## Background

An ordinal $\alpha$ is called a *recursive ordinal* if there is a recursive well-order $R$ on $\mathbb{N}$ such that ordertype($\mathbb{N},R) = \alpha$. For example, $\omega\cdot 2$ is a recursive ordinal because the ordering of $\mathbb{N}$ as 0, 2, 4, 6, 8, ... 1, 3, 5, 7, ... is computable and has order type $\omega\cdot 2$.

Kleene encoded the recursive ordinals in the natural numbers in a nifty way which is described at the Wikipedia page on Kleene's O. Now Kleene's $\mathcal{O}$ is a fairly powerful set -- given a Turing machine index for a linear order, $\mathcal{O}$ can decide whether that ordering is a well-ordering or not.

Using Kleene's $\mathcal{O}$, it is possible to describe how to iterate the Turing jump through the recursive ordinals. For each natural number $a\in\mathcal{O}$, we can define a set $H_a$ recursively as follows:

- $H_a = \emptyset$ if $a=0$
- $H_a = {H_b}'$ if $a=2^b$
- $H_a = \{\langle n, x \rangle | x \in H_{\phi_e(n)} \}$ if $a = {3\cdot 5^e}$

For each $a\in \mathcal{O}$, we have $H_a$ <$_T \ \mathcal{O}$ (strict inequality), and no $H_a$ is powerful enough to decide which recursive orders are well-orders.

## Question

Among recursive non-well-orders, some hide their descending chains better than others do.

For example, if we only wanted to flag the non-well-orders sporting a *recursive* descending chain, the full power of $\mathcal{O}$ would not be necessary -- $\emptyset'''$ would do $(\exists e [ \phi_e$ is total and $\forall n [\ \phi_e(n+1)$ <$_R\ \phi_e(n)\ ]]$?). Thus there is a recursive linear non-well-order with no recursive descending chain.

In fact (by similar reasoning), for each $a \in \mathcal{O}$ there must be a recursive linear non-well-order with no recursive-in-$H_a$ descending chain.

I wonder whether we could effectively construct these sneaky recursive non-well-orders.

Is there a recursive function $f$ such that whenever $a\in\mathcal{O}$, $f(a)$ is a Turing index for a linear non-well-order with no $H_a$ -computable descending chain?