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2 [n] b and 3 [n] b for (large) integer n, b
#1
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.
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Messages In This Thread
2 [n] b and 3 [n] b for (large) integer n, b - by dyitto - 03/08/2011, 10:56 PM

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