  $E_{\mathrm{SF}}$, $E_{\mathrm{DSF}}$ for the numbers of sets lying in exactly
  those families,
  \begin{equation}\label{eq:incexc}
    a(n)=\sum_{\mathcal F}|\mathcal F|
    -E_{\mathrm{DS}}-E_{\mathrm{DF}}-E_{\mathrm{SF}}-2E_{\mathrm{DSF}} .
  \end{equation}
\item \emph{Fibres.} The number of parameter tuples with five different shapes
  that produce a given set is $3$ for $\mathrm A$, $\mathrm V$, $6$ for
  $\mathrm C$ and $1$ for $\mathrm F$ and $\Delta$; for $\mathrm B$,
  $\mathrm D$, $\mathrm S$ it is $3$ except for finitely many accidental
  coincidences per $n$ (the quotient tuples/sets is $3.003$ for $n=40$).
\end{enumerate}
The fibres of $\mathrm A$, $\mathrm V$, $\mathrm B$, $\mathrm D$, $\mathrm S$
come from the cyclic rotation of $(u,i)$, and the extra factor $2$ for
$\mathrm C$ from the involution of Proposition~\ref{prop:mult}(a).

\begin{lemma}\label{lem:famA}
For every $n$,
\[
  |\mathrm A|=\frac13\Bigl[\binom{n-1}{4}-\sum_{j\ge1}\bigl(n-3j-1\bigr)_+\Bigr],
  \qquad\text{i.e.}\qquad
  \sum_n |\mathrm A|\,x^n=\frac{x^6}{(1-x)^5\,(1+x+x^2)} .
\]
\end{lemma}

\begin{proof}
In a set of the family $\mathrm A$ the shape with long base $n$ is unique, since
$n-a<n$ and the bases of $\mathcal P(m;u)$ are at most $m-1<n-a$; so $a$ is
determined by the set, and then so are $c$ and $m=n-a-c$, and the three
remaining shapes are those of $\mathcal P(m;u)$. Two shapes of the set coincide
only if $u_1=u_2=u_3$: indeed $\trap{m-u_2}{u_1}=\trap{m-u_3}{u_2}$ forces
$u_2=u_3$ and $u_1=u_2$, and similarly for the other two pairs. Conversely, the
multiset $\mathcal P(m;u)$ determines $(u_1,u_2,u_3)$ up to a cyclic rotation:
the multiset of legs is $\{u_1,u_2,u_3\}$, and if the $u_r$ are pairwise
different, the two cyclic classes of a multiset give different multisets of
pairs, because the shape with leg $u_1$ is $\trap{m-u_2}{u_1}$ in one and
$\trap{m-u_3}{u_1}$ in the other; if two of the $u_r$ are equal there is only
one cyclic class. Hence the tuples $(a,c,u)$ with five different shapes fall
into orbits of exactly three elements under the rotation of $u$, and
$|\mathrm A|$ is a third of their number. Finally, the number of tuples is
$\binom{n-1}{4}$ (compositions of $n$ into five positive parts), of which those
with $u_1=u_2=u_3$, i.e.\ with $m\equiv0 \pmod 3$, number
$\sum_{j\ge1}(n-3j-1)_+$. The generating function follows from
$\sum_n\binom{n-1}{4}x^n=x^5/(1-x)^5$ and
$\sum_n\sum_j (n-3j-1)_+x^n=x^5/\bigl((1-x^3)(1-x)^2\bigr)$.
\end{proof}

\begin{lemma}[family $\Delta$]\label{lem:famX}
Write $(x)_+=\max(x,0)$. For every $n$,
\[
  |\Delta|=\sum_{a\ge1}\binom{n-1-2a}{2}
  -\binom{\lfloor(n-1)/2\rfloor}{2}
  -\sum_{\mu\ge1}(n-3\mu-1)_+
  -[\,3\mid n\,]\Bigl(\tfrac n3-1\Bigr)
  +\Bigl\lfloor\tfrac{n-1}4\Bigr\rfloor .
\]
\end{lemma}

\begin{proof}
Let $P_1,\dots,P_5$ be the five shapes in the order of the table. Putting
$\mu=a$, $m=\mu+b$, $\beta=m+c$ and $d=n-\mu-1-\beta$ identifies the parameter
set with $\{(a,b,c,d):a,b,c\ge1,\ d\ge0,\ 2a+b+c+d=n-1\}$, which has
$\sum_{a\ge1}\binom{n-1-2a}{2}$ elements. Comparing the shapes pairwise, the only
possible coincidences are
\[
\begin{array}{ll}
  P_1=P_3\iff\beta=n-m, &
  P_1=P_5\iff\beta=n-2\mu,\\[2pt]
  P_2=P_4\iff\beta=n-m,\ n=3m-\mu, &
  P_2=P_5\iff\beta=n-m-\mu,\ n=3m,\\[2pt]
  P_3=P_5\iff m=2\mu,\ \beta=n-2\mu, &
  P_4=P_5\iff\beta=2m-2\mu,\ n=3m-2\mu,
\end{array}
\]
whereas $P_1=P_2$, $P_1=P_4$, $P_2=P_3$ and $P_3=P_4$ contradict
$\mu<m<\beta$. The loci of $P_2=P_4$ and of $P_4=P_5$ are contained in the locus
$\beta=n-m$, and that of $P_3=P_5$ in the locus $\beta=n-2\mu$; so the union of
the six is the union of $\{\beta=n-m\}$, $\{\beta=n-2\mu\}$ and
$\{\beta=n-m-\mu,\ n=3m\}$. On the first one the constraints reduce to
$1\le\mu<m<n/2$, which gives $\binom{\lfloor(n-1)/2\rfloor}{2}$ points; on the
second to $\mu\ge1$, $\mu<m<n-2\mu$, which gives $\sum_\mu(n-3\mu-1)_+$; the
third is non-empty only if $3\mid n$, and then $m=n/3$ and $1\le\mu<n/3$. The
first two meet exactly when $m=2\mu<n/2$, in $\lfloor(n-1)/4\rfloor$ points, and
the other intersections are empty. As the fibre is $1$, $|\Delta|$ is the number
of triples whose five shapes are different.
\end{proof}

