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+ω法序数超运算分析

来自Googology Wiki

以下是+ω法序数超运算的分析。

分析1

分析1:SCO~HCO
序数超运算 BMS
ω↑↑ω (0)(1,1) = SCO
ω↑↑(ω+1)=ωω↑↑ω+ω (0)(1,1)(1)(2)
ω↑↑(ω+2)=ωωω↑↑ω+ω (0)(1,1)(1)(2,1)(2)(3)
ω↑↑(ω+3) (0)(1,1)(1)(2,1)(2)(3,1)(3)(4)
ω↑↑ω2 (0)(1,1)(1,1)
ω↑↑(ω2+1)=ωω↑↑ω2+ω (0)(1,1)(1,1)(1)(2)
ω↑↑ω3 (0)(1,1)(1,1)(1,1)
ω↑↑ω2 (0)(1,1)(2)
ω↑↑(ω2+ω) (0)(1,1)(2)(1,1)
ω↑↑ω22 (0)(1,1)(2)(1,1)(2)
ω↑↑ω3 (0)(1,1)(2)(2)
ω↑↑ωω (0)(1,1)(2)(3)
ω↑↑↑3=ω↑↑ω↑↑ω (0)(1,1)(2)(3,1)
ω↑↑↑4 (0)(1,1)(2)(3,1)(4)(5,1)
ω↑↑↑ω (0)(1,1)(2,1) = CO
ω↑↑↑(ω+1)=ω↑↑(ω↑↑↑ω+ω) (0)(1,1)(2,1)(1,1)
ω↑↑↑(ω+2) (0)(1,1)(2,1)(1,1)(2)(3,1)(4,1)(3,1)
ω↑↑↑ω2 (0)(1,1)(2,1)(1,1)(2,1)
ω↑↑↑ω2 (0)(1,1)(2,1)(2)
ω{4}3 (0)(1,1)(2,1)(2)(3,1)(4,1)
ω{4}ω (0)(1,1)(2,1)(2,1) = LCO
ω{4}(ω+1)=ω↑↑↑(ω{4}ω+ω) (0)(1,1)(2,1)(2,1)(1,1)(2,1)
ω{4}ω2 (0)(1,1)(2,1)(2,1)(1,1)(2,1)(2,1)
ω{5}ω (0)(1,1)(2,1)(2,1)(2,1)
ω{ω}ω (0)(1,1)(2,1)(3) = HCO

分析2

分析2:HCO~FSO
序数超运算 BMS
ω{ω}ω (0)(1,1)(2,1)(3) = HCO
ω{ω}(ω+1)=ωω{ω}ω+ω (0)(1,1)(2,1)(3)(1)(2)
ω{ω}(ω+2)=ω↑↑(ω{ω}ω+ω) (0)(1,1)(2,1)(3)(1,1)
ω{ω}(ω+3)=ω↑↑↑(ω{ω}ω+ω) (0)(1,1)(2,1)(3)(1,1)(2,1)
ω{ω}ω2 (0)(1,1)(2,1)(3)(1,1)(2,1)(3)
ω{ω}ω2 (0)(1,1)(2,1)(3)(2)
ω{ω+1}3 (0)(1,1)(2,1)(3)(2)(3,1)(4,1)(5)
ω{ω+1}ω (0)(1,1)(2,1)(3)(2,1)
ω{ω+2}ω (0)(1,1)(2,1)(3)(2,1)(2,1)
ω{ω2}ω (0)(1,1)(2,1)(3)(2,1)(3)
ω{ω2}ω (0)(1,1)(2,1)(3)(3)
ω{ωω}ω (0)(1,1)(2,1)(3)(4)
ω{ω↑↑ω}ω (0)(1,1)(2,1)(3)(4,1)
ω{Ω}3=ω{ω{ω}ω}ω (0)(1,1)(2,1)(3)(4,1)(5,1)(6)
ω{Ω}4 (0)(1,1)(2,1)(3)(4,1)(5,1)(6)(7,1)(8,1)(9)
ω{Ω}ω (0)(1,1)(2,1)(3,1) = FSO

分析3

分析3:FSO~BHO
序数超运算 BMS
ω{Ω}ω (0)(1,1)(2,1)(3,1) = FSO
ω{ω{Ω}ω}(ω+1)=ω{ω}ω{Ω}ω (0)(1,1)(2,1)(3,1)(1,1)(2,1)(3)
ω{ω{Ω}ω}(ω+2)=ω{ω{ω}ω}ω{Ω}ω (0)(1,1)(2,1)(3,1)(1,1)(2,1)(3)(4,1)(5,1)(6)
ω{ω{Ω}ω}ω2 (0)(1,1)(2,1)(3,1)(1,1)(2,1)(3)(4,1)(5,1)(6,1)
ω{ω{Ω}ω+1}ω (0)(1,1)(2,1)(3,1)(1,1)(2,1)(3)(4,1)(5,1)(6,1)(2,1)
ω{Ω}(ω+1)=ω{ω{Ω}ω+ω}ω (0)(1,1)(2,1)(3,1)(1,1)(2,1)(3)(4,1)(5,1)(6,1)(2,1)(3)
ω{Ω}(ω+2)=ω{ω{ω{Ω}ω+ω}ω}ω (0)(1,1)(2,1)(3,1)(1,1)(2,1)(3)(4,1)(5,1)(6,1)(4,1)(5,1)(6)(7,1)(8,1)(9,1)(5,1)(6)
ω{Ω}ω2 (0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,1)
ω{Ω}ω2 (0)(1,1)(2,1)(3,1)(2)
ω{Ω+1}3 (0)(1,1)(2,1)(3,1)(2)(3,1)(4,1)(5,1)
ω{Ω+1}ω (0)(1,1)(2,1)(3,1)(2,1)
ω{Ω+2}ω (0)(1,1)(2,1)(3,1)(2,1)(2,1)
ω{Ω+ω}ω (0)(1,1)(2,1)(3,1)(2,1)(3)
ω{Ω2}3=ω{Ω+ω{Ω+ω}ω}ω (0)(1,1)(2,1)(3,1)(2,1)(3)(4,1)(5,1)(4,1)(5)
ω{Ω2}ω (0)(1,1)(2,1)(3,1)(2,1)(3,1)
ω{Ωω}ω (0)(1,1)(2,1)(3,1)(3)
ω{Ω2}3=ω{Ω×ω{Ωω}ω}ω (0)(1,1)(2,1)(3,1)(3)(4,1)(5,1)(5)
ω{Ω2}ω (0)(1,1)(2,1)(3,1)(3,1) = ACO
ω{Ω2+1}ω (0)(1,1)(2,1)(3,1)(3,1)(2,1)
ω{Ω22}ω (0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,1)(3,1)
ω{Ω2ω}ω (0)(1,1)(2,1)(3,1)(3,1)(3)
ω{Ω3}ω (0)(1,1)(2,1)(3,1)(3,1)(3,1)
ω{Ωω}ω (0)(1,1)(2,1)(3,1)(4) = SVO
ω{ΩΩ}3=ω{Ωω{Ωω}ω}ω (0)(1,1)(2,1)(3,1)(4)(5,1)(6,1)(7,1)(8)
ω{ΩΩ}ω (0)(1,1)(2,1)(3,1)(4,1) = LVO
ω{ΩΩ+1}ω (0)(1,1)(2,1)(3,1)(4,1)(2,1)
ω{ΩΩ+1}ω (0)(1,1)(2,1)(3,1)(4,1)(3,1)
ω{ΩΩ2}ω (0)(1,1)(2,1)(3,1)(4,1)(3,1)(4,1)
ω{ΩΩ2}ω (0)(1,1)(2,1)(3,1)(4,1)(4,1)
ω{ΩΩω}ω (0)(1,1)(2,1)(3,1)(4,1)(5)
ω{Ω↑↑ω}3=ω{ΩΩΩ}ω (0)(1,1)(2,1)(3,1)(4,1)(5,1)
ω{Ω↑↑ω}4=ω{Ω↑↑4}ω (0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,1)
ω{Ω↑↑ω}ω (0)(1,1)(2,2) = BHO

