ConwayLife.com

2^^5 [also, new page!]

ababa11e wrote: August 14th, 2024, 12:42 pm 2^^5 [also, new page!]

wth is tetration
U(n+1)=U(n)!
U(0)=3
U(6)

unname4798 wrote: August 14th, 2024, 12:45 pm

ababa11e wrote: August 14th, 2024, 12:42 pm 2^^5 [also, new page!]

wth is tetration
U(n+1)=U(n)!
U(0)=3
U(6)

n1^^n2 = n1^(n1^^n2-1), n1^^1 = n1
U(6) is roughly (10^^5)^3
so, my number is 10^^10, or 10^10^10^10^10^10^10^10^10^10
which is between 14!!!!!!!!!! and 15!!!!!!!!!!

guess I'll use my other ideas next round.

my number is 10!^^10!, which is also 10!^^^2

rutabaga wrote: August 14th, 2024, 5:01 pm guess I'll use my other ideas next round.

my number is 10!^^10!, which is also 10!^^^2

3628800^^3628800... hm what about 10^^10^^10,or 10^^^3

I don't want to speed things up too much, but adding up arrows is boring, so I'll make a kinda weak function.

boring(n) = (n-1)^^(n-1). boring(1) = 2.
[EDIT: I meant boring(n-1)^^boring(n-1).]

my number is boring(5).

rutabaga wrote: August 14th, 2024, 5:19 pm I don't want to speed things up too much, but adding up arrows is boring, so I'll make a kinda weak function.

boring(n) = (n-1)^^(n-1). boring(1) = 2.

my number is boring(5).

okay, but we can make this stronger: boringModified(N) = boringModified(N-1)^^boringModified(N-1), boringModified(1) = 3

My number is boringModified(1000), or 3^^^(2^999)

frick, that's what i meant to say.

with my accidentally MUCH weaker definition of boring(n), my number would have been 4^^4, which is smaller than the previous number. of course, boring(1000) is larger than boring(5) either way, so your number is much bigger.

alright, back to just adding up arrows. my number is 55^^^^^555.

rutabaga wrote: August 14th, 2024, 6:06 pm frick, that's what i meant to say.

with my accidentally MUCH weaker definition of boring(n), my number would have been 4^^4, which is smaller than the previous number. of course, boring(1000) is larger than boring(5) either way, so your number is much bigger.

alright, back to just adding up arrows. my number is 55^^^^^555.

My new function works like this: b_1(n) is n^n, b_2(n) = n^^n, b_3(n) = n^^^n... my number is b_6(10), or 10^^^^^^10 [this is just fgh with f_M+2(N) = b_M(N) but shhhh]

b_10(10), or 10{10}10 in BAN.

time for expansion!

NEW RULE PROPOSAL: If nobody else has posted in 12 hours, you can take two turns in a row (but you cannot take 3 turns in a row unless the thread goes completely dead).

If that's a good rule, then my number is G2. Otherwise, I didn't say anything lol

rutabaga wrote: August 15th, 2024, 7:49 am NEW RULE PROPOSAL: If nobody else has posted in 12 hours, you can take two turns in a row (but you cannot take 3 turns in a row unless the thread goes completely dead).

If that's a good rule, then my number is G2. Otherwise, I didn't say anything lol

U(U(4))

G(100)

rutabaga wrote: August 15th, 2024, 9:02 amG(100)

U[U[2024](0)](0)

unname4798 wrote: August 15th, 2024, 9:19 amU[U[2024](0)](0)

Sorry for the jump, but

Ghee(n) = G(G(...G(Ghee(n-1)) times...Ghee(n-1)...))

Ghee(64)

[EDIT: Ghee(1) = G64]

rutabaga wrote: August 15th, 2024, 10:20 am

unname4798 wrote: August 15th, 2024, 9:19 amU[U[2024](0)](0)

Sorry for the jump, but

Ghee(n) = G(G(...G(Ghee(n-1)) times...Ghee(n-1)...))

Ghee(64)

U[U[U[4](0)](0)](0)

rutabaga wrote: August 15th, 2024, 10:20 am

unname4798 wrote: August 15th, 2024, 9:19 amU[U[2024](0)](0)

Sorry for the jump, but

Ghee(n) = G(G(...G(Ghee(n-1)) times...Ghee(n-1)...))

Ghee(64)

you never defined Ghee(1), and i have NO ideas what U[n] is unname4798

unname4798 wrote: August 15th, 2024, 11:37 am U[U[U[4](0)](0)](0)

not entirely sure this is bigger but ok.

Gh(n) = Ghee[Ghee[...Ghee(n-1) times...[Ghee(n-1)](n-1)...](n-1)](n-1)

Gh(64).

Ghee(1) = G64, and Gh(1) = Ghee(64). Sorry for forgetting to fully define Ghee earlier lol

Also I believe unname defined U(n) as (U(n-1))! where U(0) = 3

rutabaga wrote: August 15th, 2024, 11:55 am

unname4798 wrote: August 15th, 2024, 11:37 am U[U[U[4](0)](0)](0)

not entirely sure this is bigger but ok.

Gh(n) = Ghee[Ghee[...Ghee(n-1) times...[Ghee(n-1)](n-1)...](n-1)](n-1)

Gh(64).

