What is the period of the fraction $1/3^k$ in base $2$, for $k=1, 2, \dots$?

Question

I am hoping for the following:

• The period increases with $k$ (on average), and eventually becomes infinite.
• The proportion of digits equal to $0$ approaches 50% as $k$ increases.
• For large values of $k$, all the first few digits are all zero. But if
  you remove these zeros, you still have a proportion of $0$ approaching 50%.

Answer 1

Yes, yes, yes.

1. If you stare at the long division $1 \div 3^k$, you will see that the $n$-th binary decimal digit of $1/{3^k}$ depends on whether $2a_{n-1} > 3^k$, where $a_{n-1}$ is the remainder of $2^{n-1}$ divided by $3^k$. What is the period of the binary representation? It will be the order of $2$ in $(\mathbb Z/3^k\mathbb Z)^{\times}$. By induction you will see that it is $2\cdot3^{k-1}$ (i.e. $2$ is indeed a generator). So as $k \to \infty$, also $\text{(the period)} \to \infty$.

2. Indeed the proportion of the digit $0$ does not approach $50\%$; it is always exactly $50\%$. From part 1, you will see that the remainder series $a_n$ indeed passed through the entire multiplicative group $(\mathbb Z/3^k\mathbb Z)^{\times}$, which is every number from $1$ to $3^k-1$, not divisible by $3$. Among these numbers, exactly half ($3^{k-1}$) are less than $3^k/2$ and the other half ($3^{k-1}$) is greater than $3^k/2$. This implies from part 1 that exactly half of the digits in a period is $0$.

3. The series is purely periodic. No matter what you removed from the beginning, it is still periodic with the same period and shifted pattern.

Answer 2

It is the multiplicative order of $2$ (mod $3^k$), which is $m = 2 \cdot 3 ^{k-1}$.

To be precise, the binary expansion of $3^{-k}$ is:

$$3^{-k} = \frac{2^m - 1}{3^k}\sum_{n=1}^\infty \frac{1}{2^{mn}}$$

$$3^{-k} = \frac{2^m - 1}{3^k}\frac{1}{2^m - 1}$$

Use Euler's theorem to see that $\frac{2^m - 1}{3^k}$ is integer and that the above expansion is correct.