Consider a model of Ordinal Turing Machines (called “$\omega_{\alpha}$-machines”) with a special oracle that provides a dynamic access to the [transfinite initial ordinals](https://mathworld.wolfram.com/InitialOrdinal.html).  
  
Any $\omega_{\alpha}$-machine is an Ordinal Turing Machine equipped with two extra tapes, the _oracle input_ tape and the _oracle output_ tape. Additionally (for convenience), the alphabet is extended to four symbols in the set $\{0,1,2,3\}$, instead of the usual $\{0,1\}$.  
  
Let $t(\alpha)$ denote the symbol written on an $\alpha$-th cell of the oracle input tape at the time when a machine goes into the ASK state.  
  
The oracle operates as follows.  
  
If there are no non-zero symbols written on the oracle input tape (that is, $t(\alpha) = 0$ for all $\alpha \in \text{Ord}$), the oracle puts a machine into the NO state and the computation continues. Otherwise, the oracle performs the following five steps in this exact order (obviously, the order of steps here is significant, although it is possible to consider steps 2 and 3 to be a single step):  
  
1. All non-zero symbols on the oracle output tape are replaced with 0;
2. For any ordinal $\alpha$ such that $t(\alpha) = 1$ and $\omega_{\alpha} \ne \alpha$, a zero symbol on the $\omega_{\alpha}$-th cell of the oracle output tape is replaced with $1$;
3. For any ordinal $\alpha$ such that $t(\alpha) = 1$ and $\omega_{\alpha} = \alpha$, a zero symbol on the $\omega_{\alpha+1}$-th cell of the oracle output tape is replaced with $2$;
4. Assuming that $T(\alpha)$ denotes the symbol written on an $\alpha$-th cell of the oracle output tape at the time when the result of step 3 is complete, let $\beta  = \sup \{\alpha : T(\alpha) \in \{ 1,2 \}\}$. Then a zero symbol on the $\beta$-th cell of the oracle output tape is replaced with $3$;
5. The oracle puts a machine into the YES state and the computation continues.  
  
If $\epsilon > 0$, then the $\epsilon$-stabilization time of a machine is the least ordinal $\gamma_0$ such that the values of all symbols written on all cells of the initial segment of length $\epsilon$ of the output tape (not the _oracle output_ tape) never change at any time $\gamma > \gamma_0$. If $\epsilon = 0$, then the $\epsilon$-stabilization time of a machine is the least ordinal $\gamma_0$ such that the values of all symbols written on all cells of the output tape (not the _oracle output_ tape) never change at any time $\gamma > \gamma_0$. That is, if $\epsilon = 0$, we take into account _all cells_ of the output tape for all $\epsilon \in \text{Ord}$. If a machine halts, then $\gamma_0$ is not greater than the halting time.  
  
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). Here we assume that if a corresponding machine diverges (i.e. does not stabilize), then $F(i) = 0$.  
  
Consider the following two ordinals: $$\begin{array}{l}
{\tau_0} = \sup \{F_{\omega}(i) : i \in \mathbb{N}, F_{\omega}(i) < {\omega _1}\} ,\\
{\tau_1} = \sup \{F_0(i) : i \in \mathbb{N}, F_0(i) \ge {\omega _1}\}.
\end{array}$$  
  
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)?
  
**EDIT**  
Made a correction to the definition of the stabilization time.  
**EDIT #2**  
Made another correction to the definition of the stabilization time.