B Squaring finite decimals of 2/3 and 1/3 - growing patterns

[bahamagreen]

TL;DR

The number of decimal repeats feeds segments of the square with same digits

I was doing some probability calculations that include squaring a number between 0 and 1.
When I approximate 2/3 using 0.6 or 0.66 or 0.666 etc. I get an interesting series of growing same digit segments...

0.6^2=0.36
0.66^2=0.4356
0.666^2=0.443556
0.6666^2=0.44435556
0.66666^2=0.4444355556
0.666666^2=0.444443555556

And (2/3)^2=0.4444444444444...

Similar thing squaring 1/3 approximated as 0.3 or 0.33 or 0.333 etc.

What is this called?
Is it an artifact of base 10?
Sometimes a long division yields a repeating remainder so a similar string of repeats forms, but this is multiplication that produces two growing strings that preserve the digits that separate the strings.

 

[andrewkirk]

You can investigate this by observing that the n-th item in your sequence is
\begin{align*}\left(\frac23 \times (1 - 10^{-n})\right)^2
&=\left(\frac23\right)^2 \times \left(1 - 2 \times 10^{-n} + 10^{-2n}\right)
\\&=0.\dot4 - 0.0...(n\mathrm{\ zeros})...0\dot8 + 0.0...(2n\mathrm{\ zeros})...0\dot4
\end{align*}
That gives a decimal with zero before the decimal point and 2n digits after the decimal point, with those digits given by:
$$0.4...(n\mathrm{\ fours})...40...(n\mathrm{\ zeros})...0 - 0.0...(n\mathrm{\ zeros})...04...(n\mathrm{\ fours})...4$$
eg the 3rd number is:
\begin{align*}
&0.444000-\\
&0.000444\\
=&0.443556\end{align*}
You can see how the regularity arises from the regularity of that subtraction.

Similarly, with 1/3 we have that the n-th item is
$$0.1...(n\mathrm{\ ones})...10...(n\mathrm{\ zeros})...0 - 0.0...(n\mathrm{\ zeros})...01...(n\mathrm{\ ones})...1$$
eg the 3rd number is:
\begin{align*}
&0.111000-\\
&0.000111\\
=&0.110889\end{align*}

The neat patterns arise for 1/3 and 2/3 because d/9 has base 10 representation of d-recurring. Other bases will have similar patterns for different numerals.