Dallime mes rishikimeve të "Përdoruesi:Armend/nënfaqe"

If <math>m=1\,</math>, <math>c=(x)\,</math> for some <math>x \in \mathbb{N}\,</math>. If <math>x>0\,</math>, <math>t(c) = (0^x,1)\,</math>, and either <math>x=1\,</math>, in which case <math>t^2(c) = (0,2), t^3(c) = (1,0,1)\,</math>, and <math>t^4(c) = (1,2) \in R\,</math>. If <math>x=0\,</math>, <math>t(c) = (1)\,</math>, and we’re in the first case.
===2===
Now let <math>m=2\,</math> and <math>c = (x,y)\,</math>. Observation: If <mathbr>n>2\,</math>, <math>t^2((n,2)) = (n-1,2)\,</math>, while <math>t((2,2)) = (0,0,2) \in B\,</math>, so <math>t^{1+2(n-2)}((n,2)) = t^{2n-3}((n,2)) \in B\,</math> whenever <math>n \ge 2\,</math>.
Observation: If <math>n>2\,</math>, <math>t^2((n,2)) = (n-1,2)\,</math>, while <math>t((2,2)) = (0,0,2) \in B\,</math>,
<br>so <math>t^{1+2(n-2)}((n,2)) = t^{2n-3}((n,2)) \in B\,</math> whenever <math>n \ge 2\,</math>.
 
#If <math>x=y \in {0,1}\,</math>, <math>t(c) = (0^x,2),t^2(c) = (x,0,1)\,</math>, and <math>t^4(c) = (0,1,1) \in R\,</math>.
#If <math>x=y>1\,</math>, <math>t(c) = (0^x,2),t^2(c) = (x,0,1),t^3(c) = (1,1,0^{x-2},1)\,</math>, and <math>t^4(c) = (x-2,3)\,</math>.
#If <math>x=y=2\,</math>, <math>t^4(c) \in B\,</math>; if <math>x=y \in {3,4}\,</math>, <math>t^6(c) = (2,2) \in B\,</math>; and if <math>x=y=5\,</math>, <math>t^{10}(c) = (2,2) \in B\,</math>.
#Ifif <math>x=y>5 \,</math>,in <math>t^5(c) = (0,0,0,1,0^\{x-63,4\},1)\,</math>, and <math>t^6(c) = (x-32,2) \,</math>,in where <math>x-3>2B\,</math>, whence;
<br>and if <math>x=y=5\,</math>, <math>t^{10}(c) = (2,2) \in B\,</math>.
#If <math>x=y>5\,</math>, <math>t^5(c) = (0,0,0,1,0^{x-6},1)\,</math>, and <br>
<<math>t^6(c) = (x-3,2)\,</math>, where <math>x-3>2\,</math>, whence
:<math>t^{6+2(x-3)-3}(c) = t^{2x-3}(c) \in B\,</math> by the Observation.
 
Now assume without loss of generality that $x Observation ensures that
:<math>t^{2+2(y-1)-3}(c) = t^{2y-3}(c) \in B\,</math>.
-- Brian M. Scott 8 hours ago
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