Expanding on my comment, here is a reason why you shouldn't find any better (in your sense) approximation.

Let $p_n/q_n$ be the $n$-th convergent of the continued fraction of $\pi$, and $R_n$ its quality as you defined it in your question.

We purposefully ignore integer parts and off-by-one errors in expressing the number of decimal digits and simply write
$$ R_n \doteq \frac{-\log_{10} \left | \pi-p_n/q_n\right |}{\log_{10} p_n+\log_{10} q_n} .$$

What we need now is that $\pi$ is a typical real number in the Khinchin-Lévy sense, which by the way holds for all real numbers but a set of measure $0$. This is an open conjecture, but the numerical evidence is very strong (check it for yourself if you wish).

This would mean, in particular,
$$ \lim_{n \rightarrow \infty} q_n^{1/n}=\lim_{n \rightarrow \infty} \left (\frac{p_n}{\pi}\right )^{1/n}=\mathrm{e}^{\pi^2/12 \log 2}$$
and
$$ -\lim_{n \rightarrow \infty} \frac 1 n \log_{10} \left | \pi-\frac{p_n}{q_n} \right |=\frac{\pi^2}{6 \log 2 \log 10}$$
(see here and here for the first equality, here, here and here for the latter).

A consequence of this would be $\lim_{n \rightarrow \infty} R_n=1$. This has of course nothing to do with base $10$ representation, Lévy's theorem answers your question in any base.

This is not a proof that $355/113$ is optimal, but you can check the first convergents with the code you were given in the comments; see also here and here for some effective results.

Just for the sake of completeness, mine was

```
a(n)={A=contfracpnqn(contfrac(Pi,n+1),n));return((1+floor(-log(abs(Pi-A[1,n]/A[2,n]))/log(10)))/(floor(log(A[1,n])/log(10)+1)+floor(log(A[2,n])/log(10)+1))+0.0);};
```

but this should be optimised.

Also note that: a) It is enough to consider convergents instead of any rational number. b) Something weaker than $\pi$ being a Khinchin-Lévy number would suffice, but this is the easiest way to see what's going on (the question is related to "how do the rational approximations to $\pi$ behave" anyways).

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