cuz why not? its a incremental game forum
rule: https://googology.fandom.com/wiki/My_number_is_bigger!
i first, 1.79e308 as being infinity in a lot of game

For people trying to go to that link, the reason it'a broken is because you need to add an exclamation mark to the end.

I'll go with 10↑10↑10

10↑↑10

1J100

JJJJ1000

K1000

TREE(3)

Rayo(10↑100)

Fish number 7 (https://googology.fandom.com/wiki/Fish_number_7)

Rayo(Fish number 7) 🤔

ω

[this post was deleted]

yo bro ω isnt finite thaats agains riuel

actually the rules have been broken since JJJJ1000 because you can't directly refer to the previous number

whar

oops my bad

Xi^g(g(64))(27) where Xi^m(n) is recursion of Xi m times, this was hard to come up with, and i don't even know if its larger than the previous (finite) numbers, heres the googology wiki page https://googology.fandom.com/wiki/Xi_function

if we're gonna go with ordinals then ig ω^2

stole most of this from the og my number is bigger forum >:)

basically let f(x) be this massive thingy:

            / X X X X X … X X X X X \
           /  X X X X X … X X X X X  \
          /   X X X X X … X X X X X   \
         /    X X X X X … X X X X X    \
        /     X X X X X … X X X X X     \
       |      | | | | | \ | | | | |      |
        \     X X X X X … X X X X X     /
         \    X X X X X … X X X X X    /
          \   X X X X X … X X X X X   /
           \  X X X X X … X X X X X  /
            \ X X X X X … X X X X X /

with x columns and rows
now let H(x) be zeta(1+(1/g(g_(x!)!)!))
next let Y(x) = x↑(x number of arrows) H^H(g_g_g_...[H(g_g_g_xkcd!) sub-levels of g_ here]...g_g(H(g_g_g_xkcd!)))(g_g_g_...[H(g_g_g_xkcd!) sub-levels of g_ here]...g_g(H(g_g_g_xkcd!))) where g_x is the xth number in the graham's number sequence, and xkcd is the xkcd number
finally, let Z = H(x)↑(Y(x) number of arrows) h(x) ↑ (H(x) number of arrows) Y(x)

my number is f(Z) applied to itself recursively TREE(TREE(TREE(Z)) times

going to the ordinal route ill just throw in cantor's ordinal

[this post was deleted]

i dont have much googology knowledge so it might not even be the slightest bit larger than the original post

tree(tree(tree(tree(tree(tree(tree6000000))))))

sscg(3)

sscg(4)

sscg(6)

sscg(10^10)

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{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^20}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^20}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^20}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^20}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}
{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}
{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}
{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}
{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

{{{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}, {{SSCG(SSCG(R^(R^(tree(TAR(5)))))),TREE(TREE(R^(SSCG(SSCG(R^(R^(TREE(TAR(3))))))))),TAR(TREE(3)),TREE(5),TREE(5),TREE(4),TREE(4),SCCG(R^(R^(R^(R^(TREE(5))))),TAR(8),TAR(8),10},{SSCG(SSCG(TAR(6)))},TAR(3),TAR(3),R^(R^(R^(TREE))),TAR(3)^2}}

Too much copy and paste

Big number

Icecreamdude says:

Big number

Big number*Big number

ω

SECTION 0: INTRODUCTION

after 3 months, i present to you... with the power of incremental game notation, and some fictional googology stupidity, the bigger number!!!!

SECTION 1: SOMEWHAT UNDERSTANDABLE EXPLANATION

part 1: the basics

first, let's start with a small number. 1000. in scientific notation this is 1e3, because there are 3 zeroes.
you can then make 1e1e3, which is a number with 1000 zeroes. that got big quick.
then there's 1e1e1e3, then 1e1e1e1e3, but theres starting to be a lot of 1e's, so we just write eeee3
but that'll still cause there to be a lot of e's so we introduce F. eeeee3 = 3F5, since there are 5 e's and those e's end in a 3. the 3 wont matter though so we'll just write F5.
and F's can be chained, just like e. FF5, FFF5, FFFF5... let's introduce G now, which counts the number of Fs. FFFFF5 = G5.
G's can also be chained. starting to see a pattern? GG5, GGG5, GGGG5, introduce H, H5, H's can be chained, blah blah blah

part 2: recursion

this will go on until we reach Z. what do we do then? we can't loop back to A, since we'll eventually reach E again. the solution: Z is the 21st layer of the recursion in part 1, so we'll notate it as...
ZZZZZ5 = {22}5. now that we can represent our current layer of recursion with a number... MORE RECURSION!!!
{{22}5}5... {{{22}5}5}5... yeah that got hard to read quickly, let's represent it with an array! {{{22}5}5}5 will be represented as [3, 22, 5] since there are 3 opening brackets, 22 inside all brackets, and 5 in none of the brackets. now...
[[3, 22, 5], 22, 5]... [[[3, 22, 5], 22, 5], 22, 5]... welp, time to make some more notation... {3, [3, 22, 5], 5} should do, since there are 3 opening brackets, a [3, 22, 5] inside all brackets, and 5 as the final item in the array. and thus,
{{3, [3, 22, 5], 5}, [3, 22, 5], 5}... {{{3, [3, 22, 5], 5}, [3, 22, 5], 5}, [3, 22, 5], 5}... when will it end? this is our third time... wait a minute...

