User blog comment:Deedlit11/Ordinal Notations V: Up to a weakly Mahlo cardinal/@comment-24509095-20140509071457/@comment-5150073-20140509155805

King, note that M plays in $$\chi$$ function the same role as $$\Omega$$ in $$\vartheta$$. So $$\chi(M,\alpha)$$ must be reduced by the same way as $$\theta(\Omega,\alpha)$$. We can define:

$$\chi(M,\alpha+1)[0] = \chi(M,\alpha)+1$$

$$\chi(M,\alpha+1)[n+1] = \chi(\chi(M,\alpha+1)[n])$$

Yet I found it is better to define $$\chi(\alpha)_{\beta+1} = \chi(\alpha,\beta)$$, in order to make $$\chi$$ more similar to $$\theta$$. So:

$$\chi(M)_2[n] = \chi(M,1)[n] = \chi(\chi(\cdots (\chi(M)+1) \cdots))$$ (with n+1 $$\chi$$'s.)