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Apr 15, 2020 at 16:57 vote accept nflswsykimi
Jun 27, 2014 at 21:23 answer added Lucia timeline score: 11
Jun 27, 2014 at 20:55 review First posts
Jun 27, 2014 at 21:26
Jun 27, 2014 at 20:43 comment added Anthony Quas Probably this is very hard. There's a well known open problem, showing that every sufficiently large power of 2 has at least one 7 in its decimal expansion. I don't see why the base 3 would be any easier. [ You can, however, show that for almost every $n$, $2^n$ has at least one 1 in its ternary expansion ].
Jun 27, 2014 at 20:35 history asked nflswsykimi CC BY-SA 3.0