I'm looking for a power of 3 close to a power of 2.

Let's say, what is $(n,m)$ such that $$\left|\frac{2^n}{3^m}-1\right| = \min\left \{\left|\frac{2^i}{3^j}-1 \right|, 1\leq i,j\leq 20\right\} \quad ?$$


The idea is to understand the intervals between musical notes. Any link on this subject is welcomed (I don't know music theory but I'm interested).

  • 4
    $\begingroup$ $\dfrac{2^1}{3^{20}}$ sounds good? (no pun intended :) But if you want a power of $3$ which is close to a power of $2$ maybe you wanted $$ \min\left \{|3^j-2^i|, 1\leq i,j\leq 20\right\} $$ $\endgroup$
    – Raffaele
    Oct 20, 2017 at 7:48
  • $\begingroup$ Thanks, I will edit!!! $\endgroup$
    – Colas
    Oct 20, 2017 at 7:51
  • $\begingroup$ @Raffaele I just understood your good (intended!) pun. Another way to see the issue is to say that's about two superpowers that should be equilibrated. $\endgroup$
    – Jean Marie
    Oct 20, 2017 at 10:37

3 Answers 3


You want $2^i \approx 3^j$ with $i,j \in \mathbb N$

(preferably $1 \leq i,j \leq 20$).

Take the logarithm (in any base) of both sides: $$i \log(2) \approx j \log(3).$$

Thus the issue is now to obtain a good rational approximation

$$\underbrace{\dfrac{\log(3)}{\log(2)}}_{N}\approx \dfrac{i}{j}$$

Such rational approximations are obtained as the so-called "convergents" of the continued fraction expansion of number $N$ (https://en.wikipedia.org/wiki/Continued_fraction):


with $(a_0; a_1, a_3, \cdots )=(1;1,1,2,2,3,1,5,2,23,2,2,1,55,1,...).$

with successive convergents:

$$ i/j = 3/2, \ 5/3, \ 8/5, \ 11/7, \ 19/12, \ 46/29, \ 65/41, \ 84/53, ...$$

$$317/200,\ 401/253, \ 485/306, \ 569/359, \ 1054/665...$$

(that can be found in (https://oeis.org/A254351/internal)).

Good convergents are obtained by (in plain terms) "stopping just before a big $a_k$".

If the "big $a_k$" (!) is 3, one gets :

$$\dfrac{19}{12} \approx 1.5833 \ \ \ \text{of} \ \ \ N = 1.5850...$$

Thus a good tradeoff is to take $i=19$ and $j=12$.

This is connected to the "Pythagorician comma". (https://en.wikipedia.org/wiki/Pythagorean_comma).

Very related : (http://www.math.uwaterloo.ca/~mrubinst/tuning/12.html)

  • $\begingroup$ I know the definition of continued fractions, but I never thought it had a link to rational approximation. $\endgroup$
    – Colas
    Oct 20, 2017 at 8:00
  • $\begingroup$ That makes senses since there are 12 musical notes. $\endgroup$
    – Colas
    Oct 20, 2017 at 8:15
  • $\begingroup$ see as well (math.stackexchange.com/q/11669) $\endgroup$
    – Jean Marie
    Oct 20, 2017 at 8:50
  • $\begingroup$ Related (andrewduncan.net/cmt) $\endgroup$
    – Jean Marie
    Oct 20, 2017 at 10:49
  • $\begingroup$ Each term in the continued fraction ($a_n$) corresponds to a convergent. You are giving the "semiconvergents" when you include $\frac53$ and $\frac{11}7$. For their denominators, these are not as close to the number being approximated as the convergents. $\endgroup$
    – robjohn
    Oct 20, 2017 at 14:11

Here I have written different things in different parts; the first and third are related to each other:

Irrelevant to the main question (For those who are interested in the connection between music theory and powers of primes): Years ago when I was writing this answer, I did not know about its connection with music theory. This morning I received a notification for an up-vote to this question (That's why I added this information). 3 or 4 months ago, I got information about the relationship between music theory (Superparticular ratios) and prime powers $2$, $3$, etc.; the links to which I will briefly share here:

  1. Størmer's theorem

  2. ABC, High Quality Commas, and Epimericity (The original link has expired)

  3. A Weaker Version of the ABC Conjecture

Just read the first link (I mean Størmer's theorem), to understand the connection between music theory (Superparticular ratios and ...) and the powers of prime numbers through Pell equations.

