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For some positive integer $k \gg 1$, the largest bit index of $3^k$ is given by

$$m \equiv \lfloor k\log_2 3 \rfloor$$

The distance between $3^k$ and $2^m$ can be written as

$$ 3^k - 2^m \equiv a_k \cdot 2^m$$ where $0 \lt a_k \lt 1$.

My questions are:

  1. Is it true that $a_k \gtrsim 10^{-c}$, where $c$ is a positive constant. How do I prove, or disprove, this?
  2. If bullet 1 is true, how do I estimate the constant $c$?

As a side note, I have computed numerically for $k$ up to 10,000. It appears that the value of $a_k$ kept in a stable range in $(0, 1)$, and it could be as small as $\sim \mathcal{O}(10^{-6})$.

J. Doe
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    I think this (https://mathoverflow.net/questions/116840/distance-between-powers-of-2-and-powers-of-3) proves that there's no such $c$ – Exodd Sep 13 '21 at 20:09
  • @Arthur There is really no obvious reason. $c$ could definitely be non-integers, I believe. I will edit and update my question to remove this limit. – J. Doe Sep 13 '21 at 20:15
  • The smallest values of $a$ will be when $m/k$ is an odd continued fraction of $\log_2 3.$ – Thomas Andrews Sep 13 '21 at 20:21
  • @Exodd Thanks for the link. Yes, it proves that the lower bound of $a$ depends on $m$. However, the estimate of the constant in that link does not match correctly what I observed numerically, if I haven't missed anything. – J. Doe Sep 13 '21 at 20:28
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    Since $a$ depends on $k.$ you should write it as $a_k$ – Thomas Andrews Sep 13 '21 at 20:31
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    You may enjoy this question on rational representations of $\log_2(3)$ https://math.stackexchange.com/q/3981318/207316 – PM 2Ring Sep 13 '21 at 20:40
  • @ThomasAndrews I re-edited to update $a$ -> $a_k$. – J. Doe Sep 13 '21 at 20:44

2 Answers2

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When $\frac mk$ is an odd continued fraction for $\log_2 3$ then:

$$0\leq \log_2 3 - \frac{m}{k}<\frac1{k^2}$$

Then $$3^k=2^{k\log_2 3}< 2^{m+\frac1k}$$

So:

$$3^k-2^m<2^{m+1/k}-2^{m}=2^{m}\left(2^{1/k}-1\right)$$

So: $a=a_k\leq 2^{1/k}-1.$ We can make that arbitrarily small.

The continued fractions are infinite, and half of them will be odd ones. (The even continued fraction terms give you examples where $2^m>3^k$ but the difference is small.)

For example, when $k=12,m=19,$ $a_k\approx .0136.$ When $k=53,m=84,$ then $a_k\approx 0.00209.$

When $m/k$ is an even continued fraction for $\log_2 3,$ then $$2^{m}>3^{k}> 2^{m-1}$$ so:

$$3^{k}=2^{k\log_2 3}>2^{m-\frac{1}{k}}$$ and: $$3^{k}-2^{m-1}>2^{m-1}(2^{1-1/k}-1)$$

So for $m/k$ an even continued fraction, like $8/5$ or $65/41,$ you get $a_k\to 1.$

So there are values of $a_k$ arbitrarily close to $0,$ and $a_k$ arbitrarily close to $1.$

Thomas Andrews
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  • I am not following the first inequality $0\leq \log_2 3 - \frac{m}{k}<\frac1{k^2}$. I am not aware of the odd continued fraction expansion for $log_2 3$, either. When I checked your values numerically. It seems that when $k log_2 3$ has the fraction less than $0.5$, the obtained $m/k$ is what you call the odd continued fraction for $log_2 3$. Could you please explain a bit about how you can get the first inequality? – J. Doe Sep 14 '21 at 00:37
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    The continued fractions for an irrational number $\alpha$ are a sequence of rational numbers $p_n/q_n$ with $|\alpha-p_n./q_n|<1/q_n^2.$ When $n$ is odd, $p_n/q_n<\alpha,$ when $n$ is even, then $p_n/q_n>\alpha.$ There's a lot of reading to do about continued fractions - these aren't typical $m,k$ but just a special set of cases where we get "good" approximations of $\log_2 3.$ @J.Doe – Thomas Andrews Sep 14 '21 at 01:29
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    In these cases, $k\log_2 3$ will be "close" to an integer, not just having fractional parts $<0.5.$ The point is, we can can always find bigger $k$ with the fractional part of $k\log_2 3$ less than $\frac{1}{k}.$ – Thomas Andrews Sep 14 '21 at 01:36
  • I have one more question. In the final result of the odd continued fraction for $log_2 3$, namely, $a=a_k\leq 2^{1/k}-1.$ Here we know there doesn't exist the lower bound for $a_k$. However, is there any way to show that the $2^{1/k}-1$, or something similar, is how fast the smallest value of $a_k \rightarrow 0$? Really appreciate your answer. – J. Doe Sep 14 '21 at 01:42
  • On the last question, there isn't really a way. It's possible that $\log_2 3-m/k<1/k^3$ or even smaller for infinite values of $k.$ @J.Doe – Thomas Andrews Sep 14 '21 at 03:47
  • Please bear my ignorance on this subject. Is the existence of infinite values of $k$ that "$log_2 3 - m/k \lt 1/k^3$ or even smaller" a proved result somewhere? – J. Doe Sep 14 '21 at 04:30
  • I said “it’s possible.” It is almost certainly unknown. – Thomas Andrews Sep 14 '21 at 04:40
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There is no $N$ such that $a_k$ is below $1/3$ for all $k>N$.

For suppose $a_k<(1/3)$ for some $k$. Then $2^m<3^k<(4/3)2^m$. Multiply by $3$ to get $(3/2)2^{m+1}<3^{k+1}<2^{m+2}$ forcing the succeeding value, $a_{k+1}$, to exceed $1/2$.

Barry Cipra
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Oscar Lanzi
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