分析4

分析4:BHO~(0)(1,1)(2,2)(3,2)(4)
序数超运算 BMS
ω{Ω↑↑ω}ω (0)(1,1)(2,2) = BHO
ω{Ω↑↑ω}(ω+1)=ω{Ω}(ω{Ω↑↑ω}ω+ω) (0)(1,1)(2,2)(1,1)(2,1)(3,1)
ω{Ω↑↑ω}(ω+1)=ω{ΩΩ}(ω{Ω↑↑ω}ω+ω) (0)(1,1)(2,2)(1,1)(2,1)(3,1)(4,1)
ω{Ω↑↑ω}ω2 (0)(1,1)(2,2)(1,1)(2,2)
ω{Ω↑↑ω}ω2 (0)(1,1)(2,2)(2)
ω{Ω↑↑ω+1}3 (0)(1,1)(2,2)(2)(3,1)(4,2)
ω{Ω↑↑ω+1}ω (0)(1,1)(2,2)(2,1)
ω{Ω↑↑ω+Ω}ω (0)(1,1)(2,2)(2,1)(3,1)
ω{Ω↑↑ω×2}ω (0)(1,1)(2,2)(2,1)(3,2)
ω{Ω↑↑ω×ω}ω (0)(1,1)(2,2)(2,1)(3,2)(3)
ω{Ω↑↑ω×Ω}ω (0)(1,1)(2,2)(2,1)(3,2)(3,1)
ω{Ω↑↑(ω+1)}ω=ω{ΩΩ↑↑ω+ω}ω (0)(1,1)(2,2)(2,1)(3,2)(3,1)(4)
ω{Ω↑↑ω2}ω (0)(1,1)(2,2)(2,2)
ω{Ω↑↑ω3}ω (0)(1,1)(2,2)(2,2)(2,2)
ω{Ω↑↑ω2}ω (0)(1,1)(2,2)(3)
ω{Ω↑↑Ω}3=ω{Ω↑↑ω{Ω↑↑Ω}ω}ω (0)(1,1)(2,2)(3)(4,1)(5,2)(6)
ω{Ω↑↑Ω}ω (0)(1,1)(2,2)(3,1)
ω{Ω↑↑(Ω+ω)}ω (0)(1,1)(2,2)(3,1)(2,2)
ω{Ω↑↑Ω2}ω (0)(1,1)(2,2)(3,1)(2,2)(3,1)
ω{Ω↑↑Ω2}ω (0)(1,1)(2,2)(3,1)(3,1)
ω{Ω↑↑Ω↑↑ω}ω (0)(1,1)(2,2)(3,1)(4,2)
ω{Ω↑↑↑3}ω (0)(1,1)(2,2)(3,1)(4,2)(5,1)
ω{Ω↑↑↑ω}ω (0)(1,1)(2,2)(3,2)
ω{Ω↑↑↑(ω+1)}ω=ω{Ω↑↑(Ω↑↑↑ω+ω)}ω (0)(1,1)(2,2)(3,2)(2,2)
ω{Ω↑↑↑ω2}ω (0)(1,1)(2,2)(3,2)(2,2)(3,2)
ω{Ω↑↑↑ω2}ω (0)(1,1)(2,2)(3,2)(3)
ω{Ω↑↑↑Ω}ω (0)(1,1)(2,2)(3,2)(3,1)
ω{Ω{4}ω}ω (0)(1,1)(2,2)(3,2)(3,2)
ω{Ω{ω}4}ω=ω{Ω{4}Ω}ω (0)(1,1)(2,2)(3,2)(3,2)(3,1)
ω{Ω{5}ω}ω (0)(1,1)(2,2)(3,2)(3,2)(3,2)
ω{Ω{ω}ω}ω (0)(1,1)(2,2)(3,2)(4)

