ConwayLife.com

ababa11e wrote: August 15th, 2024, 12:58 pm

unname4798 wrote: August 15th, 2024, 12:52 pm

ababa11e wrote: August 15th, 2024, 12:50 pm
New idea: D heriarchy: D_0(n) = n^2, D_1(n) = D_0(D_0(...D_0(n) TIMES...D_0(n))), or n^(2(n^2)) or roughly n^n^2, D_2(n) = D_1(D_1(...D_1(n) TIMES...D_1(n))) is roughly n^^3n, now, D_100(n) is roughly b_100(n), but now write D_n(n) as D_(1,0)(n), where the array goes like ...n^2, n, 1,
write D_(1,0,0,0...n times)(n) = D_(1,0/1)(n), D_(1,0,0,0...n times/1)(n) = D_(1,0/2)(n), D_(1,0/n)(n) = D_(1,0/1,0)(n) [the /1,0 can be stacked), and now, call E_0(n) = D_(1,0/1,0,0,0,0,0,0...G64 0s)(n),
E_1(n) = D_(1,0/1,0,0,0,0,0,0...E_0(n) 0s)(n), my number is E_(E_G64)(G64)

E__(n) is E_(n^[n arrows]^n)(n^[n arrows]^n)
E__(G64)

Well, i can go further F0_0(n) = E_E_E_E_...E_0(n) E's...E_0(n)(n)(n)..., F0_1 = E_E_E_E_...E_1(n) E's...E_1(n)(n)(n)... F1_0(n) = F0_F0_ F0_ F0_...F0_0(n) F0's...F0-0(n)(n)(n)... F1_1 = F0_F0_ F0_ F0_...F0_0(n) F0's...F0_1(n)(n)(n)..., F2_0 = F1_F1_ F1_ F1_...F1_0(n) F0's...F1_0(n)(n)... my number is F(G64)_G64

H(n) is F(n^[n arrows]^n)_n^[n arrows]^n
H(G64)

unname4798 wrote: August 15th, 2024, 1:02 pm

ababa11e wrote: August 15th, 2024, 12:58 pm

unname4798 wrote: August 15th, 2024, 12:52 pm
E__(n) is E_(n^[n arrows]^n)(n^[n arrows]^n)
E__(G64)

Well, i can go further F0_0(n) = E_E_E_E_...E_0(n) E's...E_0(n)(n)(n)..., F0_1 = E_E_E_E_...E_1(n) E's...E_1(n)(n)(n)... F1_0(n) = F0_F0_ F0_ F0_...F0_0(n) F0's...F0-0(n)(n)(n)... F1_1 = F0_F0_ F0_ F0_...F0_0(n) F0's...F0_1(n)(n)(n)..., F2_0 = F1_F1_ F1_ F1_...F1_0(n) F0's...F1_0(n)(n)... my number is F(G64)_G64

H(n) is F(n^[n arrows]^n)_n^[n arrows]^n
H(G64)

thats it, time to stack, I0_0(n) = Fn_n(n), I_0,1(n) = F(Fn_n(n))_(Fn_n(n))(n), and repeat this till you get I1_0(n), = I0_n(n), I1_1(n) = I0_(I0,n(n))(I0,n(n)), and keep on repeating until you get to In_n(n), whill ill shorten to J_n, J_J_J_J_J_J_J_...J_100 J's..._J_100 is my number

Hi(x) = J_↪[Hi(x-1)](Hi(x-1)). The function it "returns to" is D_0(x).

Hi(n,x) = Hi(n-1,Hi(n-1,Hi(...Hi(n-1,x) times...n-1,Hi(n-1,x)...))) where Hi(0,x) = Hi(x)

Hi(n2,n1,x) = Hi(n2-1,Hi(n2-1,n1,x),Hi(n2-1,Hi(n2-1,n1,x),Hi(...Hi(n2-1,n1,x) times...n2-1,n1,x...))) where Hi(0,n1,x) = Hi(n1,x)

Hi(nA,..n2,n1,x) = Hi(nA-1,..Hi(nA-1,..n2,n1,x),Hi(nA-1,..n2,n1,x),Hi(nA-1,..Hi(nA-1,..n2,n1,x),Hi(nA-1,..n2,n1,x),Hi(...Hi(nA-1,..n2,n1,x) times...(nA,..n2,n1,x)...))) where Hi(0, n[A-1],..n2,n1,x) = Hi(n[A-1],..n2,n1,x)

Hi(\x) = Hi(Hi(x),Hi(x),...Hi(x) times...Hi(x)).

Hi↪[Hi(\Hi(\Goobol))](Hi(\Hi(\...Hi(\Goobol) times...Hi(\Goobol)...))) is my number, where Goobol is {10,100(1)2} in BEAF, or 10(100)10 in HAN.

