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

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 meeting 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: $1, 2, 3, \dots$. What would come next? The number $4$ is a continuation of the ordinalsâ€™ pattern, so we might then say that $4$ is implied by $1, 2, 3, \dots$ or, similarly, $1, 2, 3, \dots \implies 4$. The number $4$ is not the only thing that may be implied by $1, 2, 3, \dots$, however. We can also say that $1, 2, 3, \dots \implies \omega$. If there are any other ordinal numbers that happen to be universally agreeable continuations or limits for the three-member sequence $1, 2, 3, â€¦$, we could consider them too. In this case, I am unaware of any. 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:**Let $a, b, c, \dots$ be ordinals.

$$\text{Rule 1 : }a, a+1, a+2, \dots \implies \{ a+3, a + \omega \}$$

$$\text{Rule 2 : }a, a-1, a-2, \dots \implies \{ a-3 \} \text{ where } a \geq 3$$

$$\text{Rule 3 : }a \cdot b, a \cdot (b+1), a \cdot (b+2), \dots \implies \{ a \cdot (b+3), a \cdot (b + \omega) \}$$

$$\text{Rule 4 : }0, 1, a, \dots \implies \{a+1\}$$

$$\text{Rule 5 : }0, 2, a, \dots \implies \{a+2\}$$

$$\text{Rule 6 : }a^{b}, a^{b+1}, a^{b+2}, \dots \implies \{a^{b + \omega}\}$$

$$\text{Rule 7 : }a_{b}, a_{b+1}, a_{b+2}, \dots \implies \{a_{b+\omega}\}$$

$$\text{Rule 8 : }a^{b}, a^{b \cdot 1}, a^{b \cdot 2}, \dots \implies \{a^{b \cdot \omega}\}$$

$$\text{Rule 9 : }a_{b}, a_{b \cdot 1}, a_{b \cdot 2}, \dots \implies \{a_{b \cdot \omega}\}$$

$$\text{Rule 10 : }a^b, a^{b^b}, a^{b^{b^b}}, \dots \implies \{a^{b^{b^{b^{\vdots}}}}\}$$

$$\text{Rule 11 : }a_b, a_{b_b}, a_{b_{b_b}}, \dots \implies \{a_{b_{b_{b_{\vdots}}}}\}$$

$$\text{Rule 12 : }0, 1, \omega, \dots \implies \{\omega_1\}$$

$$\text{Rule 13 : }1, \omega, \omega_1, \dots \implies \{\omega_2\}$$

$$\text{Rule 14 : }a_{0}, a_{1}, a_{b}, \dots \implies \{a_{b+1}\}$$

$$\text{Rule 15 : }a_{0}, a_{2}, a_{b}, \dots \implies \{a_{b+2}\}$$

$$\text{Rule 16 : }a^{0}, a^{1}, a^{b}, \dots \implies \{a^{b+1}\}$$

$$\text{Rule 17 : }a^{0}, a^{2}, a^{b}, \dots \implies \{a^{b+2}\}$$

$$\text{Rule 18 : }a^{0}, a^{1}, a^{\omega}, \dots \implies \{a^{\omega_1}\}$$

$$\text{Rule 19 : }a^{1}, a^{\omega}, a^{\omega_1}, \dots \implies \{a^{\omega_2}\}$$

$$\text{Rule 20 : }a^{\omega}, a^{\omega_1}, a^{\omega_2}, \dots \implies \{a^{\omega_{\omega}}\}$$

$$\text{Rule 21 : }a_{0}, a_{1}, a_{\omega}, \dots \implies \{a_{\omega_1}\}$$

$$\text{Rule 22 : }a_{1}, a_{\omega}, a_{\omega_1}, \dots \implies \{a_{\omega_2}\}$$

$$\text{Rule 23 : }a_{\omega}, a_{\omega_1}, a_{\omega_2}, \dots \implies \{a_{\omega_{\omega}}\}$$

We can now define function $f$:

$$f((a,b,c)) = \{ x : a, b, c, \dots \implies x \}, \text{ where } a, b, \text{ and } c \text{ are ordinals and } x \text{ is a set containing ordinals}$$

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

$$g(A) = \{ (a,b,c) : a,b,c \in A \}, \text{ where } A \text{ is a set containing only ordinals (or is empty).}$$

Define a sequence $T = t_1, t_2, t_3, \dots$, where:

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

2) Each $t_n$, where $n \geq 4$, is defined by the previous elements of the sequence:

a. Let $A = \{ t_i \in T : i < n \}$.

b. Let $B = \{ f((a,b,c)) : (a,b,c) \in g(A) \}$.

c. Let $C = \bigcup B \setminus A$ and let $c_1, c_2, c_3, \dots, c_j$ be a well ordering of the elements of $C$ by their natural order.

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 $|C| = |\mathbb{N}|$, then set $t_n = c_1, t_{n+2} = c_2, t_{n+4} = c_3, \dots$.

Assuming my calculations are correct (it's tricky ), the first few elements of $T$ would be:

$$T = 1,2,3,0,4, \omega, 5, \omega + 1, \omega +2, \omega_1, 6, \omega + 3, \omega \cdot 2, \omega_1 + 1, \omega_1 + 2, \omega_2, 7, \dots$$

The question arises: Is there a subsequence of $T$ that is unbounded in $\omega_1$? For this purpose, I am interested in how $T$ grows.