分析5

分析5:(0)(1,1)(2,2)(3,2)(4)~(0)(1,1)(2,2)(3,2)(4,2)
序数超运算 BMS
ω{Ω{ω}ω}ω (0)(1,1)(2,2)(3,2)(4)
ω{Ω{ω}(ω+1)}ω=ω{ΩΩ{ω}ω+ω}ω (0)(1,1)(2,2)(3,2)(4)(2,1)(3,2)(3)(4)
ω{Ω{ω}(ω+2)}ω=ω{Ω↑↑(Ω{ω}ω+ω)}ω (0)(1,1)(2,2)(3,2)(4)(2,2)
ω{Ω{ω}ω2}ω (0)(1,1)(2,2)(3,2)(4)(2,2)(3,2)(4)
ω{Ω{ω}ω2}ω (0)(1,1)(2,2)(3,2)(4)(3)
ω{Ω{Ω}ω}ω=ω{Ω{ω}Ω}ω (0)(1,1)(2,2)(3,2)(4)(3,1)
ω{Ω{ω+1}ω}ω (0)(1,1)(2,2)(3,2)(4)(3,2)
ω{Ω{ω2}ω}ω (0)(1,1)(2,2)(3,2)(4)(3,2)(4)
ω{Ω{ω2}ω}ω (0)(1,1)(2,2)(3,2)(4)(4)
ω{Ω{ω↑↑ω}ω}ω (0)(1,1)(2,2)(3,2)(4)(5,1)
ω{Ω{ω{Ω↑↑ω}ω}ω}ω (0)(1,1)(2,2)(3,2)(4)(5,1)(6,2)
ω{Ω{ω{Ω{ω}ω}ω}ω}ω (0)(1,1)(2,2)(3,2)(4)(5,1)(6,2)(7)
ω{Ω{Ω}Ω}ω (0)(1,1)(2,2)(3,2)(4,1)
ω{Ω{Ω}(Ω+1)}ω=ω{ΩΩ{Ω}Ω+ω}ω (0)(1,1)(2,2)(3,2)(4,1)(2,1)(3,2)(3)(4)
ω{Ω{Ω}(Ω+2)}ω=ω{Ω↑↑(Ω{Ω}Ω+ω)}ω (0)(1,1)(2,2)(3,2)(4,1)(2,2)
ω{Ω{Ω}(Ω+ω)}ω=ω{Ω{ω}(Ω{Ω}Ω+ω)}ω (0)(1,1)(2,2)(3,2)(4,1)(2,2)(3,2)(4)
ω{Ω{Ω}Ω2}ω (0)(1,1)(2,2)(3,2)(4,1)(2,2)(3,2)(4,1)
ω{Ω{Ω}Ωω}ω (0)(1,1)(2,2)(3,2)(4,1)(3)
ω{Ω{Ω}Ω2}ω (0)(1,1)(2,2)(3,2)(4,1)(3,1)
ω{Ω{Ω+1}3}ω (0)(1,1)(2,2)(3,2)(4,1)(3,1)(4,2)(5,3)(6,1)
ω{Ω{Ω+1}ω}ω (0)(1,1)(2,2)(3,2)(4,1)(3,2)
ω{Ω{Ω+ω}ω}ω (0)(1,1)(2,2)(3,2)(4,1)(3,2)(4)
ω{Ω{Ω2}Ω}ω (0)(1,1)(2,2)(3,2)(4,1)(3,2)(4,1)
ω{Ω{Ωω}ω}ω (0)(1,1)(2,2)(3,2)(4,1)(4)
ω{Ω{Ω2}Ω}ω (0)(1,1)(2,2)(3,2)(4,1)(4,1)
ω{Ω{Ω↑↑ω}ω}ω (0)(1,1)(2,2)(3,2)(4,1)(5,2)
ω{Ω{Ω↑↑Ω}Ω}ω (0)(1,1)(2,2)(3,2)(4,1)(5,2)(6,1)
ω{Ω{Ω↑↑↑ω}ω}ω (0)(1,1)(2,2)(3,2)(4,1)(5,2)(6,2)
ω{Ω{Ω{ω}ω}ω}ω (0)(1,1)(2,2)(3,2)(4,1)(5,2)(6,2)(7)
ω{Ω{Ω2}ω}3=ω{Ω{Ω2}3}ω=ω{Ω{Ω{Ω}Ω}Ω}ω (0)(1,1)(2,2)(3,2)(4,1)(5,2)(6,2)(7,1)
ω{Ω{Ω2}ω}4=ω{Ω{Ω2}4}ω (0)(1,1)(2,2)(3,2)(4,1)(5,2)(6,2)(7,1)(8,2)(9,2)(10,1)
ω{Ω{Ω2}ω}ω (0)(1,1)(2,2)(3,2)(4,2)

分析6

分析6:(0)(1,1)(2,2)(3,2)(4,2)~(0)(1,1)(2,2)(3,3)
序数超运算 BMS
ω{Ω{Ω2}ω}ω (0)(1,1)(2,2)(3,2)(4,2)
ω{Ω{Ω2}ω+1}ω (0)(1,1)(2,2)(3,2)(4,2)(2,1)
ω{Ω{Ω2}ω×2}ω (0)(1,1)(2,2)(3,2)(4,2)(2,1)(3,2)(4,2)(5,2)
ω{Ω↑↑(Ω{Ω2}ω+ω)}ω (0)(1,1)(2,2)(3,2)(4,2)(2,2)
ω{Ω{Ω{Ω2}ω}(ω+1)}ω=ω{Ω{Ω}(Ω{Ω2}ω+Ω)}ω (0)(1,1)(2,2)(3,2)(4,2)(2,2)(3,2)(4,1)
ω{Ω{Ω{Ω2}ω}ω2}ω (0)(1,1)(2,2)(3,2)(4,2)(2,2)(3,2)(4,1)(5,2)(6,2)(7,2)
ω{Ω{Ω{Ω2}ω}Ω}ω (0)(1,1)(2,2)(3,2)(4,2)(2,2)(3,2)(4,1)(5,2)(6,2)(7,2)(3,1)
ω{Ω{Ω{Ω2}ω+1}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(2,2)(3,2)(4,1)(5,2)(6,2)(7,2)(3,2)
ω{Ω{Ω2}(ω+1)}ω=ω{Ω{Ω{Ω2}ω+ω}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(2,2)(3,2)(4,1)(5,2)(6,2)(7,2)(3,2)(4)
ω{Ω{Ω2}ω2}ω (0)(1,1)(2,2)(3,2)(4,2)(2,2)(3,2)(4,2)
ω{Ω{Ω2}ω2}ω (0)(1,1)(2,2)(3,2)(4,2)(3)
ω{Ω{Ω2}Ω}ω (0)(1,1)(2,2)(3,2)(4,2)(3,1)
ω{Ω{Ω2}Ω{Ω2}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(3,1)(4,2)(5,2)(6,2)
ω{Ω{Ω2+1}3}ω (0)(1,1)(2,2)(3,2)(4,2)(3,1)(4,2)(5,2)(6,2)(7,1)
ω{Ω{Ω2+1}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(3,2)
ω{Ω{Ω2+2}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(3,2)(3,2)
ω{Ω{Ω2+ω}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(3,2)(4)
ω{Ω{Ω2+ω{Ω{Ω2+ω}ω}ω}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(3,2)(4)(5,1)(6,2)(7,2)(6,2)(7)
ω{Ω{Ω2+Ω}Ω}ω (0)(1,1)(2,2)(3,2)(4,2)(3,2)(4,1)
ω{Ω{Ω22}3}ω=ω{Ω{Ω2+Ω{Ω2+Ω}Ω}Ω}ω (0)(1,1)(2,2)(3,2)(4,2)(3,2)(4,1)(5,2)(6,2)(7,2)(6,2)(7,1)
ω{Ω{Ω22}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(3,2)(4,2)
ω{Ω{Ω2ω}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(4)
ω{Ω{Ω2Ω}Ω}ω (0)(1,1)(2,2)(3,2)(4,2)(4,1)
ω{Ω{Ω22}3}ω=ω{Ω{Ω2×Ω{Ω2Ω}Ω}Ω}ω (0)(1,1)(2,2)(3,2)(4,2)(4,1)(5,2)(6,2)(7,2)(7,1)
ω{Ω{Ω22}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(4,2)
ω{Ω{Ω23}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(4,2)(4,2)
ω{Ω{Ω2ω}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(5)
ω{Ω{Ω2Ω}Ω}ω (0)(1,1)(2,2)(3,2)(4,2)(5,1)
ω{Ω{Ω2Ω2}3}ω=ω{Ω{Ω2Ω{Ω2Ω}Ω}Ω}ω (0)(1,1)(2,2)(3,2)(4,2)(5,1)(6,2)(7,2)(8,2)(9,1)
ω{Ω{Ω2Ω2}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(5,2)
ω{Ω{Ω2↑↑3}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(5,2)(6,2)
ω{Ω{Ω2↑↑ω}ω}ω (0)(1,1)(2,2)(3,3)

