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\title{\texorpdfstring{The binary Collatz dynamics of the trinomials $x^n+x+1$: proof of two conjectures of Gallardo and Rahavandrainy}{The binary Collatz dynamics of the trinomials x\^{}n+x+1: proof of two conjectures of Gallardo and Rahavandrainy}}
\author{David Niedbala Giraudin\\\small ORCID 0009-0009-1526-1178}
\date{}

\begin{document}
\maketitle

\begin{abstract}
Gallardo and Rahavandrainy studied an analogue of the Collatz map on $\F_2[x]$, in which
$x(x+1)$ plays the role of $2$ and $x^2+x+1$ that of $3$: an odd polynomial $A$ is sent to the
odd part of $1+(x^2+x+1)A$. They proved that every orbit reaches $1$, and conjectured that the
orbit of the trinomial $x^n+x+1$ has length $2^s+1$, where $s$ is the largest integer with
$2^{s+1}<n$, and that for $1\le t\le s-1$ it contains $2^t$ terms of the same degree. We prove both
statements for every $n\ge3$. The proof rests on the identity $(x^2+x+1)+x(x+1)=1$: it reduces the
dynamics to a map from one plateau to the next, characterises the last plateau of any orbit,
and shows that along the orbit of $x^n+x+1$ the plateaus start at the $2^j$-th powers of the
odd parts of the polynomials $x^m(x^2+x+1)+1$, the index $m$ following an explicit binary
recursion. We also note that a later restatement of the conjecture, with a non-strict
inequality, fails exactly when $n$ is a power of $2$. A script checks all the statements
numerically.
\end{abstract}

\section{Introduction}

Throughout we work in $\F_2[x]$ and put
\[
D=x(x+1)=x^2+x,\qquad M=x^2+x+1,\qquad\text{so that}\qquad M+D=1. \tag{1}
\]
A polynomial $A$ is \emph{odd} if $\gcd(A,D)=1$, that is, $A(0)=A(1)=1$. For $B\ne0$ we write
$\odd(B)$ for $B$ with all its factors $x$ and $x+1$ removed. For odd $A$ the Collatz step of
Gallardo and Rahavandrainy \cite{GR23} is
\[
\st(A)=\odd(1+MA).
\]
In their notation $A_1=\odd(A)$, $A_{2k}=1+MA_{2k-1}$ and $A_{2k+1}=\odd(A_{2k})$, so the
\emph{odd sequence} of an odd $A$ is $A,\st(A),\st^2(A),\dots$; they proved
\cite[Theorem~1.2]{GR23} that it reaches $1$, with length at most $2^{\deg A-1}$. Its
\emph{length} $\ell_A$ is the number of its terms up to and including the first $1$; this is
the integer $\ell_A=m+1$ of \cite[Theorem~1.2]{GR23}, whose odd sequence is
$[A_1,A_3,\dots,A_{2m-1},1]$.

For the trinomials $T_n=x^n+x+1$ they made two conjectures \cite[Section~3.10]{GR23}. Let $s$
be the largest integer such that $n-2^{s+1}\ge1$.

\begin{itemize}
\item \textbf{Conjecture 3.12.} For every positive integer $t\le s-1$, the odd sequence of
$T_n$ contains $2^t$ polynomials which have the same degree $d_t$; in particular
$d_1=\deg A_5=\deg A_7$ and $d_2=\deg A_9=\dots=\deg A_{15}$.
\item \textbf{Conjecture 3.13.} $\ell_{T_n}=2^s+1$.
\end{itemize}

The family is considered for $n\ge2$; for $n=2$ one gets $s=-1$, while $T_2=M$ has length
$2$, so the conjectures are meaningful for $n\ge3$. For $n\ge3$ one has $s=k-2$, where
$k=\lceil\log_2n\rceil$, since $2^{s+1}<n\le2^{s+2}$. We prove both conjectures for every
$n\ge3$.

