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strong fffz分析Part2

来自Googology Wiki
Alice留言 | 贡献2026年7月25日 (六) 17:38的版本

ψZ(ω2)=(0)(1,1,1)

ψZ(ω2+ω)=(0)(1,1,1)(1,1)

ψZ[ω2×2](ω2×2)=(0)(1,1,1)(1,1)(2)

ψZ[ω2×2,ωω](ωω)=(0)(1,1,1)(1,1)(2)(3)

ψZ[ω2×2](ωω)=(0)(1,1,1)(1,1)(2,1)

ψZ[ω2×2](ωωω)=(0)(1,1,1)(1,1)(2,1)(3,1)

ψZ[ω2×2](ψZ(ω))=(0)(1,1,1)(1,1)(2,2)

ψZ[ω2×2](ψZ(ω2))=(0)(1,1,1)(1,1)(2,2,1)

ψZ[ω2×2](ψZ(ω2)+ω)=(0)(1,1,1)(1,1)(2,2,1)(1,1)

ψZ[ω2×2](ψZ(ω2)×2)=(0)(1,1,1)(1,1)(2,2,1)(1,1)(2,2,1)

ψZ[ω2×2](ψZ(ω2+1))=(0)(1,1,1)(1,1)(2,2,1)(2)

ψZ[ω2×2,ψZ[ω2+ω](ω2+ω)](ψZ[ω2+ω](ω2+ω))=(0)(1,1,1)(1,1)(2,2,1)(2)(3)

ψZ[ω2×2](ψZ[ω2+ω](ω2+ω))=(0)(1,1,1)(1,1)(2,2,1)(2,1)

ψZ[ω2×2](ψZ[ω2+ω](ωω))=(0)(1,1,1)(1,1)(2,2,1)(2,1)(3,1)

ψZ[ω2×2](ψZ[ω2+ω](ψZ(ω)))=(0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2)

ψZ[ω2×2](ψZ[ω2+ω](ψZ(ω2)))=(0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)

ψZ[ω2×2](ψZ[ω2+ω](ψZ(ω2)+1))=(0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(2,1)

ψZ[ω2×2](ψZ[ω2+ω](ψZ(ω2)×2))=(0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(2,1)(3,2,1)

ψZ[ω2×2](ψZ[ω2+ω](ψZ(ω2+1)))=(0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3)

ψZ[ω2×2](ψZ[ω2+ω](ψZ[ω2+ω](ω2+ω)))=(0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3,1)

ψZ[ω2×2](ψZ(ω2+ω))=(0)(1,1,1)(1,1)(2,2,1)(2,2)

ψZ[ω2×2](ψZ(ω2+ω+1))=(0)(1,1,1)(1,1)(2,2,1)(2,2)(2)

ψZ[ω2×2](ψZ[ω2+ω×2](ω2+ω×2))=(0)(1,1,1)(1,1)(2,2,1)(2,2)(2,1)

ψZ[ω2×2](ψZ(ω2+ω×2))=(0)(1,1,1)(1,1)(2,2,1)(2,2)(2,2)

ψZ[ω2×2](ψZ[ω2×2](ω2×2))=(0)(1,1,1)(1,1)(2,2,1)(2,2)(3)

ψZ[ω2×2](ψZ[ω2×2,ωω](ωω))=(0)(1,1,1)(1,1)(2,2,1)(2,2)(3,1)

ψZ[ω2×2](ψZ[ω2×2](ωω))=(0)(1,1,1)(1,1)(2,2,1)(2,2)(3,2)

ψZ[ω2×2](ψZ[ω2×2](ψZ(ω)))=(0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3)

ψZ[ω2×2](ψZ[ω2×2](ψZ(ω2)))=(0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3,1)

ψZ(ω2×2)=(0)(1,1,1)(1,1,1)

ψZ[ω3](ω3)=(0)(1,1,1)(2)

ψZ[ω3,ωω](ωω)=(0)(1,1,1)(2)(3)

ψZ[ω3](ωω)=(0)(1,1,1)(2,1)

ψZ[ω3](ωω+ω)=(0)(1,1,1)(2,1)(1,1)

ψZ[ω3,ωω+ω2](ωω+ω2)=(0)(1,1,1)(2,1)(1,1)(2)

ψZ[ω3,ωω+ω2](ψZ(ω2))=(0)(1,1,1)(2,1)(1,1)(2,2,1)

ψZ[ω3,ωω+ω2](ψZ(ω2×2))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(2,2,1)

ψZ[ω3,ωω+ω2](ψZ[ω3](ω3))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(3)

ψZ[ω3,ωω+ω2,ψZ[ω3,ωω](ωω)](ψZ[ω3,ωω](ωω))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(3)(4)

ψZ[ω3,ωω+ω2](ψZ[ω3,ωω](ωω))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)

ψZ[ω3,ωω+ω2](ψZ[ω3,ωω](ωω+1))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(3)

ψZ[ω3,ωω+ω2](ψZ[ω3,ωω](ωω×2))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(3,1)

ψZ[ω3,ωω+ω2](ψZ[ω3,ωω](ωω2))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4)

ψZ[ω3,ωω+ω2](ψZ[ω3,ωω](ωωω))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,1)

ψZ[ω3,ωω+ω2](ψZ[ω3,ωω](ψZ(ω)))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,2)

ψZ[ω3,ωω+ω2](ψZ[ω3,ωω](ψZ(ω2)))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,2,1)

ψZ[ω3,ωω+ω2](ψZ[ω3](ωω))=(0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)

ψZ[ω3](ωω+ω2)=(0)(1,1,1)(2,1)(1,1,1)