TITLE: Decidable but nonrecursive sets
QUESTION [7 upvotes]: Until recently, I believed that recursive=decidable,
subscribing to this Wikipedia quote:
"In computability theory, a set is decidable, computable, or recursive if there 
is an algorithm that terminates after a finite amount of time and correctly decides 
whether or not a given object belongs to the set."
But my confidence in equating 'recursive' and 'decidable'
has been undermined by encountering intriguing work of Benjamin Wells that seems to
point to decidable but nonrecursive theories and sets:
[1] B. Wells, 
"Is There a Nonrecursive Decidable Equational Theory?"
J. Minds Machines, Volume 12, Number 2, 2002, 301-324.
[2] B. Wells,
"Hypercomputation by definition," 
Theor. Comput. Sci., 317, 1-3 (Jun. 2004), 191-207.
He has constructed "finitely based pseudorecursive" equational
theories which Tarski (his advisor) believed are decidable.
My (shakey) understanding based on these two papers is as follows.
$T$ is the equational theory, and $T_n$ is the subset of $T$ consisting
of the equations in which no more than $n$ distinct variables occur.
Each $T_n$ is recursive, but $T$ is not recursive.
For each $T_n$, there is a procedure for deciding whether an
arbitrary equation is in $T_n$;
so there is a "catalog" of these procedures indexed by $n$.
The $T_n$ are individually recursive but they are not
"uniformly recursive" in $n$,
apparently because the catalog is too chaotically arranged for indexing.
Despite reading Wells' description[1] of the sense in which 
$T$ might or should be considered decidable, I do not understand it.
(I am well beyond my expertise here, 
pushed into this unfamiliar territory by work with a student.)
I have two concrete questions.
First, is there later work on nonrecursive but decidable theories?
Has Wells' challenge to resolve "the current impasse" been
addressed by others?
Is there acceptance that there is indeed an impasse?
Second and relatedly, I am especially interested if
other models of computation ("hypercomputation"?)
have been suggested to capture the sense in which
$T$ is decidable by extensions of Turing machines.
Wells' 2004 summary[2] is negative:
"So far, there has emerged no concrete extension of computing
models corresponding to the extension of decidability
to finitely based pseudorecursive theories."

REPLY [3 votes]: In some situations there is a distinction between decidability and computability, but it works in the other direction, with decidability being stronger than computability.
The example is this:  A (possibly partial) function $f$ from $\mathbb N$ to $\mathbb N$ is decidable if its graph is decidable.  It is computable (or recursive) if there is an algorithm that for each $n\in\mathbb N$ computes $f(n)$ as long as $n$ is in the domain of $f$ and does not terminate on input $n$, otherwise.  
Now, if $f$ is computable (or even decidable), then its domain is c.e. (recursively enumerable).  Assume that $f$ has a domain that is c.e. but not computable.  Assume further that $f$ has only finitely many values (for instance only the single value $0$).
Such computable functions exist.  An example is the function that returns $0$ on every $n$ in the halting problem and is otherwise undefined.  In this case $f$ is not decidable.