\begin{lemma}[family $\mathrm C$]\label{lem:famC}
Write $s=u_i$, $j=u_{i+1}$, $k=u_{i+2}$, so that the five shapes are
$\trap{s+j}{j}$, $\trap{j+k}{k}$, $\trap{s+k-p}{s}$, $\trap{c}{p}$ and
$\trap{s+p-c}{p}$, with $1\le p\le k-1$ and $p<c<s$. Two of them coincide
exactly in the four cases
\[
  \mathrm E_1:\ 2c=s+p,\qquad \mathrm E_2:\ s=j=k,\qquad
  \mathrm E_3:\ j=s,\ p=k-s,\qquad \mathrm E_4:\ k=s,\ p=s-j,
\]
of which the last three are pairwise disjoint, and
\[
  6\,|\mathrm C|=3\bigl(T-E_1-E_2-E_3-E_4+I_{12}+I_{13}+I_{14}\bigr)
\]
with, all sums being over the compositions $n=s+j+k$,
\begin{align*}
  T&=\#\{(\alpha,\gamma,\delta,\varepsilon,\eta)\ge0:\
     2\alpha+\gamma+\delta+\varepsilon+\eta=n-6\},\\
  E_1&=\sum\#\{p:1\le p\le\min(k-1,s-2),\ p\equiv s\ (\mathrm{mod}\ 2)\},\qquad
  E_2=[\,3\mid n\,]\binom{n/3-1}{2},\\
  E_3&=\sum_{1\le s\le(n-1)/3}(4s-n-1)_+,\qquad
  E_4=\sum_{s>n/3}(n-2s-1)_+,\\
  I_{12}&=[\,3\mid n\,]\#\{1\le p\le\tfrac n3-2:p\equiv\tfrac n3\ (2)\},\\
  I_{13}&=[\,2\mid n\,]\#\{s:\lceil\tfrac{n+2}4\rceil\le s\le\lfloor\tfrac{n-1}3\rfloor\},
  \qquad
  I_{14}=[\,2\mid n\,]\#\{s:\lceil\tfrac{n+1}3\rceil\le s\le\lfloor\tfrac{n-2}2\rfloor\}.
\end{align*}
\end{lemma}

\begin{proof}
The pairs of shapes that can coincide are found as in Lemma~\ref{lem:famX}; the
two shapes with leg $p$ coincide iff $2c=s+p$, the two pinwheel pieces iff
$s=j=k$, and $\trap{s+k-p}{s}$ equals the first or the second pinwheel piece iff
$\mathrm E_3$ or $\mathrm E_4$ holds; all other pairs are impossible because
$p<c<s$ and $p<k$. Putting $p=1+\alpha$, $c=p+1+\gamma$, $s=3+\alpha+\gamma+\delta$,
$k=2+\alpha+\varepsilon$ and $j=1+\eta$ identifies the parameters (for a fixed
$i$) with the solutions counted by $T$. On $\mathrm E_1$ the value of $c$ is
determined by $p$, and $p<c<s$ is equivalent to $p\le s-2$; on $\mathrm E_3$ one
has $j=s$, $k=n-2s$, $p=n-3s$ and $c$ ranges over $s-1-p=4s-n-1$ values; on
$\mathrm E_4$, $k=s$, $j=n-2s$, $p=3s-n$ and $c$ ranges over $n-2s-1$ values; on
$\mathrm E_2$ the pair $(p,c)$ is free. The three intersections
$\mathrm E_1\cap\mathrm E_r$ are computed in the same way, and $\mathrm E_2$,
$\mathrm E_3$, $\mathrm E_4$ are pairwise disjoint because each pair of them
forces $p=0$. The factor $3$ is the number of indices $i$, and the fibre is $6$.
\end{proof}

\begin{lemma}[family $\mathrm V$]\label{lem:famV}
Let $d(N)=\#\{(a,b,c)\ge0:3a+2b+c=N\}$ and, for $K,v\ge0$, let $e(K,v)$ be the
number of sets of three different non-negative integers with sum $K$ one of
which is $v$. Then
\[
  |\mathrm V|=\sum_{\substack{s+j+k=n\\ k\ge s+3}}
  \Bigl[\,d(k-s-6)-[\,j=s\,]\;e(k-s-3,\ s-1)\Bigr].
\]
\end{lemma}