Also, you can see the contents of the link in the "Internet Web Archive" in the image below (just to be sure, if this link becomes unavailable in the future): enter image description here

(I still do not even understand why, in the question, why we consider the upper limit to be $20$? and set this restriction? Perhaps for a practical exercise?)

Related to the main question:

Lemma: Let $\alpha, \beta$ be linearly independent over rationals $\mathbb{Q}$; i.e. $\dfrac{\alpha}{\beta} \notin \mathbb{Q}$.
Then the group generated by $\alpha, \beta$ is dense in $\mathbb{Q}$;
i.e. $ \{ i\alpha+j\beta \ | \ i,j \in \mathbb{Z} \} = \{ i\alpha-j\beta \ | \ i,j \in \mathbb{Z} \} $ is dense in $\mathbb{Q}$.

Now let $\alpha:=\log2$ and $\beta:=\log3$, note that $\dfrac{\beta}{\alpha}=\dfrac{\log3}{\log2}=\log_23 \notin \mathbb{Q}$;
so if we let $i, j$ varies arbitrary in $\mathbb{N}$,
then the phrase $i\alpha-j\beta$ will become as small as we want;

i.e. in more mathematical discipline we can say:
for every $0 < \delta \in \mathbb{R}$ there exist $i, j \in \mathbb{N}$ such that $0 < i\alpha-j\beta < \delta$.

Note that: $$ 10^{i\alpha-j\beta}= \dfrac{10^{i\alpha}}{10^{j\beta}}= \dfrac{(10^{\alpha})^{i}}{(10^{\beta})^{j}}= \dfrac{2^i}{3^j} $$

So we can conclude that:
for every $0 < \epsilon \in \mathbb{R}$ there exist $i, j \in \mathbb{N}$ such that $1 < \dfrac{2^i}{3^j} < 1+ \epsilon$.

So ignoring some mathematical disciplines we can say:
if we let $i, j$ varies arbitrary in $\mathbb{N}$,
then the phrase $\dfrac{2^i}{3^j}$ will become as close to $1$ as we want.

We can restase some similar facts about $-\delta < i\alpha-j\beta < 0$ and $-\delta < i\alpha-j\beta < \delta$;
also for $1- \epsilon < \dfrac{2^i}{3^j} < 1$ and $1- \epsilon < \dfrac{2^i}{3^j} < 1+ \epsilon$; but it does not matters.

I have seen this problem before:

unsolved problems in number theory by Richard K. Guy;
$F$(None of the above);
$F23$(Small differences between powers of $2$ and $3$.) :

Problem $1$ of Littlewood's book asks how small $3^n — 2^m$ can be in comparison with $2^m$.
He gives as an example $$ \dfrac{3^{12}} {2^{19}} = 1+ \dfrac{ 7153 } {524288} \approx 1+ \dfrac{ 1 } { 73 } \qquad \text{ (the ratio of } D^{\sharp} \text{ to } E^{\flat}). $$ [Where $\sharp$ and $\flat$ stands for dièse and Bémol.]

