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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%.
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  • $\begingroup$ The fraction $1/3^k$, for any integer $k$, is a rational number and, thus, always has a finite period. When you write the period "eventually becomes infinite", I believe you mean to say there's no upper bound on the period as $k$ keeps increasing, which is true. $\endgroup$ – John Omielan Aug 1 '19 at 21:40
  • $\begingroup$ @John Omielan: correct, period always finite but increasing to infinity as $k$ increases to infinity. $\endgroup$ – Vincent Granville Aug 1 '19 at 21:42
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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.

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    $\begingroup$ This is all I wanted to know, thanks a lot. Good starting point for my research (next step is how chaotic is the distribution of 0 and 1 in the period, when $k$ is large? How correlated are the digits in $1/3^k$ and $1/3^n$ if both $k$ and $n$ are very large, depending on $|k - n|$. And how auto-correlated are these digits for $1/3^k$ if $k$ is large? At least, it gives me some hope to get to something interesting regarding the digits of $1/\log 2$ which is why I am interested in this topic in the first place. $\endgroup$ – Vincent Granville Aug 1 '19 at 22:56
  • $\begingroup$ I cannot see yet at this moment whether they should or should not have correlations, but good luck! ;) $\endgroup$ – Hw Chu Aug 1 '19 at 22:59
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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.

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    $\begingroup$ This is true but does not answer any part of the question. $\endgroup$ – Ethan Bolker Aug 1 '19 at 21:50
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    $\begingroup$ A reference would be useful. This is part of some ongoing work to show that some mathematical constants have a binary representation with 50% zeros and 50% ones. In this case, the constant $1/\log 2$. I am looking at the problem in a backward way, looking at "randomness" properties of some fractions, and see if it leads to anything interesting. $\endgroup$ – Vincent Granville Aug 1 '19 at 21:53

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