Ndryshimi mes inspektimeve të "Përdoruesi:Armend/nënfaqe"

:<math>t(c)=(t_0,t_1,...,t_p)\,</math>
 
where <math>t_j,j\in I_{p+1}\,</math> denote number of terms of sequence c thats are equal at j, is called trace of c.<br>
Is clear that terms of trace fulfills the conditions
 
Reasons for that name are because I suppose that:
 
Claim:For each finite sequence <math>a\,</math> of natural numbers exists natural number n such that <math>t^n(a)\in H\,</math> in other words each sequence converges to H.
 
#Sequence :<math>t^n(a)\,</math>in is of type <math>BH\,</math> if its convergein toother Hwords from B for exampleeach sequence (2,3)converges isto of type B becauseH.
 
#Sequence <math>a\,</math> is of type <math>B\,</math> if its converge to H from B for example sequence
(2,3) is of type B because
 
: <math>t^3((2,3))=(0,0,2)\in B\,</math>
:<math>t^6((0))=(1,2)\in R\,</math>
 
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 algorithm.<br>
numbers, can be done any programme or algorithm.
===0===
Firstly we reduce the problem to an analysis of the cases <math>m\le2\,</math>.
I'll write <math>t^k\,</math> for the <math>k\,</math>-fold composition of <math>t\,</math> with itself, so that <mathbr>t^1=t\,</math>, <math>t^2=t\circ t\,</math>, etc.
I'llwith also write <math>|{c}|\itself,</math> for the number of distinct elements ofso thethat <math>mt^1=t\,</math>-tuple, <math>ct^2=(c_0,c_1,t\dots,c_{m-1})\in\mathbb{N}^mcirc t\,</math>, etc.
And I'll also write things like <math>(n,m^k,p)|{c}|\,</math> asfor shorthandthe fornumber <math>(n,m,m,\dots,m,p)\,</math>of wheredistinct <math>m\,</math>elements isof repeatedthe <mathbr>k\,</math> times.
<math>m\,</math>-tuple <math>c=(c_0,c_1,\dots,c_{m-1})\in\mathbb{N}^m\,</math>.
And I'll write things like <math>(n,m^k,p)\,</math> as shorthand for <math>(n,m,m,\dots,m,p)\,</math><br>
where <math>m\,</math> is repeated <math>k\,</math> times.
 
A first observation is that
:<math>t(c)=t(c')\,</math> for any permutation <math>c'\,</math> of <math>c\,</math>.
 
Lemma: <math>|t^2(c)|\geq |c|\,</math> if and only if either <math>|{c}|=1\,</math>, or <math>|{c}|=2\,</math> and <mathbr>c=(a,b^{|c|-1})\,</math> (up to permutation) where <math>a\ne b\,</math>.
and <math>c=(a,b^{|c|-1})\,</math> (up to permutation) where <math>a\ne b\,</math>.
 
Proof: We have <math>|t^2(c)|=\max\{t_0,\dots,t_{p+1}\}\,</math>, so <math>|t^2(c)|\ge |c|\,</math> if and only if <mathbr>|t^2(c)|\in \{|c|,|c|+1\}\iff \{|c|-1,|c|\}\cap \{t_0,\dots,t_p\}\ne \emptyset\,</math>, which is equivalent to the conditions above.
<math>|t^2(c)|\in \{|c|,|c|+1\}\iff \{|c|-1,|c|\}\cap \{t_0,\dots,t_p\}\ne \emptyset\,</math>, which is equivalent to the conditions above.
 
Claim: If <math>m=|c|\ge 3\,</math> then <math>|t^n(c)|<|c|\,</math> for some <math>n\ge 1\,</math>.