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葛立恒数吧 关注：1,260贴子：58,323
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萌新迭代葛立恒数挑战TREE(3)

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最近了解了葛立恒数，感觉很厉害，如果有G64个粒子,每1/G64秒随机排列一次并随机出现G64个颜色中的其中一个，那么连续排列G64年后每次排列完全一样的概率的倒数，和TREE(3)比怎么样
假的一楼钓鱼，正文发楼下
IP属地:北京1楼2023-08-23 16:54回复
    G(a,b)=GGG...b...GGG(a)
    末项为0时去掉
    G(a,b,c)=G(G(a,b-1,c),b,c-1)
    ......
    相当于BEAF的线性数阵
    G(a,b,[1<1>1])=G(a,a,a,a,a...b次...a)
    G(a,b,c,1[1<1>1])=G(a,G(a,b-1,c,1[1<1>1]1),c-1,1[1<1>1]1)
    相当于BEAF的{n,n,...n(1)2}
    ......
    G(a,b,[1<1>1],n+)=G(a,a,a,a,a...b次...a,1[1<1>1]1,n)
    G(a,b,c,[1<1>1],n)(c)=G(a,G(a,b-1,c,1[1<1>1]1,n),c-1,1[1<1>1]1,n)
    相当于BEAF的{n,n,...n(1)n}
    G(a,b,[1<1>1],1,2)=G(a,1[1<1>1]1,G(a,b-1,1[1<1>1]1,1,2))
    G(a,1,[1<1>1],1,2)=a
    G(a,b,[1<1>1],n+,c)=G(a,a,...b次...a,a,1[1<1>1]1,n,c)
    G(a,b,[1<1>1],1,c+)=G(a,1[1<1>1]1,G(a,b-1,1[1<1>1]1,1,c))
    相当于BEAF的{n,n(1)n,n,...n}
    ......
    G(a,b,[1<1>1],c,d,[1<1>1],e)=G(a,[1<1>1],c,c,c,c...d次...,c,c[1<1>1],e-1)
    G(a,b,[1<1>1],c,d,e,[1<1>1],f)=G(a,b,[1<1>1],c,G(a,b,[1<1>1],c,d-1,e,[1<1>1],f),e-1,[1<1>1],f)
    G(a,b,[1<1>1],c,d,[1<1>1],e,f)=G(a,b,[1<1>1],c,d,[1<1>1],G(a,b,[1<1>1],c,d,[1<1>1],e-1,f),f-1)
    ......
    G(a,b,[2<1>1],1)=G(a,a,[1<1>1],a,a,[1<1>1],...b次...a,a,[1<1>1],a,a)
    G(a,b,[2<1>1],c)=G(a,a,a,...b次...a[2<1>1],c-1)
    G(a,b,c,[2<1>1],d)=G(a,G(a,b-1,c,[2<1>1],d),c-1,[2<1>1],d)
    G(a,1,#)=a，#代表任意数阵部分
    G(a,b,[2<1>1],c,d,[1<1>1],n)=G(a,b,[2<1>1],c,c,c,...b次...c,[1<1>1],n-1)
    ......
    G(a,b,[3<1>1],1)=G(a,a,[2<1>1],a,a,[2<1>1],...b次...a,a,[2<1>1],a,a)
    G(a,b,[n+1<1>1],1)=G(a,a,[n<1>1],a,a,[n<1>1],...b次...a,a,[n<1>1],a,a)
    相当于BEAF的{n,n(n)n}
    G(a,b,[1,2<1>1],1)=G(a,b,[G(a,b-1,[1,2<1>1],1)<1>1],1)
    接下来规则不变
    G(a,b,[1,2<1>1][1<1>1],1)=G(a,a,[1,2<1>1],a,a,[1,2<1>1],...b次...a,a,[1,2<1>1],a,a)
    G(a,b,[1,2<1>1][2<1>1],1)=G(a,a,[1,2<1>1][1<1>1],a,a,[1,2<1>1][1<1>1],...b次...a,a,[1,2<1>1][1<1>1],a,a)
    G(a,[1,2<1>1][1,2<1>1],1)(b)=G(a,b,[1,2<1>1][G(a,b-1,[1,2<1>1][1,2<1>1],1)<1>1],1)
    ......
    G(a,b,[n+,2<1>1],1)=G(a,a,[n,2<1>1][n,2<1>1]...b次...[n,2<1>1],1)
    G(a,b,[1,n<1>1],1)=G(a,b,[1,n<1>1],1)=G(a,b,[G(a,b-1,[1,n<1>1],1)<1>1],1)
    G(a,b,[n,n,n<1>1],1)=G(a,b,[n,G(a,b,[n,n-1,n<1>1],1),n-1<1>1],1)
    相当于BEAF的{n,n(n,n,n,n,n,...n,n)n}
    ......
