User blog comment:Edwin Shade/The Grand List Of Transfinite Ordinals/@comment-30754445-20171130012046/@comment-32213734-20171130142711

I have no idea, is this correct, and will this work: I think, we can define another "Veblen" function φ1(α) = Ωα (the "ordinary" Veblen function is φ(α) = ωα), and as for the "ordinary" Veblen function for many variables. We can count Ψ of various expressions f(Ω). For example, of Ω = ωΩ; ΩΩ = ωΩ 2 ; ΩΩ Ω = ωω Ω 2 ; ΩΩ Ω Ω  = ωω ω Ω 2   ;... We can just continue beyond this sequence with our new function φ1(α). For example, limit of this sequence is φ1(1,0) = εΩ + 1. Then Ωφ1(1,0) + 1 = ωεΩ + 1 + 1; ΩΩ φ1(1,0) + 1 = ωω εΩ + 1 + 1 ; ΩΩ Ω φ1(1,0) + 1 = ωω ω εΩ + 1 + 1  ;... φ1(1,1) = εΩ + 2. Generally, φ1(1,α) = εΩ + 1 + α. Then φ1(1,φ1(1,0)) = εε Ω + 1 ; φ1(1,φ1(1,φ1(1,0))) = εε ε Ω + 1 ;... φ1(2,0) = ζΩ + 1, and, generally, φ1(2,α) = ζΩ + 1 + α. Then φ1(Ω,0) = φ(Ω,1); φ1(1,0,0) = ΓΩ + 1 etc. Then, the same way as we expressed "ordinary" Veblen function through Ψ(f(Ω)), we can express our new "Veblen" function through new function Ψ1(f(Ω2)). For example, φ1(1,0) = Ψ1(0); φ1(2,0) = Ψ1(Ω2); φ1(1,0,0) = Ψ1(Ω2Ω2); Ψ1(Ω2Ω2 Ω2 ); Ψ1(Ω2Ω2 Ω2 Ω2 );... Then we can define another "Veblen" function φ2(α) = Ω2α, then express it through another function Ψ2(f(Ω3)) etc. But this is just my thoughts, in fact, I don't undestand OCF. :)