Here is another way of studying these processes. In the sequel $k=3$ but it is clear how to generalize.

The idea is to look at the matrices formed by three consecutive triples used in the insertion process.
To this end we write the triples produced by the process into an infinite matrix $S$ with $3$ columns (indexed by 0,1,2), and rows
indexed $1,2,,\ldots$. 
The $i$-th row of $S$ consists of the triple (in order) which sums to  $x_i$. 

So the top of $S$ is 

\begin{align*}
	\begin{matrix}
	c0&c1&c2&\;\;x\\ 
	1&2&3&\;\;6\\
	4&5&6&\;\;15\\
	6&7&8&\;\;21\\
	9&10&11&\;\;30\\
	12&13&14&\;\;39\\
        \ldots
\end{matrix}\end{align*}

We let $M(n)$ be the $3\times 3$ submatrix of $S$ which consists of the rows $3n-1,3n,3n+1$ of $S$,
and let $T_0(n),T_1(n),T_2(n)$ be the $3 \times 3$ matrices
\begin{align*} 
	T_0(n):=\begin{pmatrix} 
	8n-4 & 8n-3 & 8n-2\\
	8n-2 & 8n-1 & 8n\\
	8n+1 & 8n+2 & 8n+3\\
	\end{pmatrix},\;\;
	T_1(n):=\begin{pmatrix} 
	8n-4 & 8n-3 & 8n-3\\
	8n-2 & 8n-1 & 8n\\
	8n+1 & 8n+2 & 8n+3\\
	\end{pmatrix}\,\;\end{align*}
\begin{align*}
	T_2(n):=\begin{pmatrix} 
	8n-4 & 8n-3 & 8n-2\\
	8n-1 & 8n-1 & 8n\\
	8n+1 & 8n+2 & 8n+3\\
\end{pmatrix}\;\;.
\end{align*}

Our aim is to show:

**Lemma**: for all $n$ $$M(n)\in\{T_0(n),T_1(n),T_2(n)\}$$

Auxiliary considerations:

(1) Insertion of  $x_n$.
$x_n$ is inserted after the first appearance of its value. Since  $n-1+x_n$ elements of $S$ are less or equal than $x_n$
$x_n$ will be inserted in row $\lceil{n+x_n \over 3}\rceil$ and in column $(n+x_n-1)\bmod 3$.

(2) Succession rules. Assume that $M(j)=T_i(n)$. Where will the produced $x$s be inserted?
Application of the insertion rule gives:

for $i=0$ :

$x_{3j-1}$ goes to row $8n-3+j$ , colummn $1$.
$x_{3j}$ goes to row $8n-1+j$ , colummn $2$.    
$x_{3j+1}$ goes to row $8n+3+j$ , colummn $0$.

for $i=1$ 

$x_{3j-1}$ goes to row $8n-3+j$ , colummn $1$.
$x_{3j}$ goes to row $8n-1+j$ , colummn $0$.
$x_{3j+1}$ goes to row $8n+3+j$ , colummn $0$.


for $i=2$ 

$x_{3j-1}$ goes to row $8n-3+j$ , colummn $0$.
$x_{3j}$ goes to row $8n-1+j$ , colummn $2$.
$x_{3j+1}$ goes to row $8n+3+j$ , colummn $0$.

In other words: if $S(j)$ is of type $T_0$ it generates a type sequence $120$ later, type $T_1$ generates $100$, type $T_2$ generates $020$.
.

**Proof** of the lemma:(sketch) comparison shows that $M(1)=T_0(1)$. Iterated use of the succession rules
now shows that $(M(2),M(3),M(4))=(T_1(2),T_2(3),T_0(3))$, the next block of nine matrices has 
indices $120, 100, 020$, and in general the indices of a following 3^{k+1} block are generated by substituting 
$120$ for $0$, $100$ for $1$ and and $020$ for $2$ in the previous $3^k$ block. **End** of proof.

Some corollaries:

(1) for all n $$x_n\in  \{8n-1,8n-2,8n-3\}$$
Thus $$1+\lfloor{n+x_n \over 3}\rfloor=3n$$.

(2) the row $3n+1$ is $8n+1,8n+2,8n+3$ and $x_{3n+1}=24n+6$


Open question: has the replicative sequence above a simple ternary description?