part 3: the α function

that was our third time making a recursion notation, so why not display it as a function to show how many times we've made a new notation?
{{{3, [3, 22, 5], 5}, [3, 22, 5], 5}, [3, 22, 5], 5} can be represented with our brand new function α(r, x, y). r is a little hard to explain, so just think about it as the amount of times we've made a new recursion notation. x is the number of brackets in the rth recursion notation. y is the value for the final array item. so {{{3, [3, 22, 5], 5}, [3, 22, 5], 5}, [3, 22, 5], 5} = α(2, 3, 5). with just 3 small numbers, we've described a really big number! you know what- eh let's just get into it
α(α(2, 3, 5), 3, 5)... α(α(α(2, 3, 5), 3, 5), 3, 5)... it's getting hard to read, so we'll simplify α(r, x, y) to just α(r), and use the notation for applying a function to itself recursively, which is α(^x)(r) for x = number of times to apply α(r) to itself. okay, now...
α(^3)(2)... α(^α(2))(2)... α(^α(^α(2))(2))(2)... sigh yeah we need a new function

part 4: more functions

introducing... β(x)!!! x is the amount of times we uhh we do uhh,,, lets just say β(3) is equal to α(^α(^α(2))(2))(2)
you know what's happening at this point. we're gonna recurse β(3), leading to β(^β(3))(3), then we're gonna need γ(x)...
and then δ(x)... then ε(x)... all the way down to ω(x) (NOT THE ORDINAL ω, THE FUNCTION ω)... what now?

part 5: my number?

i think it's time to end it off. let x_0 = ω(3). x_(n+1) = ω(^ω(x_n))(x_n). my number is equal to x_ω(3). but wait... why does the forum post begin with section-

SECTION 2: THE STUPIDITY BEGINS

part 6: you want more? i'll give you more...

have you ever seen one of those "numbers 0 to Ơ̷̢͉͙̫͖̞͓̝̲̗̘͆̇̃̄͌̈́͋͊͝͝M̶̙̜̫̳̝͊̌͑̐̎̚͜Ẻ̸̹͎̣͒́͆̂̀̿͑̓̄͛͛Ģ̵̧̧̛̹͈̲̙̩̞̣̻̥̬͊̃͗̉͌̒̽͝A̷̲̩̮͇̗̞̟̣̯͎̯̱͗̀̿̕͜͜ ̸̟̈͗̒̊̈́͊̓̍͛Ư̶̧͕͔̮̥̳̘͕͕͖̟̪͛̌̀͊̾͆̽̑͑̐͗̋͘̚P̴͍̕P̴̨̯̩̥͚̱̪͍̜̮͎͚̺͗̀͌̔͊͑͛̄̍̈́̌Ȩ̸͚͉̫̦̪͇͈̰͔̠̙͍͚̦͋̃̓̑̽̎̓̚R̵̢̩͔͇̖̞̪̒̈́̎̏͌̓̕̕͝ ̴̭̭̮͑̋̓̌͐̆͌͌͂̊͒̕Ļ̷̧̲̻̲̤̫̼͕̹̦̜͍̌̀I̸̢̨̹̪̞̖̱̬̩̥̦̣̰͙̤̋̈́͆̎̍̓̓͘͝M̴̨̨̛̭͇̽̈́͂́͋̀̊̿́̑̚̚ͅI̶͉̒̋̾̀́̃̄̽̎͊̿̋͝͝T̴͍̋̒" videos where they start doing recursion over and over, then the video starts going faster and faster until it turns into a bass boosted epileptic sequence (example, epilepsy warning btw: https://www.youtube.com/watch?v=0TsagtbAusA&t=203)? well, you probably know what i'm thinking...
but how exactly do we cause an explosion? well, we can introduce a new variable: ER^1, which is Explosive Recursion. every time we increase ER's exponent by 1, think of the universe exploding for extra effect. but what exactly is ER?

part 7: what is ER?

remember around 6 minutes ago in part 1 where we decided past Z we would notate layers of recursion as a number? we'll do it again. ω is the 24th letter of the greek alphabet, so the next function after ω(x) would be {25}(x). now let y_0 = {25}(3). y_(n+1) = {y_n}(y_n). now let z_0 = y_{25}(3). z_(n+1) = y_{z_n}(z_n). as you continue defining variables after z, think of that video speeding up and getting faster, to the point where it becomes a bass boosted epileptic mess. after ω(3) variable definitions, we hit ER^1.

part 8: larger ER exponents

what about ER^2? ER^2 is reached after ER^1 variable definitions. you can probably predict what ER^3 is. ER^n+1 = ER^n variable definitions. and as you can probably predict, ER^0 = 1.
oh boy, you can probably predict what happens next. ER^ER^2... ER^ER^ER^2... i'm already tired of this. hey, this somewhat resembles tetration! so...
ER^^2... ER^^ER^^2... onto pentation... ER^^^2... skip to arrow notation... ER↑(ER^2)↑2... ER↑(ER↑(ER^2)↑2)↑2... all roads lead to recursion. new variable!

