2 [n] b and 3 [n] b for (large) integer n, b

[dyitto]

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.

[dyitto]

n = 3:
b > 3 -> 2 [4] b > 3 [3] b

Proof:

For b = 4:
2 [4] 4 = 2 ^ (2 ^ (2 ^ 2)) = 2 ^ (2 ^ 16) = 2 ^ 65536 > 81 = 3 ^ 4 = 3 [3] 4

Let's assume that 2 [4] b > 3 [3] b for some b > 3.
Then we wish to prove that 2 [4] (b + 1) > 3 [3] (b + 1)


2 [4] (b + 1) = 2 ^ (2 [4] b)
> 2 ^ (3 [3] b)
> 3 * (3 [3] b)
= 3 [3] (b + 1)

[dyitto]

n = 4:
b > 3 -> 2 [5] b > 3 [4] b

Proof:

For b = 4:

Apply lemma 8 having a = 2, b = 3, c = 2, m = 4, k = 2:

2 [4] 4 > 2 * (3 + 2) is certainly true
-> 2 [4] 7 >= 3 [4] 4
2 [5] 4 = 2 [4] 65536 > 2 [4] 7 >= 3 [4] 4

Let's assume that 2 [5] b > 3 [4] b for some b > 3.
Then we wish to prove that 2 [5] (b + 1) > 3 [4] (b + 1)


2 [5] (b + 1) = 2 [4] (2 [5] b)
> 2 [4] (3 [4] b)
> 3 [3] (3 [4] b)
= 3 [4] (b + 1)