My Number is WAYYY larger [game]

[get_Snacked]

BTW, current growth rate for the message above me is f_{ω2}(f_{ω+1}(64)), i think.
edit: well, it started a new page. now i look stupid, contrary to my profile picture.
i'll go with {3,3,1,4} using BEAF, approx. f_{ω3}(3).

[rutabaga]

BTW, current growth rate for the message above me is f_{ω2}(f_{ω+1}(64)), i think.

How do you calculate the growth in FGH in general? I can see how you figured it out in this case, but it would be helpful to know lol
My number is just f_{w4}(10^100) because I don't have any good ideas :P

[get_Snacked]

BTW, current growth rate for the message above me is f_{ω2}(f_{ω+1}(64)), i think.

How do you calculate the growth in FGH in general? I can see how you figured it out in this case, but it would be helpful to know lol

it requires some pretty simple knowledge of patterns; i.e.x{{1}}x = x{x{...{x}...}x}x with 2x-1 x's is f_{ω+1}(x).
until you reach ω^2. i don't really know how to help you beyond that, but my confidence in accuracy of growth rates remains and slowly disappears until i have no idea what i'm doing at ψ_0(Ι).

[ababa11e]

BTW, current growth rate for the message above me is f_{ω2}(f_{ω+1}(64)), i think.

How do you calculate the growth in FGH in general? I can see how you figured it out in this case, but it would be helpful to know lol
My number is just f_{w4}(10^100) because I don't have any good ideas

B(n1,n2) = B(1n,B(n1^B(n1-1,n2-1),n2-1), B(0,n) = n, and B(n,0) = n^^^...^^^n with b(n-1,0) arrows, where B(0,0) = G(64)
is the base function.
B_2(n,n) = B(B(n1,n2),B(n1,n2))
B_k(n,n) = B_k-1(B_k-1(n1,n2),B_k-1(n1,n2))
my number is B_G64(G64,G64), if its smaller, ill change it

[tommyaweosme]

how about a number of 4-glider collisions

Still infinite.

The list of single objects that can be synthesized with 4 gliders is pretty much complete. If we don't know of a 4-glider synthesis, it probably doesn't exist.

[confocaloid]

how about a number of 4-glider collisions

Still infinite.

The list of single objects that can be synthesized with 4 gliders is pretty much complete. If we don't know of a 4-glider synthesis, it probably doesn't exist.

That's pretty much unsubstantiated, and (considering the date of the forum post) just plain wrong.
Check external links in 4-glider collision, in particular this: viewtopic.php?f=2&t=1394&start=150#p146998

Currently, it seems unfeasible (practically impossible) to prove any claim of the sort "we know all objects that have 4G syntheses".
(While disproving takes just one discovery of a new 4G synthesis for an object that previously wasn't known to be constructible with less than 5 gliders.)

[rutabaga]

eg no rayo's number, no TREE[TREE[TREE[G[64]], no using overused functions more than once, and NO tree(x)

so I can use overused functions that aren't TREE or RAYO exactly once? In that case, SSCG(10^100).
(this function is stronger than TREE but weaker than BB, SCG, or RAYO)

[unname4798]

so I can use overused functions that aren't TREE or RAYO exactly once?

You can't, as the game rules state.
Anyways, new game.
100

[get_Snacked]

B(n1,n2) = B(1n,B(n1^B(n1-1,n2-1),n2-1), B(0,n) = n, and B(n,0) = n^^^...^^^n with b(n-1,0) arrows, where B(0,0) = G(64)
is the base function.
B_2(n,n) = B(B(n1,n2),B(n1,n2))
B_k(n,n) = B_k-1(B_k-1(n1,n2),B_k-1(n1,n2))
my number is B_G64(G64,G64), if its smaller, ill change it

B(1,1)
B(1,B(1^B(0,0),0))
B(1,B(1^G64,0))
B(1,B(1,0))
B(1,1{B(0,1)}1)
B(1,1{1}1)
B(1,1^1)
B(1,1)
infinite loop.

Anyways, new game.
100

please don't keep doing that until nobody understands what's happening. then it's valid to restart.

my number is {10,10,1,1,2} using BEAF, approx. f_{ω^2}(10).

[unname4798]

{10,10,10,10,10}

[get_Snacked]

time to define a simple array notation.

domain: {0}∪N
where # is any string
where Z is any string of 0's
x[0] = x+1
x[a,#] (a>0) = ((x[a-1,#])...[a-1,#]) with x [a-1,#]'s
x[Z,0,a,#] = x[Z,x,a-1,#]

my number is 3[0,...,0,1] with 3[0,0,0,1] 0's, which is exactly f_{ω^{f_{ω^ω}(3)}}(3).

