MISTAKENREPAIRED ARGUMENT
I don't think that theThe memorables are strictly bounded belowbetween the leastfirst $\Sigma_2$-stable in $\omega_1$ and the first $\Sigma_3$-correct.stable in (On grounds of length I post an "answer" rather than$\omega_1$. The least non-memorable is thus a comment, which hopefully characterizes the memorables$(\Sigma_2\wedge \Pi_2)$-singleton.)
Let $\delta$$\delta_0$ be as above: the least ordinal satisfyingthis first non-memorable. The previous erroneous argument yields a characterisation or perhaps a paraphrase of $L_\delta \prec L_{\omega_1}$$\delta_0$.
Claim For $\delta$ is precisely$\beta<\omega_1$ let
(A) $H(\beta)$ be the setSkolem Hull inside $L_\beta$ of memorable ordinalsthe empty set.
Proof:(B) Let $\tau$ be$\omega_1(\beta):= (\omega_1)^{L_\beta}$ if the least non-memorable. Itlatter is easy to see the memorables are downwards closeddefined, $= \beta$ otherwise.
Then (see Joel's comment abovei), and as Joel's answer shows all memorables are less than $\delta$, it suffices to show$H(\beta)$ is the set of pointwise definable objects in $\delta = \tau$$L_\beta$. As $\tau$ is not memorable, for arbitrarily large(ii) $\beta < \omega_1$$H(\omega_1(\beta))=$ $\tau$ is not point-wise definable in$L_\tau$ for some $L_\beta$$\tau\leq \omega_1(\beta)$. Let $S$ be the set of such(iii) For unboundedly many $\beta$, $\beta = \omega_1(\beta)$.
SubclaimClaim For arbitrarily largeLet $\beta\in S$$\delta_1 =$ the least $L_\beta\models$ "Every ordinal is countable”$\delta< \omega_1$ so that for unboundedly many $\beta$ $H(\beta)=L_{\delta_1}$. Then $\delta_1=\delta_0$.
Proof: Suppose otherwiseIt is easy to argue that $\delta_1$ is defined. Then, by definition $\delta_1$ is not memorable (as $\delta_1 \notin H(\beta)$ for all sufficientlyarbitrarily large $\beta\in S$ $\omega_1(\beta):=\omega_1^{L_\beta}<\beta$$\beta$). Note So it suffices to show that $\tau$ cannot be point-wise definable in $L_{\omega_1(\beta)}$ since that definition coupled with the$\tau<\delta_1 \rightarrow \tau$ is memorable.
By definition, and countability, of $\omega_1$ would make it pointwise definable in $L_\beta$. For $\beta\in S$ as$\delta_1$:
$\exists \beta_0\forall \beta> \beta_0\forall\tau<\delta_1\,\, H(\beta)\neq L_\tau \quad (*).$
As $\beta \rightarrow \omega_1$$\beta \longrightarrow \omega_1$ so does $\omega_1(\beta)\longrightarrow \omega_1$$\omega_1(\beta) \longrightarrow \omega_1$ non-decreasingly. But of course every ordinal in $L_{\omega_1(\beta)}$Thus there is countable and this contradicts the supposition. QED (Subclaim)$\beta_1>\beta_0$ so that:
Let $T$ be the set of $\beta\in S$ satisfying the Subclaim. Then $T$ is unbounded in $\omega_1$.$\forall\beta>\beta_1\,\, \omega_1 (\beta)>\beta_0.$
We may assume that suchThen, using $\beta\in T$ are sufficiently large so that all(ii), for any $\eta<\tau$ are pointwise definable in$\beta>\beta_1\,\, H(\omega_1(\beta))=L_\gamma$ for some $L_\beta$. But then if$\gamma \leq\omega_1(\beta)$ but by $H\prec L_\beta$ is the Skolem Hull of the empty set inside$(*)$ $L_\beta$$\delta_1\leq \gamma.$ Thus any (i.e the set of$\tau< \delta_1$ is pointwise definable objects in $L_\beta$) then $H = L_\tau$$H(\omega_1(\beta))$ and so (this is why we thinned $S$ down tousing a definition of $T$$\omega_1$). But this implies $L_\tau\prec L_\beta$. And that, it is true for arbitrarily largepointwise definable in $\beta\in T$$L_\beta$. HenceSo all $L_\tau\prec L_{\omega_1}$. We have$\tau$ less than $\tau=\delta$ and the Claim$\delta_1$ are memorable as required. QED $\quad $ QED.
So something similar would work if CH holds, or in CH models. Eg let $A\subseteq\omega_1$ be such that $H_{\omega_1}=L_{\omega_1}[A]$. Consider the least $\delta$ with $L_\delta[A\cap \delta]\prec L_{\omega_1}[A]$.