葛立恒数二吧 关注:846贴子:67,232

我自创了记号和hydra是否相同

只看楼主收藏回复

n(0)n=n+1
n(1)n=n(0)n(0)n……n
n(n+1)n=n(n)n(n)n……n
n(1,0)n=n(n)n
n(1,1)n=n(1,0)n(1,0)n……n
n(1,(1,0))n=n(1,n)n
n(1,(1,0)1)n=n(1,(1,0))n(1,(1,0))n……n
n(1,(1,0)(1,0))n=n(1,(1,0)n)n
n(1,(1,0)(1,(1,0)1))n=
n(1,(1,0)(1,(1,0))(1,(1,0))……n)n
n(1,(1,0)(1,(1,0)2))n=
n(1,(1,0)(1,(1,0)1)(1,(1,0),1)……n)n
n(1,(1,0)(1,(1,0)(1,(1,0),1)))n=
n(1,(1,0)(1,(1,0)(1,(1,0))(1,(1,0))……n)n
n(1,(1,1))n=n(1,(1,0)(1,(1,0)(1,(1,0)……n)n
n(1,(1,2))n=n(1,(1,1)(1,(1,1)(1,(1,1)……n)n
n(1,(1,(1,0)))n=n(1,(1,n))n
n(1,(1,(1,0)(1,(1,(1,0)1)))n=
n(1,(1,(1,0)(1,(1,(1,0))(1,(1,(1,0))(1,(1,(1,0))……n)n
n(1,(1,(1,1)))n=
n(1,(1,(1,0)(1,(1,(1,0)(1,(1,(1,0)……n)n
n(2,0)n=n(1,(1,(1,(……n))))n
n(2,1)n=n(2,0)n(2,0)n……n
n(2,(1,0))n=n(2,n)n
n(2,(2,0))n=n(2,(1,(1,(1,(……n)))))n
n(2,(2,1))n=n(2,(2,0)(2,(2,0)(2,(2,0)……n)n
n(3,0)n=n(2,(2,(2,(……n))))n
n((1,0),0)n=n(n,0)n
n(1,0,0)n=n((((1,0),0),0)……n)n
再往后就参考φ函数


IP属地:浙江来自Android客户端1楼2024-11-10 15:23回复
    IP属地:浙江来自Android客户端2楼2024-11-10 15:24
    回复
      IP属地:浙江来自Android客户端3楼2024-11-10 15:24
      回复
        IP属地:浙江来自Android客户端6楼2024-11-12 23:46
        回复
          @jdihdib @英短蓝猫_ @贴吧用户_5Ee7ACZ


          IP属地:浙江来自Android客户端7楼2024-11-12 23:47
          回复
            @五年高考💯 最近吧里一直冷冷清清的


            IP属地:浙江来自Android客户端8楼2024-11-13 09:22
            回复
              看没人自己分析一下。


              IP属地:浙江来自Android客户端9楼2024-11-14 21:50
              回复
                n(1)n=p1
                n(1,0)n=p1(p1)
                n(1,(1,0))n=p1(p1)+p1(p1)
                n(1,(1,0)(1,1))n=p1(p1+p1)


                IP属地:浙江来自Android客户端10楼2024-11-14 21:55
                回复
                  n(1,(1,0)(1,2))n=p1(p1+p1)
                  n(1,(1,0)(1,(1,0)))n=p1(p1(p1))


                  IP属地:浙江来自Android客户端14楼2024-11-15 12:06
                  回复
                    n(1,(1,0)(1,(1,0)1))n=p1(p1(p1+p1))
                    n(1,(1,0)(1,(1,0)2))n=
                    p1(p1(p1+p1+p1))
                    n(1,(1,0)(1,(1,0)(1,0)))n=
                    p1(p1(p1(p1))))


                    IP属地:浙江来自Android客户端15楼2024-11-15 12:13
                    回复
                      n(1,(1,0)(1,(1,0)(1,1)))n=
                      p1(p1(p1(p1+p1)))
                      n(1,(1,0)(1,(1,0)(1,(1,0))))n=
                      p1(p1(p1(p1(p1))))
                      n(1,(1,0)(1,(1,0)(1,(1,0)1)))n=
                      p1(p1(p1(p1(p1+p1))))


                      IP属地:浙江来自Android客户端16楼2024-11-15 12:21
                      回复
                        n(1,(1,1))n=p1(p2)


                        IP属地:浙江来自Android客户端17楼2024-11-15 12:26
                        回复
                          n(1,(1,1)(1,(1,1))n=
                          n(1,(1,1)(1,(1,0)(1,(1,0)……n)n=
                          p1(p2+p1(p2))


                          IP属地:浙江来自Android客户端18楼2024-11-15 12:33
                          回复
                            n(1,(1,1)(1,(1,1)1))n=
                            p1(p2+p1(p2+p1))
                            n(1,(1,1)(1,(1,1)(1,(1,1))))n=
                            p1(p2+p1(p2+p1(p2)))


                            IP属地:浙江来自Android客户端19楼2024-11-15 12:49
                            回复
                              n(1,(1,2))n=p1(p2+p2)


                              IP属地:浙江来自Android客户端20楼2024-11-15 12:52
                              回复