Dallime mes rishikimeve të "Përdoruesi:Armend/nënfaqe"

My question is.Is my assumption true and 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. Thanks
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)|
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 $x 3$ then $|t^4(c)|=\max{m-3,2}+1=\max{m-2,3} y$ then $t(c)$ is a permutation of $t(y,x^{m-1})$, so the previous argument applies.
-- mac 12 hours ago
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 Observation ensures that $t^{2+2(y-1)-3}(c) = t^{2y-3}(c) \in B$.
-- Brian M. Scott 8 hours ago