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+ω法序数超运算分析:修订间差异

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无编辑摘要
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!序数超运算
!序数超运算
!BMS
!BMS
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|<math>\omega\{\Omega_2\}(\omega+1)=\omega\{\Omega\}(\omega\{\Omega_2\}\omega+\omega)</math>
|(0)(1,1,1)(1,1)(2,1)(3,1)
|-
|<math>\omega\{\Omega_2\}(\omega+2)=\omega\{\Omega\{\Omega_2\}\omega\}(\omega\{\Omega_2\}\omega+\omega)</math>
|(0)(1,1,1)(1,1)(2,2)(3,2)(4,2)
|-
|-
|<math>\omega\{\Omega_2\}\omega2</math>
|<math>\omega\{\Omega_2\}\omega2</math>
|(0)(1,1,1)(1,1)(2,2,1)
|(0)(1,1,1)(1,1)(2,2,1)
|-
|<math>\omega\{\Omega_2\}\omega^2</math>
|(0)(1,1,1)(1,1)(2,2,1)(2)
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|<math>\omega\{\Omega_2+1\}\omega</math>
|<math>\omega\{\Omega_2+1\}\omega</math>
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|<math>\omega\{\Omega_22\}\omega</math>
|<math>\omega\{\Omega_22\}\omega</math>
|(0)(1,1,1)(1,1,1)
|(0)(1,1,1)(1,1,1)
|-
|<math>\omega\{\Omega_2\omega\}\omega</math>
|(0)(1,1,1)(2)
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|<math>\omega\{\Omega_2\Omega\}\omega</math>
|<math>\omega\{\Omega_2\Omega\}\omega</math>
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|<math>\omega\{\Omega_2^2\}\omega</math>
|<math>\omega\{\Omega_2^2\}\omega</math>
|(0)(1,1,1)(2,1,1)
|(0)(1,1,1)(2,1,1)
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|<math>\omega\{\Omega_2^\omega\}\omega</math>
|(0)(1,1,1)(2,1,1)(3)
|-
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|<math>\omega\{\Omega_2^\Omega\}\omega</math>
|<math>\omega\{\Omega_2^\Omega\}\omega</math>
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|<math>\omega\{\Omega_2^{\Omega_2}\}\omega</math>
|<math>\omega\{\Omega_2^{\Omega_2}\}\omega</math>
|(0)(1,1,1)(2,1,1)(3,1,1) = [[SIO]]
|(0)(1,1,1)(2,1,1)(3,1,1) = [[SIO]]
|-
|<math>\omega\{\Omega_2^{\Omega_2\times\Omega\uparrow\uparrow\omega}\}\omega</math>
|(0)(1,1,1)(2,1,1)(3,1,1)(3,1)(4,2) = [[SRO]]
|-
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|<math>\omega\{\Omega_2^{\Omega_2^2}\}\omega</math>
|<math>\omega\{\Omega_2^{\Omega_2^2}\}\omega</math>
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|<math>\omega\{\Omega_{\Omega_2}\}\omega</math>
|<math>\omega\{\Omega_{\Omega_2}\}\omega</math>
|(0)(1,1,1)(2,2,1)(3,1,1)
|(0)(1,1,1)(2,2,1)(3,1,1)
|-
|<math>\omega\{\Omega_{\Omega_\omega}\}\omega</math>
|(0)(1,1,1)(2,2,1)(3,1,1)(4,2,1)(5)
|-
|-
|<math>\omega\{\Phi(1,0)\}\omega</math>
|<math>\omega\{\Phi(1,0)\}\omega</math>

2026年4月26日 (日) 05:50的版本

+ω法序数超运算是一种超运算型记号,作者是量子杰克,与+1法超运算类似,但区别是遇到不动点会+ω而不是+1,原因是根据分析,这样能给出更整的基本列。序数超运算可以进行许多更高级别的拓展。

定义

一个序数超运算的格式必须是β{λ}α,其中β,λ,α都是正序数。

底数β只能是正整数或超限基数。若底数是超限基数,只能是Ωx,其中x是序数。定义:Ω0=ω,对于正序数x,Ωx为第x个不可数基数(不是可数非递归序数,因为在超运算型记号中,不可数与可数非递归的效果存在本质区别)。

若底数有限,则指数α必须有限。若底数为超限基数,则指数的势必须小于等于底数的势。例如,若底数为ω,指数必须可数。若底数为Ωx,指数必须小于Ωx+1

算符序数λ的势在目前版本中最多允许比底数多不超过I(首个不可达基数)。.

