Ndryshimi mes inspektimeve të "Përdoruesi:Armend/nënfaqe"

The set
 
<math>H=B\ cup R\,</math> is called ''' black hole of sequences'''
 
Reasons for that name are because I suppose that:
-- 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.
===1===
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.
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 $n \ge 2$.
===2===
If $x=y \in {0,1}$, $t(c) = (0^x,2),t^2(c) = (x,0,1)$, and $t^4(c) = (0,1,1) \in R$.
Now let $<math>m=2$\,</math> and $<math>c = (x,y)$\,</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>, so $<math>t^{1+2(n-2)}((n,2)) = t^{2n-3}((n,2)) \in B$\,</math> whenever $<math>n \ge 2$\,</math>.
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 $<math>x=y \in {0,1}\,</math>5$, $<math>t^5(c) = (0,0,0,1,0^{x-6},12)$, and $t^62(c) = (x-3,20,1)$\, where $x-3</math>2$, whenceand $<math>t^{6+2(x-3)-3}4(c) = t^{2x-3}(c0,1,1) \in B$ by the ObservationR\,</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>.
Now assume without loss of generality that $x Observation ensures that $t^{2+2(y-1)-3}(c) = t^{2y-3}(c) \in B$.
#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>.
#If <math>x=y>5\,</math>, <math>t^5(c) = (0,0,0,1,0^{x-6},1)\,</math>, and <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 $t^{2+2(y-1)-3}(c) = t^{2y-3}(c) \in B$.
:<math>t^{2+2(y-1)-3}(c) = t^{2y-3}(c) \in B\,</math>.
-- Brian M. Scott 8 hours ago