165

## 定義

（原始定義： 發表於大数搜索吧於2014年8月18日）  

dim A(Infinity):dim B(Infinity):C=9 for D=0 to 9 for E=0 to C A(E)=E:B(E)=E next for F=C to 0 step -1 C=C*C if B(F)=0 then G=1 else G=0 for H=0 to F*G if A(F-H)<A(F) or A(F)=0 then I=H:H=F*G next for J=1 to C*G*I A(F)=A(F-I):B(F)=B(F-I):F=F+1 next G=1-G for K=1 to F*G if A(F-K)<A(F) and B(F-K)<B(F) then L=A(F)-A(F-K):M=K:K=F*G next for N=1 to C*G*M A(F)=A(F-M)+L:B(F)=B(F-M):F=F+1 next next next print C 

（更正後：作者本人於2018年5月26日更新，計算內容與原始版本不同）  

dim A[∞],B[∞]:C=9 for D=0 to 9 for E=0 to C A[E]=E:B[E]=E next for F=C to 0 step -1 C=C*C for G=0 to F if A[F]=0 | A[F-G]<A[F]-H then if B[F]=0 then I=G:G=F else H=A[F]-A[F-G] if B[F-G]<B[F] then I=G:G=F endif endif next for J=1 to C*I A[F]=A[F-I]+H:B[F]=B[F-I]:F=F+1 next H=0 next next print C 

 

#define A a[f] #define B b[f] #define M = malloc(9), #define W while ( main(f) { int *a M *b M c = 9, d = 10, h, i, k; W d--) { f = h = c + 1; W h--) a[h] = b[h] = h; W f--) { c *= c; h = f + 1; W h--) (a[h] < A && (b[h] < B || !B) || A + B == 0)? (k = B ? A - a[h]: 0, i = f - h, h = 0): 0; h = f + c * i; a = realloc(a, h); b = realloc(b, h); W f < h) A = a[f-i] + k, B = b[f-i], f++; } } return c; } 

## 數學的定義

\begin{eqnarray*} \mathrm{大数:}~K&=&\mathrm{Pair}^{10}(9)\\ \mathrm{大函数:}~\mathrm{Pair}(n)&=&\mathrm{expand}\left((0,0)(1,1)\cdots (n+1,n+1)[n]\right)\\ \mathrm{擴充規則:}~\mathrm{expand}([n])&=&n\\ \mathrm{expand}({\boldsymbol S}[n])&=&\left\{\begin{array}{ll} \mathrm{expand}((S_{00},S_{01})\cdots(S_{(X-2)0},S_{(X-2)1})[f(n)])&(\mathrm{if}~/forall yS_{(X-1)y}=0) \\ \mathrm{expand}({\boldsymbol G}{\boldsymbol B}^{(0)}{\boldsymbol B}^{(1)}{\boldsymbol B}^{(2)} \cdots {\boldsymbol B}^{(f(n))}[f(n)])&(\mathrm{otherwise}) \end{array}\right.\\ \mathrm{激活函数:}~f(n)&=&n^2\\ \mathrm{對數列:}~{\boldsymbol S}&=&(S_{00},S_{01})(S_{10},S_{11})\cdots (S_{(X-1)0},S_{(X-1)1})\\ \mathrm{好的部分:}~{\boldsymbol G}&=&(S_{00},S_{01})(S_{10},S_{11})\cdots (S_{(r-1)0},S_{(r-1)1})\\ \mathrm{壞的部分:}~{\boldsymbol B}^{(a)}&=&(B_{00}^{(a)},B_{01}^{(a)})(B_{10}^{(a)},B_{11}^{(a)})\cdots (B_{(X-2-r)0}^{(a)},B_{(X-2-r)1}^{(a)})\\ B_{x0}^{(a)}&=&\left\{\begin{array}{ll} S_{(r+x)0}+a(S_{(X-1)0}-S_{r0})&~(S_{(X-1)1}\gt 0)\\ S_{(r+x)0} &~(\mathrm{otherwise})\\ \end{array}\right.\\ B_{x1}^{(a)}&=&S_{(r+x)1}\\ \mathrm{Bad~root:}~r &=& \left\{\begin{array}{ll} P_1(X-1) & (S_{(X-1)1} \neq 0)\\ P_0(X-1) & (\mathrm{otherwise}) \end{array}\right.\\ S_{x1}~\mathrm{的親}:~P_1(x)&=&\max\{p|p\lt x \land S_{p1} \lt S_{x1} \land \exists a( p=(P_0)^a(x))\}\\ S_{x0}~\mathrm{的親}:~P_0(x)&=&\max\{p|p\lt x \land S_{p0} \lt S_{x0} \}\\ \end{eqnarray*}

