## Ordinals in googology

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### Re: Ordinals in googology

There are total functions based on intermediate Turing degrees (r.e. sets that aren't universal or recursive) that dominate every computable function but are dominated by the busy beaver function, i.e. fM(n)=(the maximum time Turing machine M takes to accept a string of length n), where M is any Turing machine recognising such a set. Do any of these functions have ordinals? (EDIT: It's probably impossible to answer this question, especially since no explicit examples of such Turing Machines or even sets have been found.)

Also, what property of an ordinal determines whether its function is computable?
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### Re: Ordinals in googology

Keeps decreasing, I guess?
How to XSS:

Code: Select all

``Function(‘a’+’lert(1)’)()``

Moosey
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### Re: Ordinals in googology

What is the decord sequence for e_0 + 1? (And what is dco(e_0 + 1))

Code: Select all

``````e_0 +1
e_0
w^w + w (*depending on the definition of the fundamental sequence of e_0)
w^w +6
w^w +5
w^w +4
w^w +3
w^w +2
w^w +1
w^w (10th term)
w^10+w^9+w^8+w^7+w^6+w^5+w^4+w^3+w^2+w
w^10+w^9+w^8+w^7+w^6+w^5+w^4+w^3+w^2+66...
...
w^10+w^9+w^8+w^7+w^6+w^5+w^4+w^3+w^2 (78th term)
w^10+w^9+w^8+w^7+w^6+w^5+w^4+w^3+w3081
w^10+w^9+w^8+w^7+w^6+w^5+w^4+w^3+w3080 + 3160 (80th term)
...
w^10+w^9+w^8+w^7+w^6+w^5+w^4+w^3+w3080 (3240th term)
w^10+w^9+w^8+w^7+w^6+w^5+w^4+w^3+w3079 + 5250420
...
w^10+w^9+w^8+w^7+w^6+w^5+w^4+w^3+w3079 (5253661th term)
w^10+w^9+w^8+w^7+w^6+w^5+w^4+w^3+w3078 + 13783457713410
...
``````
What is the magnitude of dco(e_0+1)?
Clearly, it's far far greater than anything you could write in decimal.

Of course, it's far exceeded by dco(e_0+2):

Code: Select all

``````e_0 + 2
e_0 + 1
e_0
w^w^w + w^w + w
w^w^w + w^w + 10 (term 5)
...
w^w^w + w^w (term 15)
w^w^w + w^15 + w^14 + w^13 + w^12 + w^11 + w^10 + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w
...``````
let's try dco(Γ_0+1)!

Code: Select all

``````Γ_0+1
Γ_0
z_0 + e_0
...
...
... (dco(e_0+2), give or take a few, terms)
...
...
z_0
e_e_e_e_e_... (many e's) + e_e_e_e... (one fewer e's) ... many, many, many, many terms ... e_0
e_e_e_e_e_... (many e's) + e_e_e_e... (one fewer e's) ... many, many, many, many terms ... w^^(a very large value) + w^^(Same -1) ... w
e_e_e_e_e_... (many e's) + e_e_e_e... (one fewer e's) ... many, many, many, many terms ... w^^(a very large value) + w^^(Same -1) ... something which is larger than dco(e_0+2) (though not notably; it's about the dco(e_0+2)th triangular number)
...
... ~dco(e_0+2)(dco(e_0+2)+1)/2 terms
...
e_e_e_e_e_... (many e's) + e_e_e_e... (one fewer e's) ... many, many, many, many terms ... w^^(a very large value) + w^^(Same -1) ... w^w
e_e_e_e_e_... (many e's) + e_e_e_e... (one fewer e's) ... many, many, many, many terms ... w^^(a very large value) + w^^(Same -1) ... w^(very large value) + w^(Same value -1) ... w
etc.``````
I think that dco(Γ_0+1) may be one of the largest numbers, if not THE largest number, that I have ever coined. (Specifically coined... I guess I could coin much larger numbers via the sm hierarchy). However, I don't want to overestimate its value. But I do suspect that it beats Graham's number because it's an enormous recursion thing. What's a good approximation using the fgh for dco(Γ_0 +n) ? f_w(n) ? (In which case it would not beat Graham's number, however that would be a hilariously crappy approximation since f_w(1) = 2)
f_w(n+small value)?
f_w^a(n+small value)?
f_w^w(n+small value)?
f_e_0(n+small value)?
f_(higher things in veblen hierarchy)(n+small value)?
I have a sneaking suspicion that dco(a+n) << f_a(n) for large n, and thus that dco(Γ_0 +n) is eventually dominated by f_Γ_0(n).