part 9: introduction to NR

let's define a new variable: NR, for Nuclear Recursion. let's begin some more variable definitions. let a_0 = ER↑(ER^2)↑2. a_(n+1) = ER↑(a_n)↑2.
let b_0 = a_(ER↑(ER^2)↑2). b_n+1 = a_((ER↑(b_n)↑2)). continue defining new variables with this pattern. after ER↑(ER^2)↑2 variable definitions, NR^1 is reached. further NR exponents are defined similarly to further ER exponents. now let's not sugarcoat the progression and go straight to NR↑(NR^2)↑2. this is where the final push begins.

part 10: the final push

let α_0 = NR↑(NR^2)↑2. α_n+1 = NR↑(α_n+1)↑2. let β_0 = a_(NR↑(NR^2)↑2). β_n+1 = a_(NR↑(β_n)↑2). continue defining variables as usual. after NR↑(NR^2)↑2 variable definitions, you reach my number. is there really a point in getting the number any further? it's already so massive. but hey, the 9850 forum post character limit (as of finishing this sentence) says otherwise.

SECTION 3: REPETITION SIMULATOR (FREE DOWNLOAD) (NO VIRUS)

part 11: more power

what if i told you that the entirety of section 2 is just 2 layers of recursion? yeah. i'm sure you're tired of hearing "recursion" 31 billion times already. ER^1 can be represented with our brand new function OR(x) (omega recursion) as OR(1). NR^1 can be represented as OR(2). and "my number" from part 10 is OR(3). time for more re- actually screw that

part 12: more powerful variable definitions

defining variables is more fun. wait, we don't have any alphabe-
HIRAGANA. let あ_0 = OR(^OR(^OR(3))(3))(3). あ_(n+1) = OR(あ_n). い_0 = あ_OR(^OR(^OR(3))(3))(3). い_(n+1) = あ_OR(^OR(^OR(い_n))(3))(3). continue defining variables for every hiragana character up to ん. use the variable-to-number trick to define variables all the way up to variable OR(^OR(^OR(3))(3))(3). now let's define a new explosion variable: ED^1, for Explosive Definitions.

part 13: more powerful explosions

now that we have ED^1, we can go straight to arrow notation, leading us to ED↑(ED^2)↑2!!! now for some more variable definition shenanigans...
let ア_0 = ED↑(ED^2)↑2. ア_(n+1) = ED↑(ア_n)↑2. let イ_0 = ア_ED↑(ED^2)↑2. イ_(n+1) = ア_ED↑(イ_n)↑2. define ED↑(ED^2)↑2 more variables. you have reached RD^1.
let's make a new function. OD(1) (omega definitions) = ED^1, OD(2) = RD^1.

part 14: more powerful recursions

we have another function, meaning that in order to progress we need to continue recursion. every time we try to escape it, it always comes back. well actually, just jump to OD(^OD(^OD(2))(2))(2). let's create another explosion variable to end this section off!
let ✂0 = OD(^OD(^OD(2))(2))(2).✂(n+1) = OD(^OD(^OD(✂_n))(2))(2). let ✁_0 = ✂OD(^OD(^OD(2))(2))(2). ✁(n+1) = ✂_OD(^OD(^OD(✁_n))(2))(2). define OD(^OD(^OD(2))(2))(2) more variables. we now have ΩR^1 (you can probably guess what this stands for).

part 15: end it already...

let's get this part over with. time for the fake conclusion, then the sUrPrIsInG rEvEaL that this forum post is over. my number is ΩR↑(ΩR^2)↑2. it's not over yet, we have 7736 more characters to go.

SECTION 4: THE META SECTION

part 16: the section function

we've gone a long way. how about we automate things? let S(1) = x_ω(3). S(2) = OR(3). S(3) = ΩR↑(ΩR^2)↑2. notice something? S(x) produces the final result of each section. which is why we're going to generalize sections now!

part 17: generalizing sections

each section will have a variable definition section, an explosion section, and a recursion section in that order. but what exactly does that mean? here's a step by step for each section, using S(3) -> S(4) as an example:

1. define variables with an obscure alphabet. let (char_a)0 = ΩR↑(ΩR^2)↑2, (char_a)(n+1) = ΩR↑(ΩR^(char_a)n)↑2, let (char_b)0 = (char_a)(ΩR↑(ΩR^2)↑2), (char_b)(n+1) = (char_a)_(ΩR↑(ΩR^(char_b)_n)↑2)
2. explode. define ΩR↑(ΩR^2)↑2 more variables and create a new explosion variable (AB^1 for example). jump to using arrow notation. AB↑(AB^2)↑2.
3. repeat step 1 and 2 with the result from the previous step once.
4. create a function that creates new explosion variables.
5. recurse your function. for example ΩA(3) would become ΩA(^ΩA(^ΩA(3))(3))(3)
6. the next S(x) value is the result of step 5

part 18: the fun begins

now that we have defined S(x), let's recurse it!!! we now have S(^S(^S(4))(4))(4).
let's define S(^S(^S(4))(4))(4) variables now (skipping the process since i already generalized the process) and create a new explosion variable: ES^1
now jump to arrow notation... ES↑(ES^2)↑2
and finally... create a function that creates new explosion variables. OS(1) = ES^1.
recurse that function. OS(^OS(^OS(2))(2))(2). this is actually kinda getting boring...

part 19: the fun ends?

well, we tried automating the process of defining, recursing, and exploding but we ended up continuing to define, recurse, and explode. wait...
introducing S_2(x)! this repetitively repeats the steps done in part 18! let's recurse i-
introducing S_3(x)! this repetitively defines variables, explodes functions, and re-
introducing S_4(x)! this-
introducing S_(OS(^OS(^OS(2))(2))(2))(x)! yeah.

part 20: 5665 more.

now, first things first: recurse this monstrosity of a function. for simplicity we'll rename it to ΩS(x). now we have ΩS(^ΩS(^ΩS(3))(3))(3). now, explode this function. ΩE^1.
jump to arrow notation. ΩE↑(ΩE^2)↑2. that's my number.