分析7

分析7:(0)(1,1)(2,2)(3,3)~BO
序数超运算 BMS
ω{Ω{Ω2↑↑ω}ω}ω (0)(1,1)(2,2)(3,3)
ω{Ω{Ω2↑↑ω}ω+1}ω (0)(1,1)(2,2)(3,3)(2,1)
ω{Ω{Ω2↑↑ω}ω×2}ω (0)(1,1)(2,2)(3,3)(2,1)(3,2)(4,3)
ω{Ω↑↑(Ω{Ω2↑↑ω+ω)}ω}ω (0)(1,1)(2,2)(3,3)(2,2)
ω{Ω↑↑↑(Ω{Ω2↑↑ω+ω)}ω}ω (0)(1,1)(2,2)(3,3)(2,2)(3,2)
ω{Ω{Ω2↑↑ω}(ω+1)}ω=ω{Ω{Ω2}(Ω{Ω2↑↑ω+ω)}ω}ω (0)(1,1)(2,2)(3,3)(2,2)(3,2)(4,2)
ω{Ω{Ω2↑↑ω}ω2}ω (0)(1,1)(2,2)(3,3)(2,2)(3,3)
ω{Ω{Ω2↑↑ω}ω2}ω (0)(1,1)(2,2)(3,3)(3)
ω{Ω{Ω2↑↑ω}Ω}ω (0)(1,1)(2,2)(3,3)(3,1)
ω{Ω{Ω2↑↑ω+1}ω}ω (0)(1,1)(2,2)(3,3)(3,2)
ω{Ω{Ω2↑↑ω+ω}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4)
ω{Ω{Ω2↑↑ω+Ω}Ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,1)
ω{Ω{Ω2↑↑ω+Ω2}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,2)
ω{Ω{Ω2↑↑ω×2}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,3)
ω{Ω{Ω2↑↑ω×ω}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,3)(5)
ω{Ω{Ω2↑↑ω×Ω}Ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,3)(5,1)
ω{Ω{Ω2Ω2↑↑ω+1}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,3)(5,2)
ω{Ω{Ω2↑↑(ω+1)}ω}ω=ω{Ω{Ω2Ω2↑↑ω+ω}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,3)(5,2)(6)
ω{Ω{Ω2↑↑ω2}ω}ω (0)(1,1)(2,2)(3,3)(3,3)
ω{Ω{Ω2↑↑ω2}ω}ω (0)(1,1)(2,2)(3,3)(4)
ω{Ω{Ω2↑↑Ω}Ω}ω (0)(1,1)(2,2)(3,3)(4,1)
ω{Ω{Ω2↑↑Ω2}ω}ω (0)(1,1)(2,2)(3,3)(4,2)
ω{Ω{Ω2↑↑↑3}ω}ω (0)(1,1)(2,2)(3,3)(4,2)(5,3)(6,2)
ω{Ω{Ω2↑↑↑ω}ω}ω (0)(1,1)(2,2)(3,3)(4,3)
ω{Ω{Ω2{ω}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,3)(5)
ω{Ω{Ω2{Ω}Ω}Ω}ω (0)(1,1)(2,2)(3,3)(4,3)(5,1)
ω{Ω{Ω2{Ω2}Ω2}ω}ω (0)(1,1)(2,2)(3,3)(4,3)(5,2)
ω{Ω2}4=ω{Ω{Ω2{Ω3}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,3)(5,3)
ω{Ω{Ω2{Ω32}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,3)(5,3)(4,3)(5,3)
ω{Ω{Ω2{Ω32}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,3)(5,3)(5,3)
ω{Ω{Ω2{Ω3Ω3}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,3)(5,3)(6,3)
ω{Ω{Ω2{Ω3↑↑ω}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,4)
ω{Ω{Ω2{Ω3↑↑↑ω}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,4)(5,4)
ω{Ω2}5=ω{Ω{Ω2{Ω3{Ω4}ω}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,4)(5,4)(6,4)
ω{Ω{Ω2{Ω3{Ω4↑↑ω}ω}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,4)(5,5)
ω{Ω2}ω (0)(1,1,1) = BO

