User blog comment:Deedlit11/Ordinal Notations V: Up to a weakly Mahlo cardinal/@comment-24509095-20140509071457/@comment-5150073-20140605104025

Wait, that was a typo. I meant of course, $$f_{psi_{\Omega_1}(\psi_{\chi(\alpha)}(\chi(\alpha)))}(n)$$.

Generally, I don't prefer to think about $$\chi(\alpha)$$ as the uncountable cardinal. If we shall not use $$\chi(\alpha)[\beta]$$ for $$\beta \geq \omega$$, I think we can just ignore it for our notation and resolve $$f_{psi_{\Omega_1}(\psi_{\chi(\alpha)}(\chi(\alpha)))}(n)$$ to  $$f_{psi_{\Omega_1}(\psi_{\chi(\alpha)}(\chi(\alpha)[n]))}(n)$$, where $$\chi(\alpha)[n]$$ isn't equal to n, but rather to the expression we want and expect.

As for reducing binary $$\chi$$ function to unary, it is trivial that we can reduce $$\chi(\alpha,0) = \lambda(\alpha)$$, but what must we do for $$alpha$$ to get $$\lambda(f(\alpha)) = \chi(\alpha,\beta)$$? For instance, $$\psi(\Omega^\alpha*\beta) = \theta(\alpha,\beta)$$ so $$f(\alpha) = Omega^\alpha*\beta$$ but that doesn't work for $$\chi$$ and $$\lambda$$ if we replace $$\Omega$$ to M.