SECTION 5: PROGRESSION INNOVATION

part 21: a new method of progression

let's introduce another type of progression: RP (recursion power). ΩE↑(ΩE^2)↑2 = RP {R^2/2}. there are 2 layers of recursion (marked by the /2) and the final ΩE being raised to the power of 2 (marked by the R^2) you know what we're gonna do. let's boost those values up!!! we now have RP {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2} you know what's special about this? RP {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2} is equal to RP^2 {R^2/2}. you know what's happening now.

part 22: return of arrow notation

time to abuse the notation to reach a new massive number! RP^ΩE↑(ΩE^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}... RP^RP^ΩE↑(ΩE^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2} {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2} oh god that's way too long... no worries, arrow notation! RP↑(RP^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2} yeah that's still way too long

part 23: a somewhat brand new notation

let's introduce RP_1 {R^2/2}! each time we reach RP_n↑(RP_n^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}, increase n by 1. now... RP_ΩE↑(ΩE^2)↑2↑(RP_ΩE↑(ΩE^2)↑2^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}... RP_RP_ΩE↑(ΩE^2)↑2↑(RP_RP_ΩE↑(ΩE^2)↑2^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}... huh. looks like we're gonna need a function.

part 24: a not so brand new function

introducing PR(x) (power recursion)! for a simple explanation, PR(1) = RP_ΩE↑(ΩE^2)↑2↑(RP_ΩE↑(ΩE^2)↑2^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}, and PR(2) = RP_RP_ΩE↑(ΩE^2)↑2↑(RP_RP_ΩE↑(ΩE^2)↑2^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}! let's recurse our function, leading to PR(^PR(^PR(2))(2))(2). let's do something similar to what we did to S(x), creating PR_n(x) which counts the amount of times we recursed PR_(n-1)(x)! hard to understand? basically, PR_(n+1)(3) =PR_n(^PR_n(^PR_n(2))(2))(2), since there are 3 PR_n's used in the recursion. now, we have... PR_(PR(2))(^PR_(PR(2))(^PR_(PR(2))(2))(2))(2)... PR_(PR_(PR(2))(2))(^PR_(PR_(PR(2))(2))(^PR_(PR_(PR(2))(2))(2))(2))(2)... oh god. i think this is a good time to bring in the variable definitions

part 25: the cycle continues

let's see... which unicode symbols should i use today... okay i got ⥯ and ⏻ cool so anyways
let ⥯0 = PR(PR(2))(^PR_(PR(2))(^PR_(PR(2))(2))(2))(2), ⥯(n+1) = PR(PR(⥯n))(^PR(PR(⥯n))(^PR(PR(⥯n))(⥯n))(⥯n))(⥯n), and let ⏻0 = ⥯PR(PR(2))(^PR(PR(2))(^PR(PR(2))(2))(2))(2), ⏻(n+1) = ⥯PR(PR(⏻n))(^PR(PR(⏻n))(^PR(PR(⏻n))(⏻n))(⏻n))(⏻n). define PR(PR(PR(2))(2))(^PR(PR(PR(2))(2))(^PR(PR(PR(2))(2))(2))(2))(2) more variables. new explosion variable! EP^1. now jump to using arrow notation... EP↑(EP^2)↑2. yeah. let's convert it into a power formula: EP↑(EP^2)↑2 = PE {P^2/2}
now jump to PE {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}, then to PE^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}, then to PE↑(PE^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}, then to RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}... then to RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}... yeah i'll just call that my number and get this section over with

SECTION 6: THE TRUE FINAL PUSH

part 26: return to the basics

let's start again, but simpler. let's call our previous number (RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}) "new 1." you know what we're gonna do. we're doing it all over again.

NEW SECTION 1: NEW SOMEWHAT UNDERSTANDABLE EXPLANATION

new part 1: new the basics

first, let's start with a small new numbe- you know what skip all that

new part 25: new the cycle continues

let's see... which unicode symbols should i use today... okay i got ⥯ and ⏻ cool so anyways
let new ⥯0 = new PR(PR(2))(^PR_(PR(2))(^PR_(PR(2))(2))(2))(2), new ⥯(n+1) = new PR(PR(⥯n))(^PR(PR(⥯n))(^PR(PR(⥯n))(⥯n))(⥯n))(⥯n), and let new ⏻0 = new ⥯PR(PR(2))(^PR(PR(2))(^PR(PR(2))(2))(2))(2), ⏻(n+1) = ⥯PR(PR(⏻n))(^PR(PR(⏻n))(^PR(PR(⏻n))(⏻n))(⏻n))(⏻n). define new PR(PR(PR(2))(2))(^PR(PR(PR(2))(2))(^PR(PR(PR(2))(2))(2))(2))(2) more variables. new explosion variable! EP^1. now jump to using arrow notation... new EP↑(EP^2)↑2. yeah. let's convert it into a power formula: new EP↑(EP^2)↑2 = new PE {P^2/2}
now jump to new PE {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}, then to PE^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}, then to PE↑(PE^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}, then to new RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}... then to new RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}... yeah i'll just call that my number and get this section over with

NEW SECTION 6: THE NEW TRUE FINAL PUSH

new part 26: new return to the basics

let's start again, but simpler. let's call our previous new number (new RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}) "new new 1." you know what we're gonna do. we're doing it all over again.