计算规则:

若指数为1,则值为底数。β{λ}1=β

若算符序数与指数均为后继序数,使用带跳不动点函数的简单迭代规则。β{λ+1}(α+1)=if(β{λ}β{λ+1}α=β{λ+1}α)?β{λ}j(β{λ+1}α) else β{λ}β{λ+1}α

跳跃函数j(x)取j(x)=x+ω,虽然可以取其他值,但根据分析,取x+ω能使基本列更整。

若指数为极限序数,值为对指数取基本列时得到的值的极限。β{λ}α=lim(β{λ}α[x])。共尾性Ωx的极限序数的基本列长度为Ωx

若指数为后继序数,算符序数为共尾性小于等于底数的极限序数:对于β=Ωx, cof(λ)=Ωy, yx,将指数分解为一个Ωy的倍数与一个小于Ωy的序数之和,α=Ωy×c+d。若α<Ωy即c=0,则β{λ}α=β{λ}d=β{λ[d]}β。若α>Ωy即c>0,β{λ}α=β{λ}(Ωy×c+d)=β{λ[d]}j(β{λ}(Ωy×c))

若指数为后继序数,算符序数为共尾性为底数的下一个超限基数:对于β=Ωx, cof(λ)=Ωx+1,展开为:β{λ}(α+1)=if(β{λ[β{λ}α]}β=β{λ}α)?β{λ[j(β{λ}α)]}β else β{λ[β{λ}α]}β

若指数是后继序数,算符序数的共尾性大于底数的下一个超限基数,底数必须是超限基数(不能为有限数)。对于β=Ωx, cof(λ)=Ωx+y+1,y为正序数,将指数分为ω的倍数与自然数之和:α=ω×c+d

定义:λ1=λλn(n为大于1的正整数)= 将λn1的表达式中所有x>0的Ωx全部替换为Ωx+y。若指数有限,β{λ}α = Ω_x{λ[Ω_(x+y+1)]}d;β{λ}1 = Ω_x{λ[Ω_(x+y)]}ω;β{λ}(n+1) = 将β{λ}n表达式中的λ_n[Ω_(x+y*n)]替换为λ_n[Ω_(x+y*n)]{λ_(n+1)[Ω_(x+y*(n+1))]}ω。若指数为超限后继序数,β{λ}α = Ω_x{λ[Ω_(x+y+1)]}(ω*c+d);β{λ}(ω*c+1) = Ω_x{λ[Ω_(x+y)]}j(β{λ}(ω*c));β{λ}(ω*c+n+1) = 将β{λ}n表达式中的λ_n[Ω_(x+y*n)]替换为λ_n[Ω_(x+y*n)]{λ_(n+1)[Ω_(x+y*(n+1))]}ω;若指数为极限序数,β{λ}α = lim(β{λ}(α[n]))。