[get_Snacked]

12-hour rule:
Z[x] = x string of 0's
3[Z[3[Z[3[Z[3],1]]]],1]
which is exactly f_{ω^f_{ω^f_{ω^ω}(3)}(3)}(3). (still waiting on f_{ω^(ω+1)})

[rutabaga]

7x7x7x7 tetrational BEAF array where all arguments are 77 (tetrational BEAF is around f_{w^^3}(x) but idk what x is in this case)

i don't understand the Z thing so i'll just stick to BEAF for now. i wonder how quickly we'll finally pass my mn(x) function...

[get_Snacked]

7x7x7x7 tetrational BEAF array where all arguments are 77 (tetrational BEAF is around f_{w^^3}(x) but idk what x is in this case)

i don't understand the Z thing so i'll just stick to BEAF for now. i wonder how quickly we'll finally pass my mn(x) function...

is that just 77^^7 & 7? if so, that's, like, f_{ω^^7}(77) or something. i'm gonna have to ask someone, though.

[hotdogPi]

77^^7 & 7

If I calculated correctly, this number is equal to 5, assuming & is bitwise AND.

[get_Snacked]

If I calculated correctly, this number is equal to 5, assuming & is bitwise AND.

how did you manage to calculate 77^^7? or was there some shortcut?

[hotdogPi]

77 to the power of any odd number is 5 mod 8.

[ababa11e]

12-hour rule:
Z[x] = x string of 0's
3[Z[3[Z[3[Z[3],1]]]],1]
which is exactly f_{ω^f_{ω^f_{ω^ω}(3)}(3)}(3). (still waiting on f_{ω^(ω+1)})

ill just... Z^{k}[n] = Z[1Z^{k-1}[n]]. Where Z^1[n] = Z[n]
What if I just... Z'[n] = Z[1Z^{1Z'[n-1]}[n]], where Z'[0] = 100 0's.
Z^3[1] = 10^^^...1000000001 ARROWS...^^^2
Z'[1] = Z[1Z^{10^100}[1]] = A very BIG number.
ill just 2... Z'^{k}[n] = Z'[1Z'^{k-1}[n]]. Where Z'^1[n] = Z'[n]
My number is Z'^{Z'^{Z'^{Z'^{Z'^{100}[100]}[100]}[100]}[100]}[100].

[rutabaga]

12-hour rule:
Z[x] = x string of 0's
3[Z[3[Z[3[Z[3],1]]]],1]
which is exactly f_{ω^f_{ω^f_{ω^ω}(3)}(3)}(3). (still waiting on f_{ω^(ω+1)})

ill just... Z^{k}[n] = Z[1Z^{k-1}[n]]. Where Z^1[n] = Z[n]
What if I just... Z'[n] = Z[1Z^{1Z'[n-1]}[n]], where Z'[0] = 100 0's.
Z^3[1] = 10^^^...1000000001 ARROWS...^^^2
Z'[1] = Z[1Z^{10^100}[1]] = A very BIG number.
ill just 2... Z'^{k}[n] = Z'[1Z'^{k-1}[n]]. Where Z'^1[n] = Z'[n]
My number is Z'^{Z'^{Z'^{Z'^{Z'^{100}[100]}[100]}[100]}[100]}[100].

I don't think this is larger than around f_{w^^7}(77), but then again, I have no clue how large this is.

[ababa11e]

12-hour rule:
Z[x] = x string of 0's
3[Z[3[Z[3[Z[3],1]]]],1]
which is exactly f_{ω^f_{ω^f_{ω^ω}(3)}(3)}(3). (still waiting on f_{ω^(ω+1)})

ill just... Z^{k}[n] = Z[1Z^{k-1}[n]]. Where Z^1[n] = Z[n]
What if I just... Z'[n] = Z[1Z^{1Z'[n-1]}[n]], where Z'[0] = 100 0's.
Z^3[1] = 10^^^...1000000001 ARROWS...^^^2
Z'[1] = Z[1Z^{10^100}[1]] = A very BIG number.
ill just 2... Z'^{k}[n] = Z'[1Z'^{k-1}[n]]. Where Z'^1[n] = Z'[n]
My number is Z'^{Z'^{Z'^{Z'^{Z'^{100}[100]}[100]}[100]}[100]}[100].

I don't think this is larger than around f_{w^^7}(77), but then again, I have no clue how large this is.

