[seqfan] Re: Why should A003001 be finite?
njasloane at gmail.com
Sat Oct 10 19:42:38 CEST 2015
I looked hard at this question back in 1972 - see my JRM article.
It is certain that it is finite - as soon as you have a 2 and a 5 anywhere
in the number it dies 2 steps later. We just can't prove it ...
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 at gmail.com
On Sat, Oct 10, 2015 at 1:37 PM, Heinz, Alois <alois.heinz at hs-heilbronn.de>
> Am 10.10.2015 um 17:22 schrieb Felix Fröhlich:
> > Hi Seqfans,
> > in A003001 there is comment saying the sequence is probably finite. What
> > property of n points towards that assumption? It seems to me the opposite
> > is more likely to be true. In fact, isn't it possible for any composite c
> > to construct a number k such that the digital product of k is c?
> Try it with c=77.
> There is no k such that the digital product of k is c
> because there is no digit "11" in the decimal system.
> A003001 is finite because you cannot avoid digit 0
> long enough to get a larger persistence.
> Best regards,
> Seqfan Mailing list - http://list.seqfan.eu/
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