分析8

分析8:BO~(0)(1,1,1)(2,1)
序数超运算 BMS
ω{Ω2}ω (0)(1,1,1) = BO
ω↑↑(ω{Ω2}ω+ω) (0)(1,1,1)(1,1)
ω{Ω2}(ω+1)=ω{Ω}(ω{Ω2}ω+ω) (0)(1,1,1)(1,1)(2,1)(3,1)
ω{Ω↑↑ω}(ω{Ω2}ω+ω) (0)(1,1,1)(1,1)(2,2)
ω{Ω2}(ω+2)=ω{Ω{Ω2}ω}(ω{Ω2}ω+ω) (0)(1,1,1)(1,1)(2,2)(3,2)(4,2)
ω{Ω2}ω2 (0)(1,1,1)(1,1)(2,2,1)
ω{Ω2}ω2 (0)(1,1,1)(1,1)(2,2,1)(2)
ω{Ω2+1}3 (0)(1,1,1)(1,1)(2,2,1)(2)(3,1,1)
ω{Ω2+1}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)
ω{Ω2+2}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(2,1)
ω{Ω2+ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3)
ω{Ω2+ω{Ω2}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3)(4,1,1)
ω{Ω2+Ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,1)
ω{Ω2+Ω+1}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,1)(2,1)
ω{Ω2+Ω2}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,1)(2,1)(3,1)
ω{Ω2+Ω2}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,1)(3,1)
ω{Ω2+ΩΩ}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,1)(4,1)
ω{Ω2+Ω↑↑ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2)
ω{Ω2+Ω↑↑Ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2)(4,1)
ω{Ω2+Ω↑↑↑ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2)(4,2)
ω{Ω2+Ω{ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2)(4,2)(5)
ω{Ω2+Ω{Ω}Ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2)(4,2)(5,1)
ω{Ω2+Ω{Ω2}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2)(4,2)(5,2)
ω{Ω2+Ω{Ω2↑↑ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2)(4,3)
ω{Ω2+Ω{Ω3}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)
ω{Ω2+ΩΩ{Ω3}ω+1}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3,1)
ω{Ω2+Ω{Ω3}(ω+1)}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3,1)(4,1)(5,1)
ω{Ω2+Ω{Ω2↑↑ω}(Ω{Ω3}ω+ω)}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3,1)(4,2)
ω{Ω2+Ω{Ω3}ω2}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3,1)(4,2,1)
ω{Ω2+Ω{Ω3}ω2}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3,1)(4,2,1)(5)
ω{Ω2+Ω{Ω3}Ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3,1)(4,2,1)(5,1)
ω{Ω2+Ω{Ω3}Ω{Ω3}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3,1)(4,2,1)(5,1)(6,2,1)
ω{Ω2+Ω{Ω3+1}3}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3,1)(4,2,1)(5,1)(6,2,1)(7,1)
ω{Ω2+Ω{Ω3+1}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)
ω{Ω2+Ω{Ω3+1}ω+1}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(2,1)
ω{Ω2+Ω{Ω3+1}ω+Ω{Ω3}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(2,1)(3,2,1)
ω{Ω2+Ω{Ω3+1}ω×2}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(2,1)(3,2,1)(3,2)
ω{Ω2+ΩΩ{Ω3+1}ω+1}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(2,1)(3,2,1)(3,2)(3,1)
ω{Ω2+Ω{Ω3+1}(ω+1)}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(2,1)(3,2,1)(3,2)(3,1)(4,2,1)
ω{Ω2+Ω{Ω3+1}ω2}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(2,1)(3,2,1)(3,2)(3,1)(4,2,1)(4,2)
ω{Ω2+Ω{Ω3+2}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(2,2)
ω{Ω2+Ω{Ω3+ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3)
ω{Ω2+Ω{Ω3+Ω}Ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,1)
ω{Ω22}3=ω{Ω2+Ω{Ω3+Ω2}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,2)
ω{Ω2+Ω{Ω3+Ω2↑↑ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3)
ω{Ω2+Ω{Ω3+Ω2↑↑↑ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3)(4,3)
ω{Ω2+Ω{Ω3+Ω2{Ω3}ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3)(4,3)(5,3)
ω{Ω2+Ω{Ω3+Ω2{Ω3↑↑ω}ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3)(4,4)
ω{Ω2+Ω{Ω3+Ω2{Ω4}ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3,1)
ω{Ω2+Ω{Ω3+Ω2{Ω4+1}ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3,1)(3,3)
ω{Ω22}4=ω{Ω2+Ω{Ω3+Ω2{Ω4+Ω3}ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3,1)(3,3)(4,3)
ω{Ω2+Ω{Ω3+Ω2{Ω4+Ω3↑↑ω}ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3,1)(3,3)(4,4)
ω{Ω2+Ω{Ω3+Ω2{Ω4+Ω3{Ω5}ω}ω}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3,1)(3,3)(4,4,1)
ω{Ω22}ω (0)(1,1,1)(1,1,1)
ω{Ω2}(ω{Ω22}ω+ω) (0)(1,1,1)(1,1,1)(1,1)(2,2,1)
ω{Ω22}(ω+1)=ω{Ω2+Ω}(ω{Ω22}ω+ω) (0)(1,1,1)(1,1,1)(1,1)(2,2,1)(2,1)(3,1)
ω{Ω22}ω2 (0)(1,1,1)(1,1,1)(1,1)(2,2,1)(2,2,1)
ω{Ω22+1}ω (0)(1,1,1)(1,1,1)(1,1)(2,2,1)(2,2,1)(2,1)
ω{Ω22+Ω}ω (0)(1,1,1)(1,1,1)(1,1)(2,2,1)(2,2,1)(2,1)(3,1)
ω{Ω22+Ω↑↑ω}ω (0)(1,1,1)(1,1,1)(1,1)(2,2,1)(2,2,1)(2,1)(3,2)
ω{Ω22+Ω{Ω3}ω}ω (0)(1,1,1)(1,1,1)(1,1)(2,2,1)(2,2,1)(2,1)(3,2,1)
ω{Ω22+Ω{Ω32}ω}ω (0)(1,1,1)(1,1,1)(1,1)(2,2,1)(2,2,1)(2,1)(3,2,1)(3,2,1)
ω{Ω22+Ω{Ω32+1}ω}ω (0)(1,1,1)(1,1,1)(1,1)(2,2,1)(2,2,1)(2,2)
ω{Ω23}ω (0)(1,1,1)(1,1,1)(1,1,1)
ω{Ω2ω}ω (0)(1,1,1)(2)
ω{Ω2×ω{Ω2}ω}ω (0)(1,1,1)(2)(3,1,1)
ω{Ω2Ω}3 (0)(1,1,1)(2)(3,1,1)(4)
ω{Ω2Ω}ω (0)(1,1,1)(2,1)

分析9

分析9:(0)(1,1,1)(2,1)~TFBO
序数超运算 BMS
ω{Ω2Ω}ω (0)(1,1,1)(2,1)
ω{Ω2Ω}ω2 (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)
ω{Ω2Ω+1}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,1)
ω{Ω2Ω+Ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,1)(3,1)
ω{Ω2Ω+Ω↑↑ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,1)(3,2)
ω{Ω2Ω+Ω{Ω3}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,1)(3,2,1)
ω{Ω2Ω+Ω{Ω3Ω}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,1)(3,2,1)(4,1)
ω{Ω2Ω+Ω{Ω3Ω+1}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2)
ω{Ω2Ω+Ω{Ω3Ω+Ω2}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2)(3,2)
ω{Ω2Ω+Ω{Ω3Ω+Ω2↑↑ω}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2)(3,3)
ω{Ω2Ω+Ω{Ω3Ω+Ω2{Ω4}ω}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2)(3,3,1)
ω{Ω2Ω+Ω{Ω3Ω+Ω2{Ω4Ω}ω}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2)(3,3,1)(4,1)
ω{Ω2Ω+Ω{Ω3×(Ω+1)}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2,1)
ω{Ω2Ω+Ω{Ω3×(Ω+2)}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2,1)(2,2,1)
ω{Ω2Ω+Ω{Ω3×Ω2}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2,1)(3,1)
ω{Ω2Ω+Ω{Ω3×Ω2}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(3,1)
ω{Ω2Ω+Ω{Ω3×ΩΩ}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,1)
ω{Ω2Ω+Ω{Ω3×Ω↑↑ω}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,2)
ω{Ω2Ω+Ω{Ω3×Ω↑↑↑ω}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,2)(5,2)
ω{Ω2Ω+Ω{Ω3×Ω{Ω2}ω}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,2)(5,2)(6,2)
ω{Ω2Ω+Ω{Ω3×Ω{Ω2↑↑ω}ω}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,2)(5,3)
ω{Ω2Ω+Ω{Ω3×Ω{Ω3}ω}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,2,1)
ω{Ω2×(Ω+1)}3=ω{Ω2Ω+Ω{Ω3Ω2}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)
ω{Ω2×(Ω+1)}4=ω{Ω2Ω+Ω{Ω3Ω2+Ω2{Ω4Ω3}}ω}ω (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3,1)(4,3)
ω{Ω2×(Ω+1)}ω (0)(1,1,1)(2,1)(1,1,1)
ω{Ω2×(Ω+2)}ω (0)(1,1,1)(2,1)(1,1,1)(1,1,1)
ω{Ω2×(Ω+ω)}ω (0)(1,1,1)(2,1)(1,1,1)(2)
ω{Ω2×Ω2}ω (0)(1,1,1)(2,1)(1,1,1)(2,1)
ω{Ω2×Ωω}ω (0)(1,1,1)(2,1)(2)
ω{Ω2×Ω2}ω (0)(1,1,1)(2,1)(2,1)
ω{Ω2×(Ω2+1)}ω (0)(1,1,1)(2,1)(2,1)(1,1,1)
ω{Ω2×Ωω}ω (0)(1,1,1)(2,1)(3)
ω{Ω2×ΩΩ}ω (0)(1,1,1)(2,1)(3,1)
ω{Ω2×(ΩΩ+1)}ω (0)(1,1,1)(2,1)(3,1)(1,1,1)
ω{Ω2×Ω↑↑ω}ω (0)(1,1,1)(2,1)(3,2) = TFBO

