User blog:Kel47/Illion Notation Part 3

Review from last time:

1. I(x) = 10^(3x+3)

2. I(x,1) = I(x)

3. I(x,y) = I(I(x,y-1),y-1)

Generalizing rule 3:

4. I(a,b,c.....,y,z)=I(a,b,c..... I(y,z-1),z-1)

I(2,3,2) = I(2,I(3,1),1) = I(2,I(3),1)

= I(2,10^12)

=I(I(2,10^12-1),10^12-1) = =I(I(I(2,10^12-1),10^12-2),10^12-2)

.... 1 trillion steps

=I(I(I...I(2,10^12-1),10^12-2),10^12-3)..... 2),1) = I(I(I...I(2,10^12-1),10^12-2),10^12-3)....., ) 2))

= I(I(#,1), 1)

Adds (2-1)+(2^2-1)+(2^3-1)+......+(2^(10^12-2)-1) I's

Adds  ~ approximately 5 x 10301029995663

I's.

I(I(I.....(10^9)... ) < about 10^300,000,000,000 I's

about (10^^300,000,000,000)