what loops of numbers are possible when you take the alternating sum of the digits of squared? I've heard about the happy numbers and the sad numbers. if you don't know the happy numbers are numbers when you add it's digits$^2$ and do that n times it hits the floor value of $1$. sad numbers are numbers that go into the same endless loop of sadness $(20,4,16,37,58,89,145,42,20,...)$ $$(\space2^2+1^2=4,\space4^2=16,\space1^2+6^2=37,\space3^2+7^2=58,\space5^2+8^2=89,\space8^2+9^2=145,\space1^2+4^2+5^2=42,\space4^2+2^2=20)$$
my question is slightly different instead of the sum mine uses the absolute value of the alternating sum. so far I've found $2$ loops and to $2$ points $$(\space9^2=81,\space8^2-1^2=63,\space6^2-3^2=27,\space|2^2-7^2|=45,\space|4^2-5^2|=9,\dots),(\space|1^2-6^2|=35,\space|3^2-5^2|=16)$$ $$the\space points\space are\space0\space and\space 1 \space . 1^2-0^2=1,1^2-1^2=0$$ doing this with a $3$ digit number looks like this you start with $125$ do this$|1^2-2^2+5^2|=23 $ and you get $22$.
My question is how many points and loops are there are?