NEW NEW SECTION 6: THE NEW NEW TRUE FINAL PUSH

new new part 26: new new return to the basics

let's start again, but simpler. let's call our previous new new number (new new RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}) "new new 1." you know what we're gonna do. we're doing it all over again.

RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} "new"s later...

SECTION 7: THE END

my number is (new xRP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}) RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}. you simply CANNOT beat this number. all it took was a 16384 character post. look how far we've gone.

oh yeah multiply my number by big numb

Too late I already posted infinity

∞+1

pixelperfect12 says:

SECTION 0: INTRODUCTION

after 3 months, i present to you... with the power of incremental game notation, and some fictional googology stupidity, he bigger number!!!!

SECTION 1: SOMEWHAT UNDERSTANDABLE EXPLANATION

part 1: the basics

first, let's start with a small number. 1000. in scientific notation this is 1e3, because there are 3 zeroes.
you can then make 1e1e3, which is a number with 1000 zeroes. that got big quick.
then there's 1e1e1e3, then 1e1e1e1e3, but theres starting to be a lot of 1e's, so we just write eeee3
but that'll still cause there to be a lot of e's so we introduce F. eeeee3 = 3F5, since there are 5 e's and those e's end in a 3. the 3 wont matter though so we'll just write F5.
and F's can be chained, just like e. FF5, FFF5, FFFF5... let's introduce G now, which counts the number of Fs. FFFFF5 = G5.
G's can also be chained. starting to see a pattern? GG5, GGG5, GGGG5, introduce H, H5, H's can be chained, blah blah blah

part 2: recursion

this will go on until we reach Z. what do we do then? we can't loop back to A, since we'll eventually reach E again. the solution: Z is the 21st layer of the recursion in part 1, so we'll notate it as...
ZZZZZ5 = {22}5. now that we can represent our current layer of recursion with a number... MORE RECURSION!!!
{{22}5}5... {{{22}5}5}5... yeah that got hard to read quickly, let's represent it with an array! {{{22}5}5}5 will be represented as [3, 22, 5] since there are 3 opening brackets, 22 inside all brackets, and 5 in none of the brackets. now...
[[3, 22, 5], 22, 5]... [[[3, 22, 5], 22, 5], 22, 5]... welp, time to make some more notation... {3, [3, 22, 5], 5} should do, since there are 3 opening brackets, a [3, 22, 5] inside all brackets, and 5 as the final item in the array. and thus,
{{3, [3, 22, 5], 5}, [3, 22, 5], 5}... {{{3, [3, 22, 5], 5}, [3, 22, 5], 5}, [3, 22, 5], 5}... when will it end? this is our third time... wait a minute...

part 3: the α function

that was our third time making a recursion notation, so why not display it as a function to show how many times we've made a new notation?
{{{3, [3, 22, 5], 5}, [3, 22, 5], 5}, [3, 22, 5], 5} can be represented with our brand new function α(r, x, y). r is a little hard to explain, so just think about it as the amount of times we've made a new recursion notation. x is the number of brackets in the rth recursion notation. y is the value for the final array item. so {{{3, [3, 22, 5], 5}, [3, 22, 5], 5}, [3, 22, 5], 5} = α(2, 3, 5). with just 3 small numbers, we've described a really big number! you know what- eh let's just get into it
α(α(2, 3, 5), 3, 5)... α(α(α(2, 3, 5), 3, 5), 3, 5)... it's getting hard to read, so we'll simplify α(r, x, y) to just α(r), and use the notation for applying a function to itself recursively, which is α(^x)(r) for x = number of times to apply α(r) to itself. okay, now...
α(^3)(2)... α(^α(2))(2)... α(^α(^α(2))(2))(2)... sigh yeah we need a new function

part 4: more functions

introducing... β(x)!!! x is the amount of times we uhh we do uhh,,, lets just say β(3) is equal to α(^α(^α(2))(2))(2)
you know what's happening at this point. we're gonna recurse β(3), leading to β(^β(3))(3), then we're gonna need γ(x)...
and then δ(x)... then ε(x)... all the way down to ω(x) (NOT THE ORDINAL ω, THE FUNCTION ω)... what now?

part 5: my number?

i think it's time to end it off. let x_0 = ω(3). x_(n+1) = ω(^ω(x_n))(x_n). my number is equal to x_ω(3). but wait... why does the forum post begin with section-

SECTION 2: THE STUPIDITY BEGINS

part 6: you want more? i'll give you more...

