User blog comment:P進大好きbot/What does a computable large number mean?/@comment-4224897-20180610135217/@comment-35470197-20180612015555

> Now, a number n (defined in L) is computable (relative to T) if in all models of T, n is actually finite (equivalently, T proves n=k for some fixed numeral k=0+1⋯+1).

Oh, this sounds a good definition. So for example, let \(\Sigma\) denote the set of definitions in L of computable natural numbers of length smaller than \(10^{100}\). Then, by the definition of the computability, the maximum of the natural numbers defined by a formula in \(\Sigma\) is computable for you, right?

But there is a problem, which might be not so serious. If a Turing machine \(M\) is provably halting with input \(0\) under \(T\) but any formula of the type \(M(0) = k\) is not provable under \(T\), then \(M(0)\) is not computable in your definition.