分析10

分析10:TFBO~EBO
序数超运算 BMS
ω{Ω2×Ω↑↑ω}ω (0)(1,1,1)(2,1)(3,2) = TFBO
ω{Ω2×Ω↑↑ω×2}ω (0)(1,1,1)(2,1)(3,2)(1,1,1)(2,1)(3,2)
ω{Ω2×Ω↑↑ω×ω}ω (0)(1,1,1)(2,1)(3,2)(2)
ω{Ω2×ΩΩ↑↑ω+1}ω (0)(1,1,1)(2,1)(3,2)(2,1)
ω{Ω2×Ω↑↑(ω+1)}ω=ω{Ω2×ΩΩ↑↑ω+ω}ω (0)(1,1,1)(2,1)(3,2)(2,1)(3)
ω{Ω2×ΩΩ↑↑ω×2}ω (0)(1,1,1)(2,1)(3,2)(2,1)(3,2)
ω{Ω2×ΩΩ↑↑ω×ω}ω (0)(1,1,1)(2,1)(3,2)(3)
ω{Ω2×ΩΩΩ↑↑ω+1}ω (0)(1,1,1)(2,1)(3,2)(3,1)
ω{Ω2×Ω↑↑(ω+2)}ω (0)(1,1,1)(2,1)(3,2)(3,1)(4)
ω{Ω2×Ω↑↑ω2}ω (0)(1,1,1)(2,1)(3,2)(3,2)
ω{Ω2×Ω↑↑ω2}ω (0)(1,1,1)(2,1)(3,2)(4)
ω{Ω2×Ω↑↑Ω}ω (0)(1,1,1)(2,1)(3,2)(4,1)
ω{Ω2×Ω↑↑Ω↑↑ω}ω (0)(1,1,1)(2,1)(3,2)(4,1)(5,2)
ω{Ω2×Ω↑↑↑3}ω (0)(1,1,1)(2,1)(3,2)(4,1)(5,2)(6,1)
ω{Ω2×Ω↑↑↑ω}ω (0)(1,1,1)(2,1)(3,2)(4,2)
ω{Ω2×Ω{ω}ω}ω (0)(1,1,1)(2,1)(3,2)(4,2)(5)
ω{Ω2×Ω{Ω}Ω}ω (0)(1,1,1)(2,1)(3,2)(4,2)(5,1)
ω{Ω2×Ω{Ω2}ω}ω (0)(1,1,1)(2,1)(3,2)(4,2)(5,2)
ω{Ω2×Ω{Ω2↑↑ω}ω}ω (0)(1,1,1)(2,1)(3,2)(4,3)
ω{Ω2×Ω{Ω3}ω}ω (0)(1,1,1)(2,1)(3,2,1)
ω{Ω22}3=ω{Ω2×Ω{Ω3Ω2}ω}ω (0)(1,1,1)(2,1)(3,2,1)(4,2)
ω{Ω2×Ω{Ω3×Ω2↑↑ω}ω}ω (0)(1,1,1)(2,1)(3,2,1)(4,2)(5,3)
ω{Ω2×Ω{Ω3×Ω2{Ω4}ω}ω}ω (0)(1,1,1)(2,1)(3,2,1)(4,2)(5,3,1)
ω{Ω22}4=ω{Ω2×Ω{Ω3×Ω2{Ω4Ω3}ω}ω}ω (0)(1,1,1)(2,1)(3,2,1)(4,2)(5,3,1)(6,3)
ω{Ω22}ω (0)(1,1,1)(2,1,1)
ω{Ω23}ω (0)(1,1,1)(2,1,1)(2,1,1)
ω{Ω2ω}ω (0)(1,1,1)(2,1,1)(3)
ω{Ω2ω{Ω2}ω}ω (0)(1,1,1)(2,1,1)(3)(4,1,1)
ω{Ω2Ω}3 (0)(1,1,1)(2,1,1)(3)(4,1,1)(5,1,1)(6)
ω{Ω2Ω}ω (0)(1,1,1)(2,1,1)(3,1) = BIO
ω{Ω2Ω+Ω2}ω (0)(1,1,1)(2,1,1)(3,1)(1,1,1)
ω{Ω2Ω2}ω (0)(1,1,1)(2,1,1)(3,1)(1,1,1)(2,1,1)(3,1)
ω{Ω2Ωω}ω (0)(1,1,1)(2,1,1)(3,1)(2) = EBO

