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\hypersetup{pdftitle={The Hankel determinants of the Baum-Sweet sequence are 2-automatic},
pdfauthor={David Niedbala Giraudin}}

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\title{The Hankel determinants of the Baum--Sweet sequence are $2$-automatic}
\author{David Niedbala Giraudin}
\address{ORCID 0009-0009-1526-1178}
\date{4 November 2026}

\begin{document}

\begin{abstract}
Let $b$ be the Baum--Sweet sequence and $H_n=\det(b_{i+j})_{0\le i,j<n}$ its Hankel determinants over
$\F$. We prove that the sequence $\chi_0(n)=[H_n\ne0]$ is $2$-automatic: $\chi_0(n+1)$ is computed by an
explicit automaton with $8$ states reading the binary expansion of $n$, and the minimal automaton of
$\chi_0$ reading least significant digits first has $74$ states. The same holds for the shifted
determinants $\det(b_{i+j+1})$, with the same $8$-state graph and $68$ states. This contrasts with
Mkaouar's theorem that the partial quotients of the Baum--Sweet series are not automatic. On the way
we show that $\Theta=tf^2$, a root of $Z^3+Z+t^3$ in $\F((t^{-1}))$, has all its partial quotients in
$\{t,t^3,t^3+t\}$, given by an explicit coding of the Mills--Robbins expansion, and that the linear
complexity profile of $b$ satisfies $-2\le2L(b,N)-N\le3$ for all $N$, all six values occurring. The
proofs reduce to finite verifications, which an ancillary script performs.
\end{abstract}

\maketitle

\section{Introduction}

The Baum--Sweet sequence $b=(b_n)_{n\ge0}$ over $\F$ is defined by $b_0=1$ and, for $n\ge1$,
$b_n=1$ if and only if the binary expansion of $n$ has no block of zeros of odd length; equivalently
$b_{2n+1}=b_n$, $b_{4n}=b_n$, $b_{4n+2}=0$. Its generating series $f=\sum_{n\ge0}b_nt^{-n}$ is the
unique root in $\F((t^{-1}))$ of
\[
tf^3+f+t=0 .
\]
Baum and Sweet \cite{BS76} showed that all partial quotients of $f$ have degree $1$ or $2$. Mills and
Robbins \cite{MR86} gave the expansion explicitly (Section~\ref{sec:mr}), and Mkaouar \cite{Mk95}
proved that the sequence of partial quotients is not automatic, although it is the image of a fixed
point of a non-uniform substitution on $20$ letters.

For $c\ge0$ let $H^{(c)}_n=\det(b_{c+i+j})_{0\le i,j<n}$ (with $H^{(c)}_0=1$) and
\[
\chi_c(n)=\begin{cases}1&\text{if }H^{(c)}_n\ne0,\\0&\text{otherwise.}\end{cases}
\]
Hankel determinants of automatic sequences have been studied since Allouche, Peyri\`ere, Wen and Wen
\cite{APWW98}; Hu and Han \cite{HH19} proved automaticity for the reduced Hankel determinants modulo $2$
of a family of $\pm1$ sequences, and Allouche, Han and Niederreiter \cite[Section 7]{AHN20} ask whether
the Hankel determinants of a $d$-automatic sequence are $d$-regular or, reduced modulo some integer,
$d$-automatic. For the Baum--Sweet sequence and reduction modulo $2$ (the determinants of the $0,1$
matrices modulo $2$ are the determinants over $\F$) the answer is positive.

\begin{theorem}\label{thm:main}
For $c\in\{0,1\}$ the sequence $\chi_c$ is $2$-automatic. More precisely, let $\mathcal A$ be the
automaton with states $0,\dots,7$, initial state $0$, and transitions
\begin{center}
\begin{tabular}{c|cccccccc}
\toprule
state & 0&1&2&3&4&5&6&7\\
\midrule
digit $0$ & 0&2&4&6&1&7&6&5\\
digit $1$ & 1&3&5&0&4&5&7&6\\
\midrule
output $\omega_0$ & 1&1&0&0&1&0&0&1\\
output $\omega_1$ & 1&1&0&1&1&0&1&1\\
\bottomrule
\end{tabular}
\end{center}
reading the binary expansion of $n$ from the most significant digit. Then $\chi_c(n+1)$ is the
output $\omega_c$ of the state reached, for every $n\ge0$; and $\chi_c(0)=1$. The minimal automaton
of $\chi_0$ (resp.\ $\chi_1$) reading the most significant digit first has $15$ (resp.\ $8$) states,
and the one reading the least significant digit first has $74$ (resp.\ $68$) states.
\end{theorem}

