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Zack_M_Davis comments on Rationality Quotes May 2012 - Less Wrong
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I haven't studied number theory, but I expect that someone who has would be able to answer this. Successive powers of three have final digits in the repeating pattern 1, 3, 9, 7, so if we can find N mod 4 for the N such that 3^N = 3^^^3, then we would have our answer.
3^odd = 3 mod 4
so it ends in 7.
(but I repeat myself)
I think you're mistaken. Counterexample: 3^9 = 19683.
19683 = 3 mod 4
and 3^19683 = 150 ... 859227, which ends in 7.
( The full number is 9392 digits long, which messes up the spacing in these comments. )
Oh, sorry; I agree that odd powers of three are 3 mod 4, but I had read VKS as claiming that odd powers of three had a final digit of seven; I probably misunderstood the argument. [EDIT: Yes, I was confused; I understand now.]
right, well, it's just that 3^^^3 = 3^3^3^3^3...3^3^3 = 3^(3^3^3^3...3^3^3), for a certain number of threes. So, 3^^^3 is 3^(some odd power of three).
Yes, thanks; I apologize for having misunderstood you earlier.
That is entirely ok -- I am badly in need of sleep and may have failed to optimise my messages for legibility.