\begin{theorem}\label{thm:main}
Let $n\ge3$ and $k=\lceil\log_2 n\rceil$. The odd sequence of $T_n=x^n+x+1$ has length
$2^{k-2}+1$. If $n\ge5$, it consists of $T_n$, then of $k-2$ blocks $B_1,\dots,B_{k-2}$, where
$B_j$ consists of exactly $2^{j-1}$ consecutive terms of the same degree $\delta_j$, with
$n>\delta_1>\delta_2>\dots>\delta_{k-2}$, and finally of $1$.
\end{theorem}

\begin{corollary}\label{cor:conj}
Conjectures 3.12 and 3.13 of \cite{GR23} hold for every $n\ge3$.
\end{corollary}

\begin{proof}
Theorem~\ref{thm:main} with $s=k-2$ gives Conjecture 3.13. For $1\le t\le s-1$ the block
$B_{t+1}$ consists of $2^t$ terms of degree $\delta_{t+1}$, and no other term has this degree
since the degrees $n,\delta_1,\delta_2,\dots$ are distinct; this is Conjecture 3.12, with
$d_t=\delta_{t+1}$. Moreover $T_n$ and $B_1,\dots,B_t$ have $2^t$ terms in all, so
$B_{t+1}$ occupies the positions $2^t+1,\dots,2^{t+1}$ of the odd sequence, that is, it consists
of $A_{2^{t+1}+1},A_{2^{t+1}+3},\dots,A_{2^{t+2}-1}$; for $t=1,2$ these are $A_5,A_7$ and
$A_9,\dots,A_{15}$.
\end{proof}

The blocks are made explicit in Proposition~\ref{prop:induction}: $B_j$ starts with
$V(m_j)^{2^{j-1}}$, where $V(m)=\odd(x^mM+1)$, $m_1=n-3$, and $m_{j+1}$ is obtained from $m_j$
by a binary rule (Lemma~\ref{lem:binary}). The last block starts from a power of one of the
four polynomials $x^3+x+1$, $x^3+x^2+1$, $x^4+x+1$, $x^5+x^3+x^2+x+1$.

Section~\ref{sec:general} contains three statements valid for every odd polynomial: the plateau
lemma, the plateau-to-plateau map with a necessary and sufficient criterion for the orbit to
end after a given plateau, and an infinite family of polynomials whose orbit is a single
plateau. Section~\ref{sec:trinomials} treats the trinomials. In Section~\ref{sec:erratum} we
note that the restatement of Conjecture 3.13 as Conjecture 3.3 of \cite{GR25}, with a
non-strict inequality, fails exactly at the powers of $2$.

The polynomial analogue of the $3x+1$ problem goes back to Hicks, Mullen, Yucas and Zavislak
\cite{HMYZ}, who used the map sending $f$ to $f/x$ if $x\mid f$ and to $(x+1)f+1$ otherwise;
see also Nichols \cite{Nic}. The variant
studied here, with $x(x+1)$ in the role of $2$, is that of \cite{GR23,GR25}.

\section{Three facts valid for every odd polynomial}\label{sec:general}

For odd $u$ we write $z=u+1$, which is divisible by $D$, and
\[
\mu(u)=\max\{t\ge0:\ D^t\mid u+1\}\ \ge1 .
\]
Since $D=x(x+1)$ with $x$, $x+1$ coprime, $\mu(u)=\min\bigl(v_x(u+1),v_{x+1}(u+1)\bigr)$, and
$\mu(u^{2^e})=2^e\mu(u)$ because $u^{2^e}+1=(u+1)^{2^e}$.

\begin{lemma}[Plateau lemma]\label{lem:plateau}
Let $u$ be odd of degree at least $1$, $z=u+1$ and $\mu=\mu(u)$.
\begin{enumerate}
\item[(i)] $\deg\st(u)\le\deg u$, with equality if and only if $\mu\ge2$; in that case
$\st(u)+1=Mz/D$ and $\mu(\st(u))=\mu-1$.
\item[(ii)] Consequently $u,\st(u),\dots,\st^{\mu-1}(u)$ all have degree $\deg u$, and
$\deg\st^{\mu}(u)<\deg u$.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) Since $M+1=D$, we have $1+Mu=M(z+1)+1=Mz+D$. As $M(0)=M(1)=1$, the valuation of $Mz$ at $x$
is $v_x(z)\ge1$, while $v_x(D)=1$. Hence $v_x(Mz+D)=1$ if $v_x(z)\ge2$, and $v_x(Mz+D)\ge2$ if
$v_x(z)=1$; the same holds at $x+1$. Since $\deg(Mz+D)=\deg u+2$, the degree of
$\st(u)=\odd(Mz+D)$ is $\deg u+2-v_x(Mz+D)-v_{x+1}(Mz+D)\le\deg u$, with equality exactly when
both valuations equal $1$, that is, when $D^2\mid z$, that is, when $\mu\ge2$. In that case
$\st(u)=(Mz+D)/D=Mz/D+1$, so $\st(u)+1=Mz/D$, and $\mu(\st(u))=\mu-1$ because $M$ is coprime
to $D$.

