User blog:Primussupremus/Extending my notation past the 2nd level and into the 3rd.

Last time we looked at numbers like (3,f,3#10) and (6,z,6#100) this time I am going to extend the general principles from my last entry on the notation I developed yesterday to produce even larger numbers. Suppose we have something like (3,f,3#3,f,3) or (3,f,3) recursed (3,f,3) times we can write this as (3,f,3#(2)) ,another example of this is (3,f,3#3,f,3#3,f,3)=(3,f,3#(3)). (3,f,3#(100)) would be (3,f,3) recursed (3,f,3#(99)) times and (3,f,3#(165)) would be (3,f,3) recursed (3,f,3#(164)) times. Lets move away from (3,f,3) and use some different symbols (10,g,10). (10,g,10#(200))=(10,g,10) recursed (10,g,10#(199)) times. So (n,p,n(q) is (n,p,n) recursed (n,p,n(q-1) times. A large number example is (10,g,10#(10,c,100)) or (10,g,10) recursed (10,c,10)-1 times where (10,c,10) is equal to one googol. Now that we have a firm grasp of whats happening we can move on to more powerful things how about something like (4,A,4#(4,A,4)) or (4,A,4) recursed (4,A,4)-1 times. We write this as (4,A,4#[2]) to distinguish it from the last level of the notation.

(4,A,4(4,A,4#(4,A,4)))=(4,A,4#[3]) thats (4,A,4) recursed (4,A,4#[2])-1 times.

(4,A,4#[500]) is equal to (4,A,4) recursed (4,A,4#[499])-1 times and (11,Z,22#[3333]) is equal to (11,Z,22) recursed (11,Z,22#[3332]) times.