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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Bowers' operators ： ウィキペディア英語版 Bowers' operators Let $a\backslash b$ = $H\_n(a,b)$, the hyperoperation. Invented by Jonathan Bowers, the first operator is $\backslash$ and it's defined: $m\backslash 1\; =\; m$ $m\backslash 2\; =\; m\backslash m$ $m\backslash 3\; =\; m\backslash m$ $m\backslash 4\; =\; m\backslash m\backslash \}m$ ⋮ The number inside the brackets can change. If it's two $m\backslash 1\; =\; m$ $m\backslash 2\; =\; m\backslash (m\backslash 1)$ $m\backslash 3\; =\; m\backslash (m\backslash 2)$ $m\backslash 4\; =\; m\backslash (m\backslash 3)$ ⋮ Operators beyond $\backslash$ can also be made, the rule of it is the same as hyperoperation: $m\backslash n\; =\; m\backslash (m\backslash (n-1))$ The next level of operators is $\backslash \backslash \}$, it to $\backslash$ behaves like $\backslash$ is to $\backslash$. For every fixed positive integer $q$, there is an operator $m\backslash ...\backslash \}\backslash \}n$ with $q$ sets of brackets. The domain of $(m,\; n,\; p)$ is $(\backslash mathbb^+)^3$, and the codomain of the operator is $\backslash mathbb^+$. Another function $\backslash$ means $m\backslash ...\backslash \}\backslash \}n$, where $q$ is the number of sets of brackets. It satisfies that $\backslash \; =\; \backslash$ for all integers $m\; \backslash ge\; 1$, $n\; \backslash ge\; 2$, $p\; \backslash ge\; 2$, and $q\; \backslash ge\; 1$. The domain of $(m,\; n,\; p,\; q)$ is $(\backslash mathbb^+)^4$, and the codomain of the operator is $\backslash mathbb^+$. Numbers like TREE(3) are unattainable with Bowers' operators, but Graham's number lies between $3\backslash 63$ and $3\backslash 64$. ==References== 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Bowers' operators」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.