Hn(a,b)H_n(a,b)Hn​(a,b)递归定义如下Hn(a,b){b1n0,an1, b0,1n≥2, b0,Hn−1(a, Hn(a,b−1))n≥1, b≥1. H_n(a, b) \begin{cases} b 1 n 0, \\ a n 1,\; b 0, \\ 1 n \ge 2,\; b 0, \\ H_{n-1}\big(a,\;H_n(a, b-1)\big) n \ge 1,\; b \ge 1. \end{cases}Hn​(a,b)⎩⎨⎧​b1a1Hn−1​(a,Hn​(a,b−1))​n0,n1,b0,n≥2,b0,n≥1,b≥1.​例子H0(a,b)b1(后继运算) H_0(a, b) b 1 \quad \text{(后继运算)}H0​(a,b)b1(后继运算)H1(a,b)ab(加法) H_1(a, b) a b \quad \text{(加法)}H1​(a,b)ab(加法)H2(a,b)a×b(乘法) H_2(a, b) a \times b \quad \text{(乘法)}H2​(a,b)a×b(乘法)H3(a,b)ab(幂运算) H_3(a, b) a^b \quad \text{(幂运算)}H3​(a,b)ab(幂运算)H4(a,b)a↑↑b(四运算) H_4(a, b) a \uparrow\uparrow b \quad \text{(四运算)}H4​(a,b)a↑↑b(四运算)Hn(a,b)a↑↑⋯↑↑a(超运算) H_n(a, b) a \uparrow\uparrow \dots \uparrow\uparrow a \quad \text{(超运算)}Hn​(a,b)a↑↑⋯↑↑a(超运算)用n-1层定义第n层第 n 层 对第 n-1 层做“b 次右结合折叠”写成结构就是a∘a∘a∘⋯∘a(共 b 个 a) a \circ a \circ a \circ \dots \circ a \quad (\text{共 } b \text{ 个 } a)a∘a∘a∘⋯∘a(共b个a)满足性质∘\circ∘是第 n-1 层运算右结合a,b 是两个符号n是自然数H(n,a,b):ifn1:returnabifb1:returnareturnH(n-1,a,H(n,a,b-1))print(H(3,2,8))层级 n类型运算名称表达式统一形式Hn(a,b)H_n(a,b)Hn​(a,b)解 b右逆依据方程解 a左逆本质0正向后继a1a1a1H0(a,b)b1H_0(a,b)b1H0​(a,b)b1解H0(a,b)c⇒bc−1H_0(a,b)c \Rightarrow bc-1H0​(a,b)c⇒bc−1—构造1正向加法abababH1(a,b)H_1(a,b)H1​(a,b)解abc⇒bc−aabc \Rightarrow bc-aabc⇒bc−a减法—平移1扩展减法a−ba-ba−bH1(a,−b)H_1(a,-b)H1​(a,−b)—已是逆—平移1扩展实数加法a0.5a0.5a0.5H1(a,0.5)H_1(a,0.5)H1​(a,0.5)——连续平移2正向乘法a×ba \times ba×bH2(a,b)H_2(a,b)H2​(a,b)解abc⇒bc/aabc \Rightarrow bc/aabc⇒bc/a除法—缩放2扩展除法a/ba/ba/bH2(a,1/b)H_2(a,1/b)H2​(a,1/b)——缩放2扩展分数乘a×0.5a \times 0.5a×0.5H2(a,0.5)H_2(a,0.5)H2​(a,0.5)——缩放2扩展负乘−ab-ab−abH2(a,−b)H_2(a,-b)H2​(a,−b)——方向翻转3正向幂aba^babH3(a,b)H_3(a,b)H3​(a,b)解abc⇒blog⁡aca^bc \Rightarrow b\log_a cabc⇒bloga​c对数解ac1/bac^{1/b}ac1/b开方指数3扩展负指数a−ba^{-b}a−bH3(a,−b)H_3(a,-b)H3​(a,−b)——倒数3扩展分数指数a1/ba^{1/b}a1/bH3(a,1/b)H_3(a,1/b)H3​(a,1/b)——开方3逆向对数log⁡ac\log_a cloga​c—定义解abca^bcabc—指数反演3逆向开方cb\sqrt[b]{c}bc​——定义解abca^bcabc指数反演4正向四运算a↑↑ba \uparrow\uparrow ba↑↑bH4(a,b)H_4(a,b)H4​(a,b)解a↑↑bca \uparrow\uparrow b ca↑↑bc定义 super-log解super-root迭代4逆向超对数super-log—定义解a↑↑bca \uparrow\uparrow b ca↑↑bc—迭代反演4逆向超根super-root——定义解a↑↑bca \uparrow\uparrow b ca↑↑bc迭代反演5正向五运算a↑↑↑ba \uparrow\uparrow\uparrow ba↑↑↑bH5(a,b)H_5(a,b)H5​(a,b)解H5(a,b)cH_5(a,b)cH5​(a,b)c高阶反函数解左逆超递归…………Hn(a,b)H_n(a,b)Hn​(a,b)解Hn(a,b)cH_n(a,b)cHn​(a,b)c解Hn(a,b)cH_n(a,b)cHn​(a,b)c无限层级