Trace of sequence of natural numbers Trace of sequence
Denote by $\mathbb{N}=\{0,1,2,...\},~$ the set of natural numbers, and by $I_{m}=\{0,1,...,m-1\}\,$  the set of natural numbers lesser than given natural number $m$. Let $c=(c_0,c_1,...,c_{m-1})\,$  a $m$-sequence of natural numbers, and $p=max\{c_0,c_1,...,c_{m-1}\},\,$  the greatest term of sequence $c$ 
Then the sequence 
$t(c)=(t_0,t_1,...,t_p)\,$
where $t_j,j\in I_{p+1}\,$  denote number of terms of sequence $ c$ thats are equal at $j$, is called trace of $c$. It is clear that terms of trace fulfills the conditions
$t_0+t_1+...+t_p=m\,$
$t_1+2t_2+...+pt_p=c_0+c_1+...+c_{m-1}\,$
Denote by 
$t^{0}(c)=c\,$
$t^{n}(c)=t(t^{n-1}(c))\,$
1.The set of sequences  
$B=\{(1,0,0,1),(2,2),(0,0,2),(2,0,1),(1,1,1),(0,3)\}\,$
that is cycle of length 6 is called '''bracelet of sequences''' because for each sequence $c$ from $B$ holds 
$t^6 (c)=c\,$
2.The set of sequences 
$R=\{(0,1,1),(1,2)\}\,$   
that is cycle of length 2 is called '''ring of sequences''' because for each sequence $c $ from $R$ holds 
$t^2 (c)=c\,$
The set 
$ H=B\cup R\,$  is called ''' black hole of sequences''' 
Reasons for that name are because I suppose that: 
Claim: For each finite sequence $a$ of natural numbers exists natural number $n$ such that $t^n(a)\in H\,$   in other words each sequence converges to $H$.
Sequence $a$ is of type $B$ if its converge to $H$ from $B$ for example sequence $(2,3)$ is of type $B$ because
$t^3((2,3))=(0,0,2)\in B\,$     
And sequences that converges to $H$ from $R$ are of type $R$ for example the sequence $(0)$ is of type $R$ because 
$t^6((0))=(1,2)\in R\,$
My questions are.


*

*Is my assumption true 

*If it is true how to decide of which type is any given finite sequence of natural numbers

*Can be done any programme or algorithme to determine of which type is certain sequence. Thanks

 A: This is not a complete solution, but it seems to reduce the problem to
an analysis of the cases $m\le2$.
I'll write $t^k$ for the $k$-fold composition of $t$ with itself, so
that $t^1=t$, $t^2=t\circ t$, etc.
I'll also write $|\{c\}|$ for the number of distinct elements of the
$m$-tuple $c=(c_0,c_1,\dots,c_{m-1})\in\mathbb{N}^m$.
And I'll write things like $(n,m^k,p)$ as shorthand for $(n,m,m,\dots,m,p)$ where $m$ is repeated $k$ times.
A first observation is that $t(c)=t(c')$ for any permutation $c'$ of
$c$.
Lemma: $|t^2(c)|\geq |c|$ if and only if either $|\{c\}|=1$, or
$|\{c\}|=2$ and $c=(a,b^{|c|-1})$ (up to permutation) where $a\ne b$.
Proof: We have $|t^2(c)|=\max\{t_0,\dots,t_p\}+1$, so $|t^2(c)|\ge |c|$
if and only if $|t^2(c)|\in \{|c|,|c|+1\}\iff \{|c|-1,|c|\}\cap
\{t_0,\dots,t_p\}\ne \emptyset$, which is equivalent to the conditions
above.
Claim: If $m=|c|\ge 3$ then $|t^n(c)|<|c|$ for some $n\ge1$.
Remark: This allows us to reduce to the case $m\in \{1,2\}$ to establish the
claim in the question, if it's true.
Proof of the claim: this is a case-by-case analysis.


*

*If $|\{c\}|>2$ then $n=2$ will do by the lemma.

