This blog post is a tip for new googologists who want to create a fast-growing function by recursion. It leads you to \(\psi(\Omega_\omega)\) - the limit of \(\Pi_1^1-CA_0\).

## Basic idea of "legion"

In BEAF, the symbol "X" can take on some arithmetic, such as exponentiation. Exponentiation is not very fast-growing in googoloical sense. However, when symbol "X" takes on exponentiation, "X^X&" will be a very strong operator, and result very fast-growing functions - much fast-growing than exponentiation. Also, in ExE, when symbol "#" takes on exponentiation, "#^#" will be a very strong separator, and the result functions are much faster-growing than exponentiation. And, in FGH, when symbol "ω" takes on exponentiation, "ω^ω" also results a much faster-growing function than exponentiation...

Naturally such ideas come out - why not using BEAF itself on X's, using ExE itself on #'s, using FGH itself on ω's... And that's the basic idea of "legion".

## Success or failure

### Using BEAF on X's

Jonathan Bowers worked on that idea , but unfortunately, now we all know he failed, and it turns out that BEAF is ill-defined beyond tetrational arrays.

Let go one more step to see the reason he fails. Have a look at an X^^(X+1) array (or an {X,X+1,2} array). By definition of up-arrow notation or linear BEAF, we have X^^(X+1) = X^(X^^X) (or {X,X+1,2} = {X,{X,X,2}}). The next time we reduce it, it'll become X^(X^^n) - note that the "X^^X" reduces immediately after X^^(X+1) reduces. Not only "X", but # and ω also have this sort of problem, if they take on tetration.

As a result, X^^(X+1) is even worse than X^^X+1. The latter is something equivalent to {a,b(X^^X)a+1} = {X^^X&a(X^^X)a} - where the outside separator (X^^X) will reduce a-1 more times, and every time the prime, "b", will become much larger than last time. Not only in BEAF, but also #^^#*# and ω^^ω+1 are stronger than #^^(#+1) and ω^^(ω+1) respectively.

### Using ExE on #'s

Let's try another thing - using ExE itself on #'s. f(n) = En#n have a tetrational growth rate, so let's start with that. (To avoid confusing, I use #_{2} here for the # mark of E# notation, and use # for the #'s we operate on) E##_{2}# reduces to E##_{2}n, then EE##_{2}(n-1), then EEE##_{2}(n-2), etc. We finally get 10^10^...10^10^#, then the # reduces immediately, making it even worse than the "##".

But the next step is more fatal. What about E##_{2}(#+1)? It's EE##_{2}#, and just adds a 10 to the bottom of the power tower, which still pales compared with "##". As a result, we can't go very further with that, too. Also, EX#X and Eω#ω are weaker than X^2 and ω^2 respectively.

### Using FGH on ω's

This time it works. \(f_3(\omega)\) reaches ω^^ω, and \(f_0(f_3(\omega))\), \(f_3^2(\omega)\) and \(f_4(\omega)\) are stronger, comparable to \(\varepsilon_0+1\), \(\varepsilon_{\varepsilon_0}\) and \(\zeta_0\) respectively.

## Conclusion

The point that decides the success of this sort of notations is whether the strongest part of the operator can keep enough time after one reduction. A^^(A+1) and EA#A fail (where A is X, #, ω, or this sort of symbols in your own notation) because the strongest part, the A's in EA#A and the A in A^^A, reduces immediately after EA#A and A^^(A+1) reduces. On the opposite, \(f_3^2(A)\) successes because the strongest part, \(f_3(A)\), doesn't reduce until the outside \(f_3\) in \(f_3(f_3(A))\) reduces.

One can fix the problem with a very simple way. Even simply adding 1 can make big difference. For example, if the rules are modified to {a,b+1,c+1 #} = {a,{a,b,c+1 #}+1,c #} and E@m#(n+1) = E@((E@m#n)+1), then they work. Sbiis uses a complicated way to define a ">" operator, and then define hyper-operation on #'s - that's also one way, but more complicated.