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\begin{center}
\vskip 1cm{\LARGE\bf 
On Collatz Words, Sequences, and Trees
}
\vskip 1cm
\large
Wolfdieter Lang \\
Karlsruhe \\
Germany\\
\href{mailto:wolfdieter.lang@partner.kit.edu}{\tt wolfdieter.lang@partner.kit.edu}
\end{center}

\vskip .2 in

\begin{abstract}
Motivated by recent work of  Tr\"umper, we consider the general Collatz
word (up-down pattern) and the sequences following this pattern. We
derive recurrences for the first and last sequence entries from
repeated application of the general solution of a binary linear
inhomogeneous Diophantine equation. We solve these recurrences and
also discuss the Collatz tree.
\end{abstract}

\section{Introduction}

The Collatz map $C$ for natural numbers maps an odd number $m$ to $3 m
+ 1$ and an even number to $m/2$.  The Collatz conjecture (see Lagarias
\cite{Lagarias} for original references) is the claim that every
natural number $n$ ends up, after sufficiently many iterations of the
map $C$, in the trivial cycle $(4,  2,  1)$.

Motivated by the work of Tr\"umper \cite{Truemper} we consider a
general finite Collatz word on the alphabet $\{u, d\}$, where $u$ (for
`up') indicates application of the map $C$ on an odd number, and $d$
(for `down')  for applying the map $C$ on an even number. The task is
to find all sequences which follow this word pattern (to be read from
left to right). These sequences are called $CS$ (for Collatz sequence,
also for the plural) realizing the CW (for Collatz word, also for the
plural) under consideration. This problem was solved by Tr\"umper
\cite{Truemper} under the restriction that the first and last sequence
entries are odd. Here we do not use this restriction.  The solutions
are given in terms of recurrence relations for the first and last
entries of the $CS$ for a given $CW$. This involves a repeated
application of the general solution in positive integers of the linear
inhomogeneous Diophantine equation $a x + b y = c$, with $a = 3^m$, $b
= 2^n$ and given integer $c = c(m,n)$. Because $\gcd(3,2) = 1$, one
always has a countably infinite number of solutions. This general
solution depends on a non-negative integer parameter $k$. We believe
that our solution is more straightforward than the one given by
Tr\"umper \cite{Truemper}.


\section{Collatz words, sequences and the Collatz tree}\label{section2}
The Collatz map $C:\ \mathbb{N} \to \mathbb{N},\ m \mapsto 3 m + 1$ if
$m$ is odd,  $m \mapsto m/2$
if $m$ is even, leads to an
increase $u$ (for `up') or decrease $d$ (for `down'), respectively.
Finite Collatz words over the alphabet $\{u, d\}$ are considered with
the restriction that, except for the one letter word $u$, every $u$ is
followed by a $d$, because $2 m + 1 \mapsto 2 (3 m+1)$. This is the
reason for also introducing (with Tr\"umper \cite{Truemper}) $s := ud$.
Thus $s$ stands for  $2 m + 1\mapsto 3 m + 1$. The general finite word
is encoded by an $(S+1)-$tuple $\vec n_{S} = [n_0, n_1, \ldots , n_S ]$
with $S\in \mathbb N$.

\begin{eqnarray} \label{CW}
 CW(\vec n_{S+1}) & = & d^{n_0} s d^{n_1-1} s \cdots  s d ^{n_S-1} \\
                              & =  &   (d^{n_0} s) (d^{n_1-1} s) \cdots (d ^{n_{S-1}-1} s)  d ^{n_S-1} \nonumber
\end{eqnarray}
with $n_0\in {\mathbb N}_0 := {\mathbb N} \cup  \{0\} $, $n_i \in \mathbb N$, for $i = 1, 2, \ldots,  S$. The number of  letters of $u$ (that is  of $s = ud$) in the word $CW(\vec n_{S})$, or $CW(S)$, for short, is $S$ (which is why we have used $\vec n_S$ not $\vec n_{S+1}$ for the $(S+1)-$tuple), and the number of letters of $d$ is 
$$D(S) := \sum_{j=0}^{S} n_j .$$
In the paper of Tr\"umper \cite{Truemper} $n_0 = 0$ (start with an odd number), $y = S$ and $x = D(S)$.

Some special words are not covered by this notation: first the 
one-letter word $u$ with the Collatz sequence ($CS$) of length two $CS(u;
k) = [2 k + 1,  2 (3 k +  2)]$, and $CW([n_0]) = d^{n_0}$ with the
family of  sequences $CS([n_0];k) = [2^{n_0} k, 2^{n_0 - 1} k, \ldots,
1 k] $ with  $k \in {\mathbb N}_0$.

A Collatz sequence $CS$ realizing a word $CW(\vec n_{S})$  is of length
$L = D + S + 1$, and follows the word pattern from the left to the
right: 
$$CS(\vec n_{S}) = [c_1,  c_2, \ldots , c_L].$$
For example,
$CW([1,2,1]) = dsds$ with  $S = 2$,  $D(2) = 2$, and length $L = 7$
with $SC_0(\vec n_S) = [2, 1, 4, 2, 1, 4, 2]$ is the first of these
sequences (for non-negative integers), the one with smallest starting
number $c_1$. In order to conform with the notation used by Tr\"umper
\cite{Truemper}, we use $c_1 = M$ for the starting number and 
$c_L = N$ for the last number.

However, in the paper \cite{Truemper} $M$ and $N$ are
restricted  to be odd, which is not the case here. Later one can get the
words with odd starting number $M$ by choosing $n_0 = 0$. In order to also have
$N$ odd, one has to pick only the odd numbers
from $SC_k([0, n_1, \ldots , n_S])$, for
$k \in \mathbb N_0$.

In the Tr\"umper \cite{Truemper} article the monoid of Collatz words,
with the unit element $e =$ {\it empty word} is treated. This is not
considered in this work. Also the connection to the $3 m - 1$ problem
is not pursued here.