分析11

分析11:EBO~SSO
序数超运算 BMS
ω{Ω2Ωω}ω (0)(1,1,1)(2,1,1)(3,1)(2) = EBO
ω{Ω2ΩΩ}ω (0)(1,1,1)(2,1,1)(3,1)(2,1)
ω{Ω2Ω+1}ω (0)(1,1,1)(2,1,1)(3,1)(2,1,1)
ω{Ω2Ω+ω}ω (0)(1,1,1)(2,1,1)(3,1)(2,1,1)(3)
ω{Ω2Ω2}ω (0)(1,1,1)(2,1,1)(3,1)(2,1,1)(3,1)
ω{Ω2Ω2}ω (0)(1,1,1)(2,1,1)(3,1)(3,1)
ω{Ω2ΩΩ}ω (0)(1,1,1)(2,1,1)(3,1)(4,1)
ω{Ω2Ω↑↑ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2) = JO
ω{Ω2Ω↑↑ω+1}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(2,1,1)
ω{Ω2Ω↑↑ω×2}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(2,1,1)(3,1)(4,2)
ω{Ω2Ω↑↑ω×ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(3)
ω{Ω2ΩΩ↑↑ω+1}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(3,1)
ω{Ω2ΩΩ↑↑ω+1}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(3,1)
ω{Ω2Ω↑↑(ω+1)}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(3,1)(4)
ω{Ω2ΩΩ↑↑ω×2}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(3,1)(4,2)
ω{Ω2ΩΩ↑↑ω×ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(4)
ω{Ω2ΩΩΩ↑↑ω+1}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(4,1)
ω{Ω2Ω↑↑(ω+2)}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(4,1)(5)
ω{Ω2Ω↑↑ω2}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(4,2)
ω{Ω2Ω↑↑ω2}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(5)
ω{Ω2Ω↑↑Ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(5,1)
ω{Ω2Ω↑↑↑ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(5,2)
ω{Ω2Ω{ω}ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(5,2)(6)
ω{Ω2Ω{Ω}Ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(5,2)(6,1)
ω{Ω2Ω{Ω2}ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(5,2)(6,2)
ω{Ω2Ω{Ω2↑↑ω}ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2)(5,3)
ω{Ω2Ω{Ω3}ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2,1)
ω{Ω2Ω{Ω32}ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2,1)(5,2,1)
ω{Ω2Ω{Ω3Ω}ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2,1)(5,2,1)(6,1)
ω{Ω2Ω2}3=ω{Ω2Ω{Ω3Ω2}ω}ω (0)(1,1,1)(2,1,1)(3,1)(4,2,1)(5,2,1)(6,2)
ω{Ω2Ω2}ω (0)(1,1,1)(2,1,1)(3,1,1) = SIO
ω{Ω2Ω2+1}ω (0)(1,1,1)(2,1,1)(3,1,1)(2,1,1)
ω{Ω2Ω22}ω (0)(1,1,1)(2,1,1)(3,1,1)(2,1,1)(3,1,1)
ω{Ω2Ω2ω}ω (0)(1,1,1)(2,1,1)(3,1,1)(3) = MBO
ω{Ω2Ω2Ω}ω (0)(1,1,1)(2,1,1)(3,1,1)(3,1)
ω{Ω2Ω2Ωω}ω (0)(1,1,1)(2,1,1)(3,1,1)(3,1)(2) = TBO
ω{Ω2Ω2×Ω↑↑ω}ω (0)(1,1,1)(2,1,1)(3,1,1)(3,1)(4,2) = SRO
ω{Ω2Ω2×Ω{Ω3}ω}ω (0)(1,1,1)(2,1,1)(3,1,1)(3,1)(4,2,1)
ω{Ω2Ω22}ω (0)(1,1,1)(2,1,1)(3,1,1)(3,1,1) = SMO
ω{Ω2Ω23}ω (0)(1,1,1)(2,1,1)(3,1,1)(3,1,1)(3,1,1) = SNO
ω{Ω2Ω2ω}ω (0)(1,1,1)(2,1,1)(3,1,1)(4)
ω{Ω2Ω2Ω}ω (0)(1,1,1)(2,1,1)(3,1,1)(4,1)
ω{Ω2Ω2Ω↑↑ω}ω (0)(1,1,1)(2,1,1)(3,1,1)(4,1)(5,2) = RO
ω{Ω2Ω2Ω{Ω3}ω}ω (0)(1,1,1)(2,1,1)(3,1,1)(4,1)(5,2,1)
ω{Ω2↑↑3}ω (0)(1,1,1)(2,1,1)(3,1,1)(4,1,1) = SKO
ω{Ω2↑↑3|Ω↑↑ω}ω (0)(1,1,1)(2,1,1)(3,1,1)(4,1,1)(5,1)(6,2) = DO
ω{Ω2↑↑4}ω (0)(1,1,1)(2,1,1)(3,1,1)(4,1,1)(5,1,1)
ω{Ω2↑↑ω}ω (0)(1,1,1)(2,2) = SSO

预期强度

预期强度(已进行部分分析,待填写)
序数超运算 BMS
ω{Ω2↑↑ω}ω (0)(1,1,1)(2,2) = SSO
ω{Ω2↑↑ω+Ω2}ω (0)(1,1,1)(2,2)(1,1,1)
ω{Ω2↑↑ω×2}ω (0)(1,1,1)(2,2)(1,1,1)(2,2)
ω{Ω2↑↑ω×ω}ω (0)(1,1,1)(2,2)(2)
ω{Ω2↑↑ω×Ω}ω (0)(1,1,1)(2,2)(2,1)
ω{Ω2Ω2↑↑ω+1}ω (0)(1,1,1)(2,2)(2,1,1)
ω{Ω2↑↑(ω+1)}ω=ω{Ω2Ω2↑↑ω+ω}ω (0)(1,1,1)(2,2)(2,1,1)(3)
ω{Ω2Ω2↑↑ω×2}ω (0)(1,1,1)(2,2)(2,1,1)(3,2)
ω{Ω2↑↑ω2}ω (0)(1,1,1)(2,2)(2,2)
ω{Ω2↑↑ω2}ω (0)(1,1,1)(2,2)(3)
ω{Ω2↑↑Ω}ω (0)(1,1,1)(2,2)(3,1)
ω{Ω2↑↑Ω2}ω (0)(1,1,1)(2,2)(3,1,1)
ω{Ω2↑↑Ω2↑↑ω}ω (0)(1,1,1)(2,2)(3,1,1)(4,2)
ω{Ω2↑↑↑3}ω (0)(1,1,1)(2,2)(3,1,1)(4,2)(5,1,1)
ω{Ω2↑↑↑ω}ω (0)(1,1,1)(2,2)(3,2)
ω{Ω2↑↑↑Ω2}ω (0)(1,1,1)(2,2)(3,2)(3,1,1)
ω{Ω2{4}ω}ω (0)(1,1,1)(2,2)(3,2)(3,2)
ω{Ω2{ω}ω}ω (0)(1,1,1)(2,2)(3,2)(4)
ω{Ω2{Ω}Ω}ω (0)(1,1,1)(2,2)(3,2)(4,1)
ω{Ω2{Ω}Ω×ω}ω (0)(1,1,1)(2,2)(3,2)(4,1)(2) = LSO
ω{Ω2{Ω2}Ω2}ω (0)(1,1,1)(2,2)(3,2)(4,1,1) = APO
ω{Ω2{Ω3}ω}ω (0)(1,1,1)(2,2)(3,2)(4,2)
ω{Ω2{Ω3↑↑ω}ω}ω (0)(1,1,1)(2,2)(3,3)
ω{Ω2{Ω4}ω}ω (0)(1,1,1)(2,2)(3,3,1)
ω{Ω2{Ω4↑↑ω}ω}ω (0)(1,1,1)(2,2)(3,3,1)(4,4)
ω{Ω3}ω (0)(1,1,1)(2,2,1) = BGO
ω{Ω3+Ω2}ω (0)(1,1,1)(2,2,1)(1,1,1)
ω{Ω32}ω (0)(1,1,1)(2,2,1)(1,1,1)(2,2,1)
ω{Ω3ω}ω (0)(1,1,1)(2,2,1)(2)
ω{Ω32}ω (0)(1,1,1)(2,2,1)(2,1,1)(3,2,1)
ω{Ω3↑↑ω}ω (0)(1,1,1)(2,2,1)(2,2)
ω{Ω3{Ω4}ω}ω (0)(1,1,1)(2,2,1)(2,2)(3,2)(4,2)
ω{Ω3{Ω4↑↑ω}ω}ω (0)(1,1,1)(2,2,1)(2,2)(3,3)
ω{Ω3{Ω5}ω}ω (0)(1,1,1)(2,2,1)(2,2)(3,3,1)
ω{Ω3{Ω6}ω}ω (0)(1,1,1)(2,2,1)(2,2)(3,3,1)(4,4,1)
ω{Ω4}ω (0)(1,1,1)(2,2,1)(2,2,1)
ω{Ωω}ω (0)(1,1,1)(2,2,1)(3) = SDO
ω{Ωω↑↑ω}ω (0)(1,1,1)(2,2,1)(3)(2,2)
ω{Ωω+1}ω (0)(1,1,1)(2,2,1)(3)(2,2,1)
ω{ΩΩ}ω (0)(1,1,1)(2,2,1)(3,1)
ω{ΩΩ2}ω (0)(1,1,1)(2,2,1)(3,1,1)
ω{ΩΩω}ω (0)(1,1,1)(2,2,1)(3,1,1)(4,2,1)(5)
ω{Φ(1,0)}ω (0)(1,1,1)(2,2,1)(3,2) = LDO

