User blog comment:KthulhuHimself/Plexation, TaN, and such/@comment-26428969-20151014153656/@comment-27014275-20151015055939

I've just realised that ]n {1(1)} 1[ is actually somewhat LARGER than  f(n) = ]n{n{n{...{n}...}n}n}n[ with n nestings, as with  f(n) = ]n{n{n{...{n}...}n}n}n[ with n nestings, your "base function is  Zn+1 =] n  {Z n } n [, repeated n times, whereas with ]n {1(1)} 1[ the function is  Zn+1 =] Z n  {Z n }  Z n [, repeated n times.

How did you arrive at  f[epsilon0] (n) ? That could mean that it (] {n(1)1} 1[) beats traditional BEAF, all the way to tetrational arrays. If you'd give me some proof, I'd know how to make it reach even higher ordinals.

Thanks for bringing this to my attention.