ConwayLife.com

Updated Rules:
A: NO SALAD NUMBERS! [ie no rayo's number, no TREE[TREE[TRREE[G[64]], no using overused functions more than once, and NO tree(x)]
B: As original as possible please. Ones that i consider good will be favourited by me.
C: It must be well defined, computable, and finite.
D: NO QUOTES UNLESS NESSECARY
E: Post limit of 3/hour
F: you cant increase the number too much every turn. [decided by the community on who breaks this]

I'll start: 3^(3^3).

ababa11e wrote: July 27th, 2024, 1:42 pm Rules:
A: NO SALAD NUMBERS! [ie no rayo's number, no TREE[TREE[TRREE[G[64]], no using overused functions more than once, and NO tree(x)]
B: As original as possible please. Ones that i consider good will be favourited by me.
C: It must be well defined, computable, and finite.

I'll start: 3^(3^3).

That’s 19683, I’ll go with {10,5} (or 10^5) which is 100,000.

Banananananan wrote: July 27th, 2024, 6:09 pm

ababa11e wrote: July 27th, 2024, 1:42 pm Rules:
A: NO SALAD NUMBERS! [ie no rayo's number, no TREE[TREE[TRREE[G[64]], no using overused functions more than once, and NO tree(x)]
B: As original as possible please. Ones that i consider good will be favourited by me.
C: It must be well defined, computable, and finite.

I'll start: 3^(3^3).

That’s 19683, I’ll go with {10,5} (or 10^5) which is 100,000.

Okay then, i'll go for 9^(9^9) instead, or about 1.9662705*10^77.

ababa11e wrote: July 27th, 2024, 6:21 pm

Banananananan wrote: July 27th, 2024, 6:09 pm

ababa11e wrote: July 27th, 2024, 1:42 pm Rules:
A: NO SALAD NUMBERS! [ie no rayo's number, no TREE[TREE[TRREE[G[64]], no using overused functions more than once, and NO tree(x)]
B: As original as possible please. Ones that i consider good will be favourited by me.
C: It must be well defined, computable, and finite.

I'll start: 3^(3^3).

That’s 19683, I’ll go with {10,5} (or 10^5) which is 100,000.

Okay then, i'll go for 9^(9^9) instead, or about 1.9662705*10^77.

Fine, how bout 1 quattourquadragenseptuagentillion?

Banananananan wrote: July 27th, 2024, 6:43 pm

ababa11e wrote: July 27th, 2024, 6:21 pm

Banananananan wrote: July 27th, 2024, 6:09 pm
That’s 19683, I’ll go with {10,5} (or 10^5) which is 100,000.

Okay then, i'll go for 9^(9^9) instead, or about 1.9662705*10^77.

Fine, how bout 1 quattourquadragenseptuagentillion?

A millinillion, 10^3003.

ababa11e wrote: July 27th, 2024, 8:38 pm

Banananananan wrote: July 27th, 2024, 6:43 pm

ababa11e wrote: July 27th, 2024, 6:21 pm
Okay then, i'll go for 9^(9^9) instead, or about 1.9662705*10^77.

Fine, how bout 1 quattourquadragenseptuagentillion?

A millinillion, 10^3003.

Quinmillillion
10^15003.

Banananananan wrote: July 27th, 2024, 8:51 pm

ababa11e wrote: July 27th, 2024, 8:38 pm

Banananananan wrote: July 27th, 2024, 6:43 pm
Fine, how bout 1 quattourquadragenseptuagentillion?

A millinillion, 10^3003.

Quinmillillion
10^15003.

2^4211744 or {2, 4211744}
Approximately equal to 1.89695836432202*10^1262741

This is the number of Radius 2 Isotropic Non-Totalistic rules.

Banananananan wrote: July 27th, 2024, 8:51 pm

ababa11e wrote: July 27th, 2024, 8:38 pm

Banananananan wrote: July 27th, 2024, 6:43 pm
Fine, how bout 1 quattourquadragenseptuagentillion?

A millinillion, 10^3003.

Quinmillillion
10^15003.

Millinilli-millinillion.
10^(3*((3003*(10^3003))+1))

ababa11e wrote: July 27th, 2024, 9:43 pm

Banananananan wrote: July 27th, 2024, 8:51 pm

ababa11e wrote: July 27th, 2024, 8:38 pm
A millinillion, 10^3003.

Quinmillillion
10^15003.

Millinilli-millinillion.
10^(3*((3003*(10^3003))+1))

Megillion
10^(3*((3*(10^(10^6)))+3))

Banananananan wrote: July 27th, 2024, 9:58 pm

ababa11e wrote: July 27th, 2024, 9:43 pm

Banananananan wrote: July 27th, 2024, 8:51 pm
Quinmillillion
10^15003.

