>>430
原始数列と順序数の対応はこれであってる？
(0)=ω^0=1
(0,1)=ω^1=ω
(0,1,1)=ω^2
(0,1,1,1)=ω^3
(0,1,1,1,1)=ω^4

(0)=1
(0,1)=ω
(0,1,2)=ω^ω
(0,1,2,3)=ω^ω^ω
(0,1,2,3,4)=ω^ω^ω^ω
(0,1,2,3,4,5)=ω^ω^ω^ω^ω

(0,2)=ε_0
(0,2,1)=ε_0×ω
(0,2,1,3)=ε_0×ε_0=ε_0^2
(0,2,1,3,2)=ε_0^ω
(0,2,1,3,2,4)=ε_0^ε_0
(0,2,1,3,2,4,3)=ε_0^ε_0^ω
(0,2,1,3,2,4,3,5)=ε_0^ε_0^ε_0

(0,2)=ε_0
(0,2,2)=ε_1
(0,2,2,2)=ε_2
(0,2,2,2,2)=ε_3
(0,2,2,2,2,2)=ε_4

(0,2)=ε_0
(0,2,3)=ε_ω
(0,2,3,5)=ε_ε_0
(0,2,3,5,6)=ε_ε_ω
(0,2,3,5,6,8)=ε_ε_ε_0
(0,2,3,5,6,8,9)=ε_ε_ε_ω
(0,2,3,5,6,8,9,11)=ε_ε_ε_ε_0

(0)=φ(0,0)=1
(0,2)=φ(1,0)=ε_0
(0,2,4)=φ(2,0)=ζ_0
(0,2,4,4)=φ(3,0)=η_0
(0,2,4,4,4)=φ(4,0)
(0,2,4,4,4,4)=φ(5,0)

(0,2)=ψ(0)=ε_0
(0,2,4)=ψ(Ω)=ζ_0
(0,2,4,6)=ψ(Ω^Ω)=Γ_0
(0,2,4,6,8)=ψ(Ω^Ω^Ω)
(0,2,4,6,8,10)=ψ(Ω^Ω^Ω^Ω)
(0,2,4,6,8,10,12)=ψ(Ω^Ω^Ω^Ω)