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testitemqlstudop wrote:

Moosey wrote: according to Wikipedia, the incredibly unreliable resource™.

The biggest lie perpetrated by humanity (cough. cough. TEACHERS) is that Wikipedia is unreliable. Studies show [citation needed] that for every 1 edit to a page, ten read the page.

PkmnQ wrote:Oh, so if f(x) = 3x+2 then f^3(5) = 161?

testitemqlstudop wrote:Where on the fgh is my function for, say, a function f with growth rate w^2

æ(F, x, y, z) = {
    F(y+z) when x = 0
    F(æ(F^x, x-1, æ(F^x, x-1, F(x), z+1))+z, z+1) when y = 0
    æ(F^y, x, y-1, z+1) otherwise
}

The googology wiki wrote:[salad numbers] are often coined by inexperienced googologists holding the false notions that a more complex definition is a more powerful one.

Moosey wrote:

testitemqlstudop wrote:Where on the fgh is my function for, say, a function f with growth rate w^2

Are you asking for œ(f_w^2,n,n)?
I'm not sure why I ask since I can't calculate it, but I doubt it's too much higher.

Variant of your function:

Code: Select all

æ(F, x, y, z) = {
    F(y+z) when x = 0
    F(æ(F^x, x-1, æ(F^x, x-1, F(x), z+1))+z, z+1) when y = 0
    æ(F^y, x, y-1, z+1) otherwise
}

Almost at this point, unfortunately:

The googology wiki wrote:[salad numbers] are often coined by inexperienced googologists holding the false notions that a more complex definition is a more powerful one.

testitemqlstudop wrote:S is f_w, right?

testitemqlstudop wrote:
Manually trying to bash out ae(S,4,4):

ae(S,4,4) =
ae(S^4,4,3) =
ae(S^12,4,2) =
ae(S^24,4,1) =
ae(S^24,4,0)=
ae(S^96,3,28)+24=...

testitemqlstudop wrote:How do you define F^x for multi-variable functions tho

EDIT: Whoops i meant oe

f(tvi(f,n-1,x,x),tvi(f,n-1,y,y)), n>1
f(x,y), n=1

PkmnQ wrote:

testitemqlstudop wrote:How do you define F^x for multi-variable functions tho

EDIT: Whoops i meant oe

Oh, theres no actual description yet?
It might be f(f(x, x), f(y, y)) maybe?

A while ago, I wrote:

A while before that, I wrote:This morning I created a notation that extends f^a(x). I will use ^. to notate an important part of it so it’s not confused with exponentiation, but think of it is superscript right above something.
It starts like this:
f^1^.a(x) = f^f^f^f^f^f^f^f^f...(x)(x)(x)(x)... with a f’s, all of x
so, where f = +1,
f^1^.5(1)=

Code: Select all

f^f^f^f^f(1)(1)(1)(1)(1)=
f^f^f^f^2(1)(1)(1)(1)
f^f^f^3(1)(1)(1)...
=6

then: f^b^.a(x) = f^b*f^b*f...(x)(x)(x) with a b*f’s.
then, where # represents anything, including a stack: f^#^.b^.a(x)=f^#^.b*f^#^.b*f...(x)(x)(x) with a f^#^.b*’s.
then we have this leap:
f^a|n(x)= f^a^.a^.a^.a^.a ... with n a's.

f^a|n^.#(x)=f^(f^#(a))^.(f^#(a))^.(f^#(a)) ... with n (f^#(a))^. ’s. The initial f, is, of course, still of x.
Then, the next logical extension:
f^a|n|m(x) = f^a|n^.a|n^.a|n... with m a|n ’s.
The logical continuation:
f^a|n|m^.#(x)= f^f^a|n^.#(a)^f^a|n^.#(a)^f^a|n^.#(a)... with f^n|m(a) f’s. The initial f will always be of x of course

Can somebody help me define f^a|b|c|d... with arbitrarily many arguments?
EDIT: nvm, f^a|n|m|#(x)= f^a|n|f^#(m)(x)

Then I can, a la bowers, define f^a||b(x) = f^a|a|a|a|a...(x) with b a’s.

where is S^2||x(1) in the grass-mowing hierarchy? (Probably very low--I mean the fgh of course)
Anyways, now:
f^#||b||c(x) = f^#||(f^b||c(b))(x)

GENERALIZATION TIME YAY
f^a(:b:)c(x) =

Code: Select all

f^a(:b-1:)a(:b-1:)a(x) with c a's, b > 1
f^a|c, b = 1

f^#a(:b:)c(x) =

Code: Select all

f^#a|c(x), b = 1
f^#(:b:)(f^a(:b:)c(a))(x), b > 1

I'm sure that S^3(:x:)3(1) is much higher in the fgh. At least, relatively speaking considering S = the successor function.

S^3^.3*S^3^.66(1)(1)

PkmnQ wrote:I haven’t seen n[x] before.

x+1, n = 0
f^x_(n-1)(x), n > 0, n not a limit ordinal
f_n[x](x), n a limit ordinal

testitemqlstudop wrote:So the sequence for w^w is [w, w*w, w*w*w, w*w*w*w, ...] and the sequence for w*w is [w, w*2, w*3, w*4, ...]?

testitemqlstudop wrote:Your explanation is better than the one on goowiki

testitemqlstudop wrote: October 4th, 2019, 9:30 pm Yes, that's right for unidimensional arrays.

But I don't get anything above 2d

{k #} -> {k-1 #}
{# 1,k #} -> {# b,k-1}
If this separator has another separator at its active point, reduce it the same way.

2,3,1{1}2 -> 2,3,1,1,1,2
2,3,1{3}3{4}2 -> 2,3,1{2}1{2}1{2}2{3}2{4}2
2,3,1{1,3}2 -> 2,3,1{3,2}1{3,2}1{3,2}2
2,3,1{1{3}2}2 -> 2,3,1{1{2}1{2}1{2}2}2

Macbi wrote: June 7th, 2019, 9:29 am I believe the fastest growing known computable functions are of the following form:

n ↦ The combined running time of all Turing machines (with no input) such that there exists a proof that they halt in second-order PA using fewer than n symbols

testitemqlstudop wrote: October 4th, 2019, 11:11 pm

Macbi wrote: June 7th, 2019, 9:29 am I believe the fastest growing known computable functions are of the following form:

n ↦ The combined running time of all Turing machines (with no input) such that there exists a proof that they halt in second-order PA using fewer than n symbols

FIrst order set theory is much stronger, and you might as well take the maximum running time:

n ↦ The maximum running time of a Turing machine (with no input) such that there exists a proof of halting in first order set theory (under the domain of the vN universe) using fewer than n symbols.

I think this is some weak version of the Rayo function now.