I think the same sort of trick (sticking a 1 or 0 in each decimal place according to some rule) can be played with several variants of the "Turing machine trick."
Here's one of a somewhat different flavor. Choose an enumeration of the diophantine Diophantine equations (over $\mathbb{Z}$), and define a number with decimal expansion $0.a_1a_2a_3\ldots$ where
$$ a_i=\begin{cases} 1&\text{ if the $i$-the Diophantine equation has a solution}\\ 0&\text{ if not.} \end{cases} $$
This is non-computable by the negative solution to Hilbert's 10th problem.
On the other hand
(Though to be fair, by that same solution, this is probably tantamount to permuting the digits in one of the Turing machine examples.examples.)
