Call $\alpha\in On$ *p.r.closed*, if every *p.r. set function* $f$ is total on $L_{\alpha}$. (Note (1) If $\alpha^\ast$ is the least p.r. closed ordinal $> \alpha$ then it is the next image of a point in the $\omega$'th Veblen function, so in any case is much smaller than the next admissible. Hence (2) the p.r.closed ordinals form a c.u.b. class, and in fact are c.u.b. below any admissible ordinal $>\omega$. (3) By Jensen-Karp [1] the p.r. functions on ordinals to ordinals are all the restrictions of the p.r. set functions to $On$. Hence it is easier to reason with the latter class.) *Definition* Let $\delta$ be *p.r. reflecting* if for all p.r. functions $F:On\rightarrow On$, if $F(\delta)\neq 0$ then $\exists \alpha < \delta (F(\alpha)\neq 0 )$. Then *Claim* $\delta$ *not p.r. reflecting iff $\delta$ is p.r. recognizable.* Now let $\beta$ be the least p.r. closed ordinal with a $\beta_0<\beta$ satisfying $L_{\beta_0}\prec_{\Sigma_1}L_\beta$. (So larger than the least $\Pi^1_0$-reflecting ordinal, but less than the least $\Pi^1_1$-reflecting ordinal.) By p.r. closure, and $\Sigma_1$-elementarity, the totality of the p.r. functions on $L_\beta$ ensures: *Lemma The least p.r. reflecting ordinal is less than or equal to $\beta_0$.* This gives an upper bound to the first p.r. reflecting, and so non-p.r. recognizable, ordinal, and will answer the first part of the Extra Quest. negatively, as we shall see below there are p.r. recognisable ordinals $>\beta$. As an imprecise lower bound one can take the least $\Pi^1_0$-reflecting ordinal mentioned above. [**Edit:** This is fully answered below at (C).] (If $\gamma$ is this ordinal, and so we have $L_\gamma \prec_{\Sigma_1}L_{\gamma +1}$, it is easy to see all ordinals $\tau < \gamma$ are p.r. recog. as for any such $\tau$ there is some sentence $B$ so that $L_\tau$ is the least level of the $L$-hierarchy where $B$ is true. But $\gamma+1$ is itself p.r. recog. as the first $\tau$ where $L_\tau$ has a proper $\Sigma_1$-substructure, etc., etc. (This answers the second part of the original Question) Similar statements hold for larger stretches of the ordinals $[0,\gamma']$ for $\gamma + 1 < \gamma'$.) To find the least upper bound of the p.r. recog. ordinals, one can reason as follows: If the $\Sigma_1$ sentence $A$ is first true at some stage $L_{\delta}$, then $\delta$ is p.r. recog. Consequently the p.r. recog. ordinals will be cofinal in $\sigma_1$ the least $\Sigma_1$-stable ordinal, *i.e.* the least ordinal with $L_{\sigma}\prec_{\Sigma_1}V $ (as new $\Sigma_1$-sentences become true cofinally in $\sigma_1$). Thus for example if $\delta$ is least so that $L_\delta$ is a $ZF^-$ model, it will be p.r. recog. Note then that no ordinal $\geq \sigma_1$ can be p.r.recongizable, as otherwise this would be a new $\Sigma_1$ fact true at a stage beyond $\sigma_1$. Thus: *Lemma The supremum of the p.r. recognisable ordinals is $\sigma_1$.