预期强度(超下标运算)

预期强度(超下标运算)
序数超运算 BMS
ω{ω↓↓ω}ω=ω{Φ(1,0)}ω (0)(1,1,1)(2,2,1)(3,2) = LDO
ω{(ω↓↓ω)↑↑ω}ω=ω{Φ(1,0)↑↑ω}ω (0)(1,1,1)(2,2,1)(3,2)(2,2)
ω{Ωω↓↓ω+1}ω=ω{ΩΦ(1,0)+1}ω (0)(1,1,1)(2,2,1)(3,2)(2,2,1)
ω{ω↓↓(ω+1)}ω=ω{ΩΦ(1,0)+ω}ω (0)(1,1,1)(2,2,1)(3,2)(2,2,1)(3)
ω{ω↓↓ω2}ω=ω{Φ(1,1)}ω (0)(1,1,1)(2,2,1)(3,2)(2,2,1)(3,2)
ω{ω↓↓ω2}ω=ω{Φ(1,ω)}ω (0)(1,1,1)(2,2,1)(3,2)(3)
ω{ω↓↓Ω}ω=ω{Φ(1,Ω)}ω (0)(1,1,1)(2,2,1)(3,2)(3,1)
ω{ω↓↓Ω2}ω=ω{Φ(1,Ω2)}ω (0)(1,1,1)(2,2,1)(3,2)(3,1,1)
ω{ω↓↓↓3}ω=ω{Φ(1,Φ(1,0)}ω (0)(1,1,1)(2,2,1)(3,2)(3,1,1)(4,2,1)(5,2)
ω{ω↓↓↓ω}ω=ω{Φ(2,0)}ω (0)(1,1,1)(2,2,1)(3,2)(3,2)
ω{ω{{ω}}ω}ω=ω{Φ(ω,0)}ω (0)(1,1,1)(2,2,1)(3,2)(4)
ω{ω{{Ω}}ω}ω=ω{Φ(Ω,0)}ω (0)(1,1,1)(2,2,1)(3,2)(4,1)
ω{ω{{Ω2}}ω}ω=ω{Φ(Ω2,0)}ω (0)(1,1,1)(2,2,1)(3,2)(4,1,1)
ω{ω{{I}}3}ω=ω{Φ(Φ(ω,0),0)}ω (0)(1,1,1)(2,2,1)(3,2)(4,1,1)(5,2,1)(6,2)(7)
ω{ω{{I}}ω}ω=ω{Φ(1,0,0)}ω (0)(1,1,1)(2,2,1)(3,2)(4,2)
ω{ω{{I+1}}ω}ω=ω{Φ(1,1,0)}ω (0)(1,1,1)(2,2,1)(3,2)(4,2)(3,2)
ω{ω{{I2}}ω}ω=ω{Φ(2,0,0)}ω (0)(1,1,1)(2,2,1)(3,2)(4,2)(3,2)(4,2)
ω{ω{{I2}}ω}ω=ω{Φ(1,0,0,0)}ω (0)(1,1,1)(2,2,1)(3,2)(4,2)(4,2)
ω{ω{{Iω}}ω}ω=ω{ψI(IIω)}ω (0)(1,1,1)(2,2,1)(3,2)(4,2)(5)
ω{ω{{II}}ω}ω=ω{ψI(III)}ω (0)(1,1,1)(2,2,1)(3,2)(4,2)(5,2)
ω{ω{{I↑↑ω}}ω}ω=ω{ψI(ΩI+1)}ω (0)(1,1,1)(2,2,1)(3,2)(4,3)
ω{ω{{I↑↑↑ω}}ω}ω=ω{ψI(ΩI+12)}ω (0)(1,1,1)(2,2,1)(3,2)(4,3)(5,3)
ω{ω{{I{ΩI+1}ω}}ω}ω=ω{ψI(ΩI+1ΩI+1)}ω (0)(1,1,1)(2,2,1)(3,2)(4,3)(5,3)(6,3)
ω{ω{{I{ΩI+1↑↑ω}ω}}ω}ω=ω{ψI(ΩI+2)}ω (0)(1,1,1)(2,2,1)(3,2)(4,3)(5,4)
ω{ω{{I{ΩI+2}ω}}ω}ω=ω{ψI(ΩI+ω)}ω (0)(1,1,1)(2,2,1)(3,2)(4,3,1)
ω{ω{{I{ΩI+3}ω}}ω}ω (0)(1,1,1)(2,2,1)(3,2)(4,3,1)(5,4,1)
ω{ω{{I{ΩI+ω}ω}}ω}ω (0)(1,1,1)(2,2,1)(3,2)(4,3,1)(5,4,1)(6)
ω{ω{{I{ΩI2}ω}}ω}ω (0)(1,1,1)(2,2,1)(3,2)(4,3,1)(5,4,1)(6,2)
ω{ω{{I{ω↓↓(I+ω)}ω}}ω}ω (0)(1,1,1)(2,2,1)(3,2)(4,3,1)(5,4,1)(6,4)
ω{I}ω=ω{ω{{ X→I{ω{{X}}(I+ω)}ω fp }}ω}ω=ω{ψI(X)}ω ct. ψ(X) (0)(1,1,1)(2,2,1)(3,2,1)
psd.ω{X}ω ct. ψ(X) (0)(1,1,1)(2,2,2) = pfec LRO