The first few convergents to the continued fraction (see $F20$) $$ log_23= 1+\tfrac{1} {1+\tfrac{1} {1+\tfrac{1} {2+\tfrac{1} {2+\tfrac{1} {3+\tfrac{1} {1+\tfrac{1} {\cdots} } } } } } } $$

for $\log 3$ to the base $2$ are: $$ \dfrac{ 1}{ 1}, \dfrac{ 2}{ 1}, \dfrac{ 3}{ 2}, \dfrac{ 8}{ 5}, \dfrac{19}{12}, \dfrac{65}{41}, \dfrac{84}{53}, \dots $$ so Victor Meally observed that
the $\color{Blue}{\text{octave}}$ may conveniently be partitioned into $12$, $41$ or $53$ intervals, and that the system of temperament with $53$ degrees is due to Nicolaus Mercator ($1620-1687$; not Gerhardus, $1512-1594$, of map projection fame).

Ellison used the Gel'fond-Baker method to show that

$$ |2^x-3^y| > 2^x e^{\dfrac{-x}{10}} \qquad \text{for} \qquad x > 27 $$

and Tijdeman used it to show that there is a $c \geq 1$ such that $2^x - 3^y > \dfrac{2^x}{x^c}$.

Croft asks the corresponding question for $n! — 2^m$. $ \ \ \cdots$

That was the exact text from the book, note that the pair $(m,n)$ has been used in the reversed order in that book versus this question has been asked in this site!

At the end of section $F23$; there are four refrences as follows:

  • F. Beukers, Fractional parts of powers of rationals, Math. Proc. Cambridge Philos. Soc., $90(1981) 13-20$; MR $83$g:$10028$.

  • A. K. Dubitskas, A lower bound on the value of $||(3/2)^k||$ (Russian), Uspekhi Mat. Nauk, $45(1990) 153-154$; translated in Russian Math. Surveys, $45(1990) 163-164$; MR $91$k:$11058$.

  • W. J. Ellison, Recipes for solving diophantine problems by Baker's method, S6m. Thiorie Nombms, $1970-71$, Exp. No. $11$, C.N.R.S. Talence, $1971$.

  • R. Tijdeman, On integers with many small factors, Compositio Math., $26 (1973) 319-330$.

  • I have found an article by Kurt Mahler with the same name as the first one here.

Also in page $14$ of this paper of Michel Waldschmidt; you can find something.

  • 1
    $\begingroup$ Very interesting ! $\endgroup$
    – Jean Marie
    Apr 9, 2022 at 8:14
  • $\begingroup$ As you are very competent in number theory, here is a question math.stackexchange.com/q/4418458/305862 which hasn't received a satisfying answer. Could you have a look ? $\endgroup$
    – Jean Marie
    Apr 10, 2022 at 13:14

The continued fraction for $\frac{\log(3)}{\log(2)}$ is $$ 1;1,1,2,2,3,1,5,2,23,2,2,1,1,55,1,4,3,1,1,\dots $$ The convergents are $$ \frac11,\frac21,\frac32,\frac85,\frac{19}{12},\frac{65}{41},\frac{84}{53},\frac{485}{306},\frac{1054}{665},\frac{24727}{15601},\frac{50508}{31867},\frac{125743}{79335},\frac{176251}{111202},\frac{301994}{190537},\frac{16785921}{10590737},\frac{17087915}{10781274},\frac{85137581}{53715833}, \frac{272500658}{171928773},\frac{357638239}{225644606}, \frac{630138897}{397573379},\cdots $$ This gives a number of "close" approximations. For example, the best with exponents between $1$ and $20$, is $$ \left|\frac{2^{19}}{3^{12}}-1\right|=0.01345963145485576 $$ This gives the approximation $2^{19/12}=2.9966$. Since their are $12$ half-steps to the octave, this approximation is very useful.

  • $\begingroup$ A certain number of convergents are missing in your list : $5/3$, $11/7, 46/29...$ $\endgroup$
    – Jean Marie
    Oct 20, 2017 at 10:10
  • $\begingroup$ @JeanMarie: are you talking about convergents, or semiconvergents, which are not as close as convergents for their denominators? $\endgroup$
    – robjohn
    Oct 20, 2017 at 14:03

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