How are Koepke's ordinal computability and E-recursion related?

In Koepke's paper, "Turing Computations On Ordinals", one has the following (well-known) result:

A set $x$ is ordinal computable from a finite set of ordinal parameters if an only if it is an element of the constructible universe.

In Sacks' survey article, "E-Recursive Intuitions", one finds these (possibly related) results:

Proposition 3.1. $V$=$L$ iff $L$ is E-recursively enumerable... Proposition 2.3. There exists a $\Delta_1$ definable class that is not E-recursively enumerable.

Now, since the above theorem of Koepke's characterizes the constructible sets as those ordinal computable from a finite set of ordinal parameters (note that the pure sets are sets of ordinals), one should be able to characterize the axiom $V$=$L$ as

'Every set of ordinals is ordinal computable from a finite set of ordinal parameters.'

Substituting this characterization for $V$=$L$ in Sacks' theorem, one has

'Every set of ordinals is ordinal computable from a finite set of ordinal parameters iff $L$ is E-recursively enumerable.'

It is interesting to note that Sacks' proof of Theorem 3.1 from his paper gives an indication of what must happen if $V$$\neq$$L$ (most set theorists do not believe that $V$=$L$ is an 'acceptable' axiom):

Proof of 3.1. Suppose $\forall$$x(x$$\in$$L \leftrightarrow{e}(x)\downarrow) and V$$\neq$$L. Then for some b$$\notin$$L, {e}(b)\uparrow. By Proposition 2.2 ["If A is E-recursively enumerable, then A is \Delta_1 definable."--my comment], divergence [\uparrow--my comment] is \Sigma_1 definable. Then by Levy-Shoenfield absoluteness, {e}(x)\uparrow for some x$$\in$$L. Hence the question asked in the title. Three other questions: 1. Can one 'force' L to be not E-recursively enumerable? 2. If V$$\neq$$L$, are there constructible, non-E-recursively enumerable sets?

3. Since $L$ is usually considered a 'class', is E-recursion defined for classes (since it is based on Normann's "Set recursion")? (Hope this is not too silly a question....)

• Any nontrivial forcing makes $L$ non-E-r.e.: if $V\models ZFC$ and $V[G]$ is a nontrivial forcing extension of $V$, then $V[G]\models V\not=L$, and hence $V[G]$ thinks $L$ is not $E$-r.e.
• And there are only countably many E-r.e. sets (every E-r.e. set is the domain of an E-recursive function, and there are only countably many indices for such), so even if $V=L$, there are constructible, non-E-r.e. sets - indeed, even sets of naturals. So I'm not sure if this was the question you meant to ask.