(ii) follows by applying (i) $\mu$ times: the values of $\mu$ along $u,\st(u),\dots$ are
$\mu,\mu-1,\dots,1$, and the step from the term with $\mu=1$ lowers the degree.
\end{proof}

We call $u,\st(u),\dots,\st^{\mu-1}(u)$ the \emph{plateau} issued from $u$.

\begin{lemma}[Plateau-to-plateau map]\label{lem:psi}
Let $u$ be odd of degree at least $1$, $\mu=\mu(u)$, and
\[
W=\frac{M^{\mu}(u+1)}{D^{\mu}}\in\F_2[x].
\]
Then $\deg W=\deg u$, $\st^{\mu}(u)=\odd(W+1)$, and
\[
\st^{\mu}(u)=1\iff W+1=x^i(x+1)^j\ \text{for some } i,j\ge0 .
\]
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:plateau}(i), $\st^t(u)+1=M^tz/D^t$ for $0\le t\le\mu-1$. For
$w=\st^{\mu-1}(u)$ we get, as in the proof of Lemma~\ref{lem:plateau},
\[
1+Mw=M(w+1)+D=\frac{M^{\mu}z}{D^{\mu-1}}+D=D\,(W+1),
\]
so $\st^{\mu}(u)=\odd(D(W+1))=\odd(W+1)$. The degree of $W$ is $\deg u+2\mu-2\mu$; in particular
$W+1\ne0$. Finally $\odd(W+1)=1$ if and only if $W+1$ is a product of powers of $x$ and $x+1$.
\end{proof}

In the terminology of \cite{GR19}, the orbit ends right after the plateau issued from $u$ if and
only if $W$ is a Mersenne polynomial, the exponents $i,j$ being allowed to vanish.

\begin{proposition}[Single-plateau orbits]\label{prop:seeds}
Let $i,j\ge0$ with $(i,j)\ne(0,0)$ and $i\equiv j\pmod3$. Then $M$ divides $x^i(x+1)^j+1$, and
the polynomial
\[
P_{i,j}=1+\frac{D\,\bigl(x^i(x+1)^j+1\bigr)}{M}
\]
is odd of degree $i+j$. For every $e\ge0$ the odd sequence of $P_{i,j}^{2^e}$ consists of
$2^e$ terms of degree $2^e(i+j)$, followed by $1$.
\end{proposition}

\begin{proof}
Let $\omega\in\F_4$ be a root of $M$; then $\omega+1=\omega^2$, so
$\omega^i(\omega+1)^j+1=\omega^{i+2j}+1$, which vanishes since $i+2j\equiv i-j\equiv0\pmod 3$.
The same holds for the conjugate root $\omega^2$, so $M\mid E:=x^i(x+1)^j+1$. Clearly
$P_{i,j}(0)=P_{i,j}(1)=1$ and $\deg P_{i,j}=2+(i+j)-2$. Since $(i,j)\ne(0,0)$, $\deg E=i+j$ and at least one of
$E(0)$, $E(1)$ equals $1$; as $P_{i,j}+1=DE/M$, we get
$\mu(P_{i,j})=1+\min\bigl(v_x(E),v_{x+1}(E)\bigr)=1$. For
$u=P_{i,j}^{2^e}$ we get $\mu(u)=2^e$ and
$W=M^{2^e}(P_{i,j}+1)^{2^e}/D^{2^e}=E^{2^e}$, so $W+1=(E+1)^{2^e}=x^{i2^e}(x+1)^{j2^e}$. Lemmas
\ref{lem:plateau} and \ref{lem:psi} conclude.
\end{proof}

For instance $P_{1,1}=M$, $P_{3,0}=x^3+x+1$, $P_{0,3}=x^3+x^2+1$, $P_{2,2}=x^4+x+1$ and
$P_{1,4}=x^5+x^3+x^2+x+1$.