Twelve-hour rule. I'm going to make a very big jump.

Let beef(x) equal the result of a beef(x-1)-dimensional array, with beef(x-1) rows in every direction, and every argument is beef(x-1).
beef(0) = 2. beef(1) would be
/ 2, 2 \
\ 2, 2 /
Which is already kinda big. Imagine that as the number of dimensions, as the length of every dimension, and every single argument.
Let me define some more stuff first. g↪(n)[h(x),f(x)](x) means the function g(x), which is f(x) repeated some large number of times, is "restarted" by repeating h(x) the same number of times, and then defining that as the new g(x) to do the same process n more times.
Let steak(n), or sk(n), be equal to Hi↪(beef(beef(...beef(sk(n-1)) times...(sk(n-1))...)))[beef(beef(x),D_0(x)](beef(sk(n-1))). sk(0) = 3.
Let g↪(g↪(...g_g(n)(g(n)) times...g↪(n)[h(x),f(x)](g_g(n)(g(n)))...)[h(x),f(x)](g_g(n)(g(n))))[h(x),f(x)](g_g(n)(g(n))) = g↱(n,h(x),f(x),x). This can also serve as a 2-argument function, g↱(n,x), where h(x) is assumed to be g_g(n)(g(n)), and f(x) is assumed to be the function that is repeated in order to form g(x) and can't be broken up any smaller.

Let's make one more function, mignon(x), or mn(x). mn(0) = sk(1), mn(1) = sk↱(sk(2),sk(2)), mn(2) = sk↱(sk↱(sk(3),sk(3)),sk↱(sk(3),sk(3))).
Continue this pattern. My number is mn↱(mn_[mn(G64)](G64),mn_[mn↱(G64,G64)](G64)).

I know beef(x) may be ill-defined because BEAF above tetrational level has some flaws, but the D hierarchy also has some sketchy parts so I hope we can allow this

Edit: I don't really know how big this number is. mn(x) is at least f_w^w^w(x) in the FGH because that's where tetrational BEAF is, but this is much stronger than that. Also this is the biggest number on the thread >:D ggs

rutabaga wrote: August 16th, 2024, 8:31 am Twelve-hour rule. I'm going to make a very big jump.

Let beef(x) equal the result of a beef(x-1)-dimensional array, with beef(x-1) rows in every direction, and every argument is beef(x-1).
beef(0) = 2. beef(1) would be
/ 2, 2 \
\ 2, 2 /
Which is already kinda big. Imagine that as the number of dimensions, as the length of every dimension, and every single argument.
Let me define some more stuff first. g↪(n)[h(x),f(x)](x) means the function g(x), which is f(x) repeated some large number of times, is "restarted" by repeating h(x) the same number of times, and then defining that as the new g(x) to do the same process n more times.
Let steak(n), or sk(n), be equal to Hi↪(beef(beef(...beef(sk(n-1)) times...(sk(n-1))...)))[beef(beef(x),D_0(x)](beef(sk(n-1))). sk(0) = 3.
Let g↪(g↪(...g_g(n)(g(n)) times...g↪(n)[h(x),f(x)](g_g(n)(g(n)))...)[h(x),f(x)](g_g(n)(g(n))))[h(x),f(x)](g_g(n)(g(n))) = g↱(n,h(x),f(x),x). This can also serve as a 2-argument function, g↱(n,x), where h(x) is assumed to be g_g(n)(g(n)), and f(x) is assumed to be the function that is repeated in order to form g(x) and can't be broken up any smaller.

Let's make one more function, mignon(x), or mn(x). mn(0) = sk(1), mn(1) = sk↱(sk(2),sk(2)), mn(2) = sk↱(sk↱(sk(3),sk(3)),sk↱(sk(3),sk(3))).
Continue this pattern. My number is mn↱(mn_[mn(G64)](G64),mn_[mn↱(G64,G64)](G64)).

I know beef(x) may be ill-defined because BEAF above tetrational level has some flaws, but the D hierarchy also has some sketchy parts so I hope we can allow this

estimate the size of this number, cuz i have no idea what size it is, and need to make a bigger number

New round for no reason
U(2)

unname4798 wrote: August 16th, 2024, 11:53 am New round for no reason
U(2)

hey >:(, okay for this round i'm making this a challenge: NO HEIRARCHIES [so i can calc it better].
My number is 3^^3

ababa11e wrote: August 16th, 2024, 12:04 pm

unname4798 wrote: August 16th, 2024, 11:53 am New round for no reason
U(2)

hey >:(, okay for this round i'm making this a challenge: NO HEIRARCHIES [so i can calc it better].
My number is 3^^3

9^9

(10^10^100+1)^[10^10^100+1]^(10^10^100+1)