\begin{proof}
For a fixed index $i$ the three pieces of the divided region are
$\trap{\beta_r}{s}$ with $\beta_r\ge s+1$ and $\beta_1+\beta_2+\beta_3=2s+k$;
writing $\beta_r=s+1+\gamma_r$, the $\gamma_r$ are non-negative with sum
$K=k-s-3$, and they must be pairwise different, since equal $\gamma$'s give equal
shapes. Ordering $\gamma_1<\gamma_2<\gamma_3$ and putting $\gamma_1=a$,
$\gamma_2=\gamma_1+1+b$, $\gamma_3=\gamma_2+1+c$ shows that there are $d(K-3)$
such sets. The two remaining shapes are $\trap{s+j}{j}$ and $\trap{j+k}{k}$; they
are different from each other, and $\trap{\beta_r}{s}=\trap{j+k}{k}$ would force
$k=s$, which contradicts $k\ge s+3$. The only remaining coincidence is
$\trap{\beta_r}{s}=\trap{s+j}{j}$, which holds iff $j=s$ and $\beta_r=2s$, i.e.\
$\gamma_r=s-1$; this removes $e(K,s-1)$ of the sets when $j=s$. Finally the fibre
is exactly $3$: as $k\ge s+3$, the leg $s$ occurs at least three times among the
five legs and $k$ does not equal $s$, so the three pieces of the region are
determined unless $j=s$; and if $j=s$, replacing one $\beta_r$ by the base $2s$ of
$\trap{s+j}{j}$ would force the discarded shape $\trap{\beta_r}{s}$ to play the
role of $\trap{s+j'}{j'}$ with $j'=s$, i.e.\ $\beta_r=2s$, which has just been
excluded. Hence the three rotations of $(u,i)$ are the only tuples giving the
same set, and summing over $i$ and dividing by $3$ gives the statement.
\end{proof}

\begin{proposition}[composition coordinates]\label{prop:comp}
In the families $\mathrm A$, $\mathrm B$, $\mathrm D$, $\mathrm S$ and $\mathrm F$
the parameters can be written as a composition $n=x_1+\dots+x_5$ into five
positive parts in such a way that the legs of the five pieces are exactly
$x_1,\dots,x_5$, while their long bases are the linear forms of the following
table (where $m=x_3+x_4+x_5$):
\begin{center}
\begin{tabular}{@{}llllll@{}}
\toprule
& leg $x_1$ & leg $x_2$ & leg $x_3$ & leg $x_4$ & leg $x_5$\\
\midrule
$\mathrm A$ & $n$ & $n-x_1$ & $m-x_4$ & $m-x_5$ & $m-x_3$\\
$\mathrm B$ & $n$ & $n-x_1-x_3$ & $x_2+x_3$ & $n-x_1-x_2$ & $x_3+x_5$\\
$\mathrm D$ & $n-x_3-x_4$ & $x_2+x_5$ & $n-x_5$ & $n-x_3-x_5$ & $n-x_1-x_2$\\
$\mathrm S$ & $n-x_4$ & $n-x_1-x_4$ & $x_3+x_5$ & $n-x_5$ & $x_4+x_5$\\
$\mathrm F$ & $n-x_5$ & $x_2+x_4$ & $n-x_1$ & $x_4+x_5$ & $n-x_3-x_4$\\
\bottomrule
\end{tabular}
\end{center}
In $\mathrm A$ the three cyclic rotations of $(x_3,x_4,x_5)$ give the same set;
in $\mathrm B$, $\mathrm D$, $\mathrm S$ and $\mathrm F$ the composition is in
bijection with the tiling. In all five cases the number of tuples is
$\binom{n-1}{4}$.
\end{proposition}

In the next three lemmas we write
$N_{abc}=\#\{(x,y,z)\ge1:ax+by+cz=n\}$ and $N_{ab}=\#\{(x,y)\ge1:ax+by=n\}$.

\begin{lemma}[family $\mathrm F$]\label{lem:famF}
\[
  |\mathrm F|=\binom{n-1}{4}-2N_{311}-2N_{331}-2N_{321}-N_{322}
  +2\,[\,6\mid n\,]+[\,5\mid n\,].
\]
\end{lemma}

\begin{proof}
In the coordinates of Proposition~\ref{prop:comp} two of the five shapes of
$\mathrm F$ coincide exactly on one of the seven loci
\[
\begin{array}{lll}
 (1)\ x_1=x_3=x_5, & (2)\ x_1=x_5=x_3+x_4, & (3)\ x_1=x_4,\ x_5=x_1+x_2+x_3,\\
 (4)\ x_1=x_3+x_4,\ x_5=x_3, & (5)\ x_5=x_4=x_1+x_2, & (6)\ x_5=x_2,\ x_4=x_1+x_2,\\
 (7)\ x_2=x_4=x_5, & &
\end{array}
\]
which are the loci of $S_1=S_2$, $S_1=S_3$, $S_1=S_4$, $S_2=S_3$, $S_3=S_4$,
$S_3=S_5$ and $S_4=S_5$ respectively, the three remaining pairs being
impossible. Eliminating the two equations of each locus expresses $n$ as
$3x+y+z$, $3x+3y+z$, $3x+2y+2z$, $3x+2y+z$, $3x+3y+z$, $3x+2y+z$ and $x+3y+z$
in three free positive variables, whence the cardinalities
$N_{311},N_{331},N_{322},N_{321},N_{331},N_{321},N_{311}$. Any two of the seven
loci are disjoint except $(1)\cap(6)$ and $(4)\cap(7)$, which are the single
point $x_1=\dots=x_5=n/6$ repeated, i.e.\ each of cardinality $[\,6\mid n\,]$,
and $(1)\cap(7)$, which is $x_1=\dots=x_5=n/5$, of cardinality $[\,5\mid n\,]$;
there are no triple intersections. As the fibre is $1$, the statement follows.
\end{proof}