have you ever seen one of those "numbers 0 to Ơ̷̢͉͙̫͖̞͓̝̲̗̘͆̇̃̄͌̈́͋͊͝͝M̶̙̜̫̳̝͊̌͑̐̎̚͜Ẻ̸̹͎̣͒́͆̂̀̿͑̓̄͛͛Ģ̵̧̧̛̹͈̲̙̩̞̣̻̥̬͊̃͗̉͌̒̽͝A̷̲̩̮͇̗̞̟̣̯͎̯̱͗̀̿̕͜͜ ̸̟̈͗̒̊̈́͊̓̍͛Ư̶̧͕͔̮̥̳̘͕͕͖̟̪͛̌̀͊̾͆̽̑͑̐͗̋͘̚P̴͍̕P̴̨̯̩̥͚̱̪͍̜̮͎͚̺͗̀͌̔͊͑͛̄̍̈́̌Ȩ̸͚͉̫̦̪͇͈̰͔̠̙͍͚̦͋̃̓̑̽̎̓̚R̵̢̩͔͇̖̞̪̒̈́̎̏͌̓̕̕͝ ̴̭̭̮͑̋̓̌͐̆͌͌͂̊͒̕Ļ̷̧̲̻̲̤̫̼͕̹̦̜͍̌̀I̸̢̨̹̪̞̖̱̬̩̥̦̣̰͙̤̋̈́͆̎̍̓̓͘͝M̴̨̨̛̭͇̽̈́͂́͋̀̊̿́̑̚̚ͅI̶͉̒̋̾̀́̃̄̽̎͊̿̋͝͝T̴͍̋̒" videos where they start doing recursion over and over, then the video starts going faster and faster until it turns into a bass boosted epileptic sequence (example, epilepsy warning btw: https://www.youtube.com/watch?v=0TsagtbAusA&t=203)? well, you probably know what i'm thinking...
but how exactly do we cause an explosion? well, we can introduce a new variable: ER^1, which is Explosive Recursion. every time we increase ER's exponent by 1, think of the universe exploding for extra effect. but what exactly is ER?

part 7: what is ER?

remember around 6 minutes ago in part 1 where we decided past Z we would notate layers of recursion as a number? we'll do it again. ω is the 24th letter of the greek alphabet, so the next function after ω(x) would be {25}(x). now let y_0 = {25}(3). y_(n+1) = {y_n}(y_n). now let z_0 = y_{25}(3). z_(n+1) = y_{z_n}(z_n). as you continue defining variables after z, think of that video speeding up and getting faster, to the point where it becomes a bass boosted epileptic mess. after ω(3) variable definitions, we hit ER^1.

part 8: larger ER exponents

what about ER^2? ER^2 is reached after ER^1 variable definitions. you can probably predict what ER^3 is. ER^n+1 = ER^n variable definitions. and as you can probably predict, ER^0 = 1.
oh boy, you can probably predict what happens next. ER^ER^2... ER^ER^ER^2... i'm already tired of this. hey, this somewhat resembles tetration! so...
ER^^2... ER^^ER^^2... onto pentation... ER^^^2... skip to arrow notation... ER↑(ER^2)↑2... ER↑(ER↑(ER^2)↑2)↑2... all roads lead to recursion. new variable!

part 9: introduction to NR

let's define a new variable: NR, for Nuclear Recursion. let's begin some more variable definitions. let a_0 = ER↑(ER^2)↑2. a_(n+1) = ER↑(a_n)↑2.
let b_0 = a_(ER↑(ER^2)↑2). b_n+1 = a_((ER↑(b_n)↑2)). continue defining new variables with this pattern. after ER↑(ER^2)↑2 variable definitions, NR^1 is reached. further NR exponents are defined similarly to further ER exponents. now let's not sugarcoat the progression and go straight to NR↑(NR^2)↑2. this is where the final push begins.

part 10: the final push

let α_0 = NR↑(NR^2)↑2. α_n+1 = NR↑(α_n+1)↑2. let β_0 = a_(NR↑(NR^2)↑2). β_n+1 = a_(NR↑(β_n)↑2). continue defining variables as usual. after NR↑(NR^2)↑2 variable definitions, you reach my number. is there really a point in getting the number any further? it's already so massive. but hey, the 9850 forum post character limit (as of finishing this sentence) says otherwise.

SECTION 3: REPETITION SIMULATOR (FREE DOWNLOAD) (NO VIRUS)

part 11: more power

what if i told you that the entirety of section 2 is just 2 layers of recursion? yeah. i'm sure you're tired of hearing "recursion" 31 billion times already. ER^1 can be represented with our brand new function OR(x) (omega recursion) as OR(1). NR^1 can be represented as OR(2). and "my number" from part 10 is OR(3). time for more re- actually screw that

part 12: more powerful variable definitions

defining variables is more fun. wait, we don't have any alphabe-
HIRAGANA. let あ_0 = OR(^OR(^OR(3))(3))(3). あ_(n+1) = OR(あ_n). い_0 = あ_OR(^OR(^OR(3))(3))(3). い_(n+1) = あ_OR(^OR(^OR(い_n))(3))(3). continue defining variables for every hiragana character up to ん. use the variable-to-number trick to define variables all the way up to variable OR(^OR(^OR(3))(3))(3). now let's define a new explosion variable: ED^1, for Explosive Definitions.

