User blog comment:Edwin Shade/Rank-on-Rank Turing Ordinals and Beyond/@comment-5029411-20180119221837/@comment-30118230-20180120094220

No,it would be $$\psi(\Omega^\Omega)^\text{CK}$$ since $$\langla 1@\alpha\rangle =\psi(\Omega^{1+\alpha})^\text{CK}$$ and $$\langle 1@\omega\rangle$$ would be $$\psi(\Omega^\omega)^\text{CK}$$