\begin{lemma}[families $\mathrm B$, $\mathrm D$, $\mathrm S$: tuples]\label{lem:famBDS}
The numbers $T_{\mathrm B},T_{\mathrm D},T_{\mathrm S}$ of compositions whose five
shapes are pairwise different are
\begin{align*}
 T_{\mathrm B}&=\binom{n-1}{4}-2N_{311}-2N_{321}-N_{331}
   +\Bigl\lfloor\tfrac{n-1}4\Bigr\rfloor,\\
 T_{\mathrm S}&=\binom{n-1}{4}-2N_{311}-2N_{321}-2N_{322}
   -[\,3\mid n\,]\binom{n/3-1}{2}+\Bigl\lfloor\tfrac{n-1}4\Bigr\rfloor,\\
 T_{\mathrm D}&=\binom{n-1}{4}-N_{311}-2N_{321}-N_{322}-2N_{331}-D_3-D_8
   +[\,3\mid n\,]\Bigl(\tfrac n3-1\Bigr)+2N_{34}+2N_{32},
\end{align*}
where $D_3=\sum_{3a+2b=n,\ a,b\ge1}(a+b-1)$ and
$D_8=\sum_{\sigma}(\sigma-1)$, the sum being over
$\lceil (n+1)/3\rceil\le\sigma\le\lfloor (n-1)/2\rfloor$.
\end{lemma}

\begin{proof}
As in Lemma~\ref{lem:famF}: in each family the pairs of shapes that can coincide
give five, eight and seven loci respectively, each of them defined by two linear
equations, and eliminating them writes $n$ as a positive combination of three
free variables. For $\mathrm B$ the only non-empty intersection is
$\{x_2=x_3=x_4\}\cap\{x_2=x_3=x_5\}=\{x_2=x_3=x_4=x_5\}$, of cardinality
$\lfloor(n-1)/4\rfloor$; for $\mathrm S$ it is
$\{x_1=x_4=x_5\}\cap\{x_3=x_4=x_5\}$, of the same cardinality; for $\mathrm D$
the non-empty ones are four, of cardinalities $[\,3\mid n\,](n/3-1)$, $N_{43}$,
$N_{34}$ and, three times, $N_{32}$, the last three loci meeting in a common
set counted once, which gives the stated $+2N_{32}$ (note $N_{43}=N_{34}$).
\end{proof}

\begin{lemma}[families $\mathrm B$ and $\mathrm S$: identifications]\label{lem:BSid}
Two different compositions give the same set precisely in the following cases,
and then in pairs: in $\mathrm B$, on the locus $x_3=x_4$, $x_2=x_5$, the
compositions $(x_1,x_2,x_3,x_4,x_5)$ and $(x_1,x_3,x_2,x_5,x_4)$; in
$\mathrm S$, on the locus $x_3=x_4$, $x_1=x_5$, the compositions
$(x_1,x_2,x_3,x_4,x_5)$ and $(x_3,x_2,x_1,x_5,x_4)$. Consequently
\[
 |\mathrm B|=T_{\mathrm B}-E,\qquad |\mathrm S|=T_{\mathrm S}-E,\qquad
 E=\frac12\Bigl[\binom{\lfloor (n-1)/2\rfloor}{2}
   -\Bigl\lfloor\tfrac{n-1}4\Bigr\rfloor\Bigr].
\]
\end{lemma}

\begin{proof}
On the locus of $\mathrm S$ the composition is $(x_1,x_2,x_3,x_3,x_1)$ and the
five shapes are $\trap{n-x_3}{x_1}$, $\trap{n-x_1-x_3}{x_2}$,
$\trap{x_1+x_3}{x_3}$, $\trap{n-x_1}{x_3}$ and $\trap{x_1+x_3}{x_1}$, which is
symmetric under $x_1\leftrightarrow x_3$; the two compositions are different
exactly when $x_1\ne x_3$. Since $n=2x_1+x_2+2x_3$, the number of points of the
locus is $\binom{\lfloor(n-1)/2\rfloor}{2}$ and the number of fixed points of the
involution is $\lfloor(n-1)/4\rfloor$, whence $E$. The case of $\mathrm B$ is
identical. That there are no further identifications was verified for all
$n\le40$.
\end{proof}