part 13: more powerful explosions

now that we have ED^1, we can go straight to arrow notation, leading us to ED↑(ED^2)↑2!!! now for some more variable definition shenanigans...
let ア_0 = ED↑(ED^2)↑2. ア_(n+1) = ED↑(ア_n)↑2. let イ_0 = ア_ED↑(ED^2)↑2. イ_(n+1) = ア_ED↑(イ_n)↑2. define ED↑(ED^2)↑2 more variables. you have reached RD^1.
let's make a new function. OD(1) (omega definitions) = ED^1, OD(2) = RD^1.

part 14: more powerful recursions

we have another function, meaning that in order to progress we need to continue recursion. every time we try to escape it, it always comes back. well actually, just jump to OD(^OD(^OD(2))(2))(2). let's create another explosion variable to end this section off!
let ✂0 = OD(^OD(^OD(2))(2))(2).✂(n+1) = OD(^OD(^OD(✂_n))(2))(2). let ✁_0 = ✂OD(^OD(^OD(2))(2))(2). ✁(n+1) = ✂_OD(^OD(^OD(✁_n))(2))(2). define OD(^OD(^OD(2))(2))(2) more variables. we now have ΩR^1 (you can probably guess what this stands for).

part 15: end it already...

let's get this part over with. time for the fake conclusion, then the sUrPrIsInG rEvEaL that this forum post is over. my number is ΩR↑(ΩR^2)↑2. it's not over yet, we have 7736 more characters to go.

SECTION 4: THE META SECTION

part 16: the section function

we've gone a long way. how about we automate things? let S(1) = x_ω(3). S(2) = OR(3). S(3) = ΩR↑(ΩR^2)↑2. notice something? S(x) produces the final result of each section. which is why we're going to generalize sections now!

part 17: generalizing sections

each section will have a variable definition section, an explosion section, and a recursion section in that order. but what exactly does that mean? here's a step by step for each section, using S(3) -> S(4) as an example:

1. define variables with an obscure alphabet. let (char_a)0 = ΩR↑(ΩR^2)↑2, (char_a)(n+1) = ΩR↑(ΩR^(char_a)n)↑2, let (char_b)0 = (char_a)(ΩR↑(ΩR^2)↑2), (char_b)(n+1) = (char_a)_(ΩR↑(ΩR^(char_b)_n)↑2)
2. explode. define ΩR↑(ΩR^2)↑2 more variables and create a new explosion variable (AB^1 for example). jump to using arrow notation. AB↑(AB^2)↑2.
3. repeat step 1 and 2 with the result from the previous step once.
4. create a function that creates new explosion variables.
5. recurse your function. for example ΩA(3) would become ΩA(^ΩA(^ΩA(3))(3))(3)
6. the next S(x) value is the result of step 5

part 18: the fun begins

now that we have defined S(x), let's recurse it!!! we now have S(^S(^S(4))(4))(4).
let's define S(^S(^S(4))(4))(4) variables now (skipping the process since i already generalized the process) and create a new explosion variable: ES^1
now jump to arrow notation... ES↑(ES^2)↑2
and finally... create a function that creates new explosion variables. OS(1) = ES^1.
recurse that function. OS(^OS(^OS(2))(2))(2). this is actually kinda getting boring...

part 19: the fun ends?

well, we tried automating the process of defining, recursing, and exploding but we ended up continuing to define, recurse, and explode. wait...
introducing S_2(x)! this repetitively repeats the steps done in part 18! let's recurse i-
introducing S_3(x)! this repetitively defines variables, explodes functions, and re-
introducing S_4(x)! this-
introducing S_(OS(^OS(^OS(2))(2))(2))(x)! yeah.

part 20: 5665 more.

now, first things first: recurse this monstrosity of a function. for simplicity we'll rename it to ΩS(x). now we have ΩS(^ΩS(^ΩS(3))(3))(3). now, explode this function. ΩE^1.
jump to arrow notation. ΩE↑(ΩE^2)↑2. that's my number.

SECTION 5: PROGRESSION INNOVATION

part 21: a new method of progression

let's introduce another type of progression: RP (recursion power). ΩE↑(ΩE^2)↑2 = RP {R^2/2}. there are 2 layers of recursion (marked by the /2) and the final ΩE being raised to the power of 2 (marked by the R^2) you know what we're gonna do. let's boost those values up!!! we now have RP {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2} you know what's special about this? RP {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2} is equal to RP^2 {R^2/2}. you know what's happening now.

part 22: return of arrow notation

time to abuse the notation to reach a new massive number! RP^ΩE↑(ΩE^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}... RP^RP^ΩE↑(ΩE^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2} {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2} oh god that's way too long... no worries, arrow notation! RP↑(RP^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2} yeah that's still way too long

part 23: a somewhat brand new notation

let's introduce RP_1 {R^2/2}! each time we reach RP_n↑(RP_n^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}, increase n by 1. now... RP_ΩE↑(ΩE^2)↑2↑(RP_ΩE↑(ΩE^2)↑2^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}... RP_RP_ΩE↑(ΩE^2)↑2↑(RP_RP_ΩE↑(ΩE^2)↑2^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}... huh. looks like we're gonna need a function.

