A "better" rational approximation of pi? $355/113$ is a good fractional approximation of $\pi$, because we use six digits to produce seven correct digits of $\pi$.
$$\frac{355}{113} = 3.1415929\ldots$$
Let $R$ be the ratio of the number of accurate digits produced to the number of digits used in the numerator and denominator, then
$$R\left(\frac{355}{113}\right) = \frac 7{3+3} = 1.166666\ldots\,{}$$
Can anyone find a "better" fraction such that $R > 1.16666\ldots\,{}$.
Added: Probably, a similar question would also make sense over a base other than $10$.
 A: Using the idea of the other answer in a different way, if $u$ is the irrationality measure of $\pi$, then except for finitely many $p/q$, we have
$$ \left| \pi - \frac{p}{q} \right| > \frac{1}{q^u} $$
and consequently
$$ \frac{ -\log |\pi - (p/q)| }{\log p + \log q } <  \frac{u}{2}$$
and there will be infinitely many fractions $p/q$ that come arbitrarily close to this bound. (and, of course, those finitely many exceptions which may exist that are allowed to exceed it) (and, whatever tiny excesses might arise due to the rounding error in the analysis)
If the irrationality measure if $\pi$ is greater than $2.34$, then there will be  infinitely many fractions with a better value of $R$ than the one you found. (although that is not reason to expect any of them are small enough for us to actually find)
If $u < 2.3$, then there can only be finitely many fractions with a better value of $R$. But I have no idea how you would go about checking if any exist at all.
Almost all irrational numbers have irrationality measure $2$; for $\pi$ it's known that $u \leq 7.10320534$
A: 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).
