User blog comment:King2218/FGH Things/@comment-10429372-20140304165143/@comment-5529393-20140305124555

I'm afraid these are incorrect.

f[1](a) will be the smallest strongly critical ordinal that is larger than a. So f[1](a) will be $$\Gamma_0$$ when $$a < \Gamma_0$$, it will be $$\Gamma_1$$ when $$ \Gamma_0 \le a < \Gamma_1$$, and so on.

$$f[1]_1 (\alpha) = \varphi (1, 0, \alpha)$$

$$f[1]_2 (\omega (1+\alpha)) = \varphi (1, 1, \alpha)$$

$$f[1]_3 (\omega (1+\alpha)) = \varphi (1, 2, \alpha)$$

and so on.

$$f[2](\alpha)$$ will be the smallest ordinal of the form $$\varphi (2, 0, \beta)$$ that is greater than $$\alpha$$.

$$f[3](\alpha)$$ will be the smallest ordinal of the form $$\varphi (3, 0, \beta)$$ that is greater than $$\alpha$$.

and in general,

$$f[\beta](\alpha)$$ will be the smallest ordinal of the form $$\varphi (\beta, 0, \gamma)$$ that is greater than $$\alpha$$.

The second hierarchy ordinal is then $$\varphi(1,0,0,0)$$.

$$f[\alpha][\beta](\gamma)$$ is the smallest ordinal of the form $$\varphi(\beta,\alpha, 0, \epsilon)$$ that is greater than $$\gamma$$.

$$f[\alpha]_\delta [\beta](\omega(1 + \gamma)) = \varphi(\beta,\alpha,\delta,\gamma)$$.

$$f[alpha]_\beta[\gamma][\delta](\omega (1+\epsilon)) = \varphi (\delta,\gamma, \alpha,\beta, \epsilon)$$,

and so on.

So there is a correspondence between the brackets and Extended Veblen notation, and the third hierarchy ordinal is the Small Veblen Ordinal. The fourth hierarchy ordinal is the Large Veblen Ordinal.

Happy extending!