- #1
fat f...
- 8
- 0
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 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
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)