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It is interesting to see how to go past psi(chi(Xi(K))). We can continue further with defining different functions on K: K*2, K*3, $\omega^{K+1}$, $\omega^{\omega^{K+1}}$, $\psi_{\Omega_{K+1}}(0)$, $\Omega_{K+1}$, $M_{K+1}$, $\Xi(\alpha,K+1)$ for any $\alpha$.

So, we can make Mahlo-hierarchy above K using $\Xi_{K_2}(0,\beta) = M_{K+\beta}$ and by the analogy for $K_3, K_\omega, K_\Omega, K_K, K_{K_K}, K_\alpha$. But my ideas doesn't stop here.

Note how we expressed $\Omega_\alpha$ and $M_\alpha$ in terms of larger functions?

$\Omega_\alpha = \chi(0,\alpha)$

$M_\alpha = \chi_{\Xi(1,0)}(0,\alpha)$

Then:

$K_\alpha = \chi_{\Xi_{\sigma(1,0)}(0,0)}(0,\alpha)$

Using $\chi_\alpha(\beta,\lambda)$, where $\alpha = \Xi_{\sigma(1,0)}(0,0)$, we can express the inaccessible hierarchy above K. Using $\Xi_\alpha(\beta,\lambda)$, we express the Mahlo hierarchy above K. Finally, $\sigma(\alpha,\beta)$ makes K-typed hierarchy above K. The diagonalizer of $\sigma$ itself is supposed to be the next ordinal in the sequence $\Omega, M, K, \cdots$. Notating this as D, we can have $D_2, D_\omega, D_\Omega, D_D, D_{D_D}, D_\alpha$, and then we must have some new function to go further. Let's reconsider our strategy:

We can notate $\chi_\alpha(\beta,\lambda)$ as $\chi_\alpha(0,\beta,\lambda)$.

$\Xi_\alpha(\beta,\lambda) = \chi_\alpha(1,\beta,\lambda)$

$\sigma_\alpha(\beta,\lambda) = \chi_\alpha(2,\beta,\lambda)$

It other words, we turn the sequence of functions past binary $\chi$ to the single ternary $\chi$ function. Then let n-th term in the sequence $\Omega, M, K, \cdots$ with index $\alpha$ is represented by $T(n,\alpha)$:

$T(n,\alpha) = \chi_{\chi_{\cdots_{\chi_{\chi(n-1,1,0)}(n-2,0,0)}\cdots}(1,0,0)}(0,0,\alpha)$. Keep in mind, $\chi_\alpha(0,\beta,\lambda)$ represents inaccessible hierarchy, $\chi_\alpha(1,\beta,\lambda)$ represents Mahlo hierarchy, $\chi_\alpha(2,\beta,\lambda)$ represents compact hierarchy, and so on.

I know it can be a bit unclear, so wait for formal definition for all this like that.