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Kindly comments on New Monthly Thread: Bragging - Less Wrong Discussion
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...62535796399618993967905496638003222348723967018485186439059104575627262464195387.
Boo-yah.
Edit: obviously this was not done by hand. I used Mathematica. Code:
Edit: this was all done to make up for my distress at only having an Erdos number of 3.
Impressive, I didn't think it could be automatized (and even if it could, that it could go so many digits before hitting a computational threshold for large exponentials). My only regret is that I have but 1 upvote to give.
It is not clear to me why the above code works. In particular, the 10^80 part.
A number mod 10^n yields the last n digits of the number.
Aaah. it was just saying how much to output. Phew. I was trying to figure out what 10^80 could have to do with 3^^^3 and failing, hard. So, it's a great relief that the answer is, 'nothing'.
Technically, I have not used the fact that the number in question is 3^^^3 -- I am treating it as 3^^X, where X is very large. So for huge numbers of digits, this will not give the correct answer, but I don't think it's practical to compute that many digits, in any case.
Should I explain how everything else works?
Nah, I just figured that the named functions were something relevant, and I'm satisfied with that level of knowledge at this point.