User blog comment:Deedlit11/Ordinal Notations V: Up to a weakly Mahlo cardinal/@comment-30004975-20171217042147/@comment-28606698-20171217213133

That is true. If we define $$I(\alpha,\beta)=$$ the $$(1+\beta)$$th $$\alpha$$-inaccessible cardinal i.e. $$I(\alpha,\beta)=$$ the $$(1+\beta)$$th ordinal in the set $$\{\gamma|\gamma\in R\wedge\forall\delta<\alpha:I(\delta,\gamma)=\gamma\}$$ where $$R$$ is the set of al uncountable regular cardinals

then the first fixed point $$\gamma=I(\alpha,\gamma)$$ is equal to first $$(\alpha+1)$$-inaccessible cardinal i.e. $$\text{min}\{\gamma|\gamma=I(\alpha,\gamma)\}=I(\alpha+1,0)$$, for example $$\text{min}\{\gamma|\gamma=I(0,\gamma)=\aleph_\gamma\}=I(1,0)$$

Yet for last definition in my previous comment:

$$\text{min}\{\beta|I(\alpha,\beta)=\beta\}=\psi_{I(\alpha+1,0)}(0)$$ is not a regular cardinal and is not equal to $$I(\alpha+1,0)$$