User blog:Mh314159/Natural number recursion - first 4 rule sets

Here is the latest version of my current rule set, rules 1-4 of a total of 10 currently planned, for which I do not yet have a catchy name. Compared to the previous version, there is a notation change, a correction, and two additions:

1. Notation change from f to A for the function letter. This anticipates other function letters which will make more sense if they are B, C, etc. And this also keep the system from looking like the FGH.

2. The zero replacement rule which requires the entire expression to be copied to the zero place, with modifications to the index numbers, was not carrying the functional power "m". That is now corrected.

3. A rule now generates an index string for an A function written without one. It recurses the argument to generate the string, so that A(5) for example, will recurse 5 times and each recursion greatly increase the length of the string and the size of the terms.

4. Subscripts of A are now defined and they recurse to generate large functional powers of A. The next two rules are much more powerful, the expression-copying subscripts. But first I want to be sure these rules are well defined and sufficiently strong.

1. Function with index indicated by angled brackets and argument in parentheses:

A‹0›(x) = x + 1

A‹n›(x) = A‹n-1›A[n](x-1)(A‹n›(x-1))

A‹n›(0) = A‹n-1›(n) Ex:

A‹1›(1) = A‹0›A‹1›(0)(A‹0›(1)) = A‹0›2(2) = 4

A‹1›(2) = A‹0›4(4) = 8

A‹1›(x) = 2^(x+1)

A‹2›(1) = A‹1›8(8) > 2^^10

A‹2›(2) = A‹1›2^^10(2^^10) > 2^^2^^10

A‹2›(3) = A‹1›2^^2^^10(2^^2^^10) > 2^^2^^2^^10 > 2^^2^^2^^2^^2 or 2^^^5

A‹3›(1) = A‹2›A‹2›(3)(1) > A‹2›2^^2^^2^^10(1)

A‹4›(1) = A‹3›A‹3›(4)(1)

A‹n›(x) is greater than 2^n+1x in Knuth up-arrow notation.

2. Function with index string and argument:

A‹S›(x) = A‹T›m(x) m = A‹S›(x-1)

A‹S›(0) = A‹T›(j)

S = string of one or more nonzero terms

T = S with 1st entry decreased by one

j = sum of the terms in S

Zero replacement rules for non-trailing zeroes inside brackets:

Replace the zero with the entire expression, and in the copied expression, copy the index number after the zero into the zero's place and decrease the number that was after the zero by one. Then decrease the index number after the replacement expression by one.

Initial zero:

A‹0,b,...c›m(x) = A‹A‹b,b-1,...c›m(x),b-1,...c›m(x)

Non-initial, non-trailing zero:

A‹a,0,b,...c›m(x) = A‹a,A‹a,b,b-1,...c›m(x),b-1,...c›m(x)

Drop trailing zeroes

Ex:

A‹1,1›(1) = A‹0,1›m(1) = A‹A‹1›m(1)›m(1)    m = A‹1,1›(0) = A‹0,1›(2) = A‹A‹1›(2)›(2) = A‹8›(2)

A‹1,1›(2) = A‹0,1›m(2) = A‹A‹1›m(2)›m(2)    m = A‹1,1›(1)

A‹1,1›2(1) = A‹1,1›(A‹1,1›(1))

A‹1,1›3(1) = A‹1,1›(A‹1,1›2(1))

A‹2,1›(1) = A‹1,1›m(1) m = A‹2,1›(0) = A‹1,1›(3).

A‹1,2›(1) = A‹0,2›m(1) = A‹A‹2,1›m(1),1›m(1) m = A‹1,2›(0) = A‹0,2›(3) and by zero replacement we have m = A‹(A‹2,1›(3)),1›(3) and the first of m iterations of A‹0,2›(1) = A‹(A‹2,1›(1)),1›(1)

3. Function without index generates an index string:

A(x) = A‹m,m,...m›(x) with m instances where m = A(x-1) and A(0) = 1 Ex:

A(1) = A‹1›(1) = 4

A(2) = A‹4,4,4,4›(2)

A(3) = A‹m,m,m,m›(3) where m = A‹4,4,4,4›(2) and there are A‹4,4,4,4›(2) instances

A(4) = A‹m,m,m,m›(4) where m = A(3) and there are A(3) instances

4. Subscripted and superscripted function:

A0(x) = A(x)

An(x) = An-1m(x) m = An(x-1)

An(0) = An-1(n)

Ex:

A1(1) = A0m(1) m = A1(0) = A0(1) = A(1) = 4, so A1(1) = A04(1) = A4(1) = A3(4) = A(A(A(4)))) For A(4), see above.