## FANDOM

10,824 Pages

1.we define a set BCS (bracket-comma string):

(1)any nonnegitive integer is in BCS

(2)if a_1 and a_2 and ... and a_n is in BCS,then the string a_1,a_2,...,a_n is in BCS

(3)if a_1 and a_2 and ... and a_(n+1) and x_1 and x_2 and ... and x_n is in BCS,then the string a_1[x_1]a_2[x_2...[x_n]a_(n+1) is in BCS for any integer n

2.we recursive define an order on BCS:

for BCS x and y

(1)if x has "outside" comma more than y,then x>y

(2)if both have the same number of "outside" comma,let x=u,v y=s,t if u>s then x>y if u=s and v>t then x>y

(3)if both don't have outside comma,write x as the from in 1.(3),let the biggest x_i be u,the biggest x_i of y be v.if v is empty and u is not then x>y. if u>v then x>y. if u=v , write x=a[u]b , y=c[u]d , both a and c do not have the substring [u] "in outside",if a>c then x>y,if a=c and b>d then x>y.

(4)if x,y both are integer and x>y then x>y.

3.we recursive define a map cz (means clean zero):BCS->BCS

(1)if x has outside comma,x=a_1,a_2,...,a_n and cz(a_1)=0,then cz(x)=cz(a_2,a_3,...,a_n)

(2)if a_1≠0,then cz(x)=cz(a_1),cz(a_2),...,cz(a_n)

(3)if x doesn't have outside comma,x=a_1[x_1]a_2...[x_n]a_(n+1) and a_1=0,then cz(x)=cz(a_2[x_2]...[x_n]a_(n+1))

(4)if x doesn't have outside comma , a_1≠0 and for some i>1 we have cz(x_(i-1))>cz(x_i) and a_i=0,

then cz(x)=cz(a_1[x_1]...a_(i-1)[x_(i-1)]a_(i+1)[x_(i+1)]...[x_n]a_(n+1))

(5)if x doesn't have outside comma and that kind of a_i doesn't exist,then

cz(x)=a_1[cz(x_1)]...[cz(x_n)]a_(n+1)

(6)for integer x,cz(x)=x

4.we define an operator f on BCS:

if n is positive integer,x is in cz(BCS) ,we can put f(n) into leftmost of x,or at a symbol "[" or "," 's right, then we can get another element of BCS.

if x is an element of BCS,y is a substring of x,y is in BCS such that y=x or [y] is a substring of x,then we call y to be a bracket substring of x.

for BCS x in the image of cz,let y to be a string which is f(m) put into x.

let z=a_1,a_2,...,a_n (n might be 1) to be a bracket substring of x such that the respect substring of y is

a_1,a_2,...,f(m)a_k,a_(k+1),...,a_n

if some of a_(k+1),a_(k+2),...,a_n is not 0,then define y=x.

if a_k=0,define y=x

if k=n and right most number of a_k is not 0,define y=x

if right most number of a_k is 0,let a_k =b_1[x_1]...[x_l]b_(l+1)

let i to be the largest number such that b_i ≠ 0.

if x_i=0,minus b_i by 1,and change b_(i+1) to m,define y to be this changed x.

if x_i≠0,and right most number of x_i is not 0,minus b_i by 1,and change b(i+1) to this thing

3[x_i-1]0[x_i-1]0...0[x_i-1]0 (the number of 0 is m,x_i-1 is x_i minus right most number by 1)

define y to be this changed x.

if x_i≠0,and right most number of x_i is 0,minus b_i by 1,and change b(i+1) to be 3[f(m)x_i]0

define y to be this changed x

in this step if b_i=1 and x_(i-1) doesn't exist or bigger than x_i,delete b_i-1 (=0) and [x_i] after it.

if k≠n and right most number of a_k is not 0,we find a bracket substring h of x includes z,such that

h=h_1,h_2,...,h_n=a_1,a_2,...,a_k-1,M,0,...,0

and for any bracket substring j of h includes z,either j=z or j>h.

if such h exists,if m=1,plus a_(k+1) by 1.if m>1,let M=Sf(m)T,here S and T is substrings of h,change a_(k+1) to be Sf(m-1)T.then minus right most number of a_k by 1,and define y to be this changed x.

if such h don't exist,if m=1,change z to be a_1,a_2,..,a_k-1,1[a_1,a_2,..,a_k,0,..,0]0,0,..,0 , keeps number of outside comma of z to be n-1 and here a(k+1)=1[before changed z]0.if m>1,change z to be a_1,a_2,...,a_k-1,1[f(m-1)a_1,a_2,...,a_k+1,0,...,0]0,0,...,0.

define y to be this changed x.

in this step,if k=1,a_1=1,delete first two symbol "0," of the place of z.

5. we define a function A on a subset of BCS.

now let x is BCS without outside comma,change x to cz(x).

if x is an integer,define A(x)=x+1

if right most number of x is 0,and x=M[u]N,right most number of M is not 0 and every outside number of N is 0,define A(x)=A(M-1[u]N+1).

if right most number of x is k>0,define A(x)=A(f(A(x-1))x-k)

6.some examples.

A(a[0]b)=Ackermann(a,b)

A(1[1]n) has level ω^ω

A(1[1[1,0]0]n) has level ε[0]

A(1[1[1,1[1[1,0]0[1,0]1]0]0]n) is the limit of BEAF

A(1[1[1,1[2,0]0]0]n) has level ψ(ψ[1](0))

A(1[1[2,0]0]n) has level ψ(Ω[ω])

A(1[1[1[1,0]0,0]0]n) is the limit of Bird's notation

A(1[1[1,1[1[2,0]0,0]0]n) has level ψ(I)

the growth level of this notation is too confusing.I have another version of this notation,it's more clear and faster but much more complex.