The Collatz tree $CT$ is an infinite (incomplete) ternary tree, starting
with the root, the number $8$ on top at level $l = 0$. Three  branches,
labeled $L$, $V$ and $R$ can be present.  If a node (vertex) has label
$n\equiv 4$ (mod $6$) the out-degree is $2$ with a left edge
(branch) labeled $L$ ending in a node with label $\frac{n-1}{3}$
and a right edge (label $R$) ending in the node labeled $2 n$. In the
other cases, $n\equiv 0, 1, 2, 3, 5$  (mod $6$), with out-degree $1$, a
vertical edge (label $V$) ends in the node labeled $2 n$. The root
labeled $8$ stands  for the trivial cycle $8$ repeat$(4, 2, 1)$. See
Figure 1 for $CT_7$ with only the first eight levels.

It may seem that
this tree is left-right  symmetric (disregarding the node labels), but
this is no longer the case starting at level $l = 12$. At level $l =
10$ the mod $6$ structure of the left and right part of $CT$, also
taking into account the node labels, is broken for the first time, but
the node labels $4$ (mod  $6$) are still symmetric. At the next level
$l = 11$ the left-right symmetry concerning the labels $4$ (mod $6$) is
also broken, leading at level $l = 12$ to a symmetry breaking in the
branch structure of the left and right part of $CT$. Thus at level $l =
12$ the number of nodes becomes odd for the first time: $15$ nodes on
the left side versus $14$  nodes on the right one. See rows $ l +  3$
of \seqnum{A127824} for the node labels of the first levels, and
\seqnum{A005186}$(l+3)$ for the number of nodes. The number of $4$ (mod
$6$) nodes at level $l$ is given in \seqnum{A176866}$(l+4)$.

A $CS$ is determined uniquely from its starting number $M$. Therefore
no number can appear twice in $CT$, except for the numbers $1, 2, 4$ of
the (hidden) trivial cycle. The Collatz conjecture is that every
natural number appears in $CT$ at some level ($1, 2,$ and $4$ are hidden
in the root $8$). A formula for $l = l(n)$ would prove the conjecture.

Reading $CT$ from bottom to top, beginning with some number $M$ at a
certain level $l$, recording the edge labels up to level $l = 0$, leads
to a certain $L, V, R$-sequence. For example, $M = 40$ at level $l = 5$
generates the length $5$ sequence $[V, R, V, L, V]$. This is related to
the CS starting with $M = 40$, namely  $[40, 20, 10, 5, 16, 8]$, one of
the realizations of the CW  $dddud = d^3s$, with $S = 1$ and $\vec n_1
= [3,1]$. (Later we shall  see that this is the realizations with the
third smallest starting number, the smaller once are $8$ and $24$). One
has to map $V$ and $R$ to $d$ and $L$ to $u$. This shows that the map
from a $L,V,R$-sequence to a CW is not one to one. The numbers $n\equiv
4$ (mod $6$) except $4$ (see \seqnum{A016957}) appear exactly in two
distinct CS. For example, $64 \equiv 4$ (mod $6$) shows up in all $CS$
starting at any vertex which descends from the bifurcation at $64$,
e.g., $21, 128$; $42, 256$; $84, 85, 512$; etc.  The $CS$ starting with
$512$ and ending in $64$ corresponds to the same $CW$ $ddd$ as the $CS$
starting with $84$ and ending in $64$.

\begin{figure}
\begin{center}
\includegraphics[height=8cm,width=.8\linewidth]{CollatzTree}
\end{center}
\caption{The Collatz Tree $\bf CT_7$} 
\end{figure}

\section{Solution of a certain linear inhomogeneous Diophantine equation}\label{section3}
For the derivation of the recurrence relations for the start and end numbers $M$ and $N$ of Collatz sequences ($CS$) with prescribed up-down pattern (realizing  a given  $CW$) we  need the general solution of the following linear and inhomogeneous Diophantine equation.  

\begin{equation}\label{Diophant}
D(m,n;c):\hskip 2cm 3^m x - 2^n y =  c(m,n),\ \ m \in \mathbb N_0, \ n\in \mathbb N_0,\ c(m,n) \in \mathbb Z\ .
\end{equation}

It is well known (e.g., Niven et al.~\cite[pp.~212--214]{NZM})
how to solve the equation $a x + b y = c$
for integers $a,  b$ (not $0$) 
and $c$ provided $g = \gcd(a, b)$ divides $c$ 
for integers $x$ and $y$;
otherwise there is no solution.
One finds a sequence of solutions parameterized by $t\in \mathbb Z$.
Then one has to restrict the $t$ range to obtain all positive solutions.
The procedure is to first find a special solution $(x_0,  y_0)$ of the equation with $c = g$. Then the general solution is $(x = \frac{c}{g} x_0 + \frac{b}{g} t,  y = \frac{c}{g}  y_0 - \frac{a}{g}  t)$ with $t\in Z$. The proof is found in \cite{NZM}. For our problem $g = \gcd(3^m, 2^n) = 1$ for non-negative $m, n$, and this $g$ divides any $c(m, n)$. 


\begin{lemma}\label{SolDiophant} Solution of $D(m,n;c)$
    
{\bf (a)} A special positive integer solution of  $D(m,n;1)$ is 
\begin{eqnarray}\label{x0y0}
y_0(m,n) & = & \left(\frac{3^m + 1}{2}\right)^{n + 3^{m-1}}  \pmod {3^m}\ , \nonumber\\
x_0(m,n) & = & \frac{1 + 2^n y_0(m,n)}{3^m}\ .
 \end{eqnarray}

 {\bf (b)} The general solution with positive $x$ and $y$ is
\begin{eqnarray}\label{Solxy}
x(m,n) & = & c(m,n) x_0(m,n) + 2^n t_{\min}(m,n;\sgn(c)) + 2^n k \ ,\nonumber  \\
y_0(m,n) & = & c(m,n) y_0(m,n) + 3^m t_{\min}(m,n;\sgn(c)) + 3^m k \ ,
 \end{eqnarray}
with $k \in \mathbb N_0$, and 
\begin{equation}\label{tmin}
t_{\min}(m,n;\sgn(c)) =  
\begin{cases} \ceil{\abs{c(m,n)} \frac{x_0(m,n)}{2^n}}, & \text{if\  $ c < 0$;}\\
& \\
 \ceil{-c(m,n) \frac{y_0(m,n)}{3^m}}, & \text{if\  $c \geq 0$.}
\end{cases}
\end{equation}
\end{lemma}

