The distance between power of 3 and the largest power of 2

Question

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 $\sim \mathcal{O}(10^{-6})$.

Answer 1

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 $a_k\approx 0.00209.$

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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 $1.$

Answer 2

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 $1/2$.