Why is the following sequence of $123456790$ keep repeating and why is it skipping $8$? in the following:
$91^2/90^2=8281/8100=1.0223456790123456790123456790123456790...$
$901^2/900^2=811801/810000=1.002223456790123456790123456790123456790...$
$9001^2/9000^2=81018001/81000000=1.00022223456790123456790123456790123456790...$
$90001^2/90000^2=8100180001/8100000000=1.0000222223456790123456790123456790123456790...$
The only kind of similar example occurs in
$10^2/9^2=100/81= 1.23456790123456790123456790...$
Any thoughts?
You have
$\frac{(9 \times 10^n + 1)^2}{(9 \times 10^n)2} = \frac{9^2 \times (10^n)^2 + 2 \times 9 \times 10^n + 1}{9^2 \times (10^n)^2} = 1 + \frac{2}{9} \times \frac{1}{10^n} + \frac{1}{9^2} \times \frac{1}{(10^n)^2}$.
$\frac{2}{9} = 0.22...$
and
$\frac{1}{9^2} = 0.12345679012345679...$
Your results show that $0$'s and $2$'s start the decimals, it is due to the order of magintude in difference of $10^n$ and $(10^n)^2$.
Now if we add $222222222$ to the sequence $012345679$, then we get $2345678911$, where the last two digits are not two $1$'s but $11$, so we shift 1 forward to get $2345678101$, where the last three digits are not $1$-$0$-$1$ but $10$ and $1$, so we shift once more to get $234567901$. This is what keeps repeating, and if you concatenate it at the end you get $234567901234567901...$.
Regarind $\frac{10^2}{9^2}$, $n$ happens to be $0$.