For the proof we use the following lemma:
\begin{lemma}\label{Anm}    
$ A(n,m) := {{n-1}\choose{m-1}}  \frac{\gcd(m,n)}{m}$  is a positive integer for $ m = 1, 2,  \ldots, n,\  n \in \mathbb N$.
\end{lemma}

\begin{proof} (due to {\sl Peter Bala}, see \seqnum{A107711}, history, 
February 28 2014)

This is the triangle \seqnum{A107711} with A(0,0) =1. 
By a rearrangement of factors one also has  
$A(n,m) = {{n}\choose{m}} \frac{\gcd(n,m)}{n}$. 
Use $\gcd(n,m) \lcm(m,n) =  n m$ (e.g., 
\cite[Theorem 2.2.2, pp.~15--16]{FR}, where the uniqueness of the $\lcm$ is
also shown).  Now
$$A(n,m) = \frac{a(n,m)}{\lcm(n,m)}$$
with 
$$ a(n,m) = {{n}\choose{m}} m,$$ 
a positive integer because the binomial is a combinatorial number.
Now $m | a(n,m)$ and $n | a(n,m)$ because
$$a(n,m) = n {{n-1}\choose{m-1}}$$
by a rearrangement. Hence $a(n,m) = k_1 m =  k_2 n$, i.e., $a(n,m)$ is a common multiple of $n$ and $m$ (call it $\cm(n,m)$).  Now
$\lcm(n,m) | a(n,m)$ because $\lcm(n,m)$ is the (unique) lowest $\cm(n,m)$.
Therefore $\frac{a(n,m)}{\lcm(n,m)}\in \mathbb{N}$, since only natural numbers are in the game.   
\end{proof}

Now to the proof of Lemma~\ref{SolDiophant}.
\begin{proof} 

{\bf (a)}  $x_0(m,n) = \frac{1 + 2^n y_0(m,n)}{3^m}$ is
a solution of $D(m,n;1)$ for any $y_0(m,n)$.
The given $y_0(m,n)$ is a positive integer $\in \{1, 2, \ldots , 3^{m}
- 1\} \  m\in \mathbb N$ and $y_0(0,n) = 1$ for $n\in \mathbb N_0$. One
has to prove that $x_0(m,n)$ is a positive integer. This can be done by
showing that $1 + 3^n y_0(n,m)\equiv 0$ (mod $3^m$) for $m\in \mathbb
N$. One first observes that $\frac{3^m + 1}{2}\equiv \frac{1}{2}$ (mod
$3^m$), because obviously $2 \frac{3^m + 1}{2} \equiv 1$  (mod $3^m$)
($2$ is a unit in the ring $\mathbb Z_{3^m}$). For $m = 0$ one has
$x_0(0,n) = 1 + 2^n$, $n \in \mathbb N_0$, which is positive. In the
following $m \in \mathbb N$.

\begin{equation}
1 + 2^n \left(\frac{3^m + 1}{2}\right)^{n+3^{m-1}} \equiv 1 + 2^n \left(\frac{1}{2}\right)^{n+3^{m-1}} \equiv 1 +  \left(\frac{1}{2}\right)^{3^{m-1}} \pmod {3^m}\ .
\end{equation} 
Now we show that $L(m) :=  \left(\frac{3^m + 1}{2}\right)^{3^{m-1}} \equiv 0$
(mod $3^m$) by using $\frac{3^m + 1}{2} = 3 k(m) - 1$
with  $k(m):= \frac{3^{m-1} + 1}{2}$, a positive integer.
The binomial theorem leads with $ a(m) =  3^{m -1}$ to
\begin{eqnarray}
(3 k(m) - 1)^{a(m)} & = & \sum_{j=0}^{a(m)-1} {{a(m)}\choose{j}} (-1)^j (3 k(m))^{a(m)-j} \nonumber \\
& = & 3^m \Sigma_1(m) + \Sigma_2(m),  {\rm with} \\
\Sigma_1(m) & = & \sum_{j=0}^{a(m)-m}  (-1)^j {{a(m)}\choose{j}} k(m)^{a(m)-j} 3^{a(m)-m-j}, \ {\rm and} \\
\Sigma_2(m) & = & \sum_{j=a(m)-m+1}^{a(m)-1} (-1)^j {{a(m)}\choose{j}} (3  k(m))^{a(m)-j}\ . 
\end{eqnarray}
Now $\Sigma_1(m)$ is an integer because of $a(m)-j \geq a(m)-m-j \geq 0$  and the integer binomial, hence $L(m)\equiv \Sigma_2$  (mod  $3^m$). Rewriting $\Sigma_2$ with $j' = j-a(m)+m-1$, and also using the symmetry of the binomial, one has
\begin{eqnarray}
\Sigma_2(m)& = & \sum_{j=0}^{m-2} {{a(m)}\choose{j}} (-1)^{j-m} (3 k(m))^{m-1-j}\nonumber \\
& = & \sum_{j=1}^{m-1} (-1)^{j+1} {{a(m)}\choose{j}}  (3 k(m))^j = 3^m \widehat\Sigma_2(m) \ {\rm with}\\
\widehat\Sigma_2(m)& = & \sum_{j=1}^{m-1} (-1)^{1+j} {{a(m)}\choose{j}} k(m)^j 3^{j-m}\nonumber \\
& = &  \sum_{j=1}^{m-1} (-1)^{1+j} k(m)^j {{a(m)-1}\choose{j-1}} \frac{1}{j} 3^{j-1}\ .
\end{eqnarray}
In the last step a rearrangement of the binomial has been applied, remembering that $a(m) =  3^{m-1}$. It remains to be shown that 
$A_{m,j} :=  3^{j-1} {{3^{m-1}-1}\choose{j-1}} \frac{1}{j}$
is a (positive) integer for $j = 1, 2, \ldots , m-1$. Here Lemma~\ref{Anm}  comes to help. Consider there $A(3^{m-1},j)$ for $j = 1, 2, \ldots , m-1$  ($m=0$ has been treated separately above), which is a positive integer. If $3 \nmid j$ then $3^{j-1} A(3^{m-1} j) = A_{m,j}$, hence a positive integer. If $j = 3^k J$, with $k\in \mathbb N$ the largest power of $3$ dividing $j$ then 
$\gcd(3,J) = 1$, and  $j = 3^k J \leq m-1 < 3^{m-1}$ and $\gcd(3^{m-1} 3^k J) = 3^q$ with $q = \min(k, m-1)$.
\end{proof}

