This is a little curiosity that came up in a project I am working on, and I thought someone might have a nice way to see the answer.

**Question.** Can we uniformly compute $n$ from an oracle for the $n^{\rm th}$ jump $0^{(n)}$?

Of course, if we hard-code $n$ into our representation of $0^{(n)}$, then the answer will be yes. Similarly, if we use a $\Sigma_n$-truth predicate instead of $0^{(n)}$, which is of course Turing equivalent, then we can compute $n$ by looking at the syntactic complexity of the assertions in the oracle. We could also get negative answers by making finite changes to each $0^{(n)}$, which would preserve Turing equivalence at each level but destroy the uniform algorithm.

My question, instead, is about using the usual representation of $0^{(n)}$ as sets of halting Turing machine programs. Specifically, using some standard Turing machine architecture and encoding of Turing machine programs, let $0'$ or $0^{(1)}$ be the collection of Turing machine programs (construed as a set of natural numbers) that halt on empty input. For $n\geq 1$, let $0^{(n+1)}$ be the collection of oracle Turing machine programs (construed as a set of natural numbers) that halt on empty input using oracle $0^{(n)}$.

The question is whether there is a program $e$ such that on empty input, program $e$ with oracle $0^{(n)}$ outputs $n$. In other words, is there a computer program such that if you give it some $0^{(n)}$ as an oracle, it can recover $n$?

I believe that the answer will be yes, by designing certain programs that halt or don't halt in a such a way that one can determine what $n$ must be.

If the answer is affirmative, then we can also answer affirmatively the following question:

**Question.** Is $0^{(k)}$ uniformly computable from $0^{(n)}$ for any $n\geq k$?

That is, using the specific representations of $0^{(n)}$ I defined above, is there an oracle Turing machine program $e$ such that on input $k$, given oracle $0^{(n)}$ for any $n\geq k$, will write out the contents of $0^{(k)}$?

I expect that the answer is yes.