15 Dec
2011
15 Dec
'11
2:44 p.m.
Does someone have a proof for k=72 (in decimal)? Sure seems right, but I don't think we've seen an upper bound on the x that attains the maximum k, which would justify computer-based evidence. --Michael On Thu, Dec 15, 2011 at 4:15 PM, N. J. A. Sloane <njas@research.att.com>wrote:
It is 72, I think, for x = 125 Yes, you are right. It seems likely that 72 should suffice for all integers x.
What about the sequence: a(n) = smallest k such that ... ? With a(125) = 72. Could you please submit it? (With keyword "base", of course) Neil
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