\begin{proof} 
{\bf (b)} The general integer solution of Eq.~\eqref{Diophant} is then (see 
\cite[pp.\ 212--214]{NZM}; note that there $b>0$, here $b<0$, and we have changed $t\mapsto -t$) 
\begin{eqnarray}
x = \hat x(m,n;t)& = & c(m,n) x_0(m,n) + 2^n t,\nonumber \\
y = \hat y(m,n;t)& = & c(m,n) y_0(m,n) + 3^m t, \ t \in \mathbb Z.
\end{eqnarray}
In order to find all  positive solutions for $x$ and $y$ one has to restrict the range of $t$, depending on $\sgn c$.
If $c(m,n) \geq 0$  then, because $x_0$ and $y_0$ are positive and
$$\frac{x_0(m,n)}{2^n} = \frac{y_0(m,n)}{3^m} + \frac{1}{2^n 3^m},$$
we have
$t > -\frac{c(m,n) x_0(m,n)}{2^n}$ and  
$t > -\frac{c(m,n) y_0(m,n)}{3^m}$, i.e., 
\begin{align*}
t &\geq \ceil{\max\left( - \frac{c(m,n) x_0(m,n)}{2^n}, -\frac{c(m,n) y_0(m,n)}{3^m}\right)} \\
& =
\ceil{-c(m,n) \min\left(\frac{x_0(m,n)}{2^n}, \frac{y_0(m,n)}{3^m}  \right)}  \\
&= \ceil{-c(m,n) \frac{y_0(m,n)}{3^m} }  \\
&=t_{\min}(m,n;+). 
\end{align*}
If $c(m,n) <  0$ then
\begin{align*}
t &\geq \ceil{ \abs{c(m,n)} \max\left(\frac{x_0(m,n)}{2^n}, \frac{y_0(m,n)}{3^m}\right)} \\
&= \ceil{\abs{c(m,n)} \frac{x_0(m,n)}{2^n}}   \\
&=  t_{\min}(m,n;-).
\end{align*}
Thus with $t = t_{\min}(m,n;\sgn(c)) + k$, with $k\in \mathbb N_0$ one has the desired result. Note that $(x_0(m,n), y_0(m,n))$ is the smallest positive solution of the equation  $D(m,n;1)$,  Eq.~\eqref{Diophant}, because, for $c(m,n) = 1$, $t_{\min}(m,n;+) = \ceil{-\frac{y_0(m,n)}{3^m}}$, but with $y_0(m,n) \in \{1, 2, \ldots , 3^m-1\}$ this is $0$.
\end{proof} 

A proposition on the periodicity of the solution $y_0(m,n)$ follows.

\begin{proposition} \label{y0Periodicity} Periodicity of $ y_0(m,n)$ in $n$

\medskip

{\bf (a)} The sequence $y_0(m,n)$ is periodic in $n$ with primitive period length $L_0 = \varphi(3^m)$, for $m \in \mathbb N_0 $ with Euler's totient function $\varphi(n) =$\seqnum{A000010}$(n)$, where $\varphi(1) := 1$. 

\medskip

{\bf (b)} The sequence $x_0(m,n =  L_0(m)) =  q(m) x_0(m,n) - r(m)$, $m \in \mathbb N_0$,  with $q(m) := 2^{\varphi(3^m)}$ and $r(m) 
:= \frac{2^{\varphi{(3^m)}} - 1}{3^m}$. See \seqnum{A152007}.

\medskip

{\bf (c)} The set $Y_0(m) := \{y_0(m,n) |  n = 0, 1, \ldots , \varphi(3^m) - 1\} $, is, for $m\in \mathbb N_0$, a representation of the set $RRS(3^m)$, the smallest positive restricted residue system  modulo $3^m$. See Apostol 
\cite[p.\ 113]{TApostol} for the definition. The multiplicative group
modulo $3^m$,  called ${\mathbb Z}_{3^m}^{\times} =  (\mathbb Z/3^m
\mathbb Z)^\times$ is congruent to the cyclic group
$C_{\varphi(3^m)}$.
\end{proposition}