\begin{proposition}[identifications and intersections]\label{prop:loci}
In the coordinates of Proposition~\ref{prop:comp}, two compositions of the same
family give the same set, or a composition of one family gives the same set as a
composition of another, exactly on the loci listed below; each of them is cut out
by two linear equations, so it is two-dimensional once $n$ is fixed.
\begin{center}
\begin{tabular}{@{}ll@{}}
\toprule
pieces & loci\\
\midrule
$\mathrm F$ with itself & none\\
$\mathrm B$ with itself & $x_3=x_4,\ x_2=x_5$\\
$\mathrm S$ with itself & $x_3=x_4,\ x_1=x_5$\\
$\mathrm D$ with itself & $x_3=x_1+x_2,\ x_5=x_1+x_4$;\quad
   $x_3+x_4=x_1+x_2,\ x_5=x_1+x_2$;\\
 & $x_2=x_3+x_4,\ x_3+x_5=x_1+x_2$;\quad $x_2=x_3,\ x_1=x_5$;\quad
   $x_2=x_3,\ x_4=x_5$\\
\midrule
$\mathrm D$ and $\mathrm S$ & $x_3=x_1+x_2$ with $x_5=x_1+x_2+x_4$, with $x_4=x_5$
   or with $x_5=x_1$;\\
 & $x_2=x_3+x_4$ with $x_5=x_1+x_2$, with $x_4=x_5$ or with $x_5=x_1$;\\
 & $x_4=x_5=x_1+x_2$;\quad $x_1=x_3+x_4,\ x_5=x_1$\\
$\mathrm D$ and $\mathrm F$ & $x_1=x_3+x_4,\ x_4=x_5$;\quad $x_4=x_1+x_2,\ x_5=x_1$\\
$\mathrm S$ and $\mathrm F$ & $x_4=x_2,\ x_5=x_1$;\quad $x_4=x_5=x_2$;\quad
   $x_4=x_3,\ x_5=x_2$\\
\bottomrule
\end{tabular}
\end{center}
The loci were obtained by solving, for each of the $120$ permutations of the five
parts, the linear system expressing that the two compositions have the same five
shapes, and discarding the systems without solutions having all parts positive or
on which the permutation acts trivially. The resulting description reproduces the
enumeration for every $n\le63$ (and, for the identifications of $\mathrm D$,
for every $n\le22$).
\end{proposition}

That the family $\mathrm F$ admits no identification is exactly the statement that
its fibre is $1$, used in Lemma~\ref{lem:famF}; for $\mathrm B$ and $\mathrm S$ one
recovers the loci of Lemma~\ref{lem:BSid}.

\begin{lemma}\label{lem:DF}
The number of sets belonging to both $\mathrm D$ and $\mathrm F$ is
\[
  |\mathrm D\cap\mathrm F|=2N_{321}-[\,3\mid n\,]\Bigl(\frac n3-1\Bigr).
\]
\end{lemma}

\begin{proof}
The two loci are $\{x_1=x_3+x_4,\ x_4=x_5\}$, on which
$n=x_2+2x_3+3x_4$, and $\{x_4=x_1+x_2,\ x_5=x_1\}$, on which $n=3x_1+2x_2+x_3$;
each therefore contains $N_{321}$ points, and they are disjoint because together
they force $x_2=0$. No point of either locus has two equal shapes, and two points
give the same set exactly $[\,3\mid n\,](n/3-1)$ times. The equality was checked
for every $n\le63$.
\end{proof}

\begin{lemma}\label{lem:SFu}
The union of the three loci of $\mathrm S$ and $\mathrm F$ contains
$2N_{221}+N_{131}-2\lfloor (n-1)/4\rfloor$ compositions.
\end{lemma}

\begin{proof}
On the three loci $n$ equals $2x_1+2x_2+x_3$, $x_1+3x_2+x_3$ and $x_1+2x_2+2x_3$
respectively, which gives $N_{221}$, $N_{131}$ and $N_{221}$ points. The first and
the second meet where $x_1=x_2=x_4=x_5$, and the second and the third where
$x_2=x_3=x_4=x_5$, each of these being $\lfloor(n-1)/4\rfloor$ points; the first
and the third meet exactly at the point $x_1=\dots=x_5=n/5$, which lies on all
three, so it is added back once.
\end{proof}

\begin{remark}
The point $x_1=\dots=x_5=n/5$ appears twice in this section: it is the
intersection $(1)\cap(7)$ of the loci of Lemma~\ref{lem:famF} and the triple
intersection of the loci of Lemma~\ref{lem:SFu}. It produces the terms
$[\,5\mid n\,]$ of $|\mathrm F|$ and of $|\mathrm S\cap\mathrm F|$, which are the
only terms of those two pieces whose generating function has a denominator
divisible by $1-x^5$. The third piece with the cyclotomic factor
$1+x+x^2+x^3+x^4$ is $|\mathrm V|$, where the period $5$ comes from a different
source: the condition $5s\le n-1$ in the correction term $e(k-s-3,s-1)$ of
Lemma~\ref{lem:famV}, which removes the sets in which one of the three shapes
$\trap{\beta_r}{s}$ equals the pinwheel shape $\trap{s+j}{j}=\trap{2s}{s}$ (the case
$j=s$).
So the period $60$ of A391498 is produced by the families $\mathrm F$ and
$\mathrm V$ alone, and not only by the configuration with five equal
parameters.
\end{remark}

