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First, let me say that this is a really great question.

It seems to me that any increasing base-bumping function would give the same Goodstein result that you eventually hit $0$. That is, I claim (tentatively) that for any increasing sequence of bases $b_1$, $b_2$ and so on, if we define the Goodstein sequence by starting with any number $a_1$, and then if $a_n$ is defined, we write it in complete base $b_n$, replace all instances of $b_n$ with $b_{n+1}$, subtract $1$, and call the answer $a_{n+1}$. The theorem would be that at some point $n$ in the construction, we have $a_n=0$.

The proof of the original theorem proceeded by associating any number $a$ in complete base $b$ with the countable ordinal obtained by replacing all instances of $b$ with the ordinal $\omega$ and interpreting the resulting expression in ordinal arithmetic. They key fact is that the ordinal associated with $a$ in base $b$ is strictly larger than the ordinal associated in base $b+1$ with the number obtained by replacing all $b$'s with $b+1$'s and subtracting $1$. But of course, if If we replace $b$ with some larger $b'$ and do the same thing, then I think it appears that this key fact still goes through---doesn't through, since it ? If sowas proved by observing what happens when the subtract-$1$ part causes a complex term to be broken up with coefficients below the new base. Thus, the newly associated ordinals would still be descending, so they must hit $0$, but this happens only if the numbers themselves hit $0$.

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It seems to me that any increasing base-bumping function would give the same Goodstein result that you eventually hit $0$. That is, I claim (tentatively) that for any increasing sequence of bases $b_1$, $b_2$ and so on, if we define the Goodstein sequence by starting with any number $a_1$, and then if $a_n$ is defined, we write it in complete base $b_n$, replace all instances of $b_n$ with $b_{n+1}$, subtract $1$, and call the answer $a_{n+1}$. The theorem would be that at some point $n$ in the construction, we have $a_n=0$.

The proof of the original theorem proceeded by associating any number $a$ in complete base $b$ with the countable ordinal obtained by replacing all instances of $b$ with the ordinal $\omega$ and interpreting the resulting expression in ordinal arithmetic. They key fact is that the ordinal associated with $a$ in base $b$ is strictly larger than the ordinal associated in base $b+1$ with the number obtained by replacing all $b$'s with $b+1$'s and subtracting $1$. But of course, if we replace $b$ with some larger $b'$ and do the same thing, then I think this key fact still goes through---doesn't it? If so, the associated ordinals would be descending, so they must hit $0$, but this happens only if the numbers themselves hit $0$.