Let $\Theta=tf^2$. From $tf^3+f=t$ one gets $\Theta^3+\Theta=t(tf^3+f)^2=t^3$, so $\Theta$ is a root of
$Z^3+Z+t^3$ in $\F((t^{-1}))$, and $f/t=1/(1+\Theta)$. Since $\Theta=t^3/(1+\Theta^2)$, $\Theta$ is
hyperquadratic in the sense of Lasjaunias \cite{La00}. As in Section~\ref{sec:mr} we write $X=t$ when
$t$ is used as a partial quotient.

\begin{theorem}\label{thm:theta}
$\Theta=[X;a_1,a_2,\dots]$ with every $a_i\in\{X,X^3,X^3+X\}$. Explicitly, cut the Mills--Robbins word
$\G_\infty$ of Section~\ref{sec:mr} into blocks consisting of one letter of degree $1$ or of a letter of
degree $2$ followed by a letter of degree $1$; then $a_i=\psi(\text{$i$-th block})$, where $\psi$ maps a
block of degree $1$ to $X$, a block beginning with $X^2$ to $X^3$, and a block beginning with $X^2+1$
to $X^3+X$.
\end{theorem}

\begin{corollary}\label{cor:lc}
Let $L(b,N)$ be the $N$-th linear complexity of $b$ over $\F$. For every $N\ge1$,
$2L(b,N)-N\in\{-2,-1,0,1,2,3\}$, and each of these six values occurs.
\end{corollary}

This refines the bounds $N/3\le L(b,N)\le(2N+1)/3$ of M\'erai and Winterhof
\cite[Corollary 3]{MW18}. Some bound $|2L(b,N)-N|\le C$ follows from the boundedness of the partial
quotients of $f/t$; the point is the exact window and the fact that it is attained. Allouche, Han and Niederreiter \cite[Section 7]{AHN20} ask for which automatic
sequences the profile can be computed exactly, and Theorem~\ref{thm:main} gives it by an automaton.
We also obtain the transfer relation $\chi_0(n)=\chi_1(n)\chi_1(n-1)$ for $n\ge1$
(Proposition~\ref{prop:transfer}).

The proofs reduce every infinite statement to a finite one: an induction on the Mills--Robbins
recursion closed by finitely many boundary words (Lemma~\ref{lem:fac}), a transducer with finitely many
configurations (Theorem~\ref{thm:transducer}), and a self-similarity of the dyadic blocks
(Proposition~\ref{prop:dyadic}) that a finite automaton is shown to satisfy (Section~\ref{sec:auto}).
The finite verifications are carried out by the ancillary script, described in
Section~\ref{sec:verif}.

\section{Continued fractions and Hankel determinants}\label{sec:dict}

For $\alpha=\sum_{n\ge1}s_{n-1}t^{-n}\in\F((t^{-1}))$ with convergents $P_j/Q_j$, one has
$\det(s_{i+j})_{0\le i,j<n}\ne0$ if and only if $n\in\{\deg Q_j:j\ge0\}$; this is the classical link
between Hankel determinants and the Pad\'e table, see e.g.\ \cite{AHN20}. We apply it to
\[
\alpha_0=\sum_{n\ge0}b_nt^{-n-1}=f/t,\qquad \alpha_1=\sum_{n\ge1}b_nt^{-n}=f+1 ,
\]
so that $\chi_0$ and $\chi_1$ are the indicator functions of $\Sigma_0=\{\deg Q_j(\alpha_0)\}$ and
$\Sigma_1=\{\deg Q_j(\alpha_1)\}$. If $\alpha=[0;q_1,q_2,\dots]$ then $\deg Q_j=\deg q_1+\dots+\deg q_j$.
By Niederreiter's theorem \cite[Theorem 1]{Ni88}, the linear complexity of $s_0,\dots,s_{N-1}$ is
$\deg Q_j$, where $j$ is determined by $\deg Q_{j-1}+\deg Q_j\le N<\deg Q_j+\deg Q_{j+1}$.