* If $\tau_0$ is the least p.r. reflecting ordinal, and $\tau >\tau_0$ the least p.r. closed ordinal above that, then the answer to the second question of the Extra Question, is to take $\tau +1$. The $\Sigma_1$ statement that "there exists such a pair $\tau_0<\tau$ " can be used to recognise this ordinal. Comment: if the definition of recognizable is adjusted (call it recognizable*) so that $\alpha$ is recognizable* if for some p.r. function $F$, $F$ is everywhere $0$ except that for some (unique) $\tau$ $F(\tau)=\alpha$, (and thus $\alpha$ is the sole non-zero value that $F$ takes) then the recognizable* ordinals are precisely those $< \sigma_1$. [1] R.B. Jensen and C. Karp *Primitive Recursive set functions*, in Proceedings of Symposia in Pure Mathematics,vol.13 Part 1, *"Axiomatic Set Theory"*, Ed. D. Scott, AMS, 1971, pp 143-167. ------------------- **Edit** added to address Gro-Tsen's queries (see comment below) and intended to complete both the original Question and its second reformulation: (A) The function $F(\delta)=L_\delta$ is p.r. (See Devlin, "Constructibility" but I think it is in [1] anyway.) Satisfaction is likewise p.r., hence the function: $G(\delta)=1$ if $L_\delta \vDash $ '' $A\wedge \forall\delta' L_{\delta'} \vDash \neg A$''; $G(\delta)=0$ otherwise is p.r. and gives that the least $\delta$ with $L_\delta\vDash A$ is p.r. recognizable. (B) Let $\beta<\sigma_1$ . We want that $\beta$ is p.r. recognizable*. Let $A$ be a $\Sigma_1$ sentence that is first true at some $\gamma \in (\beta, \sigma_1)$. By standard arguments the $<_L$-least bijection $f_\gamma:\omega\leftrightarrow \gamma$ is definable over $L_\gamma$. Suppose $f_\gamma(n)=\beta$. By the kind of argument in (A) we see that $ \gamma$ is p.r. recog*. But then so is $L_{\gamma+1}$ and $f_\gamma$. Put these facts together to build a p.r. function $G$ whose only non-zero value is $G(\gamma)=\beta$. (C) We answer this (and, using the first Lemma above, finish the original Question) by: *Lemma Let $\alpha < \beta_0$. Then $\alpha$ is p.r. recog. Hence the least p.r. reflecting ordinal is $\beta_0$ which in turn is the least non-p.r. recog. ordinal.* Proof: Say that $\alpha$ *begins a gap* if $\exists \delta ( L_\alpha \prec_{\Sigma_1} L_\delta)$. We say that $[\alpha,\delta]$ *is a gap*, if $\alpha$ begins a gap, and $\delta$ is maximal with $L_\alpha \prec_{\Sigma_1} L_\delta$. $\bullet$ If $\omega < \delta < \beta_0$ is not in any gap, then we are sufficiently low down in the $L$-hierarchy where there is a $\Sigma_1$ sentence $A=A(\delta)$ so that $L_{\delta+1}\vDash $ ``$A\wedge\forall \delta’<\delta L_{\delta’ + 1}\vDash \neg A$ ‘’. As argued above at (A), this makes $\delta+1$ and so $\delta$ p.r. recog. We just need the ordinals of gaps $[\alpha,\delta]$ with $\alpha <\beta_0$ to be p.r. recognizable. So let $[\alpha,\delta]$ be such a gap. We show that $\alpha$ is p.r. recog. and variants of the argument suffice for the other ordinals in the gap. Then $\alpha < \delta < \alpha^* <\beta_0$. Using (un)/pairing functions (which are p.r.) *etc.* one has that the closure of $\{\alpha\}$ under p.r. functions includes all ordinals up to $\alpha^*$. So let $F$ be a p.r. function, with $F(\alpha)=\delta’$ for some $\delta’\in (\delta,\alpha^*)$. Let $A=A(\delta’)$. Let: $G(\xi) = 1$ if $L_{F(\xi)+1}\vDash A$; $G(\xi) = 0$ otherwise. Then $G$ witnesses that $\alpha$ is p.r. recognizable, so we are done. Q.E.D. (The definition of $\beta_0$ and $\beta$ is that given in Gro-Tsen's second formulation of question, in terms of the Veblen function (if I understand the terms correctly).)