EDIT:
Moosey wrote:
October 7th, 2019, 6:27 pm
EDIT:
Since Goodstein sequences are equivalent to the length of "decay chains" starting at values < e_0, I guess dco(Γ_0+n) >> goodstein(n) for large enough n (since even if the decay "sequences" were less powerful you still have a whole lot of padding) = ~f_e_0(n), which would mean dco(Γ_0+n) > f_e_0(n)

Am I right to think this way?
If it is > f_e_0(n), are there more (significantly larger) ordinals for whom f_a(n) is exceeded by dco(Γ_0+n)?
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### Re: Ordinals in googology

dco(a) is slightly stronger than the SGH:
SGH wrote: s(a+1,n) = s(a,n)+1
s(a,n) = s(a[n],n)
dco wrote: s(a+1,n) = s(a,n+1)+1
s(a,n) = s(a[n]+a[n-1]+...+a,n+1)
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### Re: Ordinals in googology

October 7th, 2019, 8:12 pm
dco(a) is slightly stronger than the SGH:
SGH wrote: s(a+1,n) = s(a,n)+1
s(a,n) = s(a[n],n)
dco wrote: s(a+1,n) = s(a,n+1)+1
s(a,n) = s(a[n]+a[n-1]+...+a,n+1)
Isn't it also stronger than the HH?
Given:
dco(a+1,n) = dco(a,n+1)+1

In that case, dco(Γ_0+n) catches up to f_Γ_0(n), since the HH catches up to the fgh at e_0, and thus dco(Γ_0+n) = dco(Γ_0,n) >= HH(Γ_0,n) = f_Γ_0(n) (for large enough n)

In that case, dco(Γ_0+n) is the strongest function that I have specifically defined so far, and for small n it is larger than g_64

Possibly even for n = 1 or 2...

That would also mean that dco(SVO+n) >= lower bound on TREE(n), & dco(w_1ck+n) is uncomputable & comparable to BB(n)
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testitemqlstudop
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### Re: Ordinals in googology

Is it true for an ordinal ∂ ∂ + ∂ + ... + ∂[j] < ∂[j+1] for all finite j?

Then there's a non-trivial simplification of dco, and stronger:

s(a+1,n) = s(a,n+1)+1
s(a,n) = s(a[n+1],n+1)+1

Moosey
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### Re: Ordinals in googology

testitemqlstudop wrote:
October 7th, 2019, 8:27 pm
Is it true for an ordinal ∂ ∂ + ∂ + ... + ∂[j] < ∂[j+1] for all finite j?
No, if d= w, d + d + ... d[n] < d[n+1] for most n
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Also, the tree game
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### Re: Ordinals in googology

As I was taking a shower, I realized that blinkerspawn's dco hierarchy is slightly wrong-- for instance, try to calculate s_w2(5) and you'll see what I mean
The actual definition would require a bit more funny business.

In particular, if a is a limit ordinal, s(a,n) should be defined thusly:
If you can define b and c so that b+c = a & b is the smallest ordinal >w such that b+c = a, for some c, and b <= c,
s(a,n) = s(c+b[n]+b[n-1]+b[n-2]...,n)
Else
s(a,n) = s(a[n]+a[n-1],a[n-2]...,n)

This just means that the first argument can never become larger

This probably makes it less powerful, but I guess dco(Γ_0+n) is still one of the more powerful functions that I've come up with.
How strong is it compared to the fgh? Is the dco hierarchy still as powerful as the HH, or is it only about as strong as the sgh?
Anyways, how powerful is dco(Γ_0+n) compared to the fgh?
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My CA rules can be found here

Also, the tree game
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### Re: Ordinals in googology

