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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:

$$1 \le (4/3)^m * 2^n \le 2$$

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  • $\begingroup$ Are $m$ and $n$ positive? $\endgroup$ – Julián Aguirre Aug 29 '18 at 11:02
  • $\begingroup$ m and n can be either positive or negative. $\endgroup$ – MarkokraM Aug 29 '18 at 11:06
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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.

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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.

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  • $\begingroup$ I think you need the ceiling function instead of the floor function. $\endgroup$ – joriki Aug 29 '18 at 11:24
  • $\begingroup$ Right. Floor gives normalization between 0 and 1, but ceil between 1 and 2. $\endgroup$ – MarkokraM Aug 29 '18 at 14:57
  • $\begingroup$ @MarkokraM: I think you mean between $\frac12$ and $1$? $\endgroup$ – joriki Aug 31 '18 at 9:56
  • $\begingroup$ @joriki To be more precise, yes, floor seems to normalize series between half and one. $\endgroup$ – MarkokraM Aug 31 '18 at 10:11

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