The busy beaver function Σ(n) is not computable, at least, not for every n. If it were, it could be used to determine if an arbitrary Turing machine halts, solving the halting problem.
To get from this non-computable function to a non-computable number, you can construct the finite number R = 0.Σ(1)Σ(2)Σ(3)... that concatenates all busy beaver numbers. Its first few digits are R = 0.1621 ... (thanks to ChatGPT for this step).
However, if you include the number of digits in each number, that is
R = 0.|Σ(1)|Σ(1) |Σ(2)|Σ(2) |Σ(3)|Σ(3)... = 0.11 14 16 213
In your version of R, you can unambiguously get Σ(n) from R, therefore, the halting problem proves that R itself is not computable. Nice!
Actually, my version has an issue, which is that (while you can read the Σ(n) unambiguously after reading |Σ(n)|) you can't unambiguously get the |Σ(n)| without knowing their length.
The most efficient way, I believe, is to just include ||Σ(n)|| as well, and so on recursively down until you have only one digit. But then you'll also need to keep track of the length of the sequence, so it's a mess.
Better just write everything in unary. So (1, 4, 6, 13) gets encoded as
R = 10 11110 1111110 11111111111110 ...
Being rational means you’re a solution to a linear equation. sqrt(2) is irrational.
Being algebraic means you you’re a solution to an algebraic equation. Pi is transcendental, which means it’s not only irrational but non-algebraic.
Being computable means you can write a terminating program to compute the nth digit of the number. Pi is computable, I have no idea what isn’t computable.
But… just as naturals are countable, so are rationals, so are algebraic numbers, so are computable numbers. Which in turn means that the non-computable numbers that I can’t even describe are the VAST majority of numbers on the real number line.