What is the FGH of this number?[]
Adamjasher (talk) 01:22, 9 August 2023 (UTC)
is it bigger than TREE(3)[]
how big is this number ?
—Preceding unsigned comment added by AblanGG (talk • contribs)
- Consider this number is well-defined, this number would still be far smaller than TREE(3). It's also far smaller than Graham's number. But for the actual value, I need to analyse the number to see how big is the number.
- HaydenTheGoogologist2009 00:03, 12 August 2023 (UTC)
- nah. even if we give the same number , right-reverse factorial notation grows faster. So lets give it 3. 3↑3=27 but 3¡ = (3!x2!x1!)^(3!x2!x1!)....= 12^12^12^12^12^12^12....^12 (12 times power tower.). and 3↑↑3= 9.7 billion. but 3¡¡ means (3¡)¡ wich is way bigger. its already grows much faster than up arrow notation. and porbably k1>G. but the real problem is , is K (kk100) is bigger then TREE(3) ???
- AblanGG
AblanGG 23:11, 23 August 2023 (UTC)
- \(n!\) is the same as the normal factorial. FGH level \(f_2(n)\).
- \(n¡=(n!\times n-1!\times n-2!....\times1)^{(n!\times n-1!\times n-2!...\times1)^{.......^{(n!\times n-1!\times n-2!...\times1)}}})\) it has \(n!\times n-1!\times n-2!...\times1\) amount of power tower \(=n\$\uparrow\uparrow n\$\) where \(\$\) represents Sloane and Plouffe's superfactorial. FGH level \(f_3(f_2^2(n))\).
- \(n¡¡¡...(z \text{amount of} ¡\text{'s})...¡¡^m=(z\$\uparrow\uparrow m)¡¡¡...((z-1) \text{amount of} ¡\text{'s})...¡¡\). FGH level \(f_4(n)\).
- k1 = Googolplexian¡¡¡....¡¡^Googolplexian there is Googolplexian¡¡¡¡....(Googolplexian amount of ¡'s)...¡¡^Googolplexian amount of ¡'s. It's approximately equal to \(f_4^2(f_2^3(326))\).
- k2 = k1¡¡¡......¡¡^k1 there is k1¡¡¡...(k1 amount of ¡'s)...¡¡^k1 amount of ¡'s. It's approximately equal to \(f_4^3(f_2^3(326))\).
- k3 = k2¡¡¡......¡¡^k2 there is k2¡¡¡...(k2 amount of ¡'s)...¡¡^k2 amount of ¡'s. It's approximately equal to \(f_4^4(f_2^3(326))\).
- Hence, \(k_n\) has a growth rate of \(f_5(n)\).
- And therefore, \(K=k_{k_{100}}\) is approximately \(f_5^2(100)\), which is way smaller than \(G_{64}\) since \(G_{64}<f_{\omega+1}(64)\).
- And TREE(3) is obviously nothing compared to K.
- HaydenTheGoogologist2009 01:04, 24 August 2023 (UTC)
- i looked at Sloane and Plouffe's superfactorial and right-reverse factorial notation is not \(=n\$\uparrow\uparrow n\$\). for example if you give 3. \(=3\$\uparrow\uparrow 3\$\) = is something but 3¡ = 12^12^12^12...^12 (12 times power tower) they are different.
- and 3¡¡ means (3¡)¡ so [12^12^12^12...^12 (12 times power tower)]¡. means ((3¡)!x((3¡)-1)!....)^((3¡)!x((3¡)-1)!....)^.... (((3¡)!x((3¡)-1)!....) amount of power tower.) and this goes for n¡¡¡.....
- plus you can see as i show you above it grows faster than up arrow notation. so you can use grahams number's route and right-reverse notation will give bigger number. for example , 3¡>3↑↑3>3↑3. 3¡¡>3↑↑↑3>3¡. 3¡¡¡>3↑↑↑↑3. so 3¡¡¡>g1 that means k1 also >g1 already. so k2>g2 and even if you go with that , k64 > g64=G. but K= kk100. (kk100= ((kk100)-1)¡¡¡......¡¡^((kk100)-1) there is ((kk100)-1)¡¡¡...(((kk100)-1) amount of ¡'s)...¡¡^((kk100)-1) amount of ¡'s). well it can be smaller then TREE(3). but its clearly bigger then G.
- n¡¡¡...(z amount of ¡'s)...¡¡^m means that if you want to write without (^m) [n¡¡¡....¡¡¡] than you have to write (z!xz-1!...1!)^(z!xz-1!...1!)^.......^(z!xz-1!...1) (m amount of power tower) amount of ¡'s next to n. so n¡¡... have ((z!xz-1!...1!)^(z!xz-1!...1!)^.......^(z!xz-1!...1) (m amount of power tower) amount of i's) and its equal to n¡¡¡...(z amount of ¡'s)...¡¡^m. n¡¡¡...(z amount of ¡'s)...¡¡^m we write this to make it easy to write. (https://imgur.com/a/FsYuZBb) (and for k1 = https://imgur.com/a/QGsEwno)
- AblanGG
AblanGG 11:19, 24 August 2023 (UTC)