User blog comment:DrCeasium/new hyperfactorial array notation/@comment-5529393-20130416153852/@comment-5529393-20130418202733

I assume by z you mean zeta. As you have guessed, both zeta_alpha[m] = (zeta_{alpha-1}^^m and zeta_alpha[m] = epsilon_epsilon_..._epsilon_alpha are incorrect.  The former has limit epsilon_{zeta_{alpha-1} + 1}, and the limit of the latter is the smallest zeta number that is greater than or equal to alpha.

For fundamental sequences, you can take

zeta_0 [0] = 0

zeta_0 [m+1] = epsilon_{zeta_0 [m]}

zeta_{alpha+1} [0] = zeta_alpha + 1

zeta_{alpha+1} [m+1] = epsilon_{zeta_{alpha+1}[m]}

so zeta_0[m] = epsilon_epsilon_..._epsilon_0, and zeta_{alpha+1} [m] = epsilon_epsilon_..._epsilon_{zeta_alpha + 1}. When alpha is a limit ordinal, zeta_alpha [m] = zeta_{alpha [m]}.

Here are the general rules for phi(alpha, beta) [n]: (copied from a different post)

When \(\beta\) is limit, \(\Phi(\alpha, \beta) [n] = \Phi(\alpha, \beta[n]) \).

\Phi(0, 0) = 1 so it has to fundamental sequence.

\Phi(0, \beta+1) [n] = \Phi(0, \beta) * n

\Phi(\alpha+1, 0) [0] = 1\);

\Phi(\alpha+1, 0) [n+1] = \Phi(\alpha, \Phi(\alpha+1, 0) [n])\).

\Phi(\alpha+1, \beta+1) [0] = \Phi(\alpha+1, \beta) + 1;

\Phi(\alpha+1, \beta+1) [n+1] = \Phi(\alpha, \Phi(\alpha +1, \beta + 1) [n]) \).

When \(\alpha\) is limit:

\Phi(\alpha, 0) [n] = \Phi(\alpha[n], 0) \).

\Phi(\alpha, \beta+1) [n] = \Phi(\alpha[n], \Phi(\alpha, \beta) + 1) \).