I have added the formal definition of the function $address$. The question is divided in two parts. This is important because of two reasons. First due to the length of the question. Secondly if one gives a negative example for the first part, the second part is redundant in that case. I asked the first part of this question on Math.SE (https://math.stackexchange.com/questions/2396322/finite-registers-and-computable-well-orderings-for-omega). I was hoping that if someone would describe a negative example for the first part (which I might have failed to see for any reason), then there wouldn't be any need to post this particular question.

Also finally, even the question in second part is (quite likely) not as general as the title seems to suggest. Here is the question:

**First Part**

Suppose we are given some computable well-ordering on $\Bbb{N}$ whose order-type is $\omega$. It is assumed that the (total) notation function induced corresponding to the computable well-order relation is denoted as $address:\omega\rightarrow \Bbb{N}$, which is going to be bijective.

Given a computable well-ordering $\preceq$ on $\Bbb{N}$ of order-type $\alpha$, $address$ is meant to be the unique order-preserving bijection from $\alpha$ to $(\Bbb{N},\preceq)$. Thanks to Deedlit for suggesting this precise/formal defintion. My own informal desciption is under **EDIT2**.

Now suppose we are given **two** registers (generalising to more than two registers is self-evident). The values of registers are given by the functions $Value_1:\omega\rightarrow\omega$ and $Value_2:\omega\rightarrow\omega$. More specifically, the value of register-1 at some time $t\in\omega$ is denoted by $Value_1(t)$ (and of course same for the other register).

Before stating this question we need to define the representation of a function in a given well-ordering. Now suppose for some computable well-ordering of an element $p$ we are given the notation function $address:p\rightarrow \Bbb{N}$. For some function $F:p \rightarrow p$ we can write representation function of $F$ (in the given well-ordering for $p$) as $f:\Bbb{N} \rightarrow \Bbb{N}$ and defined by: $$f(address(x))=address(F(x)) \qquad for \; all \; x\in p$$

In general we might have to worry about extra cases (in defining representation of the function $F$), but here we won't (since $F$ is total and $F(x)<p$ for all values in the domain). One can also observe that since we are only concerned with case of $\omega$ here, we can replace $p$ with $\omega$ in the preceding paragraph.

And with this, we are pretty much set to ask the question. We just need to define the rules/constraints for how the register values can change. They are:

**(1)** We must have:
$$Value_1(0)=0$$
$$Value_1(1)=0$$
And similarly for second register.

**(2)** If we have:
$$Value_1(t)=a$$

Then only one of the following possibilities can hold at $t+1$:

(a) $Value_1(t+1)=a$ (value remains unchanged)

(b) $Value_1(t+1)=a+1$ (value increased by $1$)

(c) $Value_1(t+1)=0$ (value brought down to $0$)

And obviously the same rule holds for the second register.

**(3)** There must not exist two different times $t_1$ and $t_2$ at which "all" the register values repeat themselves (simultaneously that is). For two registers, we must not have two different times $t_1$ and $t_2$ (with $t_1\ne 0$ and $t_2\ne 0$) such that:
$$Value_1(t_1)=Value_1(t_2)$$
AND
$$Value_2(t_1)=Value_2(t_2)$$

Now the question asks that can one find a negative example with a computable well-ordering for $\omega$ such that:

(1) The representation functions corresponding to the functions $Value_1$ and $Value_2$ are recursive in the well-ordering. Note that the functions $Value_1$ and $Value_2$ are entirely arbitrary (apart from the stipulation that they have to satisfy the three constraints mentioned above). Suppose we denote the representation functions (in the given well-ordering) corresponding to $Value_1$ and $Value_2$ as $value_1:\Bbb{N} \rightarrow \Bbb{N}$ and $value_2:\Bbb{N} \rightarrow \Bbb{N}$. This condition just requires that the functions $value_1$ and $value_2$ must be recursive.

(2) The representation function corresponding to the successor function is not recursive in the well-ordering (in a more colloquial manner we could say that the successor function is not recursive in the well-ordering).

In the case a negative example exists that example can simply be described (and the second part of the question is redundant in that case). In case a negative example doesn't exist, the proof can be given (also generalisation to more than two resgisters would be nice of course).

**Second Part**

In this case the value function (for the i-th register) will generally be $Value_i:p\rightarrow p$ (instead of $Value_i:\omega\rightarrow\omega$). That's because now we are dealing with some computable well-ordering on $\Bbb{N}$ whose order-type is $p$ (where $p>\omega$). Other than that, on top of three rules for register values described in first part, we need to describe how we want to evaluate the register value for some limit ordinal (say $q$). So we need to add a fourth rule/constraint:

**(4)** For the i-th register we want to describe how to evaluate the value $Value_i(q)$ (where $q$ is a limit). We divide into two cases:

(a) There is no "last" value $r$ (where $r<q$) at which $Value_i(r)=0$. In other words, there always exists a fundamental sequence $r_0,r_1,r_2,r_3,..$ for $q$ such that $Value_i(r_j)=0$ for all $j \in \mathbb{N}$. In this case set $Value_i(q)=0$.

