User blog:Mh314159/A new approach

After two attempts to climb up the FGH and not getting very far, here's another approach. Hopefully I have learned something from my first two tries. Please note that there may be typography problems in this thing because I have not yet mastered nested subcripts and superscripts. Thank for reading!

fn(1) = f0(n+1)

f0(x) = x + 1

fn(x) = fn-1fn(x-1)(x)

within brackets, drop zeroes and collapse

[a] = fa(a)

[1] = f1(1) = f0(2) = 3

f1(2) = f0f1(1)(2) = f03(2) = 5

f1(3) = f05(3) = 8

f1(4) = f08(4) = 12

f1(x) = 3 + (x2 + x - 2)\2

[2] = f2(2) = f1f2(1)(1) = f14(1) = f13(3) = f12(8) = f1(38) = 728

[3] = f3(3) = f2f3(2)(3)

Now I will define strings in the brackets and recurse them using f

[a,b] = f[[a-1,b],b-1]|undefined(a)

[a,b,c] = f[[a-1,b,c],[a,b-1,c],c-1]|undefined(a)

etc.

[1,2] = f[[2],1]|undefined(1) = f[728,1](1) where [728,1] = f727,1(728) etc.

Now I will define very long strings in the brackets using rows of brackets. The next function after f will recurse these rows.

x is any number or string

y is a string

n is a number

[x][0] = [x]

[x][n] = [p,p,...,p][n-1] where p = [x][n-1] and there are p instances of p

[x][y] = [x][n] where n = [y]

[1][1] = [3,3,3]

[2][1] = [728,728,...,728] and there are 728 entries.

[2][2] = [p,p,...,p] where p = [2][1] and there are p entries.

[2][2,1] = [2][n] where n is the value of [2,1]

if n is a number

eliminate [0]s and collapse

[x][y][n] = [x] x][n-1][n [n-1] (notice that number of brackets remains the same until the final bracket decrements to zero)

[2,2][3][1] = [2,2] 2,2][2][1 [0] = [2,2]2,2][2][1

[2,2][2][1] = [2,2] 2,2][1][1

[2,2][1][1] = [2,2] 2,2][1

Therefore, [2,2][3][1] = [2,2]2,2][[2,2][[2,2][1]]

This means first find [2,2][1] and then use that as the recursion number m in [2,2][m] and then use [2,2][m] as the recursion number p in [2,2][p] and then use [2,2][p] as the recursion number r in [2,2][r] in general,

if n is a number,

[x][y]...[z][n] (s sets of brackets) = [x][y]...x][y]...[z-1][n [n-1] (s sets of brackets)

if q is a string,

[x][y]...[q] = [x][y]...[n] where n = the value of [q]

[4][3][2][1] = [4][3]4][3][1][1[0] = [4][3]4][3][1][1 where [4][3][1][1] = [4][3]4][3][0][1[0] = [4][3]4][3][1

Now the next function number. The symbol group f{N} indicates a function and should be treated as if it was simply an f or a g.

f{1}n(1) = f{1}0(n+1)

f{1}0(x) = [x,x,...,x]...[x,x,...,x] with [x] x's in each string and [x] sets of brackets

f{1}n(x) = f{1}n-1f{1}n(x-1)(x)

[a]1 = f{1}a(a)

[1]1 = f{1}1(1) = f{1}0(2) = [2,2,...,2]...[2,2,...,2] and because [2] = 728, there are 728 entries in each string and 728 sets of brackets

f{1}2(1) = f{1}0(3)

f{1}1(2) = f{1}0f{1}1(1)(2) = f{1}0f{1}0(2)(2)

[2]1 = f{1}2(2) = f{1}1f{1}2(1)(2) = f{1}1f{1}0(3)(2) = f{1}1(f{1}1...f{1}1(2)) with f{1}0(3) recursions

[a,b]1 = f{1}[[a-1,b] 1|undefined,b-1]1 (a)

[a,b,c]1 = f{1}[[a-1,b,c] 1|undefined,[a,b-1,c]1,c-1]1 (a)

etc.

[1,1]1 = f{1}[[1] 1|undefined]1 (1) = f{1}[n] 1 (1) where n = [1]1 (a large number, see above) and the subscript [n]1 = f{1}n(n) [x]1[0] = [x]1

[x]1[n] = [p,p,...,p]1[n-1] where p = [x]1[n-1] and there are p instances of p

[x]1[y] = [x]1[n] where n = [y] [x]1[0] = [x]1

[x]1[n]1 = [x]1[r] r = the value of [n]1

[x]1[y]1 = [x]1[n] where n = the value of [y]1

[2]1[1] = [p,p,...,p]1 with p = [2]1and with p entries.

[2]1[2] = [p,p,...,p]1 with p = [2]1[1]undefinedand with p entries.

[2]1[1]1 = [2]1[r] with r = [1]1

[2]1[2]2 = [2]1[r] with r = [2]2

Generalizing to any function number m

f{m}n(1) = f{m}0(n+1)

f{m}0(x) = [x,x,...,x]m-1...[x,x,...,x]m-1 with [x]m-1 entries in each string and [x]N-1 sets of brackets

f{m}n(x) = f{m}n-1f{m}n(x-1)(x)

[a]m = f{m}a(a)

[a,b]m = f{m}[[a-1,b] m|undefined,b-1]m (a)

[a,b,c]m = f{m}[[a-1,b,c] m|undefined,[a,b-1,c]m,c-1]m (a)

etc. [x]m[0] = [x]m

[x]m[n] = [p,p,...,p]m[n-1] where p = [x]m[n-1] and there are p instances of p

[x]m[y] = [x]m[n] where n = [y] [x]m[n]m = [x]m[r] where r = the value of [n]m

[x]m[y]m = [x]m[r] where r = the value of [y]m

Now expressions of the type [9][9]and even more deeply recursed subscripts are defined

The stringed function number:

↓(n) indicates that n is a subscript f{0,1}n(1) = f{0,1}0(n+1)

f{0,1}0(x) = [x]↓([x]↓(...[x])) with [x] instances of [x]

f{0,1}n(x) = f{0,1}n-1f{0,1}n(x-1)(x)

[a]0,1 = f{0,1}a(a)

f{0,1}0(1) = [1]↓([1]↓[1])) because [1] = 3

[1]0,1undefined= f{0,1}1(1) = f{0,1}0f{0,1}0(2)(1) where f{0,1}0(2) = [2]↓([2]↓(...[2])) with 728 instances of [2]

Next to do: define strings and rows for [a]0,1 and then f{1,1} recurses rows of [a]0,1 and [a]1,1= f{1,1}a(a) and repeat the process for the general function number f{m,1}

Then what might f{0,2} mean?