\section{The trinomials}\label{sec:trinomials}

For $m\ge0$ put
\[
F_m=x^mM+1,\qquad V(m)=\odd(F_m).
\]
Thus $F_0=D$ and $V(0)=1$.

\begin{lemma}[First step]\label{lem:first}
For $n\ge3$, $\mu(T_n)=1$ and $\st(T_n)=V(n-3)$.
\end{lemma}

\begin{proof}
Here $z=T_n+1=x^n+x=x(x^{n-1}+1)$, so $v_x(z)=1$ and $\mu(T_n)=1$. Then
$W=Mz/D=M(x^{n-1}+1)/(x+1)$ and, using $M+x+1=x^2$,
\[
W+1=\frac{Mx^{n-1}+M+x+1}{x+1}=\frac{Mx^{n-1}+x^2}{x+1}=\frac{x^2F_{n-3}}{x+1}.
\]
By Lemma~\ref{lem:psi}, $\st(T_n)=\odd(W+1)=V(n-3)$.
\end{proof}

\begin{lemma}[Valuations]\label{lem:val}
For $m\ge1$, $v_x(F_m)=0$ and
\[
v_{x+1}(F_m)=1,\ 3,\ 1,\ 2\qquad\text{according as}\qquad m\equiv0,\ 1,\ 2,\ 3\pmod4 .
\]
Hence $\deg V(m)=m+2-v_{x+1}(F_m)$; in particular $\deg V(m)\le m+1$, and $\deg V(m)\le m$ if
$m$ is odd. Moreover $V(1)=1$ and $\deg V(m)\ge3$ for $m\ge2$.
\end{lemma}

\begin{proof}
$F_m(0)=1$ for $m\ge1$. Put $y=x+1$; then $M=y^2+y+1$ and
\[
F_m=(1+y)^m(1+y+y^2)+1=(m+1)\,y+\Bigl(\tbinom m2+m+1\Bigr)y^2+\Bigl(\tbinom m3+\tbinom m2+m\Bigr)y^3+O(y^4),
\]
the coefficients being read modulo $2$. If $m$ is even, the coefficient of $y$ is $1$. If $m$ is
odd, it vanishes and the coefficient of $y^2$ is $\binom m2$, which by Lucas' theorem is odd if
and only if the bit of weight $2$ of $m$ is $1$, that is, $m\equiv3\pmod4$. If $m\equiv1\pmod4$,
then $\binom m2$ and $\binom m3$ are even, again by Lucas, and the coefficient of $y^3$ is
$m\equiv1$. The degree formula follows from $\deg F_m=m+2$. Finally $F_1=x^3+x^2+x+1=(x+1)^3$,
and for $m\ge2$ the bound $v_{x+1}(F_m)\le3$ gives $\deg V(m)\ge m-1\ge 3$ when $m\ge4$, while
$\deg V(2)=\deg V(3)=3$.
\end{proof}

\begin{lemma}[Renormalisation]\label{lem:renorm}
For $m\ge2$ define
\[
m'=\begin{cases} m/2-1 & m \text{ even},\\ (m-3)/2 & m\equiv1\pmod4,\\ (m-1)/2 & m\equiv3\pmod4.\end{cases}
\]
Then $m'\ge0$, $\st(V(m))=V(m')^2$, and $\mu(V(m))=1$.
\end{lemma}

\begin{proof}
Note $m'\ge0$ ($m\equiv1\pmod4$ and $m\ge2$ force $m\ge5$). Write $b=v_{x+1}(F_m)$, so
$V(m)=F_m/(x+1)^b$ by Lemma~\ref{lem:val}, and
\[
MV(m)+1=\frac{MF_m+(x+1)^b}{(x+1)^b}=\frac{M^2x^m+M+(x+1)^b}{(x+1)^b}.
\]
In characteristic $2$ one has $M+(x+1)=x^2$, $M+(x+1)^2=x$ and $M+(x+1)^3=x^3$. Hence, with
$b=1,3,2$ in the three cases of the definition of $m'$,
\[
MV(m)+1=\frac{x^2\bigl(M^2x^{m-2}+1\bigr)}{x+1},\qquad
\frac{x^3\bigl(M^2x^{m-3}+1\bigr)}{(x+1)^3},\qquad
\frac{x\bigl(M^2x^{m-1}+1\bigr)}{(x+1)^2},
\]
and in each case $M^2x^{2m'}+1=(Mx^{m'}+1)^2=F_{m'}^2$. Taking odd parts,
$\st(V(m))=\odd(F_{m'})^2=V(m')^2$.