Millinilli-millinillion.
10^(3*((3003*(10^3003))+1))

Megillion
10^(3*((3*(10^(10^6)))+3))

110^(18^(10^(3*((3*(10^(10^6)))+3))))

Haycat2009 wrote: July 28th, 2024, 4:38 am

Banananananan wrote: July 27th, 2024, 9:58 pm

ababa11e wrote: July 27th, 2024, 9:43 pm
Millinilli-millinillion.
10^(3*((3003*(10^3003))+1))

Megillion
10^(3*((3*(10^(10^6)))+3))

110^(18^(10^(3*((3*(10^(10^6)))+3))))

Fine. Decker. (10^^10)

Banananananan wrote: July 28th, 2024, 5:15 am

Haycat2009 wrote: July 28th, 2024, 4:38 am

Banananananan wrote: July 27th, 2024, 9:58 pm
Megillion
10^(3*((3*(10^(10^6)))+3))

110^(18^(10^(3*((3*(10^(10^6)))+3))))

Fine. Decker. (10^^10)

You asked for this:
A(9^9,9^9)
Yes, it is computable. Time is the only concern

Haycat2009 wrote: July 28th, 2024, 5:37 am

Banananananan wrote: July 28th, 2024, 5:15 am

Haycat2009 wrote: July 28th, 2024, 4:38 am
110^(18^(10^(3*((3*(10^(10^6)))+3))))

Fine. Decker. (10^^10)

You asked for this:
A(9^9,9^9)
Yes, it is computable. Time is the only concern

10000000000[10000000000]10000000000. x[y]z means y-operation.

unname4798 wrote: July 28th, 2024, 6:18 am

Haycat2009 wrote: July 28th, 2024, 5:37 am

Banananananan wrote: July 28th, 2024, 5:15 am
Fine. Decker. (10^^10)

You asked for this:
A(9^9,9^9)
Yes, it is computable. Time is the only concern

10000000000[10000000000]10000000000. x[y]z means y-operation.

f_φ(f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000))(f_φ(1,1,1,1)(1000))
f_[] = fast growing heirarchy
is it bigger? idk

ababa11e wrote: July 28th, 2024, 8:20 am

unname4798 wrote: July 28th, 2024, 6:18 am

Haycat2009 wrote: July 28th, 2024, 5:37 am
You asked for this:
A(9^9,9^9)
Yes, it is computable. Time is the only concern

10000000000[10000000000]10000000000. x[y]z means y-operation.

f_φ(f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000),f_φ(1,1,1,1)(1000))(f_φ(1,1,1,1)(1000))
f_[] = fast growing heirarchy
is it bigger? idk

How about we use BEAF/BAN from now on, since I don’t know FGH I’ll go with Goobol or (10,100[2]2)

f([10,10,10,10,10]) where f(n) = [f(n-1),f(n-1),f(n-1),f(n-1),f(n-1)] and f(1) = 3
hopefully this is larger
(i know a thing or two about googology)

eRroR_6o6 wrote: August 4th, 2024, 1:23 pm f([10,10,10,10,10]) where f(n) = [f(n-1),f(n-1),f(n-1),f(n-1),f(n-1)] and f(1) = 3
hopefully this is larger
(i know a thing or two about googology)

okay, to start this: using the same function, f([x,x,x,x,x)], where x is f([10,10,10,10,10]), which i'll shorten to f_2([10,10,10,10,10)]
now, define f_w([x,x,x,x,x...]) where it is equal to f_y([x,x,x,x,x...]) where y = f_1([x,x,x,x,x...]).
now, my number is f_w([100,100,100,100,100]), of this modified version of ErRoR's function

ababa11e wrote: August 4th, 2024, 8:35 pm

eRroR_6o6 wrote: August 4th, 2024, 1:23 pm f([10,10,10,10,10]) where f(n) = [f(n-1),f(n-1),f(n-1),f(n-1),f(n-1)] and f(1) = 3
hopefully this is larger
(i know a thing or two about googology)

okay, to start this: using the same function, f([x,x,x,x,x)], where x is f([10,10,10,10,10]), which i'll shorten to f_2([10,10,10,10,10)]
now, define f_w([x,x,x,x,x...]) where it is equal to f_y([x,x,x,x,x...]) where y = f_1([x,x,x,x,x...]).
now, my number is f_w([100,100,100,100,100]), of this modified version of ErRoR's function

I'm awful at googology, but I'll try to come up with something that isn't too cheese nor too confusing.

