User blog comment:Mh314159/Alpha numbers (and beyond)/@comment-35470197-20191007022858/@comment-39585023-20191015213204

Hi. Thanks! I am having trouble getting the formatting to display at the moment, so it looks like just a bunch of code. I see the final estimate of w^2, though. I hope you enjoy doing these things because they look like considerable effort to me!

I have a faster growing set of rules, and possibly more compact also. Here they are with some examples and estimates. I think that in these new rules even two small terms in the string f‹2,1›(1) is equivalent to tetrating the Graham's sequence, if it makes sense to tetrate a function. By this I mean if G is Graham's sequence such that G(64) is the Graham-Gardner number, I'm talking about G^(G^(...G^n(n))(n))(n) where the number of G's is itself a function of G. I don't fully understand the FGH, but isn't each power at least a +1 to w so doesn't this already reach w^2?

Here are the rules and examples. And then following them, without examples, are three added lines that implement subscripts that iterate functional powers and def. of f‹S›(x) is changed to generate a subscript instead of a power. These changes, I think, make much bigger numbers but leads to another question I have that I will post later.

Please go ahead and see if you agree with my estimate of f‹2,1›(1) but only do so if you are interested or if you enjoy the evaluating and teaching. I'm having fun with this but I don't want anyone else to start to find it to be a chore. I must be one of the few if not only people on here that want to do things with natural numbers and not resort back to ordinals in my definitions. Thank you!

f‹S›(x) = f‹T›m(x) m = f‹S›(x-1)

f‹S›(0) = f‹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 apply

f‹0,b,...›(x) = f‹(f‹b,b-1,...›(x)),b-1,...›(x)

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

Drop trailing zeroes Examples:

f‹1,1›(1) = f‹0,1›m(1) m = f‹1,1›(0) = f‹0,1›(2) = f‹f‹1›(2)›(2) = f‹8›(2) = approx 2^^^^^^^^^5 and let's call this n. So we are iterating f‹0,1› that many times where each f‹0,1›(x) = f‹f‹1›(x)›(x) and therefore each iteration pulls the argument strongly into the index and a number in the index is at least equal to the hyperoperator number. This means that f‹1,1›(1) is greater than iterating Graham's sequence n = 2^^^^^^^^^5 times so I will call it G(n) where G(64) is the Graham-Gardner number.

f‹1,1›(2) = f‹0,1›m(2) m = f‹1,1›(1) so this = G(G(n))

f‹1,1›(3) = f‹0,1›m(3) m = f‹1,1›(2) = G3(n)

f‹1,1›2(1) = f‹1,1›(f‹1,1›(1)) would equal Gn(n)

f‹1,1›3(1) = f‹1,1›(Gn(n)) would equal G^(Gn(n))(n)

Now we can look at f‹2,1›(1)

f‹2,1›(1) = f‹1,1›m(1) m = f‹2,1›(0) = f‹1,1›(3). Now f‹1,1›(3) is the functional power on the f‹1,1› function, so are we tetrating the G function?

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

to implement subscripts:

f‹S›n(x) = f‹S›n-1p(x)   p = f‹S›n(x-1)

f‹S›n(0) = f‹S›n-1(j)

f‹S›0(x) = f‹S›(x) f‹S›(x) = f‹T›m(x)   m = f‹S›(x-1)

f‹S›(0) = f‹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 apply

f‹0,b,...›(x) = f‹(f‹b,b-1,...›(x)),b-1,...›(x)

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

Drop trailing zeroes