\begin{proof} {\bf (a)} By Euler's theorem 
(e.g., \cite[Theorem 2.4.4.3, p.\ 32 ]{FR}) $a^{\varphi(n)} \equiv 1$  (mod $n$), provided $\gcd(a,n) = 1$. Now 
$\gcd\left(\frac{3^m+1}{2},3^m\right) = \gcd\left(\frac{3^m+1}{2},3\right) = 1$
because $\frac{3^m+1}{2} \equiv \frac{1}{2}$
(mod $3^m$) (see above), and hence
$\frac{3^m+1}{2} \not\equiv 0$  (mod $3^m$).
This shows that $L_0(m)$ is a period length,
but we have to show that it is in fact the length of the primitive period,
i.e., we have to prove that the order of  $\frac{3^m+1}{2}$
modulo $3^m$ is $L_0(m)$. 
(See e.g., 
\cite[Definition 2.4.4.1, p.\ 31]{FR}, for the order definition.) In other words we want to show that  $\frac{3^m+1}{2}$
is a primitive root (of $1$) modulo $3^m$. Assume that $k(m)$ is this order (the existence is certain due to Euler's theorem). Hence $(\frac{1}{2})^{k(m)} \equiv 1$  (mod $3^m$) and $k(m) | L_0(m)$. It is known that the modulus $3^m$ possesses primitive roots, and the theorem on the primitive roots says that there are precisely $\varphi(\varphi(3^m))$ incongruent ones (e.g., 
Niven et al.~\cite[pp.\ 205, 207] {NZM}, 
or Nagell \cite[Theorem 62.3, p.\ 104 and Theorem 65, p.\ 107]{Nagell}). In our case this number is  $\varphi(2\cdot 3^{m-1}) = 2\cdot 3^{m-2}$ if $m \geq 1$. The important point, proven in Nagell \cite[Theorem 65.3, p.\ 107]{Nagell}, is that if we have a primitive root $r$ modulo an odd prime, here $3$, then, if $r^{3-1} - 1$ is not divisible by $3^2$, it follows that $r$ is in fact a primitive root for any modulus $3^q$, with $q \in \mathbb N_0 $. One of the primitive roots modulo $3$ is $2$, because $2^2 =  4\equiv 1$  (mod $3$) and $2^1 \not\equiv 1$  (mod $3$). 
Also $2^{3-1} - 1 = 3$ is not divisible by $3^2$, hence $2$ is a primitive root of any modulus  $3^q$ for $q\in \mathbb N_0$. From this we proof that 
$\frac{3^m+1}{2} \equiv \frac{1}{2}$ (mod  $3^m$) is a primitive root modulo $3^m$.  Consider  $\left(\frac{3^m+1}{2}\right)^k \equiv \frac{1}{2^k}$
(mod $3^m$) for $k = 1, 2, \ldots , \varphi(3^m)$.
In order to have $\left(\frac{1}{2}\right )^k \equiv 1$
(mod $3^m$) one needs $2^k \equiv 1$  (mod  $3^m$). But due to Nagell \cite[Theorem 65.3, p.\ 107]{Nagell}, for $p = 3$, a primitive root modulo $3^m$ is $2$, and the smallest positive $k$ is therefore $\varphi(3^m)$,  and
hence $\frac{3^m+1}{2}$
is a primitive root (of $1$) of modulus $3^m$. 

\medskip

{\bf (b)} $x_0(m,n + \varphi(3^m)) =  \frac{1 + 2^n 2^{\varphi(3^m)} y_0(m,n)}{3^m}$ from the periodicity of $y_0$. Rewritten, this is
\begin{align*}
\frac{2^{\varphi(3^m)}\,\left( ( 2^{-\varphi(3^m)} - 1) + (1 + 2^n\,y_0(m,n) \right)} {3^m} &=  -\frac{1}{3^m}\,(2^{\varphi(3^m)} - 1) + 2^{\varphi(3^m)} x_0(m,n) \\
&=   q(m)  x_0(m,n) -  r(m)
\end{align*}
with the  values given in the proposition.

\medskip

{\bf (c)} This follows from the reduced residue system modulo $3^m$ for  $m \in \mathbb N_0$, 
 $$ \left\{ \left(\frac{1}{2}\right)^0, \left(\frac{1}{2}\right)^1,  \ldots ,  \left(\frac{1}{2}\right)^{\varphi(3^m) - 1} \right\},$$ 
 because $\frac{1}{2}$
 is a primitive root modulo $3^m$ (from part {\bf b)}).
 With $a(m) := \frac{3^m + 1}{2}$  one has $1 = \gcd(a(m),3) = \gcd(a(m),3^m) =  \gcd(a(m)^{b(m)} 3^m)$ with $b(m) := 3^{m - 1}$, and 
$$ \left\{a(m)^{b(m)} \left(\frac{1}{2}\right)^0, a(m)^{b(m)}  \left(\frac{1}{2}\right)^1, \ldots , a(m)^{b(m)} \left(\frac{1}{2}\right)^{\varphi(3^m) - 1}\right\}$$
is a reduced residue system modulo $3^m$ (see Apostol 
 \cite[Theorem 5.16, p.\ 113]{TApostol}).
 Thus $$Y_0(m)\equiv \{a(m)^{b(m)} 1, a(m)^{b(m)+1}, \ldots , 
 a(m)^{b(m) + \varphi(3^m) - 1} \}$$
 is a reduced residue system modulo $3^m$. Therefore this gives a permutation of the reduced residue system modulo $3^m$ with the smallest positive integers sorted increasingly.
\end{proof}     

\begin{example}  For $m = 3$, $\varphi(3^3) = 2\cdot 3^2 = 18 =
L_0(3)$,  and $$\{y_0(3,n)\}_{n=0}^{17} = \{26, 13, 20, 10, 5, 16, 8,
4, 2, 1, 14, 7, 17, 22, 11, 19, 23, 25\}$$ is a permutation of the
standard reduced residue system modulo $27$, obtained by re-sorting the
found system in increasing order. See \seqnum{A239125}. For $m=1, 2,$ and $4$
see \seqnum{A007583}, \seqnum{A234038} and \seqnum{A239130}  for the
solutions $(x_0(m,n), y_0(m,n))$.
\end{example}

\section{Recurrences and their solution}\label{section4}
After these preparations it is straightforward to derive the recurrence for the start and end numbers $M$ and $N$ for any given $CW(\vec n_S)$, for $S \in \mathbb N$. 

\medskip

{\bf (A)} We first consider the case of words with $n_S = 1$,
i.e., $\vec n_S = [n_0, n_1, \ldots, n_{S-1}, 1]$. This is the word $CW(\vec n_S ) = \rprod_{j=0}^{S-1}  d^{n_j}  s$ (with an ordered product, beginning with $j=0$ at the left-hand side). In order to simplify the notation we use $M(S)$, $N(S)$, $y_0(S)$, $x_0(S)$, and $c(S)$ for $ M(\vec n_S )$, $N(\vec n_S)$ ,  $y_0(S,n_S)$,  $x_0(S,n_S)$ and $c(S,n_S)$,  respectively. 
For $S=1$, the input for the recurrence, one has
\begin{equation}
M(1;k) =  2^{n_0} (2\,k+1)\  \text{and}\ N(1;k) = 3 k + 2,\ {\text for}\ k \in \mathbb N\ ,
\end{equation}
because there are $n_0$ factors of $2$ from $d^{n_0}$, and then an odd number $2 k + 1$ leads after application of $s$ to $3 k + 2$. Thus $M(1) = 2^{n_0}$ and $N(1) = 2$.

\begin{proposition} {Recurrences for $M(S)$ and $N(S)$ with $n_S = 1$.}
\label{Rec}