part 24: a not so brand new function

introducing PR(x) (power recursion)! for a simple explanation, PR(1) = RP_ΩE↑(ΩE^2)↑2↑(RP_ΩE↑(ΩE^2)↑2^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}, and PR(2) = RP_RP_ΩE↑(ΩE^2)↑2↑(RP_RP_ΩE↑(ΩE^2)↑2^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}! let's recurse our function, leading to PR(^PR(^PR(2))(2))(2). let's do something similar to what we did to S(x), creating PR_n(x) which counts the amount of times we recursed PR_(n-1)(x)! hard to understand? basically, PR_(n+1)(3) =PR_n(^PR_n(^PR_n(2))(2))(2), since there are 3 PR_n's used in the recursion. now, we have... PR_(PR(2))(^PR_(PR(2))(^PR_(PR(2))(2))(2))(2)... PR_(PR_(PR(2))(2))(^PR_(PR_(PR(2))(2))(^PR_(PR_(PR(2))(2))(2))(2))(2)... oh god. i think this is a good time to bring in the variable definitions

part 25: the cycle continues

let's see... which unicode symbols should i use today... okay i got ⥯ and ⏻ cool so anyways
let ⥯0 = PR(PR(2))(^PR_(PR(2))(^PR_(PR(2))(2))(2))(2), ⥯(n+1) = PR(PR(⥯n))(^PR(PR(⥯n))(^PR(PR(⥯n))(⥯n))(⥯n))(⥯n), and let ⏻0 = ⥯PR(PR(2))(^PR(PR(2))(^PR(PR(2))(2))(2))(2), ⏻(n+1) = ⥯PR(PR(⏻n))(^PR(PR(⏻n))(^PR(PR(⏻n))(⏻n))(⏻n))(⏻n). define PR(PR(PR(2))(2))(^PR(PR(PR(2))(2))(^PR(PR(PR(2))(2))(2))(2))(2) more variables. new explosion variable! EP^1. now jump to using arrow notation... EP↑(EP^2)↑2. yeah. let's convert it into a power formula: EP↑(EP^2)↑2 = PE {P^2/2}
now jump to PE {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}, then to PE^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}, then to PE↑(PE^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}, then to RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}... then to RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}... yeah i'll just call that my number and get this section over with

SECTION 6: THE TRUE FINAL PUSH

part 26: return to the basics

let's start again, but simpler. let's call our previous number (RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}) "new 1." you know what we're gonna do. we're doing it all over again.

NEW SECTION 1: NEW SOMEWHAT UNDERSTANDABLE EXPLANATION

new part 1: new the basics

first, let's start with a small new numbe- you know what skip all that

new part 25: new the cycle continues

let's see... which unicode symbols should i use today... okay i got ⥯ and ⏻ cool so anyways
let new ⥯0 = new PR(PR(2))(^PR_(PR(2))(^PR_(PR(2))(2))(2))(2), new ⥯(n+1) = new PR(PR(⥯n))(^PR(PR(⥯n))(^PR(PR(⥯n))(⥯n))(⥯n))(⥯n), and let new ⏻0 = new ⥯PR(PR(2))(^PR(PR(2))(^PR(PR(2))(2))(2))(2), ⏻(n+1) = ⥯PR(PR(⏻n))(^PR(PR(⏻n))(^PR(PR(⏻n))(⏻n))(⏻n))(⏻n). define new PR(PR(PR(2))(2))(^PR(PR(PR(2))(2))(^PR(PR(PR(2))(2))(2))(2))(2) more variables. new explosion variable! EP^1. now jump to using arrow notation... new EP↑(EP^2)↑2. yeah. let's convert it into a power formula: new EP↑(EP^2)↑2 = new PE {P^2/2}
now jump to new PE {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}, then to PE^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}, then to PE↑(PE^2)↑2 {R^ΩE↑(ΩE^2)↑2/ΩE↑(ΩE^2)↑2}, then to new RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}... then to new RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}... yeah i'll just call that my number and get this section over with

NEW SECTION 6: THE NEW TRUE FINAL PUSH

new part 26: new return to the basics

let's start again, but simpler. let's call our previous new number (new RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}) "new new 1." you know what we're gonna do. we're doing it all over again.

NEW NEW SECTION 6: THE NEW NEW TRUE FINAL PUSH

new new part 26: new new return to the basics

let's start again, but simpler. let's call our previous new new number (new new RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}) "new new 1." you know what we're gonna...

my number is (moderately) bigger in 2100:

is that number well-defined? that's my only question other than "uhhh gimme Veblen approximation" and "uhhhhh loader.c?"

probably, not sure though because I’m not sure about the variable definition and the “new number” section

what does 'new 2' mean

Royal says:

what does 'new 2' mean

new 2 is after new 1, or in more nerdy terms it successes new 1
new new 1 = new RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} so basically for every new multiply RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} by itself
the final number is actually RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} so basically for every new multiply RP^RP^EP↑(EP^2)↑2 {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2} {P^EP↑(EP^2)↑2/EP↑(EP^2)↑2}^^2 (or (new 1)^^2) so uhh
i feel like writing another 16384 character post for some reason now,,,

i know that it succeeds new 1 but what does it equal

OmegaNum (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjxh-_t8JOJAxUFFDQIHQDvI-gQFnoECAwQAQ&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FOmega_function&usg=AOvVaw0ZuxUMQzUCC95sdUkzjuRz&opi=89978449)