Memorable ordinals Suppose $V=L$ + reasonable hypotheses (e.g. "ZFC has a countable transitive model"). Call a countable ordinal $\alpha$ memorable if for some countable $\beta$, $\alpha$ is definable without parameters in every $L_\gamma$ with $\beta<\gamma<\omega_1$.
My question is:

Are there uncountably many memorable ordinals?


Some comments, right off the bat:


*

*The answer is trivially "no" for uniformly memorable ordinals, that is, ordinals which are coboundedly definable without parameters by the same formula. However, since the defining formula is allowed to vary with $\gamma$, this doesn't work.

*Towards a positive answer, note that there are uncountably many countable $\theta$ such that $L_\theta$ is pointwise definable (call such an ordinal "insightful"); this was proved by Hamkins, Linetsky, and Reitz. However, this doesn't actually resolve the issue: for a fixed $\alpha$, there may be many countable $\eta>\alpha$ with no insightful ordinals $>\alpha$ which are definable without parameters in $L_\eta$ (e.g. if there is no greatest insightful ordinal $<\eta$, this seems a distinct possibility).

Interestingly, this question has possibly interesting variants even if $L$ is a very tiny subclass of $V$! Given any hierarchy $(M_\alpha)_{\alpha\in\omega_1}$ with $M_\alpha\cap ON=\alpha$, we can ask whether uncountably many $\alpha$ satisfy "For all sufficiently large $\beta<\omega_1$, $\alpha$ is parameter-free definable in $M_\beta$." Now, we can trivially construct examples where the answer is "yes": namely, let $M_\alpha$ be $L_{\eta_\alpha}$ where $\eta_\alpha$ is the $\alpha$th insightful ordinal; so the general existence question isn't interesting. However, I would be curious if there are any "natural" hierarchies of $M_\alpha$s which do satisfy this property; especially if their union is $H_{\omega_1}$ in some $V$ which is very far from $L$.
 A: REPAIRED ARGUMENT
The memorables are strictly bounded between the first $\Sigma_2$-stable in $\omega_1$ and the first $\Sigma_3$-stable in $\omega_1$. The least non-memorable is thus a $(\Sigma_2\wedge \Pi_2)$-singleton.
Let $\delta_0$ be this first non-memorable. The previous erroneous argument yields a characterisation or perhaps a paraphrase of $\delta_0$.
For $\beta<\omega_1$ let 
(A) $H(\beta)$ be the Skolem Hull inside $L_\beta$ of the empty set.  
(B) Let $\omega_1(\beta):= (\omega_1)^{L_\beta}$ if the latter is defined,
              $= \beta$ otherwise.
Then (i) $H(\beta)$ is the set of pointwise definable objects in $L_\beta$. (ii) $H(\omega_1(\beta))=$ $L_\tau$ for some $\tau\leq \omega_1(\beta)$. (iii) For unboundedly many $\beta$, $\beta = \omega_1(\beta)$. 
Claim Let $\delta_1 =$ the least $\delta< \omega_1$ so that for unboundedly many $\beta$ $H(\beta)=L_{\delta_1}$.  Then $\delta_1=\delta_0$.
Proof: It is easy to argue that $\delta_1$ is defined. Then, by definition $\delta_1$ is not memorable (as $\delta_1
\notin H(\beta)$ for arbitrarily large $\beta$).  So it suffices to show that $\tau<\delta_1 \rightarrow \tau$ is memorable.
By definition, and countability, of $\delta_1$:
$\exists \beta_0\forall \beta> \beta_0\forall\tau<\delta_1\,\, H(\beta)\neq L_\tau \quad (*).$
As $\beta \longrightarrow \omega_1$ so does $\omega_1(\beta) \longrightarrow \omega_1$ non-decreasingly. Thus there is $\beta_1>\beta_0$ so that:
$\forall\beta>\beta_1\,\, \omega_1
(\beta)>\beta_0.$
Then, using (ii), for any $\beta>\beta_1\,\, H(\omega_1(\beta))=L_\gamma$ for some $\gamma \leq\omega_1(\beta)$ but by $(*)$ $\delta_1\leq
\gamma.$   Thus any $\tau< \delta_1$ is pointwise definable in $H(\omega_1(\beta))$ and so (using a definition of $\omega_1$), it is pointwise definable in $L_\beta$. So all $\tau$ less than $\delta_1$ are memorable as required.   $\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 $L_\delta[A\cap \delta]\prec L_{\omega_1}[A]$.
A: It is a very nice question, but unfortunately, the answer is no. 
Theorem. There are only countably many memorable ordinals.
Proof. Let $\delta$ be a countable ordinal with $L_\delta\prec
L_{\omega_1}$. I claim that every memorable ordinal is less than
$\delta$.
To see this, suppose that $\alpha$ is memorable, as witnessed by
$\beta$. Fix an ordinal $\gamma$ above both $\delta$ and $\beta$ with $L_\gamma\prec
L_{\omega_1}$. It follows that $L_\delta\prec L_\gamma\prec
L_{\omega_1}$. But since $\beta\leq\gamma$, we know $\alpha$ is
definable in $L_\gamma$ and hence (since there are no parameters)
definable in $L_\delta$. So $\alpha<\delta$, as desired. QED