\section{The Mills--Robbins expansion and its dyadic structure}\label{sec:mr}

Write $T=\{X,X+1,X^2,X^2+1\}$ (with $X=t$), let $\tau$ exchange $X\leftrightarrow X+1$ and
$X^2\leftrightarrow X^2+1$, and for a word $w$ let $\widetilde w$ be its reversal. Put
$\G_0=X+1$, $\G_1=X+1,X^2+1$, $\G_2=X+1,X^2+1,X,X^2,X+1$ and, for $n\ge3$,
\[
\G_n=\G_{n-1}\,\tau'(\widetilde\G_{n-3})\,\Lambda_n\,\G_{n-3},
\]
where $\tau'$ applies $\tau$ to the first letter only, and
\begin{align*}
\Lambda_n&=X,X{+}1,X^2,(X,X^2)^k,X{+}1,X, &&k=(2^{n-1}-4)/3,\ n\text{ odd},\\
\Lambda_n&=X{+}1,X^2{+}1,(X,X^2)^k,X,X^2{+}1,X{+}1, &&k=(2^{n-1}-5)/3,\ n\text{ even}.
\end{align*}
Each $\G_n$ is a prefix of $\G_{n+1}$; let $\G_\infty$ be the limit. Mills and Robbins proved
\cite{MR86}, see \cite[p.~363]{Mk95}, that
\[
f=[1;X,\G_\infty].
\]
For a word $w$ over $T$ let $D(w)$ be the sum of the degrees of its letters, and let $\varphi$ be one
of the codings $\varphi_1$, $\varphi_0$ defined by $\varphi_e(c)=1$ if $\deg c=1$ and
$\varphi_1(c)=01$, $\varphi_0(c)=00$ if $\deg c=2$. Both preserve lengths: $|\varphi(w)|=D(w)$.

\begin{lemma}\label{lem:len}
$D(\G_n)=2^{n+1}-1$ for all $n\ge0$; for $n\ge3$, $D(\Lambda_n)=2^{n-1}+2$ and $\Lambda_n$ is a palindrome.
\end{lemma}

\begin{proof}
$\Lambda_n$ has $(2^n+7)/3$ letters, of which $(2^{n-1}-1)/3$ have degree $2$, if $n$ is odd, and
$(2^n+5)/3$ letters, of which $(2^{n-1}+1)/3$ have degree $2$, if $n$ is even; in both cases
$D(\Lambda_n)=2^{n-1}+2$. Since $\tau$ and reversal preserve degrees,
$D(\G_n)=D(\G_{n-1})+2D(\G_{n-3})+2^{n-1}+2$, and induction from $D(\G_0)=1$, $D(\G_1)=3$, $D(\G_2)=7$
gives the claim.
\end{proof}

Put $A_n=\varphi(\G_n)$, $B_n=\varphi(\widetilde\G_n)$, $L_n=\varphi(\Lambda_n)$ and, for $m\ge5$,
\[
G_m=B_{m-3}\,L_m\,A_{m-3}.
\]
By the recursion, the palindromy of $\Lambda_n$ and $\deg\tau(c)=\deg c$, we have $A_n=A_{n-1}G_n$
and $B_n=G_nB_{n-1}$ for $n\ge5$, and $|G_m|=2^m$ by Lemma~\ref{lem:len}.

\begin{proposition}\label{prop:dyadic}
Let $\omega=1\,\varphi(\G_\infty)$ (an infinite word indexed from $0$). Then, for every $m\ge5$,
$\omega[2^m,2^{m+1})=G_m$; moreover $\omega[1,2^{n+1})=A_n$ for all $n$. Consequently, for $m\ge8$,
$q=2^{m-2}$ and $0\le x<2^m$:
\begin{itemize}
\item[(R1)] if $x\ge3q+1$, then $\omega(2^m+x)=\omega(x-3q)$;
\item[(R2)] if $q-2^{j+1}\le x<q-2^j$ with $5\le j\le m-3$, then $\omega(2^m+x)=\omega(2^j+x-q+2^{j+1})$;
\item[(R3)] if $q-32\le x\le q-2$, then $\omega(2^m+x)=B_4[x-q+32]$;
\item[(R4)] if $q-1\le x\le3q$, then $\omega(2^m+x)=L_m[x-q+1]$.
\end{itemize}
\end{proposition}

\begin{proof}
$\omega=1A_4G_5G_6\cdots$ with $|1A_4|=32$, and $|G_m|=2^m$; this gives the first two assertions. In
$G_m=B_{m-3}L_mA_{m-3}$ the three factors have lengths $q-1$, $2q+2$, $q-1$, which gives (R4), and (R1)
since $A_{m-3}[i]=\omega(i+1)$. For $m\ge8$, $B_{m-3}=G_{m-3}G_{m-4}\cdots G_5B_4$, in which $G_j$
occupies the positions $[q-2^{j+1},q-2^j)$ and $B_4$ the positions $[q-32,q-1)$; with
$G_j=\omega[2^j,2^{j+1})$ this gives (R2) and (R3).
\end{proof}

