This is related to the second part of this old question of mine.
For $A\subseteq\mathbb{N}$ let $\mathfrak{N}_A=(\mathbb{N};0,1,+,A)$ be the expansion of Presburger arithmetic with (a predicate naming) $A$. Say that $A$ is inductively self-defining iff there is a finite theory $S$ in the language $\{+,A\}$ such that $\mathfrak{N}_A\models S$ and the characteristic function of $A$ is strongly representable in the theory $S^\star$ = $S$ + the full induction for $\{+,A\}$-formulas; that is, there is an $\{+,A\}$-formula $\varphi(x,y)$ such that $S^\star\vdash$ "$\varphi$ defines a function" and for each $n$ we have $S^\star\vdash \varphi(n,A(n))$.
Question: What are the inductively self-defining sets?
Clearly any such set must be computable; however, I don't see how to narrow this down further. I don't even see how to show that not all computable sets are inductively self-defining, although in the absence of a definable pairing function it seems that any sufficiently sparse set should be a counterexample (e.g. $\{2^{2^x}: x\in\mathbb{N}\}$).