## 相應的序數

### 最多 $$\epsilon_0$$

\begin{array}{ll} (0,0) &=& 1 \\ (0,0)(0,0) &=& 2 \\ (0,0)(0,0)(0,0) &=& 3 \\ (0,0)(1,0) &=& \omega \\ (0,0)(1,0)(0,0)(0,0) &=& \omega+2 \\ (0,0)(1,0)(0,0)(1,0) &=& \omega \cdot 2 \\ (0,0)(1,0)(1,0) &=& \omega^2 \\ (0,0)(1,0)(1,0)(0,0)(1,0) &=& \omega^2+\omega \\ (0,0)(1,0)(2,0) &=& \omega^\omega \\ (0,0)(1,0)(2,0)(3,0) &=& \omega^{\omega^\omega} \\ (0,0)(1,0)(2,0)(3,0)(4,0) &=& \omega^{\omega^{\omega^\omega}} \\ \end{array}

\begin{array}{ll} (0,0)(1,1)[1] &=& (0,0)(1,0)[1] \\ (0,0)(1,1)[2] &=& (0,0)(1,0)(2,0)[2] \\ (0,0)(1,1)[3] &=& (0,0)(1,0)(2,0)(3,0)[3] \\ (0,0)(1,1)[4] &=& (0,0)(1,0)(2,0)(3,0)(4,0)[4] \\ \end{array}

\begin{array}{ll} (0,0)(1,1) &=& \epsilon_0 \\ \end{array}

### 最多 $$\epsilon_1$$

\begin{array}{ll} (0,0)(1,1) &=& \epsilon_0 \\ (0,0)(1,1)(0,0)(1,1) &=& \epsilon_0 \cdot 2 \\ (0,0)(1,1)(0,0)(1,1)(0,0)(1,1) &=& \epsilon_0 \cdot 3 \\ (0,0)(1,1)(0,0)(1,1)(0,0)(1,1)(0,0)(1,1) &=& \epsilon_0 \cdot 4 \\ (0,0)(1,1)(0,0)(1,1)(0,0)(1,1)(0,0)(1,1)(0,0)(1,1) &=& \epsilon_0 \cdot 5 \\ \end{array}

\begin{array}{ll} (0,0)(1,1)(1,0) &=& \epsilon_0 \cdot \omega \\ (0,0)(1,1)(1,0)(0,0)(1,1)(1,0) &=& \epsilon_0 \cdot \omega \cdot 2 \\ (0,0)(1,1)(1,0)(0,0)(1,1)(1,0)(0,0)(1,1)(1,0) &=& \epsilon_0 \cdot \omega \cdot 3 \\ \end{array} 它變成了 $(0,0)(1,1)(1,0)(1,0) = \epsilon_0 \cdot \omega^2$ 因此，通過在數字序列的末尾添加(1,0)，序數乘以 $$\omega$$。 接下來，考慮在數字序列的末尾添加 (1,0)(2,0)。