    G(a,b,[n,[1<1>1],1<1>1],1)=G(a,b,[n,n,n,...b次...n<1>1],1)
    G(a,b,[n,[1<1>1],2<1>1],1)=G(a,b,[n,n,n,...b次...n,n[1<1>1],1<1>1],1)
    G(a,b,[1,d,[1<1>1],2<1>1],1)=G(a,b,[G(a,b-1,[1,d,[1<1>1],2<1>1],1),d,[1<1>1],2<1>1],1)
    ......
    懒得写了，套的BEAF皮，基础部分极限G(n,n,[n,[n,[...[n,[n<1>1],n<1>1]...,n<1>1],n<1>1],n<1>1],n)
    psi(0)
    (a,b,[1<1>2],c)=(a,b,[a,[a,[...[a<1>1]b次...,b<1>1],b<1>1],b<1>1],c)
    老套路
    后面进位规则是一样的,接下来我只讲比较新的部分
    (a,b,[1<1>2][1<1>1],1)=(a,a,[1<1>2],a,a,...b次...[1<1>2],a,a)
    (a,b,[1<1>2][2<1>1],1)=(a,a,[1<1>2][1<1>1],a,a,...b次...[1<1>2][1<1>1],a,a)
    (a,b,[1<1>2][3<1>1],1)
    (a,b,[1<1>2][4<1>1],1)
    (a,b,[1<1>2][1,2<1>1],1)
    (a,b,[1<1>2][1,1,2<1>1],1)
    (a,b,[1<1>2][1,1,2<1>1],1)
    (a,b,[1<1>2][2,[2<1>1],2<1>1],1)
    (a,b,[1<1>2][1<1>2],1)
    ......
    (a,b,[1,[1<1>2]1<1>1],1)=(a,b,[1<1>2][1<1>2][1<1>2]...b次...[1<1>2][1<1>2],1)
    (a,b,[n+,[1<1>2]1<1>1],1)=(a,b,[n,[1<1>2]<1>1][n,[1<1>2]<1>1]...b次...[n,[1<1>2]<1>1],1)
    (a,b,[c,d,[1<1>2]1<1>1],1)=(a,b,[(a,b,[c-1,d,[1<1>2]<1>1],1),d-1,[1<1>2]<1>1],1)
    ......
    (a,b,[c,d,[1<1>2]2<1>1],1)=(a,b,[c,c,...b次...c[1<1>2]2<1>1],1)
    (a,b,[c,d,[1<1>2]2,2<1>1],1)
    (a,b,[c,d,[1<1>2]2,2,2<1>1],1)
    (a,b,[c,d,[1<1>2]2,2,[1<1>1],1,1<1>1],1)
    (a,b,[c,d,[1<1>2]2,2,[1<1>2],1,1<1>1],1)
    (a,b,[c,d,[1<1>2][1<1>1]2,1<1>1],1)
    (a,b,[c,d,[1<1>2][1,1,2<1>1]2,1<1>1],1)
    (a,b,[c,d,[1<1>2][1,[1<1>1],1<1>1]2,1<1>1],1)
    (a,b,[c,d,[1<1>2][1<1>2]2,1<1>1],1)
    (a,b,[c,d,[1,[1<1>2],1<1>1],1,1<1>1],1)
    (a,b,[c,d,[1,[1,[1<1>2],1<1>1],1<1>1],1<1>1],1)
    (a,b,[2<1>2],c)=(a,b,[a,[a,[...[1<1>2]b次...,b<1>1],b<1>1],b<1>1],c)
    似曾相识，跟(a,b,[1<1>2],c)时一样的，只不过最里层多了个[1<1>2]
    (a,b,[n<1>2],c)=(a,b,[a,[a,[...[n-1<1>2]b次...,b<1>1],b<1>1],b<1>1],c)
    (a,b,[1,2<1>2],c)
    (a,b,[1,[1<1>1],1<1>2],c)
    (a,b,[1,[1,[1<1>1],1<1>1],<1>2],c)
    (a,b,[c,d,[1<1>2],1<1>2],c)=(a,b,[c,[c,[...[c<1>1]d次...d<1>1],d<1>1],1<1>2],c)
    (a,b,[1,[1,[1<1>2],1<1>2],1<1>2],c)
    (a,b,[1<1>3],c)=(a,b,[a,[a,[...[a<1>2]b次...,b<1>2],b<1>2],b<1>2],c)
    (a,b,[1<1>4],c)
    (a,b,[1<1>5],c)
    (a,b,[1<1>1,2],c)=(a,b,[1<1>b],c)
    (a,b,[1<1>1,3],c)=(a,b,[1<1>b,2],c)
    (a,b,[1<1>1,m,n],c)=(a,b,[1<1>1,(a,b,[1<1>1,m-1,n],c),n-1],c)
    (a,b,[1<1>1,1,1,2],c)
    (a,b,[1<1>c,d,[1<1>1],n],c)=(a,b,[1<1>c,c,c,c,c...d次...c,[1<1>1],n-1],c)
    (a,b,[1<1>c,d,[1<1>1],n，n],c)
    (a,b,[1<1>c,d,[1<1>1],n，n,[1<1>1]],c)
    (a,b,[1<1>c,d,[2<1>1],c)
    (a,b,[1<1>c,d,[1,2<1>1],c)
    (a,b,[1<1>c,d,[2,[1<1>1],2<1>1],c)
