In music theory notes generated by the consequencing interval of $4/3$ generates harmonic series. Series can be normalized by multiplicating the fraction with a $2$ in power $n$.
What is a formula for $n$ depending on $m$ such that the ratio is always between $1$ and $2$?
I'm looking for integer solutions for n when m is a whole number:
$$1 \le (4/3)^m * 2^n \le 2$$
On
The inequality can be rewritten as $$ \Bigl(\frac{3}{4}\Bigr)^m\le 2^n\le2\,\Bigl(\frac{3}{4}\Bigr)^m. $$ Taging logarithms an dividing by $\log2>0$, we get $$ \frac{\log(3/4)}{\log2}\,m\le n\le1+\frac{\log(3/4)}{\log2}\,m. $$ From here, it follows that $$ n=\Bigl\lfloor\frac{\log(3/4)}{\log2}\,m\Bigr\rfloor, $$ where $\lfloor\ \rfloor$ is the floor function.
Take logarithms on both sides:
$$ 0\le m\log\frac43+n\log2\le\log2\;. $$
This is an area in the $(m,n)$ plane that lies between two parallel lines. Solving for $n$ yields
$$ -m\log_2\frac43\le n\le1-m\log_2\frac43\;, $$
so
$$ n=\left\lceil-m\log_2\frac43\right\rceil\approx\left\lceil-0.415m\right\rceil\;, $$
where $\lceil\cdot\rceil$ is the ceiling function.