{\bf (a)} The coupled recurrences for $M(S,t)$ and $N(S,t)$, the first and last entry of the Collatz sequences $CS(\vec n_S;t)$ for the word $ CW(\vec n_S)$ with $\vec n_S = [n_0, n_1, \ldots,  n_{S-1},1]$  ($n_S = 1$) are 

\begin{eqnarray}
M(S,t) = M(S)& + & 2^{\hat D(S)} t, \nonumber \\
N(S,t) = N(S)& + & 3^S t, \  {\rm with}\ t \in \mathbb Z\ , 
\end{eqnarray}
 where $\hat D(S) := \sum_{j=0}^{S-1} n_j$
 (we prefer to use a new symbol for the $n_S = 1$ case), and the recurrences for $M(S)$ and $\widetilde N(S) = N(S) - 2$ are
\begin{eqnarray} 
M(S) = M(S-1) & + & 2^{\hat D(S-1)} c(S-1) x_0(S-1) \ , \nonumber \\
 \widetilde N(S) & = & 3 y_0(S-1) c(S-1) ,
\end{eqnarray}
with
\begin{equation}
c(S-1) = 2 (2^{n_{S-1}-2} - 1) - \widetilde N(S-1) =: A(S-1) -  \widetilde N(S-1)\ .
\end{equation}
The recurrence for $c(S)$ is
\begin{equation}
c(S) = -3 y_0(S-1) c(S-1) + A(S), \   S \geq 2, 
\end{equation}
and the input is $M(1) = 2^{n_0}$, $\widetilde N(1) = 0$ and $c(1) = A(1)$.

\medskip

{\bf (b)} The general positive integer solution is
\begin{eqnarray} \label{eq18}
M(S;k) & = & M(S) + 2^{\hat D(S)} t_{\min}(S-1) +  2^{\hat D(S)} k, \nonumber \\
N(S;k) & = & 2 + \widetilde N(S) + 3^S t_{\min}(S-1) +  3^S k, \ k\in \mathbb N_0 ,
\end{eqnarray}
 where 
\begin{equation}
t_{\min}(S) = t_{\min}(S, n_S,\sgn(c(S))) =
\begin{cases}
\ceil{\abs{c(S)} \frac{x_0(S)}{2^{n_S}}}, &\text{if\ $c(S) < 0$;} \\
\ceil{-c(S) \frac{y_0(S)}{3^S}}, &\text{if\ $c(S) \geq 0 $.} 
\end{cases}
\end{equation}
\end{proposition}
\begin{corollary}\label{cor}
\begin{eqnarray} 
M(S;k) &\equiv& M(S) + 2^{\hat D(S)} t_{\min}(S-1) \pmod {2^{\hat D(S)}}, \nonumber \\
N(S;k) &\equiv& \widetilde N(S) +  3^S t_{\min}(S-1) \pmod { 3^S}.
\end{eqnarray}
\end{corollary} 

In Terras' article \cite{Terras} the first congruence corresponds to  
Theorem 1.2, where the encoding vector $E_k(n)$ refers to the modified Collatz tree using only $d$ and $s$ operations.
\begin{proof} 

{\bf (a)} By induction over $S$. For $S = 1$  the input $M(1) = 2^{n_0}$, $N(1) = 2$ or $\widetilde N(1) = 0$ provides the start of the induction. Assume that part {\bf (a)} of the proposition is true for $S$ values $1, 2, \ldots , S-1$. To find $M(S)$ one has to make sure that $d^{n_{S-1}} s$ can be applied to $N(S-1;k)$, the end number of step $S-1$ sequence $CS(\vec n_{S-1};t)$ which is $N_{int}(S-1,t) =  N(S-1) +3^{S-1}  t$, with integer $t$,  by the induction hypothesis. This number has to be of the form $2^{n_{S-1}-1} (2  m + 1)$ (one has to have an odd number after $n_{S-1}$ $d-$steps such that $s$ can be applied). Thus 
$3^{S-1} t - 2^{n_{S-1}} m =  2^{n_{S-1}-1} - N(S-1) =  A(S-1) - \widetilde N(S-1) =: c(S-1)$,
where $\widetilde N(S-1) = N(S-1) - 2$ and $ A(S-1) = 2 (2^{n_{S-1}-2} - 1)$. Due to  Lemma~\ref{SolDiophant} the general solution, with $t \to x(S-1,n_{S-1};t) \hat = x(S-1;t)$, $m \to y(S-1,n_{S-1};t) \hat = y(S-1;t)$, to shorten the notation, is
\begin{eqnarray} 
t\to x(S;t) & = & c(S-1) x_0(S-1) + 2^{n_{S-1}} t, \nonumber \\ 
m\to  y(S;t) & = & c(S-1) y_0(S-1) + 3^{S-1} t, \ t \in \mathbb Z\ . 
\end{eqnarray}
Therefore the first entry of the sequence $CS(\vec n_S;t)$ is 
$M(S;t) = M(s-1,x(S-1,t))$ which is
\begin{equation}
M_{int}(S;t) =  M(S-1) + 2^{\hat D(S-1)} c(S-1) x_0(S-1) + 2^{\hat D(S)} t,
\end{equation}
hence $M(S) = M(S-1) + 2^{\hat D(S-1)} c(S-1) x_0(S-1)$,
the claimed  recurrence for $M(S)$.

The last member of $CS(\vec n_{S-1};t)$ is $3 m+2$ (after applying $s$ on $2 m + 1$ from above). Thus $N_{int}(S;t) = 3 y(S;t) + 2$,
or $N_{int}(S;t) - 2 = 3 c(S-1) y_0(S-1) + 3^S t$.
Therefore, $\widetilde N(S) =  N(S) - 2  = 3 c(S-1) y_0(S-1)$ which is  the claim for the $\widetilde N$ recurrence. Note that the remainder structure of eqs.~\ref{eq18}, expressed also in the  Corollary~\ref{cor}, has also been verified by this inductive proof. The recurrence for $c(S) = A(S) - \widetilde N(S)$  follows from the one for $\widetilde N(S)$. 

