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The Wikipedia page on $\log 2$ mentions a trick to compute $\log 2$ based on the identity

$$2 = \left(\frac{16}{15}\right)^7\left(\frac{81}{80}\right)^3\left(\frac{25}{24}\right)^5$$

It's not too hard to come up with similar-looking formula, for instance:

$$2 = \left(\frac{9}{8}\right)\left(\frac{8}{7}\right)^2\left(\frac{7}{6}\right)^2$$

Consider a formula to have the form

$$2 = \prod_{i=1}^m \left(1+\frac{1}{n_i}\right)^{k_i}$$ where $n_i$ are positive integers and $k_i$ are (possibly negative) integers.

For a given $m$, the best formula maximizes the smallest $n_i$. How would one go about finding such formulas? Is there a trick besides brute force?


Edit: neat, finding consecutive pairs of p-smooth numbers and solving a lattice problem gives us things like

$$2 = \left(\frac{126}{125}\right)^{72}\left(\frac{225}{224}\right)^{27}\left(\frac{2401}{2400}\right)^{-19}\left(\frac{4375}{4374}\right)^{31}$$

and

$$\frac{144}{251} \sum_{k=0}^{\infty} \frac{1}{(2k+1)63001^k} + \frac{54}{449}\sum_{k=0}^{\infty} \frac{1}{(2k+1)201601^k}-\frac{38}{4801}\sum_{k=0}^{\infty} \frac{1}{(2k+1)23049601^k}+\frac{62}{8749}\sum_{k=0}^{\infty} \frac{1}{(2k+1)76545001^k} = \log 2$$

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  • $\begingroup$ It's not so easy, since you have to find consecutive large numbers that have only small prime factors. There's a paper by Lehmer, about Stormer's problem, that might give you some idea of what's involved. $\endgroup$ Feb 10 '20 at 2:50
  • $\begingroup$ Oh neat, the first equation is 7 minor seconds, 3 syntonic commas, and 5 minor semitones. Those all come up in 2-3-5 smooth numbers. $\endgroup$
    – Arthur B.
    Feb 10 '20 at 2:58
  • $\begingroup$ Smooth numbers are the key. I'm not sure Lehmer used that terminology, but that's what he was looking for. $\endgroup$ Feb 10 '20 at 3:00
  • $\begingroup$ Ok so smooth number, and then looks like solving a lattice problem $\endgroup$
    – Arthur B.
    Feb 10 '20 at 3:09
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    $\begingroup$ projecteuclid.org/euclid.ijm/1256067456 see also oeis.org/A002072 $\endgroup$ Feb 10 '20 at 3:10
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I think I figured it out, thanks to the comment. The key is to pay attention to the prime decomposition of the $n_i$'s and $n_i+1$'s. Let $p$ be the largest prime factor of any integer in $\{n_i\}_i \cup \{n_i+1\}_i$, so that for all i, $n_i$ and $n_{i+1}$ are p-smooth.

Commenter Gerry Myerson usefully points to Stormer's problem and the theorem that, for all $p$ there is a finite set of consecutive integers which are both $p$-smooth.

This yields the following algorithm: fix the value of $p$, find the $p$-smooth pairs of consecutive integers, pick a subset (typically one might need \Pi(p) pairs) and solve the linear system for the values of $k_i$.

This yield the identity mentioned in the edited question.

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