User blog comment:Hl3 or bust/A Newcomer Tries His Hand At a Thing/@comment-25601061-20180614162134

Your analysis is, indeed, a bit off.

As you said, AA is a synonym for Z, which means that AB will be the next letter after AA. Then, we have AB, AC, and so on until AZ = BA. Then BB, CA, DA, and so on until ZZ = AAA. However, we are only counting further letters after Z with this extension, yet we are not diagonalizing over the amount of them.

A[1](a) = AAA...(a) is the point where we first diagonalize over the amount of letters, giving us a growth rate of ω2. Then:

So the limit is actually ωω, not ωω2.
 * A[1](a) ~ fω2(a)
 * A[1]a(a) ~ fω2+1(a)
 * A[1]2(a) ~ fω2+2(a)
 * A[1]a(a) ~ fω2+ω(a)
 * B[1](a) ~ fω2+ω+1(a)
 * C[1](a) ~ fω2+ω2+1(a)
 * A[1][1](a) ~ fω22(a)
 * A[1][1][1](a) ~ fω23(a)
 * A[2](a) ~ fω3(a)
 * B[2](a) ~ fω3+ω+1(a)
 * A[1][2](a) ~ fω3+ω2(a)
 * A[1][1][2](a) ~ fω3+ω22(a)
 * A[2][2](a) ~ fω32(a)
 * A[3](a) ~ fω4(a)
 * A[3][3](a) ~ fω42(a)
 * A[4](a) ~ fω5(a)
 * A[a](a) ~ fωω(a)