分析1

分析1:SCO~HCO
序数超运算 BMS
ωω (0)(1,1) = SCO
ω(ω+1)=ωωω+ω (0)(1,1)(1)(2)
ω(ω+2)=ωωωω+ω (0)(1,1)(1)(2,1)(2)(3)
ω(ω+3) (0)(1,1)(1)(2,1)(2)(3,1)(3)(4)
ωω2 (0)(1,1)(1,1)
ω(ω2+1)=ωωω2+ω (0)(1,1)(1,1)(1)(2)
ωω3 (0)(1,1)(1,1)(1,1)
ωω2 (0)(1,1)(2)
ω(ω2+ω) (0)(1,1)(2)(1,1)
ωω22 (0)(1,1)(2)(1,1)(2)
ωω3 (0)(1,1)(2)(2)
ωωω (0)(1,1)(2)(3)
ω3=ωωω (0)(1,1)(2)(3,1)
ω4 (0)(1,1)(2)(3,1)(4)(5,1)
ωω (0)(1,1)(2,1) = CO
ω(ω+1)=ω(ωω+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)
ωω2 (0)(1,1)(2,1)(1,1)(2,1)
ωω2 (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{ωω}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{Ωω}3=ω{ΩΩΩ}ω (0)(1,1)(2,1)(3,1)(4,1)(5,1)
ω{Ωω}4=ω{Ω4}ω (0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,1)
ω{Ωω}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{Ωω}ω (0)(1,1)(2,2) = BHO
ω{Ωω}(ω+1)=ω{Ω}(ω{Ωω}ω+ω) (0)(1,1)(2,2)(1,1)(2,1)(3,1)
ω{Ωω}(ω+1)=ω{ΩΩ}(ω{Ωω}ω+ω) (0)(1,1)(2,2)(1,1)(2,1)(3,1)(4,1)
ω{Ωω}ω2 (0)(1,1)(2,2)(1,1)(2,2)
ω{Ωω}ω2 (0)(1,1)(2,2)(2)
ω{Ωω+1}3 (0)(1,1)(2,2)(2)(3,1)(4,2)
ω{Ωω+1}ω (0)(1,1)(2,2)(2,1)
ω{Ωω+Ω}ω (0)(1,1)(2,2)(2,1)(3,1)
ω{Ωω×2}ω (0)(1,1)(2,2)(2,1)(3,2)
ω{Ωω×ω}ω (0)(1,1)(2,2)(2,1)(3,2)(3)
ω{Ωω×Ω}ω (0)(1,1)(2,2)(2,1)(3,2)(3,1)
ω{Ω(ω+1)}ω=ω{ΩΩω+ω}ω (0)(1,1)(2,2)(2,1)(3,2)(3,1)(4)
ω{Ωω2}ω (0)(1,1)(2,2)(2,2)
ω{Ωω3}ω (0)(1,1)(2,2)(2,2)(2,2)
ω{Ωω2}ω (0)(1,1)(2,2)(3)
ω{ΩΩ}3=ω{Ωω{ΩΩ}ω}ω (0)(1,1)(2,2)(3)(4,1)(5,2)(6)
ω{ΩΩ}ω (0)(1,1)(2,2)(3,1)
ω{Ω(Ω+ω)}ω (0)(1,1)(2,2)(3,1)(2,2)
ω{ΩΩ2}ω (0)(1,1)(2,2)(3,1)(2,2)(3,1)
ω{ΩΩ2}ω (0)(1,1)(2,2)(3,1)(3,1)
ω{ΩΩω}ω (0)(1,1)(2,2)(3,1)(4,2)
ω{Ω3}ω (0)(1,1)(2,2)(3,1)(4,2)(5,1)
ω{Ωω}ω (0)(1,1)(2,2)(3,2)
ω{Ω(ω+1)}ω=ω{Ω(Ωω+ω)}ω (0)(1,1)(2,2)(3,2)(2,2)
ω{Ωω2}ω (0)(1,1)(2,2)(3,2)(2,2)(3,2)
ω{Ωω2}ω (0)(1,1)(2,2)(3,2)(3)
ω{ΩΩ}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{ωω}ω}ω (0)(1,1)(2,2)(3,2)(4)(5,1)
ω{Ω{ω{Ωω}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{Ωω}ω}ω (0)(1,1)(2,2)(3,2)(4,1)(5,2)
ω{Ω{ΩΩ}Ω}ω (0)(1,1)(2,2)(3,2)(4,1)(5,2)(6,1)
ω{Ω{Ωω}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{Ω23}ω}ω (0)(1,1)(2,2)(3,2)(4,2)(5,2)(6,2)
ω{Ω{Ω2ω}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{Ω2ω}ω}ω (0)(1,1)(2,2)(3,3)
ω{Ω{Ω2ω}ω+1}ω (0)(1,1)(2,2)(3,3)(2,1)
ω{Ω{Ω2ω}ω×2}ω (0)(1,1)(2,2)(3,3)(2,1)(3,2)(4,3)
ω{Ω(Ω{Ω2ω+ω)}ω}ω (0)(1,1)(2,2)(3,3)(2,2)
ω{Ω(Ω{Ω2ω+ω)}ω}ω (0)(1,1)(2,2)(3,3)(2,2)(3,2)
ω{Ω{Ω2ω}(ω+1)}ω=ω{Ω{Ω2}(Ω{Ω2ω+ω)}ω}ω (0)(1,1)(2,2)(3,3)(2,2)(3,2)(4,2)
ω{Ω{Ω2ω}ω2}ω (0)(1,1)(2,2)(3,3)(2,2)(3,3)
ω{Ω{Ω2ω}ω2}ω (0)(1,1)(2,2)(3,3)(3)
ω{Ω{Ω2ω}Ω}ω (0)(1,1)(2,2)(3,3)(3,1)
ω{Ω{Ω2ω+1}ω}ω (0)(1,1)(2,2)(3,3)(3,2)
ω{Ω{Ω2ω+ω}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4)
ω{Ω{Ω2ω+Ω}Ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,1)
ω{Ω{Ω2ω+Ω2}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,2)
ω{Ω{Ω2ω×2}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,3)
ω{Ω{Ω2ω×ω}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,3)(5)
ω{Ω{Ω2ω×Ω}Ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,3)(5,1)
ω{Ω{Ω2Ω2ω+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{Ω2Ω2ω+ω}ω}ω (0)(1,1)(2,2)(3,3)(3,2)(4,3)(5,2)(6)
ω{Ω{Ω2ω2}ω}ω (0)(1,1)(2,2)(3,3)(3,3)
ω{Ω{Ω2ω2}ω}ω (0)(1,1)(2,2)(3,3)(4)
ω{Ω{Ω2Ω}Ω}ω (0)(1,1)(2,2)(3,3)(4,1)
ω{Ω{Ω2Ω2}ω}ω (0)(1,1)(2,2)(3,3)(4,2)
ω{Ω{Ω23}ω}ω (0)(1,1)(2,2)(3,3)(4,2)(5,3)(6,2)
ω{Ω{Ω2ω}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{Ω3ω}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,4)
ω{Ω{Ω2{Ω3ω}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{Ω4ω}ω}ω}ω}ω (0)(1,1)(2,2)(3,3)(4,4)(5,5)
ω{Ω2}ω (0)(1,1,1) = BO