\medskip

{\bf (b)} Positive integer solutions from $M_{int}(S;t)$ and $N_{int}(S;t)$ of part {\bf (a)} are found from the second part of  Lemma~\ref{SolDiophant} applied to the equation $3^{S-1} x - 2^{n_{S-1}} y =  c(S-1)$, determining $t_{\min}(S-1)$ as claimed. This leads finally to the formulae for $M(S;k)$ and $N(S;k)$ with $k \in \mathbb N_0$.  
\end{proof}

\begin{example} $(sd)^{S-1} s$ Collatz sequences

Here $n_0 = 0 =  n_S $ and $n_{j} = 2$ for $j = 1, 2, \ldots , S-1$.
The first entries $M(S;k)$ and the last entries $N(S;k)$ of the Collatz sequence $CS([0, 2, \ldots , 2];k)$  (with $S-1$ times a $2$), whose length is $3 S$,  are $M(S;k) = 1 + 2^{2 S-1} k$ and $N(S;k) = 2 +  3^S k$. For $S=3$ a complete Collatz sequence $CS([0,2,2];3)$ of length $9$ is $[97, 292, 146, 73, 220, 110, 55, 166, 83]$ which is a special realization of the word $sdsds$ with starting number $M(3;3) = 97$ ending in $N(3;3) = 83$. Note that for this $u-d$ pattern the start and end numbers have remainders $ M(S;0) = M(1;0) = 1$  and $N(S;0) = N(1;0) = 2$. See the tables \seqnum{A240222} and \seqnum{A240223}.
\end{example}

The recurrences for $M(S) \hat = M(\vec n_{S-1})$,  $\tilde N(S) \hat = \tilde N(\vec n_{S-1})$ or $N(S) \hat = N(\vec n_{S-1})$ and $c(S) \hat =$ $c(\vec n_{S-1})$ are solved by iteration with the given inputs $M(1) = 2^{n_0}$, $\tilde N(1) = 0$ and $c(1) = A(1) = 2 (2^{n_0-2} - 1)$. 

\begin{proposition} Solution of the recurrences for $n_s = 1$ \label{SolRec}

The solution of the recurrences of Proposition~\ref{Rec} with the given inputs are, for $S\in \mathbb N$:

\begin{eqnarray}
c(S)& = &  A(S) + \sum_{j=1}^{S-1} (-3)^j A(S-j) \prod_{l=1}^j y_0(S-l), \nonumber \\
\tilde n(S)& = & A(S) -  c(S) =  -\sum_{j=1}^{S-1} (-1)^j A(S-j) \prod_{l=1}^{j} y_0(S-l), \nonumber \\
N(s)& = & \tilde N(S) + 2, \nonumber \\
M(S) & =& 2^{n_0} + \sum_{j=1}^{S-1} R(S-j), 
\end{eqnarray}
with $\hat D(S) := 1 + \sum_{j=0}^{S-1}  n_j$,  
$A(S) := 2 (2^{n_S-2} - 1)$,
$R(S) := 2^{\hat D(S)}  x_0(S) c(S)$,
and $y_0(S) \hat = y_0(S,n_S)$, $x_0(S) \hat = x_0(S,n_S)$,
given in Lemma~\ref{SolDiophant}.
\end{proposition}
\begin{proof}

This is obvious.
\end {proof} 

{\bf (B)} The general case $n_S = 1$ can now be found by appending the operation $d^{n_S-1}$ to the above result. This leads to the following theorem.

\begin{theorem} \label{GenSol} The general case ${\vec n}_S$.

 For the Collatz word $CW(\vec n_S) = d^{n_0} \rprod_{j=1}^{S} (s d^{n_j-1}) =  d^{n_0} s\,\rprod_{j=1}^{S} (d^{n_j-1} s) d^{n_S-1}$ (the ordered product begins with $j=1$ on the left-hand side) with  $n_0 \in \mathbb N_0$ , $n \in \mathbb N$, the first and last entries of the corresponding Collatz sequences $\{CS(\vec n_S;k)\}$, of length $L(S) = n_0 + 2 \sum_{j=1}^S n_j$,  for $k \in \mathbb N_0$, are

\begin{eqnarray}
M(\vec n_S;k) & = & M(S) - 2^{\hat D(S)} N(S) x_0(S,n_S-1) + 2^{D(S)} t_{\min}(S,n_S-1, \sgn(c_{new}(S)) \nonumber\\
&&  + 2^{D(S)}k, \nonumber \\
N(\vec n_S;k) & = & c_{new}(S) y_0(S,n_S-1) + 3^S  t_{\min}(S,n_S-1,\sgn(c_{new}(S)) + 3^S k,
\end{eqnarray}
\end{theorem}

with $c_{new}(S) := -N(S)$, $\hat D(S) = 1 + \sum_{j=0}^{S-1} n_j$, $D(S) = \sum_{j=0}^S n_j$.
\begin {proof}

In order to be able to apply to the Collatz sequences $CS([n_0,n_1,
\ldots ,n_S-1,1])$ (with the results from part (A) above) the final
$d^{n_S-1}$ operation one needs for the last entries  $N_{int}(S;t) =
N(S) + 3^S t =  2^{n_S-1} m$ with some (even or odd) integer m. The new
last entries of $CS([n_0,n_1, \ldots ,n_S];t)$ are then $m$. The
general solution of $3^S - 2^{n_S-1} m =  -N(S) =: c_{new}(S)$ is,
according to Lemma~\ref{SolDiophant}, given by
\begin{eqnarray}
t \to x(S;t)& = &  c_{new}(S)  x_0(S,n_S-1) + 2^{n_S-1}t, \nonumber \\
m \to y(S;t)& = &  c_{new}(S)  y_0(S,n_S-1) + 3^S t, \ t \in \mathbb Z . 
\end{eqnarray}
This leads to positive integer solutions after the shift $t \to
t_{\min} + k$, with $t_{\min} =  t_{\min}(S,n_S-1,\sgn(c_{new}(S)))$ to
the claimed result $N(S;k)$ for the new last number of $CS(\vec n;k)$,
with $k \in \mathbb N_0$. The new start value $M(S;k)$ is obtained by
replacing $t \to x(S;t)$ in the old $M_{int}(S;t)$ (with $n_S = 1$).
$M(S;k) = M_{int}(S,x(S;t))$ with $t \to t_{\min} + k$, also leading to
the claimed formula.
\end{proof}

The remainder structure modulo $2^{D(S)}$ for $M(\vec n_S,k)$ and modulo $3^S$ for $N(\vec n_S,k)$ is manifest.

