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The answer (highlight it): the final repeating population of a 2*(n-2) rectangle in Life (generation 1 of a 2*n zigzag, but I had more difficulty explaining that)
Moosey wrote: July 20th, 2020, 9:46 amIt matches that for the first two entries but no farther
It's sort of hard (read: impossible) to guess a sequence based off of 2 numbers, so I just gave a semi-arbitrary sequence that fits.
This is 0*2^0, 1*2^1, 2*2^2, 402653184*2^402653184, so we can look at the exponents and get a simpler sequence. However, we still need to figure out where 402653184=3*2^27 comes from.
gameoflifemaniac wrote: July 22nd, 2020, 11:42 am
i have no idea what all of those sequences are.
That's the point, isn't it?
Any sufficiently advanced software is indistinguishable from malice.
Alright, my next guess is TREE(n).
TREE(1)=1, TREE(2)=3, and there is no way to write TREE(3) without giving the answer away or collapsing into a black hole.
Any sufficiently advanced software is indistinguishable from malice.
toroidalet wrote: July 22nd, 2020, 10:56 pm
Alright, my next guess is TREE(n).
TREE(1)=1, TREE(2)=3, and there is no way to write TREE(3) without giving the answer away or collapsing into a black hole.
right
κ is measurable iff there is a nontrivial elementary embedding j:V→M (M transitive) with critical point κ
If we write out the prime factorization of n as p_1^a*p_2^b*......., element n-1 (since 1 doesn't have any prime factors) is 1^a2^b.... (treating numbers as strings rather than numbers).
Any sufficiently advanced software is indistinguishable from malice.
toroidalet wrote: July 24th, 2020, 10:05 pm
If we write out the prime factorization of n as p_1^a*p_2^b*......., element n-1 (since 1 doesn't have any prime factors) is 1^a2^b.... (treating numbers as strings rather than numbers).
