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Mar 11, 2014 at 19:11 vote accept Michael Albert
S Mar 11, 2014 at 14:54 history suggested Ben Barber
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Mar 11, 2014 at 14:51 review Suggested edits
S Mar 11, 2014 at 14:54
Mar 11, 2014 at 14:04 answer added Lucia timeline score: 22
S Mar 11, 2014 at 7:52 history suggested F. C.
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Mar 11, 2014 at 7:49 review Suggested edits
S Mar 11, 2014 at 7:52
Mar 11, 2014 at 5:43 comment added Greg Martin well ... that's quite a puzzler! I will say that I computed $(\log a(n))/\log^2 n$ for $n$ out to $10^5$; after the first few integers it's an essentially increasing quantity (other than a minor parity difference) which does pass 0.5 between 57800 and 57900. In fact, the data seems to suggest that the power of $\log n$ should be between 2.1 and 2.2, rather than 2.
Mar 11, 2014 at 1:26 history asked Michael Albert CC BY-SA 3.0