Moosey wrote:
October 7th, 2019, 8:18 pm
Isn't it also stronger than the HH?
Given:
dco(a+1,n) = dco(a,n+1)+1

In that case, dco(Γ_0+n) catches up to f_Γ_0(n), since the HH catches up to the fgh at e_0, and thus dco(Γ_0+n) = dco(Γ_0,n) >= HH(Γ_0,n) = f_Γ_0(n) (for large enough n)
Incorrect. Hardy and FGH are related in that if f_a(n) = H_b(n) then a = w^b, which is why all epsilon numbers are "catching points".
For Hardy, one would have dco(a+b+c+...,n) = dco(a,dco(b,dco(c,...,n))).
The n+1 just moves you along the fundamental sequence, it doesn't change the ordinal.

EDIT: Ah, but it does give you the requisite nesting, so you would be close to Hardy, the only difference being your collapsing of limits.
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### Re: Ordinals in googology

October 7th, 2019, 10:33 pm
Moosey wrote:
October 7th, 2019, 8:18 pm
Isn't it also stronger than the HH?
Given:
dco(a+1,n) = dco(a,n+1)+1

In that case, dco(Γ_0+n) catches up to f_Γ_0(n), since the HH catches up to the fgh at e_0, and thus dco(Γ_0+n) = dco(Γ_0,n) >= HH(Γ_0,n) = f_Γ_0(n) (for large enough n)
Incorrect. Hardy and FGH are related in that if f_a(n) = H_b(n) then a = w^b, which is why all epsilon numbers are "catching points".
For Hardy, one would have dco(a+b+c+...,n) = dco(a,dco(b,dco(c,...,n))).
The n+1 just moves you along the fundamental sequence, it doesn't change the ordinal.

EDIT: Ah, but it does give you the requisite nesting, so you would be close to Hardy, the only difference being your collapsing of limits.
So how does dco(a+n) compare to the hierarchies? And where is dco(gamma_0+n) in the fgh?
Does it eventually exceed f_a(n) for all n < gamma_0 or something? Or is it only fgh_(something in the Veblen hierarchy)-ish?
As I've stated before, I think Goodstein sequences are less powerful than dco, making my function dco(gamma_0+n) > f_e_0(n)
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My CA rules can be found here

Also, the tree game
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### Re: Ordinals in googology

Moosey wrote:
October 8th, 2019, 6:11 am
So how does dco(a+n) compare to the hierarchies? And where is dco(gamma_0+n) in the fgh?
Does it eventually exceed f_a(n) for all n < gamma_0 or something? Or is it only fgh_(something in the Veblen hierarchy)-ish?
As I've stated before, I think Goodstein sequences are less powerful than dco, making my function dco(gamma_0+n) > f_e_0(n)
Testitem's variant is very close to Hardy, which is related to FGH as I stated above, so dco's relation to Hardy is essentially dco's relation with that variant.
dco(G0+n) is certainly weaker than f_(G0+1)(n), probably between H_(G0) and H_(G0+w).
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### Re: Ordinals in googology

October 8th, 2019, 3:41 pm
Moosey wrote:
October 8th, 2019, 6:11 am
So how does dco(a+n) compare to the hierarchies? And where is dco(gamma_0+n) in the fgh?
Does it eventually exceed f_a(n) for all n < gamma_0 or something? Or is it only fgh_(something in the Veblen hierarchy)-ish?
As I've stated before, I think Goodstein sequences are less powerful than dco, making my function dco(gamma_0+n) > f_e_0(n)
Testitem's variant is very close to Hardy, which is related to FGH as I stated above, so dco's relation to Hardy is essentially dco's relation with that variant.
dco(G0+n) is certainly weaker than f_(G0+1)(n), probably between H_(G0) and H_(G0+w).
so dco(G0+n) would be between f_G0(n) and f_(G0+1)(n)?
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### Re: Ordinals in googology

Moosey wrote:
October 8th, 2019, 5:02 pm
October 8th, 2019, 3:41 pm
Testitem's variant is very close to Hardy, which is related to FGH as I stated above, so dco's relation to Hardy is essentially dco's relation with that variant.
dco(G0+n) is certainly weaker than f_(G0+1)(n), probably between H_(G0) and H_(G0+w).
so dco(G0+n) would be between f_G0(n) and f_(G0+1)(n)?
Either that or slightly weaker than G0, in the sense that dco(n) < f_G0(n) < dco(n+k) for some positive k.
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### Re: Ordinals in googology

So dco(a+n) for a limit oridinal a and a finite value n is roughly H_a(n)?