\begin{lemma}[$\mathrm S\cap\mathrm F$]\label{lem:SF}
Writing $N_{ab}=\#\{(x,y)\ge1:ax+by=n\}$,
\[
 |\mathrm S\cap\mathrm F|=2N_{221}+N_{131}
 -4\Bigl\lfloor\tfrac{n-1}4\Bigr\rfloor-N_{23}+2\,[\,5\mid n\,].
\]
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:SFu} the three loci contain $2N_{221}+N_{131}-2\lfloor(n-1)/4
\rfloor$ compositions. Two shapes of such a composition coincide only on
$\{x_1=x_2=x_4=x_5\}\cup\{x_2=x_3=x_4=x_5\}$: on the first locus the only possible
coincidence is between the shapes with bases $n-x_4$ and $n-x_5$, which needs
$x_1=x_2$; on the second these two and the shapes with bases $x_3+x_5$ and
$x_4+x_5$ can coincide, which needs $x_1=x_2$ or $x_2=x_3$; on the third only the
latter. Each of the two sets has $\lfloor(n-1)/4\rfloor$ points and they meet in
the point $x_1=\dots=x_5=n/5$, so $2\lfloor(n-1)/4\rfloor-[\,5\mid n\,]$ points are
removed. Finally, two of the remaining compositions give the same set exactly when
they lie on the identification locus $\{x_3=x_4,\ x_1=x_5\}$ of $\mathrm S$ of
Lemma~\ref{lem:BSid}, whose intersection with the three loci is
$\{x_2=x_3=x_4,\ x_1=x_5\}\cup\{x_1=x_2=x_5,\ x_3=x_4\}$, with
$2N_{23}-[\,5\mid n\,]$ points; discarding the all-equal one, which has already been
removed, this gives $N_{23}-[\,5\mid n\,]$ identified pairs.
\end{proof}

\begin{lemma}[triple intersection]\label{lem:DSF}
$E_{\mathrm{DSF}}=2N_{34}$.
\end{lemma}

\begin{proof}
Intersecting the eight loci of $\mathrm D\cap\mathrm S$ with the two of
$\mathrm D\cap\mathrm F$ leaves only two non-empty ones, namely
$\{x_1=x_2=x_3+x_4,\ x_4=x_5\}$ and $\{x_3=x_4=x_1+x_2,\ x_5=x_1\}$, on which
$n=3x_3+4x_4$ and $n=4x_1+3x_2$ respectively. They are disjoint, no point of them
has two equal shapes and no two points give the same set.
\end{proof}

Since $E_{\mathrm{DF}}=|\mathrm D\cap\mathrm F|-E_{\mathrm{DSF}}$ and
$E_{\mathrm{SF}}=|\mathrm S\cap\mathrm F|-E_{\mathrm{DSF}}$, Lemmas~\ref{lem:DF},
\ref{lem:SF} and~\ref{lem:DSF} settle three of the four intersection terms. For the
remaining family we record the following.

\begin{lemma}[the loci of $\mathrm D$]\label{lem:Dloci}
On the five identification loci of $\mathrm D$ listed in
Proposition~\ref{prop:loci}, $n$ equals respectively
\[
 3x_1+2x_2+2x_4,\quad 3x_1+3x_2,\quad 2x_1+2x_3+3x_4,\quad
 2x_1+2x_2+x_4,\quad x_1+2x_2+2x_4 ,
\]
so they contain $N_{322}$, $[\,3\mid n\,](n/3-1)^2$, $N_{322}$, $N_{221}$ and
$N_{221}$ compositions. No point of their union is fixed by the corresponding
permutation, so that
$|\mathrm D|=T_{\mathrm D}-\frac12(|L_1\cup\dots\cup L_5|-\rho)$, where $\rho$
counts the points of the union with two equal shapes.
\end{lemma}

\begin{lemma}[$\mathrm D$: identifications]\label{lem:Did}
The five loci of $\mathrm D$ are pairwise disjoint except for
$L_4\cap L_5=\{x_1=x_4=x_5,\ x_2=x_3\}$, which has $N_{32}$ points, and the points
of their union with two equal shapes form the four disjoint sets
\[
\begin{gathered}
 \{x_1=x_4=x_5,\ x_2=x_3\},\qquad \{x_2=x_4,\ x_3=x_5=x_1+x_2\},\\
 \{x_1=x_3,\ x_2=x_5=x_1+x_4\},\qquad \{x_1=x_3,\ x_2=x_4,\ x_5=x_1+x_2\},
\end{gathered}
\]
of $N_{32}$, $N_{34}$, $N_{34}$ and $[\,3\mid n\,](n/3-1)$ points. Hence
\[
 |\mathrm D|=T_{\mathrm D}-N_{322}-N_{221}+N_{32}+N_{34}
 -[\,3\mid n\,]\binom{n/3-1}{2}.
\]
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:Dloci} the union has
$2N_{322}+[\,3\mid n\,](n/3-1)^2+2N_{221}-N_{32}$ points. Removing the
$N_{32}+2N_{34}+[\,3\mid n\,](n/3-1)$ points with two equal shapes and halving,
which is legitimate because no point is fixed by its permutation, gives
$N_{322}+N_{221}-N_{32}-N_{34}+[\,3\mid n\,]\binom{n/3-1}{2}$ identified pairs,
and this is what must be subtracted from $T_{\mathrm D}$.
\end{proof}

\begin{lemma}[$\mathrm D\cap\mathrm S$]\label{lem:DS}
\[
 |\mathrm D\cap\mathrm S|=2N_{332}+2N_{331}+2N_{321}
 +2\,[\,2\mid n\,]\binom{n/2-1}{2}-4\,[\,4\mid n\,]\Bigl(\frac n4-1\Bigr)
 -4\,[\,2\mid n\,]\Bigl\lfloor\frac{n/2-1}{2}\Bigr\rfloor-N_{34}.
\]
\end{lemma}

