User blog:Mh314159/FOX notation

Here is a simple and strong set of recursions that for now I'm calling FOX notation. At these initial levels, at least, it grows faster than anything else I have done. If there are any mistakes, let me know

for all functions F, Fn(x) = Fn-1x(x) and F0(x) = F(x)

f0(x) = f(x) = x+1

f‹0›(x) = fx(x) (ω growth rate)

f‹n› and f‹S› and f‹n›mand f‹S›m are functions and are iterated by powers; n and m are natural numbers and S is a comma-separated string of natural numbers.

f‹n›(x) = f‹n-1›xf‹n›(x-1)(x) where f‹n›(0) = f‹n-1›(n)

f‹S›(x) = f‹T›xf‹S›(x-1)(x) where T is S with its first term reduced by 1.

f‹S›(0) = f‹T›(s) where s is the final term in S.

and zero replacement rules where Y and Z are any string of terms:

f‹0,a,Y›(x) = f‹(f‹x,a-1,Y›(x)),(a-1),Y›(x)

f‹Y,0,a,Z›(x) = f‹Y,(f‹Y,x,a-1,Z›(x)),(a-1),Z›(x)

drop trailing zeroes

Examples:

f‹0›(1) = f1(1) = 2

f1(2) = f2(2) = 4

f‹0›(2) = f2(2) = f12(2) = f1(4) = 8

f‹0›1(1) = f‹0›(1) = f1(1) = 2

f‹0›1(2) = f‹0›2(2) = f‹0›((f‹0›(2)) = f‹0›(8) = f8(8)

f‹1›(1) = f‹0›1f‹1›(0)(1) = f‹0›1f‹0›(1)(1) = f‹0›12(1) = f‹0›1(2) = f8(8)

f‹1›(2) = f‹0›2f‹1›(1)(2) therefore f8(8) iterations

where f‹0›2(2) = f‹0›12(2)

where f‹0›1(2) = f8(8)

so f‹0›2(2) = f‹0›1(f8(8)) = f‹0›p(f8(8)) where p = f8(8) and therefore f‹0›2 grows faster than ω+1

f‹1›(2) = f‹0›2p(2)

f‹1›1(2) = f‹1›(f‹1›(2))

f‹1›2(2) = f‹1›1(f‹1›1(2))

f‹0,1›(2) = f‹f‹2›(2)›(2)

f‹0,1›1(2) = f‹0,1›(f‹0,1›(2))

f‹1,1›(1) = f‹0,1›1f‹1,1›(0)(1) = f‹0,1›1f‹0,1›(1)(1) = f‹0,1›1f‹f‹1›(1)›(1)(1)

f‹1,1›(2) = f‹0,1›2f‹1,1›(1)(2)

f‹0,2›(1) = f‹f‹1,1›(1),1›(1)