This question is based on the assumption that $V \ne L$ and we have $\omega\_1^L < \omega\_1$ (here $\omega\_1^L$ is equal to the supremum of ordinals accidentally writable by no-oracle Ordinal Turing Machines).


Consider Ordinal Turing Machines (called “$\omega\_{\alpha}$-machines”) with an oracle that provides access to all [transfinite initial ordinals](https://mathworld.wolfram.com/InitialOrdinal.html).


Any $\omega\_{\alpha}$-machine is an Ordinal Turing Machine equipped with an extra tape (the *oracle tape*). We may assume that this tape is read-only.


Let $t(\alpha)$ denote the symbol written on an $\alpha$-th cell of the oracle tape. Then $t(\alpha) = 1$ if and only if $\alpha$ is an initial ordinal.


If $\epsilon > 0$, then the $\epsilon$-stabilization time of a machine is the successor of $\gamma\_0$, where $\gamma\_0$ is the least ordinal such that the values of all symbols written on all cells of the initial segment of length $\epsilon$ of the output tape never change at any time $\gamma > \gamma\_0$. If $\epsilon = 0$, then the $\epsilon$-stabilization time of a machine is the successor of $\gamma\_0$, where $\gamma\_0$ is the least ordinal such that the values of all symbols written on *all cells* (i.e. cells indexed by any element of the class of all ordinals) of the entire output tape never change at any time $\gamma > \gamma\_0$. If a machine halts, then $\gamma\_0$ is not greater than the halting time. If an ordinal $\gamma\_0$ does not exist, then the corresponding $\epsilon$-stabilization time is $0$.


Let $F\_{\epsilon}(i)$ denote the $\epsilon$-stabilization time of an $i$-th $\omega\_{\alpha}$-machine, assuming that all computations start with no ordinal parameters (i.e. empty input).


Assuming that we have fixed a particular way to encode a countable ordinal by an infinite binary sequence of length $\omega\_0=\omega$, the ordinal $\tau\_0$ is defined as the supremum of ordinals *eventually writable* by $\omega\_{\alpha}$-machines with empty input on the initial segment of length $\omega\_0=\omega$ of the output tape. The reasoning behind this definition of $\tau\_0$ is that there may be $\omega\_{\alpha}$-machines whose initial output segments of length $\omega$ stabilize at a time $\ge \omega\_1$ (i.e. $F\_{\omega}(i) \ge \omega\_1$), yet all other output segments are irrelevant: they stabilize at an arbitrarily large time or *even diverge*. That is, I suppose that there may exist a machine $M\_n$ such that, for example, $F\_0(n) = 0$, yet $F\_{\omega}(n) \ge \omega\_1$. In this case, if the eventually stable content on the initial $\omega$-segment of the output tape encodes an ordinal, this countable ordinal is eventually writable by $M\_n$.


The ordinal $\tau\_1$ is defined as follows: $$\tau\_1 = \sup \{F\_0(i) : i \in \mathbb{N}\}.$$


Is it possible to estimate how large are $\tau\_0$ and $\tau\_1$ (at least, give a “reasonably accurate” estimate for the lower/upper bounds)? In particular, is $\tau\_0$ larger than the least ordinal $\delta$ such that $L\_{\delta} \prec\_{\Sigma\_3} L\_{\omega\_1}$ (the latter is mentioned in [this comment](https://mathoverflow.net/questions/259100/memorable-ordinals#comment639579_259105) and [this answer](https://mathoverflow.net/a/259500) on Mathoverflow)?



How large are the stabilization times of Ordinal Turing Machines with an oracle for the transfinite initial ordinals?
For ordinal Turing machines with an oracle $S⊂Ord$,  

- the set/class of output locations that are written at some point is $Σ^{L[S]}\_{1,S}$ (i.e. $Σ\_1$ in $L[S]$ with a predicate for $S$) and can be arbitrary such,  

- the set/class of output locations that are eventually 1 is $Σ^{L[S]}\_{2,S}$ and can be arbitrary such,  

- any set in $L[S]$ (but no set not in $L[S]$) can be 'accidentally' written at some point.  



The supremum of eventually writable (as subsets of $ω$) ordinals $τ\_0$ is the least ordinal that is not $Δ^{L[S]}\_{2,S}$ definable. The supremum of stabilization times $τ\_1$ is the supremum of $Δ^{L[S]}\_{2,S}$ definable ordinals (possibly uncountable). Similarly, the supremum of writable (as subsets of $ω$) ordinals is the least ordinal not $Δ^{L[S]}\_{1,S}$ definable, and the supremum of halting times is the supremum of $Δ^{L[S]}\_{1,S}$ definable ordinals.


Your question corresponds to $S=\mathrm{Card}$ (cardinals). I actually considered $S=\mathrm{Card}$ in the question [Cardinal Register Machines](https://mathoverflow.net/questions/280638/cardinal-register-machines), but it is worth it to elaborate it further.


*Case 1:* There is no sharp for a proper class of measurable cardinals. Then:  

- $K^{L[\mathrm{Card}]}$ is an iterate of the core model $K$. ($K^{L[\mathrm{Card}]}$ does not have a sharp since otherwise enough cardinals will be indiscernible to generate a proper of class of measurables $K^{L[\mathrm{Card}]}$.)  

- If $V=K$ (which holds if $V=L$), then for sets below the least measurable cardinal (or arbitrarily if there are none), $Δ^{L[\mathrm{Card}]}\_{1,\mathrm{Card}}=Δ\_2$ and $Δ^{L[\mathrm{Card}]}\_{2,\mathrm{Card}}=Δ\_3$. I think the only difference between $K$ and $L[\mathrm{Card}]$ (if $V=K$) is that every measurable cardinal $κ$ (and its measure) in $K$ is converted into a Prikry sequence $κ,κ^{+},κ^{++},...$ for $κ^{+ω}$.  

- Non-absoluteness: There is an ordinal Turing machine $M$ such that if $V$ is a (set) generic extension of $K$ (not sure if this condition is needed), then for every set of ordinals $s$, there is a generic extension $V[G]$ such that in $V[G]$ (and using $\mathrm{Card}^{V[G]}$), $M$ outputs $s$ and halts. (For example, $M$ can check which cardinal successors are computed correctly for the least $L\_α[\mathrm{Card}]$ that correctly computes enough cardinals.) I think such $M$ (if it works for all such $s$ and $V$) cannot halt if the above sharp exists: In every case, $\mathrm{Card}$ should in a sense give indiscernibles (and thus be unavailable for coding) until $L[\mathrm{Card}∩α]$ contains all reals in $K$, and if the sharp exists, $M$ should be stuck in the pre-decoding phase.


*Case 2:* There is a sharp for a proper class of measurable cardinals. Then, $K^{L[\mathrm{Card}]}$ is an iterate of the minimal inner model $K'$ with a proper class of measurable cardinals. Writable (as subsets of $ω$) countable ordinals are exactly $Δ\_2^{K'}∩ω\_1^{K'}$, and eventually writable — $Δ\_3^{K'}∩ω\_1^{K'}$. Note that these are $Δ^1\_3$ (in $V$) and independent of forcing. As an aside, with stronger large cardinal axioms, more structure becomes forcing-independent; for example, see my questions [Complexity of L[cf]](https://mathoverflow.net/questions/279724/complexity-of-l-mathrmcf) and [Inner models with all sets generic](https://mathoverflow.net/questions/356597/inner-models-with-all-sets-generic).