*If $|\{c\}|=1$, either $c=(0^m)\implies t(c)=(m)$, or
$c=(1^m)\implies t(c)=(0,m)$, or $c=(2^3)\implies t^6(c)=(0,3)$, or
$c=(2^m)$ where $m\ge4$, so that $t(c)=(0,0,m)$, or $c=(x^m)$ where
$x\ge3$. In this case, $t(c)=(0^x,m)$, $t^2(c)=x,0^{m-1},1),
t^3(c)=(m-1,1,0^{x-2},1)$, $t^4(c)=(x-2,2,0^{m-3},1)$. If $m\ge4$ or
$x\ge5$ then $|\{t^4(c)\}|>2$, so $|t^4(c)|=m$ and $|t^6(c)|<m$ by
the lemma. If $m=3$ and $x\in \{3,4\}$ then $t^7(c)=(0,3)$.

*If $|\{c\}|=2$ and $c$ is not of the form $(x,y^{m-1})$ (up to
permutation) then $|t^2(c)|<|c|$ by the lemma.

*If $|\{c\}|=2$ and $c=(x,y^{m-1})$, suppose first that $x<y$. We
have $t(c)=(0^x,1,0^{y-x-1},m-1)$ so $t^2(c)=(y-1,1,0^{m-3},1)$ and if
$y\ge3$ then $t^3(c)=(m-3,2,0^{y-3},1)$, so if $m>3$ then
$|t^4(c)|=\max\{m-3,2\}+1=\max\{m-2,3\}<m$, whereas if $m=3=y$ then
checking the three possibilities $c=(x,3,3)$ for $x\in \{0,1,2\}$
gives $t^{11}(c)=(0,3)$, and if $m=3<y$ then
$t^4(c)=t(0,2,0^{y-3},1)=(y-2,1,1)$ so $t^5(c)=(0,1,0^{y-4},1),
t^6(c)=(y-3,1,1),\dots, t^{2(y-1)}(c)=(1,1,1), t^{2y-1}(c)=(0,3)$.  If
$y\in \{1,2\}$ then $|t^3(c)|=2$.  If $c=(x,y^{m-1})$ where $x>y$ then
$t(c)$ is a permutation of $t(y,x^{m-1})$, so the previous argument
applies.
A: Following on from mac’s partial answer, the assumption is true for $m \in \{1,2\}$, so it appears to be true in general.
If $m=1$, $c=(x)$ for some $x \in \mathbb{N}$. If $x>0$, $t(c) = (0^x,1)$, and either $x=1$, in which case $t^2(c) = (0,2), t^3(c) = (1,0,1)$, and $t^4(c) = (1,2) \in R$. If $x=0$, $t(c) = (1)$, and we’re in the first case.
Now let $m=2$ and $c = (x,y)$. Observation: If $n>2$, $t^2((n,2)) = (n-1,2)$, while $t((2,2)) = (0,0,2) \in B$, so $t^{1+2(n-2)}((n,2)) = t^{2n-3}((n,2)) \in B$ whenever $n \ge 2$.
If $x=y \in \{0,1\}$, $t(c) = (0^x,2),t^2(c) = (x,0,1)$, and $t^4(c) = (0,1,1) \in R$. 
If $x=y>1$, $t(c) = (0^x,2),t^2(c) = (x,0,1),t^3(c) = (1,1,0^{x-2},1)$, and $t^4(c) = (x-2,3)$. 
If $x=y=2$, $t^4(c) \in B$; if $x=y \in \{3,4\}$, $t^6(c) = (2,2) \in B$; and if $x=y=5$, $t^{10}(c) = (2,2) \in B$.
If $x=y>5$, $t^5(c) = (0,0,0,1,0^{x-6},1)$, and $t^6(c) = (x-3,2)$, where $x-3>2$, whence $t^{6+2(x-3)-3}(c) = t^{2x-3}(c) \in B$ by the Observation.
Now assume without loss of generality that $x<y$; then $t(c) = (0^x,1,0^{y-x-1},1)$, and $t^2(c) = (y-1,2)$, which is in $R$ if $y=2$. If $y=1$, $t^4(c) = (1,2) \in R$. And if $y \ge 3$, the Observation ensures that $t^{2+2(y-1)-3}(c) = t^{2y-3}(c) \in B$.