EDIT:
After being forced to fix my past mistakes by someone who rather unconvincingly claims that they're a bot, I found that dco(gamma_0+1) >> than I expected, since gamma_0's fundamental sequence is not {w,e0,z0,h0,...} but {0,w,phi_w(0), phi_(phi_w(0))(0),...}
Th fundamental sequence I thought gamma_0 had was the one phi_w(0), or gamma_0 had.

Code: Select all

``````gamma_0+1
gamma_0
phi_w(0)+w
phi_w(0)+6
phi_w(0)+5
phi_w(0)+4
phi_w(0)+3
phi_w(0)+2
phi_w(0)+1
phi_w(0)
phi_10(0) + phi_9(0) + phi_8(0) ... + z_0 + e_0
phi_10(0) + phi_9(0) + phi_8(0) ... + z_0 + w^^11 + w^^10 +... w^w + w
...``````
Last edited by Moosey on October 10th, 2019, 10:19 am, edited 1 time in total.
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### Re: Ordinals in googology

I'm at school, and they block anywikithatyouwanttovisit.wikia.org; can someone copy off and post the 13 fundamental sequence rules that I had in the new goowiki blog post? I need them for copying into my math stuff
I have a 14th rule I need to add: (a+b)[n] = a+(b[n]), where a >= b

EDIT:
Goowiki post is here:
https://googology.wikia.org/wiki/User_b ... d_sequence
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### Re: Ordinals in googology

Code: Select all

``````The first three rules are derived from the Wainer hierarchy; due to the fact that I'm only taking the fundamental sequences of 1-"term" ordinals I don't need all the Wainer hierarchy rules.
Rule 1. w[n] = n
Rule 2. w^(a+1)[n] = (w^a)(n)
Rule 3. If a is a limit ordinal, (w^a)[n] = w^(a[n])
The next rule is a modified version of the corresponding Wainer hierarchy rule, but rule 5 is from the Wainer hierarchy again.
Rule 4. e_0 = w
Rule 5. e_0[n+1] = e_0[n]
The next rules are stolen from the Veblen hierarchy.
Rule 6. phi_0(a) = w^a
Rule 7: phi_(a+1)(0)[n] = phi^n_a(0) (which means phi_a(0) iterated n times, not phi^(n_a)(0))
Rule 8: phi_(a+1)(b+1)[n] = phi^n_a(phi_(a+1)(b)+1)
Rule 9: if b is a limit ordinal smaller than phi_a(b) phi_a(b)[n] = phi_a(b[n])
Rule 10: if a is a limit ordinal smaller than gamma_0, phi_a(0)[n] = phi_(a[n])(0)
Rule 11: if a is a limit ordinal, phi_a(b+1)[n] = phi_a[n](phi_a(b)+1)``````
(newlines by me)
I like making rules

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### Re: Ordinals in googology

...Why are you implying the fundamental sequence of e_0 is [w, w, w, ...]? (4 and 5)

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### Re: Ordinals in googology

testitemqlstudop wrote:
October 10th, 2019, 6:57 pm
...Why are you implying the fundamental sequence of e_0 is [w, w, w, ...]? (4 and 5)
Those were a typo.
The current version of the rules is this:

Code: Select all

``````  Rule 1. w[n] = n
Rule 2. w^(a+1)[n] = (w^a)(n)
Rule 3. If a is a limit ordinal and is less than e_0, (w^a)[n] = w^(a[n])
The next rule is a modified version of the corresponding Wainer hierarchy rule, but rule 5 is from the Wainer hierarchy again.
Rule 4. e_0 = w
Rule 5. e_0[n+1] = w^e_0[n]
The next rules are stolen from the Veblen hierarchy.
Rule 6. All rules after rule 6 have a colon rather than a period after the number
Rule 7: phi_(a+1)(0)[n] = phi^n_a(0) (which means phi_a(0) iterated n times, not phi^(n_a)(0)), for all a > 0
Rule 8: phi_(a+1)(b+1)[n] = phi^n_a(phi_(a+1)(b)+1)
Rule 9: if b is a limit ordinal smaller than phi_a(b) phi_a(b)[n] = phi_a(b[n])
Rule 10: if a is a limit ordinal smaller than gamma_0, phi_a(0)[n] = phi_(a[n])(0)
Rule 11: if a is a limit ordinal, phi_a(b+1)[n] = phi_a[n](phi_a(b)+1)
Rule 12: gamma_(0)[n+1] = phi_(gamma_0[n])(0)
Rule 13: gamma_(0) = w
Rule 14: (a+b)[n] = a+(b[n]), a >=b>0, a & b both limit ordinals, a < a+b, b < a+b where a is the minimal ordinal satisfying these conditions``````
Apparently the 14th one isn't specific enough yet so I'm asking the googology wiki user P進大好きbot (who is almost certainly not a bot) how to do it right

Really, to make fundamental sequences start at 0 rather than 1, it could alternatively be

Code: Select all

``````Changed rules are indented further
Rule 1. w[n] = n
Rule 2. w^(a+1)[n] = (w^a)(n)
Rule 3. If a is a limit ordinal and is less than e_0, (w^a)[n] = w^(a[n])
The next rule is a modified version of the corresponding Wainer hierarchy rule, but rule 5 is from the Wainer hierarchy again.
Rule 4. e_0 = 1
Rule 5. e_0[n+1] = w^e_0[n]
The next rules are stolen from the Veblen hierarchy.
Rule 6. All rules after rule 6 have a colon rather than a period after the number
Rule 7: phi_(a+1)(0)[n] = phi^n_a(0) (which means phi_a(0) iterated n times, not phi^(n_a)(0)), for all a > 0
Rule 8: phi_(a+1)(b+1)[n] = phi^n_a(phi_(a+1)(b)+1)
Rule 9: if b is a limit ordinal smaller than phi_a(b) phi_a(b)[n] = phi_a(b[n])
Rule 10: if a is a limit ordinal smaller than gamma_0, phi_a(0)[n] = phi_(a[n])(0)
Rule 11: if a is a limit ordinal, phi_a(b+1)[n] = phi_a[n](phi_a(b)+1)
Rule 12: gamma_(0)[n+1] = phi_(gamma_0[n])(0)
Rule 13: gamma_(0) = w
Rule 14: (a+b)[n] = a+(b[n]), a >=b>0, a & b both limit ordinals, a < a+b, b < a+b where a is the minimal ordinal satisfying these conditions``````
By starting the sequences at 0 rather than 1; however this would make the dco sequence even stronger since gamma_0 would be phi_w(0)
Last edited by Moosey on October 11th, 2019, 12:51 pm, edited 1 time in total.
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### Re: Ordinals in googology

Is the diagonalization omega, epsilon, delta, gamma, etc. literally just w^^...n times...^w?

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### Re: Ordinals in googology

testitemqlstudop wrote:
October 13th, 2019, 11:39 pm
Is the diagonalization omega, epsilon, delta, gamma, etc. literally just w^^...n times...^w?
Which ordinal is delta? Phi_w(0) = sup{w,e0,z0,h0...}?
Regardless, no-- none of those are w^w...
I believe the diagonalisation w^2,w^w,e_0,(everybody disagrees here, but Saibian and Bowers both say it's gamma_0. Basically, it's just the first fixed point of w^^n),etc... is
the sequence w{n}w (starting at n=0)
Obviously sup{w^2,w^w,e_0,...} = "w{w}w" which could be any of a number of ordinals but even the most conservative bounds would put it at phi_w(0)
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### Re: Ordinals in googology