\begin{proof}
On the eight loci of Proposition~\ref{prop:loci}, $n$ equals $3x_1+3x_2+2x_4$,
$2x_1+3x_3+3x_4$, $3x_1+3x_2+x_3$, $x_1+2x_3+3x_4$, $x_2+3x_3+3x_4$,
$2(x_1+x_3+x_4)$, $2(x_1+x_2+x_4)$ and $3x_1+2x_2+x_4$, so that they contain
$N_{332}$, $N_{332}$, $N_{331}$, $N_{321}$, $N_{331}$,
$[\,2\mid n\,]\binom{n/2-1}{2}$, $[\,2\mid n\,]\binom{n/2-1}{2}$ and $N_{321}$
compositions. Of the twenty-eight pairs only four meet: the third with the seventh
and the fifth with the sixth, in $\{x_3=x_4=x_5=x_1+x_2\}$ and
$\{x_1=x_2=x_3+x_4,\ x_5=x_1\}$, with $[\,4\mid n\,](n/4-1)$ points each; and the
fourth with the sixth and the seventh with the eighth, in
$\{x_1=x_4=x_5,\ x_2=x_3+x_4\}$ and $\{x_1=x_4=x_5,\ x_3=x_1+x_2\}$, with
$[\,2\mid n\,]\lfloor(n/2-1)/2\rfloor$ points each; there is no triple
intersection. These same four sets are exactly the points of the union having two
equal shapes. Finally two compositions of the union give the same set exactly when
they lie on an identification locus of $\mathrm D$, which occurs on
$\{x_2=x_3,\ x_1=x_2+x_4,\ x_5=x_1\}$ and $\{x_2=x_3,\ x_4=x_1+x_2,\ x_5=x_4\}$,
i.e.\ in $N_{34}$ pairs.
\end{proof}

\begin{theorem}\label{thm:formula}
With the twelve terms given by Lemmas~\ref{lem:famA}, \ref{lem:famX},
\ref{lem:famC}, \ref{lem:famV}, \ref{lem:famF}, \ref{lem:famBDS}, \ref{lem:BSid},
\ref{lem:DF}, \ref{lem:SF}, \ref{lem:DSF}, \ref{lem:Did} and~\ref{lem:DS},
\[
 a(n)=|\mathrm A|+|\mathrm B|+|\mathrm D|+|\mathrm S|+|\mathrm C|+|\mathrm V|
 +|\mathrm F|+|\Delta|-|\mathrm D\cap\mathrm S|-|\mathrm D\cap\mathrm F|
 -|\mathrm S\cap\mathrm F|+E_{\mathrm{DSF}} ,
\]
and this closed expression equals the generating function conjectured for
A391498.
\end{theorem}

\begin{proof}
The displayed identity is \eqref{eq:incexc} rewritten by means of
$E_{\mathrm{DS}}=|\mathrm D\cap\mathrm S|-E_{\mathrm{DSF}}$ and the two analogous
equalities. Each of the twelve terms is a finite integral combination of
$\binom{n-1}{4}$, of the quantities $N_{abc}$ and $N_{ab}$, of binomials in
$\lfloor n/d\rfloor$ and of periodic terms, hence has a rational generating
function whose denominator divides
$D(x)=(1-x)^5(1-x^2)^2(1-x^3)^3(1-x^4)^2(1-x^5)(1-x^6)$, of degree $37$; the
conjectured generating function has the same property. Two power series
$P_1/D$ and $P_2/D$ with $\deg P_i\le 60$ which agree in their first $127$
coefficients are equal, and the first $127$ values of both sides do agree
(script \texttt{formula.py}). As a further check, the closed formula agrees with the
conjectured generating function up to $n=200$, and $D(x)\sum_n a(n)x^n$ has no non-zero
coefficient between $x^{38}$ and $x^{200}$ (script \texttt{verify\_formula\_200.py}).
\end{proof}

\begin{remark}
The proof of Theorem~\ref{thm:formula} rests on three statements that were
established by computation rather than by hand: the list of families and their
shape formulas (Proposition~\ref{prop:comp}), obtained from the classification;
the fact that among the eight families only $\mathrm D$, $\mathrm S$ and
$\mathrm F$ meet (checked for every $n\le63$); and the completeness of the lists
of loci of Proposition~\ref{prop:loci}, obtained by solving the $120$ linear
systems attached to the permutations of the five parts and discarding those
without admissible solutions. Everything else is elementary counting.
\end{remark}

All these formulas, together with Lemma~\ref{lem:famA}, were checked against the
enumeration for every $n\le63$ (for $\mathrm C$, $\mathrm B$, $\mathrm D$ and
$\mathrm S$, against the number of parameter tuples up to $n=40$); the scripts are
\texttt{lemmas.py}, \texttt{collisions.py} and \texttt{excess.py}. Seven of the
twelve pieces of \eqref{eq:incexc} are now given by closed formulas, all of which
were checked against the enumeration for every $n\le63$, and their combination
reproduces A391498 for every $n\le127$ (Theorem~\ref{thm:formula}).

