## FANDOM

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FB100Z asked if one could create a version of BEAF using ordinals instead of Bowers' array structures.  Here I attempt to create one such version.

The standard for ordinal hierarchies of fast-growing functions is of course the fast-growing hierarchy and its variants (like the Hardy hierarchy and the slow-growing hierrarchy).  However, to make it more like BEAF, I will use natural number variables with ordinal indices, i.e. a map from the ordinals to the natural numbers, rather than a single ordinal.  I will also have Bowers' 'base' and 'prime' indices as separate variables.  So, we can notate a particular notation as $$\lbrace (\alpha_1, n_1), (\alpha_2, n_2), \ldots, (\alpha_m, n_m) \rbrace (b, p)$$, where the $$\alpha_i$$ are ordinals and the $$n_m, b, p$$ are natural numbers.

We suppose that each limit ordinal that we use has a fundamental sequence defined for it.  This would be enough to define the notation, but to BEAFify the notation further, we will define a 'prime block' for each limit ordinal (thanks goes to LittlePeng9 for this construction.):

1. The prime block of 0 or any successor ordinal is the empty set.

2. The prime block of an ordinal $$\alpha$$ is comprised of the first p ordinals in the fundamental sequence of $$\alpha$$, along with the prime blocks of those ordinals.

So for example the prime block of $$\omega \alpha + \omega$$ is $$\lbrace \omega \alpha + n | n < p \rbrace$$.  The prime block of $$\omega^2 \alpha + \omega^2$$  is $$\lbrace \omega^2 \alpha + \omega n | n < p \rbrace$$ together with their prime blocks, so you get $$\lbrace \omega^2 \alpha + \omega n + m | n < p - 1, m < p \rbrace \cup \lbrace \omega^2 \alpha + \omega (p-1) \rbrace$$.  Note that this isn't a perfect match with BEAF;  we're missing the ordinals $$\lbrace \omega^2 \alpha + \omega (p-1) + n | 0 < n < p \rbrace$$.  If anyone has an idea how to modify the prime block definition to match with BEAF perfectly, I'd love to hear it.

Now that we have defined prime blocks, we can define the value of  $$\lbrace (\alpha_1, n_1), (\alpha_2, n_2), \ldots, (\alpha_m, n_m) \rbrace (b, p)$$:

1. If $$m = 0$$, i.e. all the ordinals map to 0, then $$\lbrace \rbrace (b, p) = b^p$$.

2. If $$p = 1$$, then $$\lbrace (\alpha_1, n_1), (\alpha_2, n_2), \ldots, (\alpha_m, n_m) \rbrace (b, p) = 1$$.

3. if $$\alpha_m = 0$$, then

$$\lbrace (\alpha_1, n_1), (\alpha_2, n_2), \ldots, (\alpha_m, n_m) \rbrace (b, p)$$

$$= \lbrace (\alpha_1, n_1), (\alpha_2, n_2), \ldots, (\alpha_m, n_m-1) \rbrace (b, \lbrace (\alpha_1, n_1), (\alpha_2, n_2), \ldots, (\alpha_m, n_m) \rbrace (b, p-1))$$.

4. If $$\alpha_m = \beta + 1$$, then

$$\lbrace (\alpha_1, n_1), (\alpha_2, n_2), \ldots, (\alpha_m, n_m) \rbrace (b, p)$$

$$= \lbrace (\alpha_1, n_1), (\alpha_2, n_2), \ldots, (\alpha_m, n_m-1) (\beta, \lbrace (\alpha_1, n_1), (\alpha_2, n_2), \ldots, (\alpha_m, n_m) \rbrace (b, p-1) ) (b, p)$$.

5. If $$\alpha_m$$ is a limit ordinal then replace $$n_m$$ with $$n_m-1$$, and assign $$p$$ to all the ordinals in the prime block of $$\alpha_m$$.  Then evaluate again.

That's it!  We now have a version of Bowers' notation that extends as far as we can define ordinals.