User blog comment:Vel!/Ordinal BEAF/@comment-5150073-20130315133136/@comment-5150073-20130315202201

I believe that. We know that f_e_0(n) ~ n^^n & n. Also we know that (n^^n)^(n^^n) ~ n^^(n+1).

So, if e_0 is n^^n space, then (e_0)^(e_0) is n^^(n+1) space. Then notice that (n^^(n+1))^(n^^(n+1)) ~ n^^(n+2), so n^^(n+2) space is (e_0)^(e_0)^(e_0)^(e_0). We can observe that every increment to n doubles the number of e_0's in the power tower. In general, n^^(n+m) space is (e_0)^(e_0)...(e_0)^(e_0) with 2^m e_0's.

Therefore, f_e_1(n) is about n^^(2n) & n. Analogously, n^^(2n+1) space is (e_1)^(e_1) space, n^^(2n+2) sp. is (e_1)^(e_1)^(e_1)^(e_1) sp. In general, e_2 sp. is n^^(3n) sp.

Under that, every increment to the alpha in the e_alpha adds n to the right of ^^ in the BEAF. In would be more convenient if it adds 1 instead of n, but it is obvious that problem becomes erased when we reach e_(w^w) sp. = n^^(n^n) sp, since n lags from 1 only by the one increment to exponent.

Therefore, e_(e_0) sp. = n^^(n^^n) sp., and e_(e_(e_0)) sp. = n^^(n^^(n^^n)) sp. The ordinal sequence leads to z_0, and operator sequence leads to pentation.