ConwayLife.com

CARuler wrote: September 15th, 2024, 6:06 pm {BB(10^[10]10),B(10^[10]10),...,BB(10^[10]10)} BB(10^[10]10) times in BEAF

brother, that is way too big of a jump.
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]
reset: 1

can everyone slow down the pace a bit?

1.000001

CARuler wrote: September 15th, 2024, 9:22 pm1.000001

too slow, we aren't 'My Number is Larger' after all.
100

a b (a^b + b^a),
where a = 2, b = 3.

confocaloid wrote: September 15th, 2024, 10:53 pm a b (a^b + b^a),
where a = 2, b = 3.

STILL TOO slow, 3^^3

ababa11e wrote: September 16th, 2024, 12:28 am

confocaloid wrote: September 15th, 2024, 10:53 pm a b (a^b + b^a),
where a = 2, b = 3.

STILL TOO slow, 3^^3

2^^4.

unname4798 wrote: September 16th, 2024, 1:22 am

ababa11e wrote: September 16th, 2024, 12:28 am

confocaloid wrote: September 15th, 2024, 10:53 pm a b (a^b + b^a),
where a = 2, b = 3.

STILL TOO slow, 3^^3

2^^4.

2^^^4

My number is X^^^X,
where X = (a * b * K * (M + L + L) - M) * (a^a * b^b * (N * L * (a^b * b^a * (a + b + a * b) * (K + a * b) + K) - M) - M),
where K = a * a + b * b, L = (a + b) * (a + b), M = a^b + b^a, N = a * a * b * b,
where a = 2, b = 3.

ababa11e wrote: September 15th, 2024, 6:02 pm {2,2,2,2,2,2... {2,2,2,2} long...,2} in BEAF

also, you do know that TREE(2) to {2,2,2,2,2} was a jump by 1, right? TREE(2) = 3, and {2,2,anything} = 4 in BEAF. and hence your post was also 4.

confocaloid wrote: September 16th, 2024, 3:10 am My number is X^^^X,
where X = (a * b * K * (M + L + L) - M) * (a^a * b^b * (N * L * (a^b * b^a * (a + b + a * b) * (K + a * b) + K) - M) - M),
where K = a * a + b * b, L = (a + b) * (a + b), M = a^b + b^a, N = a * a * b * b,
where a = 2, b = 3.

(3^^3)^^^(3^^(3^3))

10^[10]10

10^[1000]10

10^[10^[10]10]10

10{{10}}10 (sorry for the jump)

100{{100}}100

Goobol

1000 & 10 (& is Array of) which is {10,...,10} with 1000 10's which is approx. f_{ω^1000}(10)

1000&[1000]1000

get_Snacked wrote: September 16th, 2024, 9:53 am 1000 & 10 (& is Array of) which is {10,...,10} with 1000 10's which is approx. f_{ω^1000}(10)

f_{ω^ω^ω^ω}(10000)

CARuler wrote: September 16th, 2024, 10:22 am 1000&[1000]1000

that's actually a HUGE jump because c&b&a is different from (c&b)&a with the Array of notation.

ababa11e wrote: September 16th, 2024, 10:40 am f_{ω^ω^ω^ω}(10000)

+(0(1)){6} in the Buchholz hydra notation i used again because it's cool. equivalent to H_{ε0}(6), which is H_{ω^ω^ω^ω^ω^ω}(6), approx. f_{ω^ω^ω^ω^ω}(6)

f_{ω^^ω}(10^[10]10)

CARuler wrote: September 16th, 2024, 10:46 am f_{ω^^ω}(10^[10]10)

which is equal to f_{ε0}(10^[10]10).
+(0(1))(0(1)){10}, equiv. H_{ε0*2}(10), approx. f_{ε0+ω^^9}(10)

f_{ε2^ε2}(10^10^100+1)

guess we're going this speed now.
+(0(1)(1)(1)(1)){10}, equiv. H_{ε3}(10), approx. f_{ω^ω^ω^ω^ω^ω^ω^ω^(ε2*2)}(10)

H_{ε_ω}(10)

X^^(X+2) & 10 using the climbing method, approx. f_{ε(ω^ω)}(10)