**This post makes substantial revisions to the OP thanks to insights provided by Maschke and Micrm@ss (hopefully they let me know if it's still "not even wrong"):**
1) The "Brief Explanation of Function $f$" has been shortened and all ellipses have been eliminated from the original paragraph.

2) Each rule has been substantially rewritten for clarity.

3) The definition of function $f$ has been substantially rewritten for clarity.

4) Step 2, part c, of "Define the Sequence $T$" was slightly modified for clarity.

5) Concluding remarks were added that discuss $T^{\alpha}$ for any ordinal $\alpha$, where $\alpha$ represents the number of rules employed by the $T$ sequence.

**Introduction:** Making a Sequence $T$ based on â€œThe Rule of Threeâ€

The goal of this thread is to develop the basic model for what I refer to as a "$T$ sequence" and then expand on the basic model so as to create an enumeration of the ordinal $\epsilon_0$.

Each $T$ sequence is a listing of ordinals that is iteratively generated based on a list of rules. After developing a basic $T$ sequence model with only three rules that happens to be an enumeration of $\omega^2$, I will then add additional rules so as to (try to) create the explicit enumeration of $\epsilon_0$.

The primary means of generating the sequence $T$ is through the use of a function $f$. In general, function $f$ is going to be a function that takes as input a three-member sequence of ordinal numbers (an ordered triplet) and returns the set of all ordinals that are a continuation of any recognizable pattern or limit formed by the triplet. The term â€˜recognizable pattern or limitâ€™ is defined for this purpose as one that appears in a sequence of rules (see â€œRulesâ€ below).

**Brief Explanation of Function $f$:**
To get started, consider the ordered triplet $1, 2, 3$ as the initial segment of some pattern of ordinals that could be continued or a limit that could be approached. The number $4$ is a continuation of the ordinalsâ€™ pattern, so we might then say that $1, 2, 3 \implies 4$. The number $4$ is not the only thing that may be implied by $1, 2, 3$, however. We can also say that $1, 2, 3 \implies \omega$. The triplet $1, 2, 3$ is being used as an example because it implies two ordinals, $4$ and $\omega$ (and so $f((1,2,3)) = \{4, \omega \}$), whereas other triplets may imply one or none. The order of the triplet is important too. If we instead consider $3, 2, 1$ we find that $0$ is a continuation of the ordinalsâ€™ pattern (instead of $4$), so we say $3, 2, 1 \implies 0$. To summarize, we are starting to form some basic rules that collectively are going to define the term â€˜recognizable pattern or limitâ€™.

**Rules: ** Where $a, b, c, \dots$ are ordinals that are local variables with respect to each rule, the following rules are used to define function $f$:

$$\text{Rule 1 : }\mu = 3, \beta = 2, \gamma = 1 \implies \{ 0 \}$$

$$\text{Rule 2 : }\mu = 0, \beta = 1, \gamma = a \implies \{a + 1\}$$

$$\text{Rule 3 : }\mu = a, \beta = a+1, \gamma = a+2 \implies \{ a+3, a + \omega \}$$

$$\text{-Note that while we start with these three rules, we will look to add more later so as to enumerate } \epsilon_0.$$

**Define function $f$:**
$$f((\mu,\beta,\gamma)) = \bigcup \{ x : \mu,\beta,\gamma \implies x \}, \text{ where } \mu,\beta, \text{ and } \gamma \text{ are ordinals and } x \text{ is a set defined by the above rules}$$

**Define function $g$:**
For any set of ordinals, $A$, let $g(A)$ be the set of all ordered triplets that can be made from the elements of $A$:

$$g(A) = \{ (a,b,c) : a,b,c \in A \}$$

**Define $X_{Ord}$:**
$$X_{Ord} = X \setminus \{ x \in X : x \text{ is not an ordinal} \} \text{ for any set } X$$

**Define the sequence $T$:**
Define a sequence $T = t_1, t_2, t_3, \dots$ via iterations where:

**Step 1)** $t_1 = 1, t_2 = 2,$ and $t_3 = 3$.

**Step 2)** Each $t_n$, where $n \geq 4$, is defined by the previous elements of the sequence. Starting with $n = 4$:

a) Let $A = \{ t_i \in T : i < n \}$. E.g., $A = \{1, 2, 3 \}$ on the first iteration.

b) Let $B = \{ f((a,b,c))_{Ord} : (a,b,c) \in g(A) \}$. Using the previous elements of the sequence $A$, this step creates a set $B$ of all the new sets of ordinals implied by letting function $f$ range over $g(A)$. The use of the $X_{Ord}$ function in the definition of $B$ may be unnecessary for this particular $T$ sequence.

c) Let $C = (\bigcup B) \setminus A$ and let $c_1, c_2, c_3, \dots$ be an enumeration of $C$ that is also well ordered if $|C| \in \mathbb{N}$. This step removes any redundant elements from $\bigcup B$ before potentially well ordering them so that we can add them to $T$.

d) If $|C| \in \mathbb{N}$, then set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots, t_{n+j-1} = c_j$.

e) If $|C| = |\mathbb{N}|$ (not applicable for this particular $T$ sequence), then let $Tâ€™ = tâ€™_1, tâ€™_2, tâ€™_3, \dots$, be a subsequence of the remaining undefined elements of $T$ and set $tâ€™_1 = c_1, tâ€™_3 = c_2, tâ€™_5 = c_3, \dots$.

