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Busybeaver(Busybeaver(10^20^30^40^50^60^70^80^90^100))

CARuler's post is still a LUDICROUSLY big jump, but I don't care. I'll continue the pace.

BB{5}(10^100), where the 5 is used as defined in Rayo's "Oracle Turing Machines." Basically, a certain Turing machine (its function is denoted as BB{1}) would have some 'oracle' device attached to it that tells it when a normal Turing machine would halt, and then have it keep on going instead. Then, you could have a next-level oracle that tells the machine when THAT machine would stop, denoted as BB{2}, and so on. I don't remember any sort of a formal definition, but Rayo himself has written a good explanation that isn't too hard to understand. Just google "Oracle Turing Machine."

Anyway, yeah. I'm back. Beat BB{5}(10^100) and please don't call it invalid :<

let Λ(n,m,k) = Λ(n-1,m+nk,Λ(n-1,m-1,|k-m|))*m, Λ(0,m,k) = m

my number is BB{10}(Λ(100,100,99))

CARuler wrote: February 7th, 2025, 8:55 pm let Λ(n,m,k) = Λ(n-1,m+nk,Λ(n-1,m-1,|k-m|))*m, Λ(0,m,k) = m

my number is BB{10}(Λ(100,100,99))

Trivial extension, given that the Λ function is quite slow in comparison to BB, but that's okay. My next number isn't particularly interesting either. All it does is reintroduce ordinals.

BB{w+1}(10^100)

how does that work?
edit:
BB{Λ(ℵ_0+1,(ℵ_0)!,ℵ_0)}(100^[100]100)

Whelp, f(n,m) = BB{f(n-1,m+n)}(m), f(0,m)=m

My number is BB{omega}(f(1000,10000))

GrAhAmS nUmBeRrRr ^ GoOgOlPlExPLeX
(we should restart now I think)

0

0!

10!

11111111110000000000!!!!!!!!!!

70!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

LWSSONHWSSONLWSS wrote: September 25th, 2025, 10:51 am 70!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

define your number as "boring"

6{boring}7 in BAN

YN = your number
16{{YN}}11

LHL = your number
Af = Ackerman's function
Af(LHL,LHL)

R1INT = number of possible evolution rules in Range-1 INT
R2INT = number of possible evolution rules in Range-2 INT
R3INT = number of possible evolution rules in Range-3 INT
R4INT = number of possible evolution rules in Range-4 INT

My number is {R1INT,R2INT,R3INT,R4INT} using BAN

This is approximately {5070602400912917605986812821504,{2,4211874},{2,{2,49}},{2,{2,81}} }

(sorry, accidentally posted before finishing)

mine is R{R1INT,R2INT,R3INT,R4INT,R5INT}INT

CAR = CARuler's number
R(CAR)INT

R{R{R5INT}INT}INT

R(R{R(CAR)INT,R999INT}INT)INT

ThePlayzr wrote: September 26th, 2025, 6:44 am R{R{R5INT}INT}INT

(this is vastly smaller than the posts that preceded it)

ababa11e wrote: July 27th, 2024, 1:42 pm C: It must be [...] computable [...]

rip my malicious schemes

himoosy

unname4798 wrote: September 26th, 2025, 5:16 am CAR = CARuler's number
R(CAR)INT

{10, 10 [2] 2}

R{10[10]10,10[10]10}INT

CARuler wrote: September 29th, 2025, 1:50 pm R{10[10]10,10[10]10}INT

correct me if i'm wrong but that's literally just R10INT