The explicit sum versions of the results for case $n_S = 1$,
given in Proposition~\ref{SolRec}, can be inserted here.
\begin{example} $ ud^{m} =  sd^{m-1}$

For $m =  1, 2, 3$ and $ k = 0, 1, \ldots , 10$ one finds for
$N([0,m],k)$:
\begin{align*}
&[2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32],  \\
&[1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31], \\
& [2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32],
\end{align*}
and for $M([0,m],k)$:
\begin{align*}
[1,3,5,7,9,11,13,15,17,19,21], \\
[1,5,9,13,17,21,25,29,33,37,41], \\
[5,13,21,29,37,45,53,61,69,77,85].
\end{align*}
Only the odd members of $N([0,m],k)$, that is,
the odd-indexed entries, and the corresponding $M([0,m],k)$  appear in the article of Tr\"umper \cite{Truemper} as Example 2.1. See \seqnum{A238475} for $M([0,2 n],k)$  and  \seqnum{A238476} for  $M([0,2 n - 1],k)$. The odd $N([0,2 n],k)$ values are the same for all $n$, namely $5 + 6 k$, and  $N([0,2 n-1],k) =  1 + 6 k$ for all $n \in \mathbb N$.
\end{example}  

\begin{example} $(ud)^{n} =  s^{S}, S \in \mathbb N$

$\vec n_S = [0,1,\ldots,1]$ with $S$ times a $1$. For $S =  1, 2, 3$
and $ k = 0, 1, \ldots , 10$ one finds for $N(\vec n_S,k)$:
\begin{align*}
[5, 8, 11, 14, 17, 20, 23, 26, 29, 32], \\
[17, 26, 35, 44, 53, 62, 71, 80, 89, 98], \\
[53, 80, 107, 134, 161, 188, 215, 242, 269, 296],
\end{align*}
and for
$M(\vec n_S,k)$:
\begin{align*}
[3, 5, 7, 9, 11, 13, 15, 17, 19, 21], \\
[7, 11, 15, 19, 23, 27, 31, 35, 39, 43], \\
[15, 23, 31, 39, 47, 55, 63, 71, 79, 87].
\end{align*}
For odd $N$ entries, and corresponding $M$ entries this is in
the article of Tr\"umper \cite[Example 2.1]{Truemper}.

See \seqnum{A239126} for these $M$ values, and \seqnum{A239127} for these $N$ values, which are here $S$ dependent.
\end{example} 

In conclusion the author does not think that the knowledge of all Collatz sequences with a given up-down pattern (a given Collatz word) helps to prove the Collatz conjecture. Nevertheless the problem considered in this paper is a nice application of the solution of a simple Diophantine equation.

\section{Acknowledgment}
Thanks go to Peter Bala who answered the author's question for a proof
that all triangle \seqnum{A107711} entries are non-negative integers.
See the history there,  dated February 28 2014.

\section{Note added}
The referee commented that ``... the line of reasoning of the author shows resemblance with the derivation of cycles for the general Collatz problem
($T(n) = \frac{pn+q}{2}$ if $n$ is odd) based on Lyndon words 
\cite{Laarhoven} and the extra equality $x = y$" (in the present work $x = M$ and $y = N$). 



\begin{thebibliography}{9}
 \bibliographystyle{plain}
\bibitem{TApostol} T. M. Apostol, {\it Introduction to Analytic Number Theory}, Springer-Verlag, 1976.
\bibitem{FR} B. Fine and G. Rosenberger, {\it Number Theory}, Birkh\"auser, 2007.
\bibitem{Laarhoven} T. Laarhoven and B. de Weger, The Collatz conjecture and De Bruijn graphs, \emph{Indag. Math. (N.S.)} \textbf{24} (2013) 971--983.  
Preprint available at  \newline
\url{http://arxiv.org/abs/1209.3495}.

\bibitem{Lagarias}  J. Lagarias, The $3x+1$ problem and its generalization, \emph{Amer. Math. Monthly} \textbf{92} (1985), 3--23.
\bibitem{Nagell} T. Nagell, {\it Introduction to Number Theory}, Chelsea Publishing Company, 1964.
\bibitem{NZM}  I. Niven, H. S. Zuckerman, and H. L. Montgomery, {\it An Introduction to the Theory Of Numbers}, Fifth Edition, John Wiley and Sons, Inc., 1991.
\bibitem{OEIS} The On-Line Encyclopedia of Integer Sequences (2010), published electronically at \url{http://oeis.org}.
\bibitem{Terras} R. Terras, A stopping time problem on the positive integers, \emph{Acta Arith.} \textbf{30} (1976), 241--252.  
\bibitem{Truemper} M. Tr\"umper, The Collatz problem in the light of an
infinite free semigroup, \emph{Chinese J. Mathematics},
{\bf 2014} (2014), Article ID 756917, \newline
 \url{http://www.hindawi.com/journals/cjm/aip/756917/}.
\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary 11D04;
Secondary 32H50.

\noindent {\it Keywords}:  Collatz problem, Collatz sequences, Collatz
tree, recurrence, iteration, linear Diophantine equation.

\bigskip
\hrule 
\bigskip

\noindent (Concerned with sequences
\seqnum{A000010},
\seqnum{A005186},
\seqnum{A007583},
\seqnum{A016957},
\seqnum{A107711},
\seqnum{A127824},
\seqnum{A152007},
\seqnum{A176866},
\seqnum{A234038},
\seqnum{A238475},
\seqnum{A238476},
\seqnum{A239125},
\seqnum{A239126},
\seqnum{A239127},
\seqnum{A239130},
\seqnum{A240222}, and
\seqnum{A240223}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  April 10 2014;
revised versions received October 2 2014; November 5 2014; November 11 2014.  
Published in {\it Journal of Integer Sequences}, December 12 2014.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                