For the last assertion, by Lemma~\ref{lem:plateau}(i) it suffices to show
$\deg\st(V(m))<\deg V(m)$, that is, $2\deg V(m')<\deg V(m)$. If $m$ is even,
$2\deg V(m')\le2(m'+1)=m<m+1=\deg V(m)$ (for $m'=0$, $V(0)=1$). If $m$ is odd, then $m'$ is odd,
since $m-3\equiv2\pmod4$ when $m\equiv1$ and $m-1\equiv2\pmod4$ when $m\equiv3\pmod4$; hence
$\deg V(m')\le m'$ by Lemma~\ref{lem:val}, and $2\deg V(m')\le m-3<m-1=\deg V(m)$ if
$m\equiv1\pmod4$, $2\deg V(m')\le m-1<m=\deg V(m)$ if $m\equiv3\pmod4$.
\end{proof}

\begin{proposition}[Frobenius induction]\label{prop:induction}
Let $m\ge2$, $e\ge0$ and $u=V(m)^{2^e}$. Then $\mu(u)=2^e$ and $\st^{2^e}(u)=V(m')^{2^{e+1}}$.
\end{proposition}

\begin{proof}
By Lemma~\ref{lem:renorm}, $\mu(u)=2^e\mu(V(m))=2^e$. Put $H=M(V(m)+1)/D$; then
$W=M^{2^e}(u+1)/D^{2^e}=H^{2^e}$ and $W+1=(H+1)^{2^e}$. By (1),
\[
H+1=\frac{M(V(m)+1)+D}{D}=\frac{MV(m)+1}{D},
\]
so $\odd(H+1)=\st(V(m))=V(m')^2$ by Lemma~\ref{lem:renorm}. By Lemma~\ref{lem:psi},
$\st^{2^e}(u)=\odd(W+1)=\odd(H+1)^{2^e}=V(m')^{2^{e+1}}$.
\end{proof}

\begin{lemma}[A binary recursion]\label{lem:binary}
For $m\ge2$ put $N=m+2$ and $N'=m'+2$. Then $N'$ is obtained from $N$ by deleting its two
lowest binary digits $b_1b_0$ and appending the digit $b_1\vee b_0$:
\[
N'=2\lfloor N/4\rfloor+(b_0\vee b_1).
\]
In particular, if $N$ has $L\ge3$ binary digits then $N'$ has $L-1$, and after $t\le L-2$
iterations the value has digits $b_{L-1}\dots b_{t+1}\,(b_t\vee\dots\vee b_0)$.
\end{lemma}

\begin{proof}
If $m$ is even, $N$ is even, $b_0=0$, and $N'=m/2+1=N/2=2\lfloor N/4\rfloor+b_1$. If
$m\equiv1\pmod4$, then $N=4q+3$ and $N'=(m+1)/2=2q+1$. If $m\equiv3\pmod4$, then $N=4q+1$ and
$N'=(m+3)/2=2q+1$. The other assertions follow by induction.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
If $n\in\{3,4\}$, then $T_3=P_{3,0}$ and $T_4=P_{2,2}$, so $\ell_{T_n}=2=2^{k-2}+1$ by
Proposition~\ref{prop:seeds}.

Let $n\ge5$, so $m_1=n-3\ge2$, and define $m_{j+1}=m_j'$ as long as $m_j\ge2$. Put
$N_j=m_j+2$. Since $2^{k-1}<n\le2^k$, the integer $N_1=n-1$ has exactly $k$ binary digits, and
by Lemma~\ref{lem:binary} $N_j$ has $k-j+1$ digits. Hence $m_j\ge2$ (i.e. $N_j\ge4$) exactly
for $1\le j\le k-2$, and $m_{k-1}\in\{0,1\}$, so $V(m_{k-1})=1$.

