What about A3001? The smallest number of multiplicative persistence n. 0, 10, 25, 39, 77, 679, 6788, 68889, 2677889, 26888999, 3778888999, 277777788888899 See A121111 for the last term: 277777788888899, 4996238671872, 438939648, 4478976, 338688, 27648, 2688, 768, 336, 54, 20, 0 (read it backwards to get your problem) I found this in 1973, and conjectured that 11 steps is the largest possible. The conjecture has is still open (but certainly true) Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Fri, Jul 26, 2019 at 2:41 PM W. Edwin Clark <wclark@mail.usf.edu> wrote:
6 terms: [6, 32, 48, 168, 27216, 666792] 7 terms: [20, 54, 336, 768, 2688, 27648, 826686] altermative 7 terms: [20, 54, 336, 768, 2688, 27648, 2239488] 8 terms: [20, 54, 336, 768, 2688, 27648, 338688, 4478976] Obtained using Maple's GraphTheory Package. I think this is the best one can do using only 7-smooth numbers <http://oeis.org/A002473> up to 10 million
On Fri, Jul 26, 2019 at 2:30 AM Éric Angelini <eric.angelini@skynet.be> wrote:
5 terms (with my smartphone)! 6,16,144,1944,3899,... Best, É.
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