(b) There is some "last" value $r$ (where $r<q$) at which $Value_i(r)=0$. In this case we assume that the value of register increases smoothly. That is, we define: $$Value_i(q)=sup(A) $$ where the set $A$ is defined as: $$A=\{\, Value_i(\alpha)\, |\, r<\alpha<q \}$$

There is "only one" exception to rule in (b), which always has to be observed. If according to rule in (b) $Value_i(q)=q$, then we always set $Value_i(q)=0$.

And now we can ask the following two questions:

(A) Suppose the number of registers used for some specific ordinal was equal to $n$. Then what is the smallest ordinal (presumably recursive) for which there exists no computable well-ordering (and a possible assignment of register values) such that the representation functions corresponding to $Value_1, Value_2,...Value_n$ are all recursive.

As an example, it seems "apparently" that highest one can go with two registers is $\varepsilon_0$. But how to prove/disprove it and what about arbitrary (but finite) number of registers?

(B) Suppose we are given two different computable well-orderings $N_A$ and $N_B$ (for same ordinal $p$). For the first well-ordering we describe a system of register values (satisfying contraints/rules (1) to (4) described in question), such that the representation functions (in $N_A$) corresponding to register values are all recursive. For the second well-ordering once again we describe a system of register values (presumably entirely distinct from the previous one), such that the representation functions (in $N_B$) corresponding to register values are again all recursive. Can we describe a case where isomorphism between $N_A$ and $N_B$ is not recursive (or show otherwise).

**EDIT:**

I will add one example so that question is a little clearer(unfortunately I don't know how I can shorten the main question any further). Suppose the $address:\omega \rightarrow \Bbb{N}$ function is: $$address(x)=x+1 \qquad if \; x \; is \; even$$ $$address(x)=x-1 \qquad if \; x \; is \; odd$$ The well-order relation corresponding to this is trivially recursive/computable.

Now if we are assuming two registers we have full choice in choosing our functions $Value_1:\omega \rightarrow \omega$ and $Value_2:\omega \rightarrow \omega$. We may choose: $$Value_1(x)=floor(x/2)$$ and $$Value_2(0)=0$$ $$Value_2(1)=0$$ $$Value_2(x)=0 \qquad if \; x \; is \; even \; and \; x\ne 0$$ $$Value_2(x)=1 \qquad if \; x \; is \; odd \; and \; x\ne 1$$ Now we have a computable well-ordering where the representation of successor functions and the functions $value_1:\Bbb{N}\rightarrow\Bbb{N}$ and $value_2:\Bbb{N}\rightarrow\Bbb{N}$ (which are representation functions of $Value_1$ and $Value_2$ respectively in the given well-ordering) are recursive. Writing them in full detail would be a little cubersome.

But the question is that whether there exists ANY choice for these two functions $Value_1$ and $Value_2$ such that there exists a computable well-ordering for $\omega$ for which the representation of successor function is non-recursive and yet the functions $value_1:\Bbb{N}\rightarrow\Bbb{N}$ and $value_2:\Bbb{N}\rightarrow\Bbb{N}$ will still be recursive.

Now actually thinking a bit more about it, one might ask whether the "computable" part in "computable well-ordering" is absolutely necessary or not. At any rate, I hope this makes the question a little clearer.

**EDIT2:**

Explanation for function $address:\alpha\rightarrow \Bbb{N}$. If we have a well-ordering on $\Bbb{N}$ whose order-type is $\alpha$, then position of each element of $\Bbb{N}$ is given by some $\beta$ (where $\beta < \alpha$). Furthermore for any two distinct numbers $a_1,a_2 \in \Bbb{N}$ ($a_1 \ne a_2$), if we denote the positions corresponding to $a_1$ and $a_2$ as $\alpha_1$ and $\alpha_2$ respectively then we have $\alpha_1 \ne \alpha_2$.

So essentially we have a function (say $position:\Bbb{N} \rightarrow \alpha$) which arises due to the well-ordering. The function $address:\alpha\rightarrow \Bbb{N}$ is just the inverse of the function $position$.

"However, note that this is the second condition in the question (towards the end). The first condition before that must also be satisfied."What I meant with this (more clearly) is that both of the conditions must be satisfied simultaneously by a given well-ordering. $\endgroup$ – SSequence Aug 21 '17 at 16:37