Moosey wrote:
October 14th, 2019, 6:25 am
h0
People usually use n0 for that
Moosey wrote:
October 14th, 2019, 6:25 am
(everybody disagrees here, but Saibian and Bowers both say it's gamma_0. Basically, it's just the first fixed point of w^^n)
Actually almost all non-Saibian ordinal hyperoperators have w^^^w=z0
I like making rules

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### Re: Ordinals in googology

testitemqlstudop wrote:
October 13th, 2019, 11:39 pm
Is the diagonalization omega, epsilon, delta, gamma, etc. literally just w^^...n times...^w?
By most definitions, w ^{n} w is phi(n-1,0).
Gamma_0 is then w ^{w ^{w^{ ... } w} w} w, in keeping with the "standard" result f_(w+1) = s_(G0)
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### Re: Ordinals in googology

x-post of the "array hierarchy", or ah
(Copied from a personal blog of mine)

Note: ah, like all hierarchies, is technically illdefined without fundamental sequences. Just use the fundamental sequence rules of, say, the extended Veblen hierarchy, unless you need others (in which case, use 'em!)
Rules:
I define a#b to be the concatenation of the arrays a and b, and n@m to be an array of m n’s — specifically,
n@m =
n, m = 1
{n}#(n@(m-1))
Additionally, I define \$_n as any entries (including no entries) in an array– it’s my symbol for we-don’t-care entries
In any one use of any one rule, if n is the same, \$_n is the same

1. If ah is iterated: ah^a_n {\$_0} is ah_ah_ah_ah…_n {\$_0} {\$_0} {\$_0} {\$_0}… with a ah’s and {\$_0}’s
Formally, define g(a,n,B) =
ah_g(a-1,n,B) {B}, a > 1;
n, a = 0.
ah^a_n {\$_0} = g(a,n,\$_0)
2. ah_n{\$_1,z} = ah_n{\$_1}, z = 0
alternatively, ah_n ({\$_1}#{z}) = ah_n {\$_1}, z = 0
3. ah_n{} = n+1
4. ah_n{a+1,\$_2} = ah^n_n{a,\$_2}
5. ah_n{a,\$_3} = ah_n{a[n],\$_3}, a a lim ord
6. ah_n ((0@b)#{a+1,\$_4}) = ah_(n+1) (((a+1)@b)#{a,\$_4}), b > 0
7. ah_n ((0@b)#{a,\$_5}) = ah_n ((a@b)#{a[n],\$_5}), a a lim ord & b > 0

Even ah_1(0,w+1) = ah_ah_ah_ah_2{w,2}{0,w}{1,w}{w,w}, which is probably larger than g_64 (welcome to hilarious understatements, episode w+1!)

How powerful is ah_n((0@n)#{w+1}) ?

EDIT: partially extended:

8a. ah_n((0@a)#{b,\$_6}) = ah_n((0@(a[n])#{b,\$_6}), a a lim ord

Anyone wanna add an 8b so we can have a sort of "transfinite amount of entries" thing?

(Note: ah_n((0@n)#{w+1}) = ah_n((0@w)#{w+1}))
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

testitemqlstudop
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### Re: Ordinals in googology

And to say something constructive I'm going to use p-adic's fundamental sequence argument
8a. ah_n((0@a)#{b,\$_6}) = ah_n((0@(a[n])#{b,\$_6}), a a lim ord
That does not work with your recursive definition at all.

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### Re: Ordinals in googology

testitemqlstudop wrote:
October 24th, 2019, 8:45 pm
You never had that problem before.
LaTeX has confused me, and we've never cared much about whether one used LaTeX.
testitemqlstudop wrote:
October 24th, 2019, 8:45 pm
And to say something constructive I'm going to use p-adic's fundamental sequence argument
8a. ah_n((0@a)#{b,\$_6}) = ah_n((0@(a[n])#{b,\$_6}), a a lim ord
That does not work with your recursive definition at all.
That is not really the type of response that I wanted, given that what I wrote at the top was essentially hinting at
"FSes are gonna cause some illdefinedness and stuff, but can we just ignore that for our purposes?"
Yes, I am well aware that FSes have some problems with computability (and, without a definition, definedness) but P-bot has already told me that, so for the moment can we put aside those problems?
Ergo: you just told me something I already knew.
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"