User blog comment:Wythagoras/Dollar function part 1/@comment-5529393-20130613074552

Okay, my first quibble with your evaluations is at

a$[[0]_2] ~ f_e_1(a)

The pattern goes:

a$[[0]_2]  ~ f_(e_0)+1(a)

a$[0]_2 ~ f_w^(e_0 + 1) (a)

a$[0]_2 ~ f_(w^w^(e_0 + 1) (a)

a$[[0]_2] ~ f_(w^w^w^(e_0 + 1) (a)

a$[[[0]_2]] ~f_(w^w^w^w^(e_0 + 1) (a)

a$[[[[0]_2]]]] ~f_(w^w^w^w^w^(e_0 + 1) (a)

and so on, so

a$[1]_2 ~ f_e_1(a)

and in general

a$[b]_2 ~ f_e_B (a), where B is the ordinal for b

so

a$[[0]_2]_2 ~ f_e_e_0 (a)

a$[[[0]_2]_2]_2 ~ f_e_e_e_0 (a)

and

a$[0]_3 ~ f_zeta_0 (a)

Again we have

a$[[0]_3]  ~ f_(z_0)+1(a)

a$[0]_3 ~ f_w^(z_0 + 1) (a)

a$[0]_3 ~ f_(w^w^(z_0 + 1) (a)

a$[[0]_3] ~ f_(w^w^w^(z_0 + 1) (a)

and

a$[[0]_3]_2 ~ f_e_(z_0 + 1) (a)

a$[[[0]_3]_2]_2 ~ f_e_e_(z_0 + 1) (a)

so

a$[1]_3 ~ f_zeta_1 (a)

and in general

a$[b]_3 ~ f_zeta_B (a) where B is the ordinal for b

so

a$[[0]_3]_3 ~ f_zeta_zeta_0 (a)

a$[[[0]_3]_3]_3 ~ f_zeta_zeta_zeta_0 (a)

and

a$[0]_4 ~ f_phi(3,0) (a)

Extrapolating, we have

a$[0]_(c+1) ~ f_phi(c,0) (a)

a$[0]_b ~ f_phi(B,0) (a)

So it does look like you reach Gamma_0, it's just that some of the intermediate ordinals are wrong.