I have another question which is too long to fit into a comment: how do you even know that $f(n)$ is increasing?

If you have two Turing Machines $M$ and $M'$ that realize the same ordinal $\alpha$, there is no guarantee (as far as I can see) that $BB_\alpha^M(n) = BB_\alpha^{M'}(n)$, because (if $\alpha > \omega$) the Turing machine that computes $BB_\alpha^M$ needs to use the encoding defined by $M$ to index into ordinals less than $\alpha$. With $M$ and $M'$, there may not even be a computable map from the index generated by $M$ to the index generated by $M'$. You might be able to compute this map using $BB_\alpha^M$, but I don't even see how to do that. Thus, $BB_\alpha(n)$ doesn't seem well-defined; you need to specify the encoding into ordinals less than $\alpha$ for it to be well-defined. So even if $\alpha > \beta$, it's not clear that $BB^M_\alpha(n)$ grows faster than $BB^{M'}_\beta(n)$. It's possible that there are some computable ordinals where the index function is so complicated that you can't use it to compute anything useful.

You should be able to fix this by defining $f(n)$ to be $BB_\alpha^M(n)$ for the Turing machine $M$ with $n$ states so that $BB_\alpha^M(n)$ takes the maximum value over all such Turing machines.

UPDATE: and now I have what may be an answer to Scott's question. Is there any reason you have to have the Turing machine $M$ that defines the oracle for $\alpha$ be a vanilla Turing machine. Couldn't you let it have access to an oracle for BB, as well. This way, you can define classes using machines like $T_\alpha^{M_\beta^{M'}}$. Now, just let $f(n)$ be the maximum value for $BB_\alpha^M(n)$ where $M$ is a machine defined in this recursive manner using $n$ symbols.

Question: can you define ordinals that are strictly larger than any computable ordinal in this way? Or does this just define the same class of ordinals in much more complicated ways?

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I have another question which is too long to fit into a comment: how do you even know that $f(n)$ is increasing?

If you have two Turing Machines $M$ and $M'$ that realize the same ordinal $\alpha$, there is no guarantee (as far as I can see) that $BB_\alpha^M(n) = = BB_\alpha^{M'}(n)$, because (if $\alpha > \omega$) the Turing machine that computes $BB_\alpha^M$ needs to use the encoding defined by $M$ to index into ordinals less than $\alpha$. With $M$ and $M'$, there may not even be a computable map from the index generated by $M$ to the index generated by $M'$. You might be able to compute this map using $BB_\alpha^M$, but I don't even see how to do that. Thus, $BB_\alpha(n)$ doesn't seem well-defined; you need to specify the encoding into ordinals less than $\alpha$ for it to be well-defined. So I don't see how you can say even if $\alpha > \beta$, it's not clear that $BB_\alpha(n)$ BB^M_\alpha(n)$grows faster than$BB_\beta(n)$if$\alpha > \beta$. BB^{M'}_\beta(n)$. It's possible that there are some computable ordinals where the index function is so complicated that you can't use it to compute anything useful.

You should be able to fix this by defining $f$ f(n)$to be$BB_\alpha^M$BB_\alpha^M(n)$ for the Turing machine $M$ with $n$ states so that $BB_\alpha^M(n)$ takes the maximum value over all such Turing machines.

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I have another question which is too long to fit into a comment: how do you even know that $f(n)$ is increasing?

If you have two Turing Machines $M$ and $M'$ that realize the same ordinal $\alpha$, there is no guarantee (as far as I can see) that $BB_\alpha^M(n) == BB_\alpha^{M'}(n)$, because (if $\alpha > \omega$) the Turing machine that computes $BB_\alpha^M$ needs to use the encoding defined by $M$ to index into ordinals less than $\alpha$. With $M$ and $M'$, there may not even be a computable map from the index generated by $M$ to the index generated by $M'$. You might be able to compute this map using $BB_\alpha^M$, but I don't even see how to do that. Thus, $BB_\alpha(n)$ doesn't seem well-defined; you need to specify the encoding into ordinals less than $\alpha$ for it to be well-defined. So I don't see how you can say that $BB_\alpha(n)$ grows faster than $BB_\beta(n)$ if $\alpha > \beta$. It's possible that there are some computable ordinals where the index function is so complicated that you can't use it to compute anything useful.

You should be able to fix this by defining $f$ to be $BB_\alpha^M$ for the Turing machine $M$ with $n$ states so that $BB_\alpha^M(n)$ takes the maximum value over all such Turing machines.