- #1

fat f...

- 8

- 0

when somebody takes a sum of 2^3^4^5^6^7^8^9 and then this sum will "^" itself, so much times as the sum's value is, is it at the end greater than grahams number? and if, then why?(if graham loses then because it ends on g64?)

i knew i shouldn't let myself influence myself by infesting my mind again with grahams number to "show off" that my number is bigger, now i can't think anything because I am also mentally in hard times where i also forget vocabulary from foreign languages and to confirm something like this isn't what I am now 100% capable of

i won't say why its concepted as it is, i only need confirmation

here is original script

classical

2^3^4^5^6^7^8^9

=

i0 (basis value number)

i0^i0

=

i1

i1^i1

=

i2

.

.

.

result of highest value of procedures in i0^result of highest value of procedures in i0

=

. number (classical)

progressive

9^9^9^9^9^9^9^9^9

=

in9 (basis value number)

in9^in9

=

in91

in91^in91

=

in92

.

.

.

result of highest value of procedures in in9^result of highest value of procedures in in9

=

number (progressive)