It's been a while since a number was posted, so...
f_{w^^n}(n) = F[n]
F[1] = 2
F[2] ≈ 10^^10
F[3] ≈10^^^10
F[4] ≈ 10^^^...10^10^153 arrows...^^^10
F[5] ≈ 10^^^...10^^4 arrows...^^^10
F[7] > my previous number.
F'[n] = F[n] iterated F'[n-1] times, where F'[1] = F[10^100]
My Number is F'[F'[F'[F'[10^100]]]]

[rutabaga]

ill just... Z^{k}[n] = Z[1Z^{k-1}[n]]. Where Z^1[n] = Z[n]
What if I just... Z'[n] = Z[1Z^{1Z'[n-1]}[n]], where Z'[0] = 100 0's.
Z^3[1] = 10^^^...1000000001 ARROWS...^^^2
Z'[1] = Z[1Z^{10^100}[1]] = A very BIG number.
ill just 2... Z'^{k}[n] = Z'[1Z'^{k-1}[n]]. Where Z'^1[n] = Z'[n]
My number is Z'^{Z'^{Z'^{Z'^{Z'^{100}[100]}[100]}[100]}[100]}[100].

I don't think this is larger than around f_{w^^7}(77), but then again, I have no clue how large this is.

It's been a while since a number was posted, so...
f_{w^^n}(n) = F[n]
F[1] = 2
F[2] ≈ 10^^10
F[3] ≈10^^^10
F[4] ≈ 10^^^...10^10^153 arrows...^^^10
F[5] ≈ 10^^^...10^^4 arrows...^^^10
F[7] > my previous number.
F'[n] = F[n] iterated F'[n-1] times, where F'[1] = F[10^100]
My Number is F'[F'[F'[F'[10^100]]]]

F[n] = f_{w^^w}(n) = f_{e_0}(n)
F'[n] = f_{e_0 + 2}(n) because there are two additional levels of recursivity (is that a word lol).

Since this is a FGH battle now, I'll go with f_{Γ_0}(10^^10).
We have now passed all other rounds in the thread.
If it's too big, I'll go with f_{w+e0}(10^^10).

[ababa11e]

I don't think this is larger than around f_{w^^7}(77), but then again, I have no clue how large this is.

It's been a while since a number was posted, so...
f_{w^^n}(n) = F[n]
F[1] = 2
F[2] ≈ 10^^10
F[3] ≈10^^^10
F[4] ≈ 10^^^...10^10^153 arrows...^^^10
F[5] ≈ 10^^^...10^^4 arrows...^^^10
F[7] > my previous number.
F'[n] = F[n] iterated F'[n-1] times, where F'[1] = F[10^100]
My Number is F'[F'[F'[F'[10^100]]]]

F[n] = f_{w^^w}(n) = f_{e_0}(n)
F'[n] = f_{e_0 + 2}(n) because there are two additional levels of recursivity (is that a word lol).

Since this is a FGH battle now, I'll go with f_{Γ_0}(10^^10).
We have now passed all other rounds in the thread.
If it's too big, I'll go with f_{w+e0}(10^^10).

too big of a jump, ill continue from f_{w+e0}(10^^10) with f_{e0*w}(10)

[get_Snacked]

7x7x7x7 tetrational BEAF array where all arguments are 77 (tetrational BEAF is around f_{w^^3}(x) but idk what x is in this case)

i don't understand the Z thing so i'll just stick to BEAF for now. i wonder how quickly we'll finally pass my mn(x) function...

hi guys! i finally found it out!
so tetrational arrays are any arrays below the level of X^^X, which basically means below the FGH level of epsilon naught. so that basically does nothing and this is actually 7^4 & 77, which is, like, f_{ω^4}(7).
so congratulations, that did not do anything.

HOWEVER, if you had just said a 7^^7 array of 77's, that would've done what you were wishing for, giving f_{ω^^7}(7). so i know you probably don't want to restart from here now, but here's what i found out about it, i guess.

[rutabaga]

Fine. I'm sorry, but it has to be done: TREE(3).

Ideas to progress: You can make some recursive version of TREE(n), you can whip out another function like SSCG or BB (not RAYO quite yet, please), or you could shove TREE(n) into some sort of thing involving FGH.

Please don't start another round just because I mentioned TREE. I just wanted to change things up a bit because right now it's just a battle of FGH and pure confusion. TREE is not remotely close to unbeatable, it's just unoriginal-- and so is the FGH.

[CARuler]

TREE(4)