By Lemma~\ref{lem:first} the odd sequence starts with $T_n$, a plateau of length $1$, followed
by $V(m_1)$. By Proposition~\ref{prop:induction} and induction on $j$, for $1\le j\le k-2$ the
term following $T_n$ and the blocks $B_1,\dots,B_{j-1}$ is $V(m_j)^{2^{j-1}}$, and the plateau
issued from it, $B_j$, has exactly $2^{j-1}$ terms; the term after $B_{k-2}$ is
$V(m_{k-1})^{2^{k-2}}=1$. By Lemma~\ref{lem:plateau}(ii) the degree drops strictly after $T_n$
and after each block. The length is $1+(1+2+\dots+2^{k-3})+1=2^{k-2}+1$.
\end{proof}

\begin{remark}[The last block]
The last block starts from $V(m_{k-2})^{2^{k-3}}$, where $N_{k-2}=m_{k-2}+2$ has the three
digits $1\,b_{k-2}\,(b_{k-3}\vee\dots\vee b_0)$ of $n-1=\sum b_i2^i$. Thus
$V(m_{k-2})$ is $V(2)=x^3+x+1$, $V(3)=x^3+x^2+1$, $V(4)=x^5+x^3+x^2+x+1$ or $V(5)=x^4+x+1$
according as $N_{k-2}=4,5,6,7$; these are $P_{3,0}$, $P_{0,3}$, $P_{1,4}$ and $P_{2,2}$ of
Proposition~\ref{prop:seeds}. For example, $n=33$ gives $m=30,14,6,2$ and $x^3+x+1$.
\end{remark}

\section{A remark on the restatement in the 2025 paper}\label{sec:erratum}

In the arXiv version of \cite{GR25}, Conjecture 3.13 of \cite{GR23} is restated as
Conjecture 3.3, for both the even and the odd sequences of $A=x^n+x+1$, $n\ge4$, with $s$ the
greatest positive integer such that $n-2^{s+1}\ge0$. With this
non-strict inequality the predicted length $2^s+1$ differs from the true length
$2^{k-2}+1$ exactly when $n$ is a power of $2$: for $n=2^k$ it predicts $2^{k-1}+1$. Thus it
fails for $n=4$ and $n=8$ (predicting $3$ and $5$ instead of $2$ and $3$), and for instance, for
$n=32$ the length is $9$, as in the table of \cite{GR25}, whereas the restated
formula gives $17$. The original statement of \cite{GR23}, with the strict inequality
$n-2^{s+1}\ge1$, is the one proved here.

\section{Verification}

The ancillary file \texttt{verify\_collatz\_binary.py} (Python~3, no dependency) recomputes, by
exact arithmetic on bit-encoded polynomials, the identities used above and the statements of
Lemmas~\ref{lem:plateau}--\ref{lem:binary}, Propositions~\ref{prop:seeds}
and~\ref{prop:induction} and Theorem~\ref{thm:main} on the ranges it prints (for instance the
length formula for $3\le n\le1199$, and the blocks and the last block for $5\le n\le699$), and
the comparison of Section~\ref{sec:erratum} for $4\le n\le2099$. It runs in less than two
minutes.
These computations check the statements; the proofs above do not depend on them.

\paragraph{Acknowledgements.} Technical assistance: Claude (Anthropic).

\raggedright
\begin{thebibliography}{9}

\bibitem{GR19} L. H. Gallardo and O. Rahavandrainy, \emph{On Mersenne polynomials over
$\F_2$}, Finite Fields Appl. \textbf{59} (2019), 284--296.

\bibitem{GR23} L. H. Gallardo and O. Rahavandrainy, \emph{On Collatz conjecture for binary
polynomials}, arXiv:2308.01181, 2023.

\bibitem{GR25} L. H. Gallardo and O. Rahavandrainy, \emph{A variant of Collatz's conjecture
over binary polynomials}, arXiv:2510.07530, 2025; Rev. Un. Mat. Argentina \textbf{69} (2026),
no.~1, 295--302, doi:10.33044/revuma.5555.

\bibitem{HMYZ} K. Hicks, G. L. Mullen, J. L. Yucas and R. Zavislak, \emph{A polynomial
analogue of the $3N+1$ problem}, Amer. Math. Monthly \textbf{115} (2008), no.~7, 615--622.

\bibitem{Nic} D. Nichols, \emph{Analogues of the $3x+1$ problem in polynomial rings of
characteristic $2$}, Experiment. Math., doi:10.1080/10586458.2016.1227734; arXiv:1610.02545.

\end{thebibliography}

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