In binary, with $n=2^m+x$, the four cases read: (R1) $n=111y$ with $y\ne0$, and then $\omega(n)=\omega(y)$;
(R2) $n=100\,1^a\,0\,z$ with $|z|=j\ge5$, and then $\omega(n)=\omega(1z)$; (R3) $n=100\,1^{m-7}w$ with
$|w|=5$, $w\ne11111$, and then $\omega(n)=B_4[\mathrm{val}(w)]$; (R4) all remaining $n$, namely $1001^{m-2}$, $101u$, $110u$ and $1110^{m-2}$. In
(R4), $L_m$ is explicit: it is $\varphi$ of the palindrome $\Lambda_m$, whose middle part
$\varphi((X,X^2)^k)$ is periodic of period $3$.

\section{The factors of length three}

\begin{lemma}\label{lem:fac}
The set $F$ of factors of length $3$ of $\G_\infty$ has $20$ elements and is closed under reversal. In
$\G_\infty$ no two letters of degree $2$ are adjacent.
\end{lemma}

\begin{proof}
Let $F$ be the set of factors of length $3$ of $\G_{16}$; it has $20$ elements and is closed under
reversal. We show $\operatorname{Fac}_3(\G_n)\subseteq F$ for all $n\ge3$ by induction; for $3\le n\le8$
this is a direct check. Let $n\ge9$. A factor of length $3$ of
$\G_n=\G_{n-1}\,\tau'(\widetilde\G_{n-3})\,\Lambda_n\,\G_{n-3}$ either lies in one of the four blocks or
meets a junction. Factors of $\G_{n-1}$ and $\G_{n-3}$ lie in $F$ by induction, factors of
$\widetilde\G_{n-3}$ by closure under reversal, and factors of $\Lambda_n$ by inspection: for $n\ge5$,
$\operatorname{Fac}_3(\Lambda_n)$ depends only on the parity of $n$, and it is contained in $F$. The
remaining factors are the factor of $\tau'(\widetilde\G_{n-3})$ containing its first letter and the six
factors meeting the three junctions. They depend only on the first two letters of $\G_{n-3}$, the last
three letters of $\G_{n-1}$ and of $\G_{n-3}$, and the first two and last two letters of $\Lambda_n$.
The first two letters of $\G_k$ are $X+1,X^2+1$ for all $k\ge1$; the last three letters of $\G_k$ for
$k\ge6$ are those of $\G_{k-3}$, hence depend only on $k\bmod3$; the ends of $\Lambda_n$ depend only on
the parity of $n$. So these seven factors depend only on $n\bmod6$, and the six cases are checked to
lie in $F$. The last statement is read off $F$.
\end{proof}

\section{A transducer for \texorpdfstring{$t/f$}{t/f}}

For a matrix $M=\begin{pmatrix}p&q\\r&s\end{pmatrix}$ over $\F[t]$ and a series $y$ write
$M(y)=(py+q)/(ry+s)$. To compute the continued fraction of $z=t/f$ from $f=[1;X,\G_\infty]$ we keep a
matrix $M$ with $z=M(y)$, where $y$ is the remaining tail of $f$. Reading a partial quotient $a$ of $f$
replaces $M$ by $M\begin{pmatrix}a&1\\1&0\end{pmatrix}$; emitting a partial quotient $\kappa$ of $z$
replaces $M$ by $\begin{pmatrix}0&1\\1&\kappa\end{pmatrix}M$. Both preserve $\det M$ up to sign, hence
$\det M=t$ throughout (we start from $M=\begin{pmatrix}0&t\\1&0\end{pmatrix}$).