$(0,0)(1,1)(1,0)(2,0)[4] = (0,0)(1,1)(1,0)(1,0)(1,0)(1,0)(1,0)[4]$

，這對應於將序數乘以 $$\omega^\omega$$，因為有一個基本序列，其中逐個添加(1,0),獲得

$(0,0)(1,1)(1,0)(2,0) = \epsilon_0 \cdot \omega^\omega$

。接下來，考慮(0,0)(1,1)(1,1)、 $(0,0)(1,1)(1,1)[4] = (0,0)(1,1)(1,0)(2,1)(2,0)(3,1)(3,0)(4,1)(4,0)(5,1)[4]$ ，獲得以下基本序列。

\begin{array}{ll} (0,0)(1,1) &=& \epsilon_0 \\ (0,0)(1,1)(1,0)(2,1) &=& \epsilon_0^2 \\ (0,0)(1,1)(1,0)(2,1)(2,0)(3,1) &=& \epsilon_0^{\epsilon_0} \\ (0,0)(1,1)(1,0)(2,1)(2,0)(3,1)(3,0)(4,1) &=& \epsilon_0^{\epsilon_0^2} \\ (0,0)(1,1)(1,0)(2,1)(2,0)(3,1)(3,0)(4,1)(4,0)(5,1) &=& \epsilon_0^{\epsilon_0^{\epsilon_0}} \\ \end{array} 因此，它變成了 $(0,0)(1,1)(1,1) = \epsilon_1 = \psi(1)$ 。

### 最多 巴克曼霍华德序數 $$\psi(\epsilon_{\Omega+1})$$

\begin{array}{ll} (0,0)(1,1)(2,1)(3,1)(4,1)(4,0) &=& \psi(\Omega^{\Omega^{\Omega \cdot \omega}}) \\ (0,0)(1,1)(2,1)(3,1)(4,1)(4,1) &=& \psi(\Omega^{\Omega^{\Omega^2}}) \\ (0,0)(1,1)(2,1)(3,1)(4,1)(5,0) &=& \psi(\Omega^{\Omega^{\Omega^\omega}}) \\ (0,0)(1,1)(2,1)(3,1)(4,1)(5,1) &=& \psi(\Omega^{\Omega^{\Omega^\Omega}}) \\ (0,0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,1) &=& ψ(\Omega^{\Omega^{\Omega^{\Omega^\Omega}}}) \\ (0,0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,1)(7,1) &=& ψ(\Omega^{\Omega^{\Omega^{\Omega^{\Omega^\Omega}}}}) \\ (0,0)(1,1)(2,2) &=& \psi(\epsilon_{\Omega+1}) = \psi(\psi_1(0)) \\ \end{array}

### 最多 $$\vartheta(\Omega_\omega)$$

\begin{array}{ll} (0,0)(1,1)(2,2)(0,0) &=& \psi(\psi_1(0))+1 \\ (0,0)(1,1)(2,2)(1,0) &=& \psi(\psi_1(0)) \omega \\ (0,0)(1,1)(2,2)(2,0) &=& \psi(\psi_1(0) \omega) \\ (0,0)(1,1)(2,2)(3,0) &=& \psi(\psi_1(\omega)) \\ (0,0)(1,1)(2,2)(3,0)(4,0) &=& \psi(\psi_1(\omega^\omega)) \\ (0,0)(1,1)(2,2)(3,0)(4,1) &=& \psi(\psi_1(\psi(0)))=\psi(\psi_1(\epsilon_0)) \\ (0,0)(1,1)(2,2)(3,1) &=& \psi(\psi_1(\Omega)) \\ (0,0)(1,1)(2,2)(3,2) &=& \psi(\psi_1(\Omega_2)) \\ (0,0)(1,1)(2,2)(3,3) &=& \psi(\psi_1(\psi_2(0))) \\ (0,0)(1,1)(2,2)(3,3)(4,4) &=& \psi(\psi_1(\psi_2(\psi_3(0)))) \\ (0,0)(1,1)(2,2)(3,3)...(9,9) &=& \psi(\psi_1(\psi_2(\psi_3(\psi_4(\psi_5(\psi_6(\psi_7(\psi_8(0))))))))) \\ \end{array}

## 終止證明

P進大好きbot[7]定義標準形式的概念（比普通意義上的正常形式更寬）， 而 Buchholz 的 $$\psi$$ 用於證明對序列系統的終止。[8]

## 九頭蛇表示法

Bashicu 顯示該對序列由標記的九頭蛇(hydra)表示。[9]

koteitan[10] 還將 Buchholz 的 hydra (也用標記的 hydra 代表)與一對標記的水印進行了比較。 [11]我還創建了一個程序來繪製帶有標籤的九頭蛇[12]