**Step 3)** Proceed to the next undefined index $j$ in $T$, set $n = j$, and repeat step 2.

**The first few elements of $T$ with the above three rules would be:**
$$T = 1,2,3,0,4,\omega,5, \omega + 1, 6, \omega + 2,7,\omega+3,\omega \cdot 2,8,\omega+4,\omega \cdot 2 + 1, \dots$$

Where $T$ is generated one iteration at a time, the first iteration takes each ordered triplet that may be comprised from the ordinals $1, 2,$ and $3$ (there are six possible triplets) and tests each triplet to see if its ordering matches one of the rules. We have $f((3,2,1)) = \{0\}$ by Rule 1 and $f((1,2,3)) = \{4, \omega \}$ by Rule 3. The iteration process then takes the union of $\{0\}$ and $\{4, \omega \}$, orders the union to produce a countable sequence that is also well ordered (because it is a finite sequence), and adds the sequence to the initial undefined elements of $T$. We then start the second iteration by taking each ordered triplet that may be comprised from the ordinals $1,2,3,0,4,\omega$ and seeing which new ordinals the rules imply. Here we get the same old $f((3,2,1)) = \{0\}$ and $f((1,2,3)) = \{4, \omega \}$ in addition to $f((0,1,4)) = \{5\}, f((2,3,4)) = \{5\},$ and $f((0,1,\omega)) = \{\omega + 1\}$. Any duplicate ordinals are removed during the iteration process so as to refrain from adding them to $T$ more than once.

If we keep going with these three rules, the $T$ sequence will become an enumeration of $\omega^2$. Note that no rule exists that is capable of taking three ordinals less than $\omega^2$ and implying a set containing an ordinal greater than or equal to $\omega^2$. Rule #1 could produce $\omega^2$ given the triplet $(\omega^2 + 3, \omega^2 + 2, \omega^2 + 1)$, but since there are no ordinals greater than $\omega^2$, Rule 1 is not capable of inserting $\omega^2$ into the sequence either.

**Adding Rules**
Rule 4 will lead to $T$ becoming an enumeration of $\omega^2 \cdot 2$ (assuming my calculations are correct for all of these):

$$\text{Rule 4 : }\mu = a \cdot b, \beta = a \cdot (b+1), \gamma = a \cdot (b+2) \implies \{a \cdot (b + \omega) \}$$

Rule 5 will lead to $T$ becoming an enumeration of $\omega^{\omega}$

$$\text{Rule 5 : }\mu = a+b \cdot c, \beta = a+b \cdot (c+1), \gamma = a+b \cdot (c+2) \implies \{a+b \cdot (c+\omega)\}$$

Rule 6 will lead to $T$ becoming an enumeration of $\omega^{\omega} \cdot 2$:

$$\text{Rule 6 : }\mu = a^b, \beta = a^{b+1}, \gamma = a^{b+2} \implies \{a^{b+\omega}\}$$

Rule 7 will lead to $T$ becoming an enumeration of $\omega^{\omega^2}$:

$$\text{Rule 7 : }\mu = a+b^c, \beta = a+b^{c+1}, \gamma = a+b^{c+2}\implies \{a+b^{c+\omega}\}$$

Rule 8 will lead to $T$ becoming an enumeration of $\omega^{\omega^{\omega}} \cdot 2$:

$$\text{Rule 8 : }\mu = a + b^{c + d \cdot e}, \beta = a + b^{c + d \cdot (e+1)}, \gamma = a + b^{c + d \cdot(e+2)}\implies \{a + b^{c + d \cdot(e+\omega)}\}$$

Rule 9 will lead to $T$ becoming an enumeration of $\epsilon_0 = \omega^{\epsilon_0}$:

$$\text{Rule 9 : }\mu = a + b^{c+d^e}, \beta = a + b^{c+d^{e+1}}, \gamma = a + b^{c+d^{e+2}}\implies \{ a + b^{c+d^{e+w}}\}$$

**Concluding Remarks:**
Assuming the above calculations are correct, the goal is to discuss a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ compiled by adding rules to the above (or some similar) $T$ sequences. Each $T^{\alpha}$ will be a $T$ sequence generated using $\alpha$ rules, where $\alpha$ is some ordinal. An uncountable $\alpha$ would require some clarification and for $T^{\alpha}$ to be a transfinite sequence, presumably. Also, let $T^0 = 1,2,3$.

Each additional rule must meet four specific criteria:

**1)** Each $T$ sequence may contain only ordinals, but not every $T$ sequence is an enumeration of an ordinal. Equivalently, $\{ x : x \in T \}$ may or may not be an ordinal based on the rules used to generate $T$. I wish to create a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ where $\{ x : x \in T^{\alpha} \}$ is an ordinal for each element of the sequence. To do this, it must be ensured that each rule, when added to the previous rules, results in a new $T$ sequence that again enumerates some ordinal.

**2)** It must hold true that $\bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \} = \{x : x \in T^{\alpha - 1} \}$ whenever $\alpha$ is not a limit ordinal.

**3)** A new rule $\alpha$ should imply an ordinal that is equal to $\bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \}$ so as to extend the sequence:

$$\text{Rule } \alpha \text{ : } \mu = a, \beta = b, \gamma = c \implies \{ \bigcup_{\delta < \alpha} \{x : x \in T^{\delta} \} \}, \text{ where } a,b,c < \bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \}$$

**4)** No new rule should should result in $|f((\mu, \beta, \gamma))| \geq \aleph_0$ for any particular triplet $(\mu, \beta, \gamma)$ in the class of all triplets.

**Of particular interest is what $T^{\omega_1}$ would equate to after $\omega_1$ rules are added in a fashion that meets the above criteria and assuming no modifications are made to the basic $T$ sequence model to accommodate a transfinite $T$ (note that we could still start with $T = 1,2,3$ and let the iterative process run, so $T^{\omega_1}$ must equate to something despite making no alterations to the $T$ sequence model to accommodate a transfinite $T$, and it may be that $\{x : x \in T^{\omega_1} \} = \omega_1$ ??? ).**