Ghee(1) = G64, and Gh(1) = Ghee(64). Sorry for forgetting to fully define Ghee earlier lol

Also I believe unname defined U(n) as (U(n-1))! where U(0) = 3

G_n[N] G(G(...n times...G(G(n)))
G_(n,2)[N] = G_n[G_n[...G_n[N] times...G_n[N]]]
G_(n,3)[N] = G_(n,2)[G_(n,2)[...G_(n,2)[N] times...G_(n,2)[N]]]
.
.
.
My number is G_(G_(100!,100!)[64],G_(100!,100!)[64])[64]

ababa11e wrote: August 15th, 2024, 12:25 pm

rutabaga wrote: August 15th, 2024, 11:55 am

unname4798 wrote: August 15th, 2024, 11:37 am U[U[U[4](0)](0)](0)

not entirely sure this is bigger but ok.

Gh(n) = Ghee[Ghee[...Ghee(n-1) times...[Ghee(n-1)](n-1)...](n-1)](n-1)

Gh(64).

Ghee(1) = G64, and Gh(1) = Ghee(64). Sorry for forgetting to fully define Ghee earlier lol

Also I believe unname defined U(n) as (U(n-1))! where U(0) = 3

G_n[N] G(G(...n times...G(G(n)))
G_(n,2)[N] = G_n[G_n[...G_n[N] times...G_n[N]]]
G_(n,3)[N] = G_(n,2)[G_(n,2)[...G_(n,2)[N] times...G_(n,2)[N]]]
.
.
.
My number is G_(G_(100!,100!)[64],G_(100!,100!)[64])[64]

G___(n) is G__(n,n) repeated U[2^n](0) times.
G__(4798!)

unname4798 wrote: August 15th, 2024, 12:29 pm

ababa11e wrote: August 15th, 2024, 12:25 pm

rutabaga wrote: August 15th, 2024, 11:55 am
not entirely sure this is bigger but ok.

Gh(n) = Ghee[Ghee[...Ghee(n-1) times...[Ghee(n-1)](n-1)...](n-1)](n-1)

Gh(64).

Ghee(1) = G64, and Gh(1) = Ghee(64). Sorry for forgetting to fully define Ghee earlier lol

Also I believe unname defined U(n) as (U(n-1))! where U(0) = 3

G_n[N] G(G(...n times...G(G(n)))
G_(n,2)[N] = G_n[G_n[...G_n[N] times...G_n[N]]]
G_(n,3)[N] = G_(n,2)[G_(n,2)[...G_(n,2)[N] times...G_(n,2)[N]]]
.
.
.
My number is G_(G_(100!,100!)[64],G_(100!,100!)[64])[64]

G___(n) is G__(n,n) repeated U[2^n](0) times.
G__(4798!)

I have a genius idea: im going to bring back b_m(n), but make it more powerful. b_(m,A)(n) is equal to b_(b_(m-1,A),A-1)(G(n)), and then define b_(0,A)(n) = b(G(A),G(A)-1)(n), and b_(A,0)(n) = G___(A^^^n), my number is b_(G__(4798!),G__(4798!))(G__(4798!))

ababa11e wrote: August 15th, 2024, 12:35 pm

unname4798 wrote: August 15th, 2024, 12:29 pm

ababa11e wrote: August 15th, 2024, 12:25 pm
G_n[N] G(G(...n times...G(G(n)))
G_(n,2)[N] = G_n[G_n[...G_n[N] times...G_n[N]]]
G_(n,3)[N] = G_(n,2)[G_(n,2)[...G_(n,2)[N] times...G_(n,2)[N]]]
.
.
.
My number is G_(G_(100!,100!)[64],G_(100!,100!)[64])[64]

G___(n) is G__(n,n) repeated U[2^n](0) times.
G__(4798!)

I have a genius idea: im going to bring back b_m(n), but make it more powerful. b_(m,A)(n) is equal to b_(b_(m-1,A),A-1)(G(n)), and then define b_(0,A)(n) = b(G(A),G(A)-1)(n), and b_(A,0)(n) = G___(A^^^n), my number is b_(G__(4798!),G__(4798!))(G__(4798!))

G____(n)=b_(n^[n arrows]^n,n^[n arrows]^n)(n^[n arrows]^n)
G____(4798)

unname4798 wrote: August 15th, 2024, 12:37 pm

ababa11e wrote: August 15th, 2024, 12:35 pm

unname4798 wrote: August 15th, 2024, 12:29 pm
G___(n) is G__(n,n) repeated U[2^n](0) times.
G__(4798!)

I have a genius idea: im going to bring back b_m(n), but make it more powerful. b_(m,A)(n) is equal to b_(b_(m-1,A),A-1)(G(n)), and then define b_(0,A)(n) = b(G(A),G(A)-1)(n), and b_(A,0)(n) = G___(A^^^n), my number is b_(G__(4798!),G__(4798!))(G__(4798!))

G____(n)=b_(n^[n arrows]^n,n^[n arrows]^n)(n^[n arrows]^n)
G____(4798)

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,
stack example: D_(3,2,1)(n) = D_n^2+2n+3(n)
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)

ababa11e wrote: August 15th, 2024, 12:50 pm

unname4798 wrote: August 15th, 2024, 12:37 pm

ababa11e wrote: August 15th, 2024, 12:35 pm
I have a genius idea: im going to bring back b_m(n), but make it more powerful. b_(m,A)(n) is equal to b_(b_(m-1,A),A-1)(G(n)), and then define b_(0,A)(n) = b(G(A),G(A)-1)(n), and b_(A,0)(n) = G___(A^^^n), my number is b_(G__(4798!),G__(4798!))(G__(4798!))

G____(n)=b_(n^[n arrows]^n,n^[n arrows]^n)(n^[n arrows]^n)
G____(4798)

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)

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

ababa11e wrote: August 15th, 2024, 12:50 pm

unname4798 wrote: August 15th, 2024, 12:37 pm
G____(n)=b_(n^[n arrows]^n,n^[n arrows]^n)(n^[n arrows]^n)
G____(4798)

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