03/08/2011, 10:56 PM

There's one thing I can easily proof about this:

n > 2 -> 3 [n] 2 > 2 [n] 3

Proof:

n = 3:

3 [3] 2 = 3 ^ 2 = 9 > 8 = 2 ^ 3 = 2 [3] 3

Suppose for some n > 2 : 3 [n] 2 > 2 [n] 3

Then we wish to prove that 3 [n+1] 2 > 2 [n+1] 3

3 [n+1] 2 = 3 [n] 3

= 3 [n-1] (3 [n] 2)

> 3 [n-1] (2 [n] 3)

> 2 [n-1] (2 [n] 3)

= 2 [n] 4

= 2 [n] (2 [n+1] 2)

= 2 [n+1] 3

But now I also suspect that for each n:

2 [n+1] b > 3 [n] b

will be true for sufficiently large b

n = 1:

b > 3 -> 2 * b > 3 + b

n = 2:

b > 3 -> 2 ^ b > 3 * b

But haven't yet proved it for any n > 2.

n > 2 -> 3 [n] 2 > 2 [n] 3

Proof:

n = 3:

3 [3] 2 = 3 ^ 2 = 9 > 8 = 2 ^ 3 = 2 [3] 3

Suppose for some n > 2 : 3 [n] 2 > 2 [n] 3

Then we wish to prove that 3 [n+1] 2 > 2 [n+1] 3

3 [n+1] 2 = 3 [n] 3

= 3 [n-1] (3 [n] 2)

> 3 [n-1] (2 [n] 3)

> 2 [n-1] (2 [n] 3)

= 2 [n] 4

= 2 [n] (2 [n+1] 2)

= 2 [n+1] 3

But now I also suspect that for each n:

2 [n+1] b > 3 [n] b

will be true for sufficiently large b

n = 1:

b > 3 -> 2 * b > 3 + b

n = 2:

b > 3 -> 2 ^ b > 3 * b

But haven't yet proved it for any n > 2.