Only considering the last four digits, for all values of i between 0 and 10000 not divisible by 10, shows 125 at 72 as the highest (next best is 3375 at 56). This proves that the limit is k=72 at 125, and there is no solution that does not end in "0125" that is not 125 itself. On Thu, Dec 15, 2011 at 1:43 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- -- http://cube20.org/ -- http://golly.sf.net/ --