# Product formula for 2

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

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

• 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. Feb 10 '20 at 2:50
• 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. Feb 10 '20 at 2:58
• Smooth numbers are the key. I'm not sure Lehmer used that terminology, but that's what he was looking for. Feb 10 '20 at 3:00
• Ok so smooth number, and then looks like solving a lattice problem Feb 10 '20 at 3:09
• Feb 10 '20 at 3:10

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 $$k_i$$.