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+1)=ω{Ω}(ω{Ω2}ω+ω) (0)(1,1,1)(1,1)(2,1)(3,1)
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}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)
ω{Ω2+Ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1)(3,1)
ω{Ω2+Ωω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)
ω{Ω2+Ω{Ω3}ω}ω (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3,1)
ω{Ω22}ω (0)(1,1,1)(1,1,1)
ω{Ω2ω}ω (0)(1,1,1)(2)
ω{Ω2Ω}ω (0)(1,1,1)(2,1)
ω{Ω2×(Ω+1)}ω (0)(1,1,1)(2,1)(1,1,1)
ω{Ω2×Ωω}ω (0)(1,1,1)(2,1)(3,2) = TFBO
ω{Ω2×Ω{Ω3}ω}ω (0)(1,1,1)(2,1)(3,2,1)
ω{Ω22}ω (0)(1,1,1)(2,1,1)
ω{Ω2ω}ω (0)(1,1,1)(2,1,1)(3)
ω{Ω2Ω}ω (0)(1,1,1)(2,1,1)(3,1) = BIO
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)(4,2) = JO
ω{Ω2Ω2}ω (0)(1,1,1)(2,1,1)(3,1,1) = SIO
ω{Ω2Ω2×Ωω}ω (0)(1,1,1)(2,1,1)(3,1,1)(3,1)(4,2) = SRO
ω{Ω2Ω22}ω (0)(1,1,1)(2,1,1)(3,1,1)(3,1,1) = SMO
ω{Ω2Ω2Ωω}ω (0)(1,1,1)(2,1,1)(3,1,1)(4,1)(5,2) = RO
ω{Ω23}ω (0)(1,1,1)(2,1,1)(3,1,1)(4,1,1) = SKO
ω{Ω2ω}ω (0)(1,1,1)(2,2) = SSO
ω{Ω2ω}ω (0)(1,1,1)(2,2)(3,2)
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{Ω3ω}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{Ω3}ω (0)(1,1,1)(2,2,1) = BGO
ω{Ω3ω}ω (0)(1,1,1)(2,2,1)(2,2)
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 (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
ω{ΩΦ(1,0)+1}ω (0)(1,1,1)(2,2,1)(3,2)(2,2,1)
ω{Φ(1,1)}ω (0)(1,1,1)(2,2,1)(3,2)(3,2)
ω{Φ(2,0)}ω (0)(1,1,1)(2,2,1)(3,2)(4,2)
ω{Φ(1,0,0)}ω (0)(1,1,1)(2,2,1)(3,2)(4,2)(5,2)
ω{ψI(ΩI+1)}ω (0)(1,1,1)(2,2,1)(3,2)(4,3)
ω{ψI(ΩI+ω)}ω (0)(1,1,1)(2,2,1)(3,2)(4,3,1)
psd.ω{I}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