LPrSSψ分析:修订间差异
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2026年4月26日 (日) 08:17的版本
以下为长初等序列序数折叠函数(LPrSSψ)的分析。LPrSSψ与BOCF在BHO前完全一致,但强度大得多,可达(0)(1,1,1)(2,2,1)(3)以上。
分析1:BHO~BO
此阶段分析很简单,仅需将每一层从对应的ψ拆出来即可。
| LPrSSψ | BOCF | BMS |
|---|---|---|
| (0)(1,1)(2,2) | ||
| (0)(1,1)(2,2)(1) | ||
| (0)(1,1)(2,2)(1,1) | ||
| (0)(1,1)(2,2)(1,1)(2,2) | ||
| (0)(1,1)(2,2)(1,1)(2,2)(1,1)(2,2) | ||
| (0)(1,1)(2,2)(2) | ||
| (0)(1,1)(2,2)(2)(3) | ||
| (0)(1,1)(2,2)(2,1) | ||
| (0)(1,1)(2,2)(2,1)(3,2) | ||
| (0)(1,1)(2,2)(2,1)(3,2)(3,1)(4,2) | ||
| (0)(1,1)(2,2)(2,2) | ||
| (0)(1,1)(2,2)(2,2)(1,1)(2,2)(2,2) | ||
| (0)(1,1)(2,2)(2,2)(2) | ||
| (0)(1,1)(2,2)(2,2)(2,2) | ||
| (0)(1,1)(2,2)(3) | ||
| (0)(1,1)(2,2)(3,1) | ||
| (0)(1,1)(2,2)(3,1)(4,2) | ||
| (0)(1,1)(2,2)(3,2) | ||
| (0)(1,1)(2,2)(3,2)(1,1)(2,2)(3,2) | ||
| (0)(1,1)(2,2)(3,2)(2,2) | ||
| (0)(1,1)(2,2)(3,2)(2,2)(3,2) | ||
| (0)(1,1)(2,2)(3,2)(3,2) | ||
| (0)(1,1)(2,2)(3,2)(4) | ||
| (0)(1,1)(2,2)(3,2)(4,1) | ||
| (0)(1,1)(2,2)(3,2)(4,1)(5,2) | ||
| (0)(1,1)(2,2)(3,2)(4,2) | ||
| (0)(1,1)(2,2)(3,2)(4,2)(5,2) | ||
| (0)(1,1)(2,2)(3,3) | ||
| (0)(1,1)(2,2)(3,3)(1,1)(2,2)(3,3) | ||
| (0)(1,1)(2,2)(3,3)(2) | ||
| (0)(1,1)(2,2)(3,3)(2,2) | ||
| (0)(1,1)(2,2)(3,3)(2,2)(3,3) | ||
| (0)(1,1)(2,2)(3,3)(3) | ||
| (0)(1,1)(2,2)(3,3)(3,2) | ||
| (0)(1,1)(2,2)(3,3)(3,2)(4,3) | ||
| (0)(1,1)(2,2)(3,3)(3,3) | ||
| (0)(1,1)(2,2)(3,3)(4) | ||
| (0)(1,1)(2,2)(3,3)(4,3) | ||
| (0)(1,1)(2,2)(3,3)(4,3)(5,3) | ||
| (0)(1,1)(2,2)(3,3)(4,4) | ||
| (0)(1,1)(2,2)(3,3)(4,4)(5,5) | ||
| (0)(1,1,1) |
分析2:BO~(0)(1,1,1)(2,1)
到(0)(1,1,1)(2,1)的分析仍较简单,仅需找层即可,但此后,LPrSSψ的行为更类似BMS,而不是BOCF,因为具有(0)(1,1,1)(2,1)(1,1,1)提升。
| LPrSSψ | BOCF | BMS |
|---|---|---|
| (0)(1,1,1) | ||
| (0)(1,1,1)(1) | ||
| (0)(1,1,1)(1,1) | ||
| (0)(1,1,1)(1,1)(2,2) | ||
| (0)(1,1,1)(1,1)(2,2)(3,3) | ||
| (0)(1,1,1)(1,1)(2,2,1) | ||
| (0)(1,1,1)(1,1)(2,2,1)(1,1)(2,2,1) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2,1) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2,1)(3,2,1) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2,2) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2,2)(2,2) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,2) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3,1) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3,1)(3,3) | ||
| (0)(1,1,1)(1,1)(2,2,1)(2,2)(3,3,1)(3,3)(4,4,1) | ||
| (0)(1,1,1)(1,1,1) | ||
| (0)(1,1,1)(1,1,1)(1,1)(2,2,1) | ||
| (0)(1,1,1)(1,1,1)(1,1)(2,2,1)(2,2,1) | ||
| (0)(1,1,1)(1,1,1)(1,1)(2,2,1)(2,2,1)(2,2) | ||
| (0)(1,1,1)(1,1,1)(1,1,1) | ||
| (0)(1,1,1)(2) | ||
| (0)(1,1,1)(2)(1,1,1) | ||
| (0)(1,1,1)(2)(3,1) | ||
| (0)(1,1,1)(2)(3,1,1) | ||
| (0)(1,1,1)(2,1) |
分析3:(0)(1,1,1)(2,1)~(0)(1,1,1)(2,1)(1,1,1)
分析进入提升阶段。可见LPrSSψ中的一个Ω_2对应的不是BOCF中的一个Ω_ω,而是BMS中的一个(1,1,1)。这里注意LPrSSψ一般不单独定义Ω_0,展开方式,Ω_2展开出ψ_1时,把外面的Ω变成了里面的Ω_2,以此类推,但ω不提升。因为阶差为d时,Ω_(n+1)变出的ψ_n只把外面的x>=n的Ω_x提升为Ω_(x+d)。不然,若定义Ω_0=ω,并在这种位置将其提升为Ω,会与没有【特别地,如果坏根中元素不是坏部中某项的祖先项,则该项在复制过程中将保持不变。】规则的BMS一样,发生无穷降链。此外,若无此种提升,可能可以得到类似IBMS的ILPrSSψ。
| LPrSSψ | BOCF | BMS |
|---|---|---|
| (0)(1,1,1)(2,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,1)(3,2,1)(4,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2)(3,3,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2)(3,3,1)(4,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2)(3,3,1)(4,1)(3,3) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2,1)(2,2,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2,1)(3) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(2,2,1)(3,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(3) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(3,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,2) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,2,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,1)(4,2,1)(5,1)(6,2,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3,1)(4,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3,1)(4,2) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3,1)(4,2)(3,3) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3,1)(4,2)(3,3,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3,1)(4,2)(3,3,1)(4,2) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3,1)(4,2)(5,3,1) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3,1)(4,3) | ||
| (0)(1,1,1)(2,1)(1,1)(2,2,1)(3,2)(2,2)(3,3,1)(4,3)(3,3)(4,4,1)(5,4) | ||
| (0)(1,1,1)(2,1)(1,1,1) |
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