For all twelve pieces we have determined the generating functions
numerically: each was fitted as a rational function using the terms up to
$n=45$ and then verified against the terms up to $n=63$ (the data were produced
by \texttt{families.c}, which generates the eight families directly from the
formulas of the table above). They are
\[
\renewcommand{\arraystretch}{2.3}
\begin{array}{ll}
|\mathrm A| & \dfrac{x^6}{(1-x)^5(1+x+x^2)}\\
|\mathrm B| & \dfrac{x^7(6+14x+23x^2+22x^3+11x^4+2x^5-6x^6)}
  {(1-x)^5(1+x)^2(1+x^2)(1+x+x^2)^2}\\
|\mathrm D| & \dfrac{x^6(1+5x+19x^2+35x^3+59x^4+56x^5+51x^6+9x^7-2x^8-17x^9)}
  {(1-x)^5(1+x)^2(1+x^2)(1+x+x^2)^3}\\
|\mathrm S| & \dfrac{x^7(5+21x+40x^2+62x^3+55x^4+38x^5+6x^6-3x^7-8x^8)}
  {(1-x)^5(1+x)^2(1+x^2)(1+x+x^2)^3}\\
|\mathrm C| & \dfrac{x^7(1+2x+5x^2+6x^3+8x^4+8x^5+8x^6+9x^7+4x^8+5x^9-2x^{10})}
  {(1-x)^5(1+x)^2(1+x^2)(1-x+x^2)(1+x+x^2)^3}\\
|\mathrm V| & \dfrac{x^{10}(1+x+x^2+3x^3+2x^4+4x^6-2x^7)}
  {(1-x)^5(1+x)^2(1+x^2)(1-x+x^2)(1+x+x^2)(1+x+x^2+x^3+x^4)}\\
|\mathrm F| & \dfrac{x^6(1+3x+14x^2+23x^3+30x^4+31x^5+37x^6+23x^7+14x^8+7x^9-3x^{11})}
  {(1-x)^5(1+x)^2(1-x+x^2)(1+x+x^2)^2(1+x+x^2+x^3+x^4)}\\
|\Delta| & \dfrac{x^7(2+6x+7x^2+11x^3+7x^4+3x^5)}
  {(1-x)^4(1+x)^2(1+x^2)(1+x+x^2)^2}\\
E_{\mathrm{DS}} & \dfrac{x^7(1+8x+22x^2+45x^3+63x^4+64x^5+42x^6+19x^7)}
  {(1-x)^3(1+x)^3(1+x^2)^2(1+x+x^2)^2}\\
E_{\mathrm{DF}} & \dfrac{x^6(1+4x^2+2x^3+5x^4)}
  {(1-x)^3(1+x)(1+x^2)(1+x+x^2)^2}\\
E_{\mathrm{SF}} & \dfrac{2x^7(1+2x+2x^2)(1+2x+2x^2+2x^3+3x^4)}
  {(1-x)^3(1+x)^2(1+x^2)(1+x+x^2)(1+x+x^2+x^3+x^4)}\\
E_{\mathrm{DSF}} & \dfrac{2x^7}{(1-x)^2(1+x)(1+x^2)(1+x+x^2)}
\end{array}
\]

\begin{proposition}\label{prop:sum}
The combination \eqref{eq:incexc} of these twelve rational functions is
identically equal to the conjectured generating function of A391498,
\[
  \frac{x^{6}\,N(x)}{(1-x)^{5}(1+x)^{3}(1+x^{2})^{2}(1-x+x^{2})(1+x+x^{2})^{3}
  (1+x+x^{2}+x^{3}+x^{4})},
\]
where
\begin{align*}
N(x)={}&2+22x+105x^{2}+300x^{3}+678x^{4}+1216x^{5}+1887x^{6}+2510x^{7}+2961x^{8}\\
&+3074x^{9}+2816x^{10}+2264x^{11}+1522x^{12}+846x^{13}+281x^{14}\\
&-2x^{15}-147x^{16}-102x^{17}-73x^{18}.
\end{align*}
\end{proposition}

\begin{proof}
A computation with exact rational arithmetic (\texttt{fit\_families.py}).
\end{proof}

\begin{remark}
The factor $1+x+x^2+x^3+x^4$, which is responsible for the period $60$ of the
quasi-polynomial of A391498, occurs in exactly three of the twelve pieces:
$|\mathrm V|$, $|\mathrm F|$ and $E_{\mathrm{SF}}$. These come from the forms
(V) and (F), that is, from the two configurations of Theorem~\ref{thm:k5II}
which cannot be obtained from a $4$-tiling by the operations of A389392
(Remark~\ref{rem:ops}(2)). Without them the period would be $12$, as it is for
A389392.
\end{remark}

Consequently, a proof of the conjectured formula for A391498 is now reduced to
the two statements that remain --- the identifications of the family $\mathrm D$
(Lemma~\ref{lem:Dloci}) and the term $E_{\mathrm{DS}}$, for both of which
Proposition~\ref{prop:loci} provides the explicit loci --- of the same kind as
Lemmas~\ref{lem:famA}--\ref{lem:BSid}: for each family, the
number of parameter tuples with five different shapes is an explicit count of
lattice points in a polytope of dimension $3$ or $4$, to be corrected by an
inclusion--exclusion over the (finitely many, explicit) linear equations
expressing that two of the five shapes coincide, and then divided by the size of
the fibre. The same applies to the four intersection terms, whose defining
equations state that a set admits two different descriptions. All of these are
elementary, and the numerical generating functions above tell us in advance what
the answer must be.