- #1

- 8

- 0

## Main Question or Discussion Point

is this bigger than grahams number, i again "accidentaly" checked grahams number out and felt dumb then lost/forgot innovation for classical progressive/ progressive progressive versions of this model

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 shouldnt let myself influence myself by infesting my mind again with grahams number to "show off" that my number is bigger, now i cant think anything because im also mentally in hard times where i also forget vocabulary from foreign languages and to confirm something like this isnt what im now 100% capable of

i wont say why its concepted as it is, i only need confirmation

here is original script

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 shouldnt let myself influence myself by infesting my mind again with grahams number to "show off" that my number is bigger, now i cant think anything because im also mentally in hard times where i also forget vocabulary from foreign languages and to confirm something like this isnt what im now 100% capable of

i wont 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)