So f_y([x,x,x,x,x]) is just a recursive function with y iterations. I'm going to make a copy of this function f and call it jeff (don't ask me why), except it's a single-argument function where the number of x's is determined by the value of x. jeff(1) is 3, jeff(2) is [3,3], jeff(3) is [jeff(2), jeff(2), jeff(2)], jeff(4) is [jeff(3),jeff(3),jeff(3),jeff(3)], and so on.

jeff_n(x) is jeff(jeff(...jeff(x)...)) with n iterations of jeff.
jeff__n(x) is jeff_[jeff_...[jeff_n(x)]...](x) with jeff_n(x) jeffs. The square brackets are just for grouping.
jeff___n(x) is jeff__[jeff__...[jeff__n(x)]...](x) with jeff__n(x) jeffs. You can probably see where this is going.
Define jeff__...__n(x) for all natural numbers of underscores. I want to stop right here, but I have some ideas to keep going...

jeff__...__n(x) with jeff_n(x) underscores is jeff_1_n(x).
jeff__...__n(x) with jeff__n(x) underscores is jeff_2_n(x), and if it had jeff___n(x) underscores it would be jeff_3_n_(x).
jeff__...__n(x) with jeff_a_n(x) underscores is jeff_a__n(x), and if it had jeff_a__n(x) underscores it would be jeff_a___n(x).
jeff_a__...__n(x) with jeff_a_n(x) underscores is jeff__a_n(x).
jeff_a__...__n(x) with jeff__a_n(x) underscores is jeff__a__n(x). You know the drill by now.
jeff__a__...__n(x) with jeff__a_n(x) underscores is jeff___a_n(x).

jeff__...__a_n(x) with jeff_a_n(x) underscores is joe(x) where a, n, and x have the same value.
My number is joe(3800).

what about, Ababa's T function. T[n] = the smallest prime number you cannot make by adding 2 smaller primes, below 2n.
T[2]= 11, T[3] = ...something,
T[T[2]] is probably higher than 10 billion,
T_3[2] = probably higher than 10^^3
T_10[2] is probably higher than 10^^^2
now one above that is T_T[2][2], which i'm gonna call T_w[2], cuz its T[2] as the iteration amount
T_w+n[2] = T_T_n-1[2][2]
T_w2[2] = T_T_T_n-1[2][2][2]
now if you do this T[n] times you get T_w^2[2]
you you can keep doing this infinitely...until you get T_e0[2]
you can swap e0 for any ordinal, and my number is T_φ(10,0)[2], where φ(10,0) is Veblen's function, a way of making large ordinals.

ababa11e wrote: August 5th, 2024, 12:40 am what about, Ababa's T function. T[n] = the smallest prime number you cannot make by adding 2 smaller primes, below 2n.
T[2]= 11, T[3] = ...something,
T[T[2]] is probably higher than 10 billion,
T_3[2] = probably higher than 10^^3
T_10[2] is probably higher than 10^^^2
now one above that is T_T[2][2], which i'm gonna call T_w[2], cuz its T[2] as the iteration amount
T_w+n[2] = T_T_n-1[2][2]
T_w2[2] = T_T_T_n-1[2][2][2]
now if you do this T[n] times you get T_w^2[2]
you you can keep doing this infinitely...until you get T_e0[2]
you can swap e0 for any ordinal, and my number is T_φ(10,0)[2], where φ(10,0) is Veblen's function, a way of making large ordinals.

Can you help me understand why this number is finite?

number of 9001-glider collisions

tommyaweosme wrote: August 5th, 2024, 9:10 pm number of 9001-glider collisions

Infinite due to RCT.

hotdogPi wrote: August 5th, 2024, 9:16 pm

tommyaweosme wrote: August 5th, 2024, 9:10 pm number of 9001-glider collisions

Infinite due to RCT.

how about a number of 4-glider collisions

ababa11e wrote: August 5th, 2024, 12:40 am what about, Ababa's T function. T[n] = the smallest prime number you cannot make by adding 2 smaller primes, below 2n.
T[2]= 11, T[3] = ...something,
T[T[2]] is probably higher than 10 billion,
T_3[2] = probably higher than 10^^3
T_10[2] is probably higher than 10^^^2
now one above that is T_T[2][2], which i'm gonna call T_w[2], cuz its T[2] as the iteration amount
T_w+n[2] = T_T_n-1[2][2]
T_w2[2] = T_T_T_n-1[2][2][2]
now if you do this T[n] times you get T_w^2[2]
you you can keep doing this infinitely...until you get T_e0[2]
you can swap e0 for any ordinal, and my number is T_φ(10,0)[2], where φ(10,0) is Veblen's function, a way of making large ordinals.

correct me if i'm wrong, but since that involves prime numbers, it should be uncomputable.

unname4798 wrote: August 6th, 2024, 12:02 am how about a number of 4-glider collisions

Still infinite.