The jump was way too large.
24!

unname4798 wrote: August 16th, 2024, 12:53 pm The jump was way too large.
24!

hey, im the op, and there's no rule saying there's too high of a jump, 3^3^5

ababa11e wrote: August 16th, 2024, 1:04 pm

unname4798 wrote: August 16th, 2024, 12:53 pm The jump was way too large.
24!

hey, im the op, and there's no rule saying there's too high of a jump, 3^3^5

U[U(3)](0) :eviltrollface:

unname4798 wrote: August 16th, 2024, 1:18 pm

ababa11e wrote: August 16th, 2024, 1:04 pm

unname4798 wrote: August 16th, 2024, 12:53 pm The jump was way too large.
24!

hey, im the op, and there's no rule saying there's too high of a jump, 3^3^5

U[U(3)](0) :eviltrollface:

you fool, i say 4^^^^4 [if smaller, i will correct]

Ack(7)

rutabaga wrote: August 16th, 2024, 1:22 pmAck(7)

3^^^^^^3 time!!!

50{50}50

rutabaga wrote: August 16th, 2024, 2:21 pm50{50}50

Ack(53,53) > 50{50}50

f_[w+1](3).

Graham's Function has an approximate growth rate of f_w+1(x).

Edit: I have no clue what multi-argument ack function does but I didn't see that

rutabaga wrote: August 16th, 2024, 3:26 pm f_[w+1](3).

Graham's Function has an approximate growth rate of f_w+1(x).

Edit: I have no clue what multi-argument ack function does but I didn't see that

it is this: https://googology.fandom.com/wiki/Ackermann_function
My number is this:
A[n,1] = n
A[n,2] = n^n
A[n,3] = n^^n
.
.
A#[n] = A[n,n]
my number is A#[100]

ababa11e wrote: August 16th, 2024, 3:42 pm

rutabaga wrote: August 16th, 2024, 3:26 pm f_[w+1](3).

Graham's Function has an approximate growth rate of f_w+1(x).

Edit: I have no clue what multi-argument ack function does but I didn't see that

it is this: https://googology.fandom.com/wiki/Ackermann_function
My number is this:
A[n,1] = n
A[n,2] = n^n
A[n,3] = n^^n
.
.
A#[n] = A[n,n]
my number is A#[100]

A++[n] = A#[n] repeated n times
A++[100]

unname4798 wrote: August 16th, 2024, 11:51 pm

ababa11e wrote: August 16th, 2024, 3:42 pm

rutabaga wrote: August 16th, 2024, 3:26 pm f_[w+1](3).

Graham's Function has an approximate growth rate of f_w+1(x).

Edit: I have no clue what multi-argument ack function does but I didn't see that

it is this: https://googology.fandom.com/wiki/Ackermann_function
My number is this:
A[n,1] = n
A[n,2] = n^n
A[n,3] = n^^n
.
.
A#[n] = A[n,n]
my number is A#[100]

A++[n] = A#[n] repeated n times
A++[100]

A'[1]= 1000
A'[n] = A#[n] repeated A'[n-1] times
A'[1000]

ababa11e wrote: August 17th, 2024, 2:12 am

unname4798 wrote: August 16th, 2024, 11:51 pm

ababa11e wrote: August 16th, 2024, 3:42 pm

it is this: https://googology.fandom.com/wiki/Ackermann_function
My number is this:
A[n,1] = n
A[n,2] = n^n
A[n,3] = n^^n
.
.
A#[n] = A[n,n]
my number is A#[100]

A++[n] = A#[n] repeated n times
A++[100]

A'[1]= 1000
A'[n] = A#[n] repeated A'[n-1] times
A'[1000]

A''[0] = 2024
A''[n] = A++[n] repeated A''[n-1] times
A''[1000]

unname4798 wrote: August 14th, 2024, 12:39 pm

ababa11e wrote: August 14th, 2024, 12:02 pm

eRroR_6o6 wrote: August 14th, 2024, 10:56 am20!

10^30

2^102 (number of INT rules)

aight time to go big

one googol

eRroR_6o6 wrote: August 17th, 2024, 12:36 pm

unname4798 wrote: August 14th, 2024, 12:39 pm

ababa11e wrote: August 14th, 2024, 12:02 pm
10^30

2^102 (number of INT rules)

aight time to go big

one googol

U(n) = U(n-1)!
U(0) = 3
U(3)
2...

A'''[1] = 1000
A'''[n] = A''[n] repeated A'''[n-1] times
A''''[1] = 1000
A''''[n] = A'''[n] repeated A''''[n-1] times
Repeat pattern

A[n] with A[n] apostrophes = AA[n]

AA[G64]

Edit: gl on 1000 posts Unname!