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The sequence $F_{\alpha}(n)$ indeed does not depend on $\alpha$ when $\alpha \geq \omega$. This question is closely related to this other question, and I will refer to my answer in that question.

The sequence $F_{\alpha}(n)$ indeed does not depend on $\alpha$ when $\alpha \geq \omega$. This question is closely related to this other question, and I will refer to my answer in that question.

Let $S_\alpha(n)$ be the set of ordinals defined by $\alpha ^ {\alpha ^ {\cdots ^ \alpha}}$ with all possible arrangements of parentheses. Since $\alpha = \alpha^{\alpha^0}$ and $(\alpha^{\alpha ^ \beta})^{\alpha^{\alpha^\gamma}} = \alpha^{\alpha ^ {\beta + \alpha^\gamma}}$, every element of $S_\alpha(n)$ is of the form $\alpha^{\alpha ^ \beta}$. So we define $T_\alpha (n) = \{ \beta | \alpha^{\alpha^\beta} \in S_\alpha(n) \}$. We observe that $T_\alpha(0) = \{ 0 \}$ and $T_\alpha(n) = \{ \beta + \alpha^\gamma | \beta \in T_\alpha(i), \gamma \in T_\alpha(j), i + j = n - 1 \}$.

Let $S_\alpha(n)$ be the set of ordinals defined by $\alpha ^ {\alpha ^ {\cdots ^ \alpha}}$ with all possible arrangements of parentheses. Since $\alpha = \alpha^{\alpha^0}$ and $(\alpha^{\alpha ^ \beta})^{\alpha^{\alpha^\gamma}} = \alpha^{\alpha ^ {\beta + \alpha^\gamma}}$, every element of $S_\alpha(n)$ is of the form $\alpha^{\alpha ^ \beta}$. So we define $T_\alpha (n) = \lbrace \beta | \alpha^{\alpha^\beta} \in S_\alpha(n) \rbrace$. We observe that $T_\alpha(0) = \lbrace 0 \rbrace$ and $T_\alpha(n) = \lbrace \beta + \alpha^\gamma | \beta \in T_\alpha(i), \gamma \in T_\alpha(j), i + j = n - 1 \rbrace$.

We claim that, for any n and any $\alpha, \beta \geq \omega$, $T_\alpha(n)$, when written in iterative normal form, is the same set as $T_\beta(n)$, except the base $\beta$ is replaced everywhere it appears by $\alpha$. If we let $f(\gamma)$ be the ordinal obtained by replacing the base $\beta$ by $\alpha$ everywhere in the iterative normal form of $\gamma$, our claim amounts to $T_{\alpha}(n) = f(T_\beta(n))$. From our above expression for $T_\alpha(n)$, we see that it suffices to show that $f(\gamma + \beta^\delta) = f(\gamma) + \alpha^{f(\delta)}$ for any $\gamma, \delta \in E(\alpha)$.

We claim that, for any n and any $\alpha, \beta \geq \omega$, $T_\alpha(n)$, when written in iterative normal form, is the same set as $T_\beta(n)$, except the base $\beta$ is replaced everywhere it appears by $\alpha$. If we let $f(\gamma)$ be the ordinal obtained by replacing the base $\beta$ by $\alpha$ everywhere in the iterative normal form of $\gamma$, our claim amounts to $T_{\alpha}(n) = f(T_\beta(n))$. From our above expression for $T_\alpha(n)$, we see that it suffices to show that $f(\gamma + \beta^\delta) = f(\gamma) + \alpha^{f(\delta)}$ for any $\gamma, \delta \in E(\beta)$.