    (a,b,[1<1>c,d,[1<1>2],2],c)
    (a,b,[1<1>c,d,[1<1>1,2],2],c)
    (a,b,[1<1>c,[1<1>1,[1<1>2],1],2],c)
    (a,b,[1<2>1],c)=(a,b,[a<1>a,[a<1>a,[a<1>b,...[a<1>b,[a<1>b],b]b次...],b],b],c)
    FSO
    (a,b,[1<2>1],c)
    按照同样的方法套娃
    (a,b,[1<2>1],c,d,[1<2>1],e)
    (a,b,[1<2>1][1<1>1],c)
    (a,b,[1<2>1][2<1>1],c)
    (a,b,[1<2>1][1,2<1>1],c)
    (a,b,[1<2>1][1,[1<1>1],2<1>1],c)
    (a,b,[1<2>1][1<1>2],c)
    (a,b,[1<2>1][1<1>3],c)
    (a,b,[1<2>1][1<1>1,2],c)
    (a,b,[1<2>1][1<1>1,[1<1>2],2],c)
    (a,b,[1<2>1][1<2>1],c)
    (a,b,[1<2>1][1<2>1][1<2>1],c)
    (a,b,[1,[1<2>1],1<1>1],c)
    (a,b,[2,[1<2>1],1<1>1],c)
    (a,b,[1,2,[1<2>1],1<1>1],c)
    (a,b,[1,[1<1>1],1,1,[1<2>1],1<1>1],c)
    (a,b,[1,[1<2>1],1,1,[1<2>1],1<1>1],c)
    (a,b,[1,1,[1<2>1][1<2>1],1<1>1],c)
    (a,b,[1,[1,[1<2>1],1<1>1],1,,1<1>1],c)
    (a,b,[1,[1<2>1],1<1>2],c)
    (a,b,[1,[1<2>1],1<1>3],c)
    (a,b,[1,[1<2>1],1<1>2,3],c)
    (a,b,[n<2>1],c)=(a,b,[a<1>a,[a<1>a,[a<1>b,...[a<1>b,[n-1<2>1],b]b次...],b],b],c)
    (a,b,[1,2<2>1],c)
    (a,b,[1,[1<1>1],2<2>1],c)
    (a,b,[2,[2<2>1],2<2>1],c)
    (a,b,[2,[2,[2<2>1],2<2>1],2<2>1],c)
    (a,b,[1<2>2],c)
    (a,b,[2,[2,[2,[2<2>1],2<2>1],2<2>2],c)
    (a,b,[2<2>3],c)
    (a,b,[2<2>4],c)
    (a,b,[2<2>5],c)
    (a,b,[2<2>5],c)
    (a,b,[2<2>1,2],c)
    (a,b,[2<2>1,[1<2>1],2],c)
    (a,b,[1<n+1>1],c)=(a,b,[a<n>a,[a<n>a,[a<n>b,...[a<n>b,[a<n>b],b]b次...],b],b],c)
    (a,b,[1<1,2>1],c)=(a,b,[1<b>1],c)
    (a,b,[1<1,3>1],c)=(a,b,[1<b,2>1],c)
    (a,b,[1<1,m,n>1],c)=(a,b,[1<(a,b,[1<1>1,m-1,n],c),n-1>1,],c)
    似曾相识，后面不会用我说了
    (a,b,[1<1,[1<1>1],2>1],c)
    (a,b,[1<1,[1<2>1],2>1],c)
    (a,b,[1<1,[1<3>1],2>1],c)
    (a,b,[1<1,[1<1,2>1],2>1],c)
    (a,b,[1<1,[1<1,[1<1>1],2>1],2>1],c)
    limit(n,n,[n<n,[n<n,[n<n,...[n<n,[n<n>n],n>n]...,n>n],n>n],n>n],n)
    我只能分析到FSO,希望大佬分析下能不能到TREE3，如果能的话TREE3如何用我这个记号表示
    IP属地:北京2楼2023-08-23 16:56
    回复
      2026-01-19 15:37:57
      广告
      不感兴趣
      开通SVIP免广告
      我这个有点复杂，争取优化一下
      IP属地:北京3楼2023-08-23 16:58
      回复
        用的弱φ吧，那极限应该是FSO
        IP属地:湖南来自Android客户端4楼2023-08-23 17:20
        收起回复
          规则还是不清晰，我到时候争取发个改良版，预计极限SVO
          IP属地:北京来自iPhone客户端5楼2023-08-23 19:05
          回复
            尊敬的先生，我告诉你:没法比
            IP属地:江苏来自Android客户端6楼2023-08-23 19:53
            收起回复
              已经在重置了，目前刚重置到e0^e0^e0
              IP属地:北京来自iPhone客户端7楼2023-08-23 23:40
              收起回复
                等于零
                来自Android客户端8楼2023-08-24 17:21
                收起回复

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