If $y=a+1/y'$ with $\deg a\ge1$, $\deg y'\ge1$, then
$M(y)-p/r=\det M/(r(ry+s))$, and $\deg(ry+s)=\deg(ra+s)$ as soon as $\deg(ra+s)\ge\deg r$. Hence, if
$\deg(ra+s)\ge\deg r$ and $\deg r+\deg(ra+s)>\deg\det M$ for every letter $a$ that can follow, the
polynomial part $\kappa$ of $p/r$ is the polynomial part of $z$, and if $\deg\kappa\ge1$ it is a
correct partial quotient of $z$: we \emph{emit} it. We emit greedily as long as this holds for every letter allowed by $F$ after the last
two letters read.

\begin{theorem}\label{thm:transducer}
Run the transducer on $1,X,\G_\infty$. The triples (matrix, last two letters read, $E-D$), where $D$ is the
degree read in $\G_\infty$ and $E$ the degree emitted, form a set of $12$ configurations closed under
reading every letter allowed by $F$. In each of them every emission is certain, and: reading a letter of
degree $1$ emits exactly one partial quotient, equal to $X$, $X^3$ or $X^3+X$ according as the previous
letter has degree $1$, is $X^2$, or is $X^2+1$, after which $E=D+1$; reading a letter of degree $2$ emits
nothing, after which $E=D-1$. The first emissions are $X+1$ (after reading $X$) and $X$ (after reading
the first letter $X+1$ of $\G_\infty$).
\end{theorem}

\begin{proof}
A finite computation over the $12$ configurations and the letters allowed by $F$ (Lemma~\ref{lem:fac}),
performed by the script.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:theta}]
By Theorem~\ref{thm:transducer}, $t/f=[X+1;a_1,a_2,\dots]$ with $a_i=\psi(\text{$i$-th block})$, and
$\Theta=t/f+1=[X;a_1,a_2,\dots]$.
\end{proof}

\begin{proposition}\label{prop:transfer}
$\chi_1=11\,\varphi_1(\G_\infty)$ and $\chi_0=11\,\varphi_0(\G_\infty)$. Consequently
$\chi_0(n)=\chi_1(n)\chi_1(n-1)$ for $n\ge1$.
\end{proposition}

\begin{proof}
$\alpha_1=f+1=[0;X,\G_\infty]$, so $\Sigma_1$ consists of $0$, $1$ and the numbers $1+D_i$, where $D_i$ is
the total degree of the first $i$ letters of $\G_\infty$; this is $\chi_1=11\varphi_1(\G_\infty)$. Next
$\alpha_0=f/t=[0;t/f]$, so $\Sigma_0$ consists of $0$ and the cumulative degrees of the partial
quotients of $t/f$. By Theorem~\ref{thm:transducer} these are $1$ and the numbers $D_i+1$ for which the
$i$-th letter of $\G_\infty$ has degree $1$; this is $\chi_0=11\varphi_0(\G_\infty)$. For the last
statement: $\chi_0(n)=1$ means that position $n$ ends a letter of degree $1$, $\chi_1(n)=1$ that it ends a
letter; a letter of degree $2$ is coded $01$ by $\varphi_1$, so $n$ ends a letter of degree $1$ if and
only if both $n$ and $n-1$ end letters.
\end{proof}

\begin{remark}
Equivalently, $\Sigma_1=\Sigma_0\cup(\Sigma_0-1)$ (restricted to $n\ge0$). One inclusion holds for
every sequence: by the Desnanot--Jacobi identity
$H^{(0)}_{n+1}H^{(2)}_{n-1}=H^{(0)}_nH^{(2)}_n-(H^{(1)}_n)^2$, if $H^{(0)}_n=H^{(0)}_{n+1}=0$ then
$H^{(1)}_n=0$, that is $\Sigma_1\subseteq\Sigma_0\cup(\Sigma_0-1)$. The reverse inclusion is specific:
it fails for most random binary sequences.
\end{remark}

