Show that $3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other. 
$3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other.  Prove it!

$3^{-1}=\overline{0\color{red}{1}}_2$
$3^{-2}=\overline{000\color{red}{111}}_2$
$3^{-3}=\overline{000010010\color{red}{111101101}}_2$
$\ldots$
These are the binary representations - add a minus sign to the left hand side and the right hand side is the 2-adic representation.

In general, the period of $2$ in $\Bbb Z/(3^{N+1})^\times$ is $2\cdot3^N$. And we expect that the $2$-adic expansion of $-3^{N+1}$ should be purely periodic, period $2\cdot3^N$.
Indeed, since $3^{N+1}|(2^{2\cdot3^N}-1)$, say with quotient $Q_N$, we get the results
\begin{align}
Q_N&=\frac{2^{2\cdot3^N}-1}{3^{N+1}}\\
-\frac1{3^{N+1}}&=\frac{Q_N}{1-2^{2\cdot3^n}}\,,
\end{align}
in which the second line says that the number of binary digits in the repeating block of the $2$-adic expansion of $-3^{-N-1}$ is $2\cdot3^N$, and what’s in the block is the number $q_N$.
What we know is that $2^{2\cdot3^N}-1\equiv0\pmod{3^{N+1}}$, so we can factor
$$ \left(2^{3^N}-1\right)\left(2^{3^N}+1\right)\equiv0\pmod{3^{N+1}}\,, $$
but please note that since $3^{N+1}$ is odd, we see that the left-hand factor above is $\equiv1\pmod3$, in particular relatively prime to $3$, and thus to $3^{N+1}$ as well. Thus $3^{N+1}$ divides the right-hand factor, i.e. $3^{N+1}\mid(2^{3^N}+1)$, and once again to make typing easier for
myself, I’ll call the quotient $\Omega$. Thus we have:
\begin{align}
\Omega&=\frac{2^{3^N}+1}{3^{N+1}}\\
0&<\Omega<2^{3^N}\\
Q_N&=\Omega\left(2^{3^N}-1\right)\\
&=2^{3^N}(\Omega-1)+\left(2^{3^N}-\Omega\right)\\
\text{where we note }0&<2^{3^N}-\Omega<2^{3^N}\,.
\end{align}
And that gives us our expression for $Q_N=2^{3^N}a+b$ with both $a$ and $b$ in the interval $\langle0,2^{3^N}\rangle$, namely $a=\Omega-1$ and $b=2^{3^N}-\Omega$. And surenough, $a+b=2^{3^N}-1$, as we desired.
