User blog comment:Unknown95387/Help please/@comment-30754445-20171126162320

The third step is actually much easier than the second, because the approximations get easier as the numbers get larger.

As a general rule:

Ackerman(n,n) ≈ fω(n) for large n

So:

mag(3) = gag(gag(gag(3))) = gag(gag(61)) ≈ fω(60)  ≈ fω(fω(60))

Yes, its that easy.

In BEAF, you can approximate this as {59,3,1,2}.

Of course, neither of these numbers are exactly equal to mag(3). Usually there aren't exact equivalences between different notations... but actually, if you insist, it is possible to write mag(3) precisely in BEAF:

First, note that you can write mag(3) precisely in arrow notation, because:

Ackerman(n,n) = n [n-2 arrows] (n+3) - 3

So:

gag(61) = (61↑5964) - 3 exactly

and:

mag(3) = [(61↑5964) - 3 ]↑(61↑ 5964) - 5 (61↑5964) - 3 exactly

An you can also write this precisely in BEAF:

mag(3) = {{61,64,59}  - 3, {61,64,59), {61,64,59}  - 5} exactly

Which is, indeed, between {59, 3, 1, 2} and {60, 3, 1, 2} (and much much closer to the former).

As for the FGH, it is impossible to write Ackerman numbers precisely with the FGH. You could have better approximations, though. Again, most of the work here is in approximating gag(61). After all, it isn't exactly fω(60). It's actually much close to f60(61). So a pretty good approximation would be:

gag(61)  ≈ f60(61)

and

mag(3)  ≈ f'ω(f60(61))'

(you could do even better, but the expressions will get very complicated very quickly, and you'll never be able to write mag(3) precisely as an FGH expression.