\section{Proof of Theorem~\ref{thm:main}}\label{sec:auto}

By Proposition~\ref{prop:transfer}, $\chi_c(n+1)=\omega(n)$ with $\omega=1\varphi_c(\G_\infty)$ for
$n\ge0$. The word $\omega$ is determined by its first $256$ letters and the rules (R1)--(R4) of
Proposition~\ref{prop:dyadic}, since each rule expresses $\omega(n)$, $n\ge256$, through a smaller index
or an explicit word. It therefore suffices to show that the output $\widehat\omega(n)$ of $\mathcal A$ on
the binary expansion of $n$ satisfies the same rules and agrees with $\omega$ on $[0,256)$. Since
$\delta(0,0)=0$, leading zeros do not matter. Each rule is a statement about all words of a regular
shape, and is checked in a finite product automaton:
\begin{itemize}
\item (R1): for all $y$ with $|y|\ge6$, $y\ne0$, the states reached from $\delta(0,111)$ and from $0$
by reading $y$ have the same output (reachability in $\mathcal A\times\mathcal A\times\{$length
$\le6\}\times\{y=0?\}$);
\item (R2): the same for the states reached from $\delta(0,1001^a0)$, $a\ge0$ (a finite set of
states), and from $\delta(0,1)$, by words of length at least $5$;
\item (R3): $\widehat\omega(1001^cw)=B_4[w]$ for $c\ge1$ and the $31$ words $w$;
\item (R4): $L_m[i]$ depends only on the parity of $m$, on $i\bmod3$, and on whether $i$ is one of the
first four or last four positions; in the regions $101u$ and $110u$ the index $i$ is $\mathrm{val}(u)+1$
and $2^{|u|}+\mathrm{val}(u)+1$, so the expected value is computed by a small automaton tracking
$\mathrm{val}(u)\bmod3$, the parity of $|u|$ and whether $u\in0^*$, $0^*1$, $0^*10$ (resp.\ $1^*$,
$1^*0$, $1^*01$); the two remaining words $1001^{m-2}$, $1110^{m-2}$ have value $1$.
\end{itemize}
All these checks succeed for $c=0$ and $c=1$ (script, Part 5). Hence $\widehat\omega=\omega$, and
$\chi_c(n+1)=\widehat\omega(n)$ for all $n$.

From $\mathcal A$ one builds an automaton $\mathcal B_c$ for $n\mapsto\chi_c(n)=\widehat\omega(n-1)$
whose states are the pairs (state of $\mathcal A$ on $n$, state of $\mathcal A$ on $n-1$), plus an initial
state for $n=0$: reading a digit updates both, since $2n+1-1=2n$ and $2n-1=2(n-1)+1$. All automata here
read the most significant digit first and accept leading zeros. Minimising $\mathcal B_c$ gives $15$
and $8$ states. For the other reading direction, the minimal automaton reading least significant digits
first has one state per element of the $2$-kernel $\{n\mapsto\chi_c(2^kn+r):k\ge0,\,0\le r<2^k\}$
(by the Myhill--Nerode theorem; see \cite[Section 6.6]{AS03}). The kernel element of $(k,r)$ is the
sequence $n\mapsto o(\delta(\sigma_n,w))$, where $\sigma_n$ is the state of $\mathcal B_c$ on $n$ and
$w$ the $k$ digits of $r$. Every state of $\mathcal B_c$ is some $\sigma_n$, so two kernel elements are
equal if and only if the maps $\sigma\mapsto o(\delta(\sigma,w))$ coincide on the states of
$\mathcal B_c$. Enumerating these maps over all words $w$ gives $74$ and $68$. \qed

\section{Proof of Corollary~\ref{cor:lc}}

By Proposition~\ref{prop:transfer} and Theorem~\ref{thm:transducer}, the consecutive elements of
$\Sigma_0=\{d_0<d_1<\dots\}$ differ by the degrees of the partial quotients of $t/f$, that is by $1$
or $3$, both occurring. By Niederreiter's theorem, $L=L(b,N)=d_j$ with $d_{j-1}+d_j\le N<d_j+d_{j+1}$,
so $2L-N\le d_j-d_{j-1}\le3$ and $2L-N\ge1-(d_{j+1}-d_j)\ge-2$. If $d_j-d_{j-1}=3$, the values
$3,2,1,0$ occur on this interval of $N$; if $d_{j+1}-d_j=3$, the values $1,0,-1,-2$ occur. \qed

\section{Verification}\label{sec:verif}

The ancillary file \texttt{verify\_baum\_sweet\_hankel.py} (Python~3, no dependency, exact arithmetic
over $\F$, under a minute) performs all the finite checks used above: Lemma~\ref{lem:fac} (the set $F$,
the base cases and the seven boundary factors of each class $n\bmod6$); Theorem~\ref{thm:transducer}
(the $12$ configurations, the certainty of every emission, the bookkeeping of $E-D$); and the rules
(R1)--(R4) for $\mathcal A$, the base table, the minimal automata and the kernels. It also checks the
statements against first principles: the Hankel determinants by Gaussian elimination for $n\le300$; the
continued fractions of $\alpha_0$ and $\alpha_1$ by Euclid's algorithm, against $\G_\infty$ and against
the automaton, for $n<2^{16}$; and the linear complexity by the Berlekamp--Massey algorithm for
$N\le3000$.

\subsection*{Acknowledgement}
Technical assistance: Claude (Anthropic).

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