[seqfan] Re: A195264
njasloane at gmail.com
Tue Jun 6 07:25:03 CEST 2017
I wish I understood that construction! I can see he is looking
for a number n = x*p which is fixed
under our map f() = A080670(). Here p is a prime which is greater than any
prime dividing x, so
f(n) = f(x)*10^y + p,
where y is the length of p.
We want f(n) = n, so f(x)*10^y = p*(x-1),
10^y*f(x)/(x-1) = p
Now we are told that x = 1047*10^y + 1 is a good choice..
But I didn't find any value of y (or p) that would work, so I must
be missing something.
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 Mon, Jun 5, 2017 at 10:56 PM, Hans Havermann <gladhobo at bell.net> wrote:
> > That's pretty amazing! Where can one read more about James Davis's work?
> He contacted me yesterday by way of a comment underneath my October 2014
> blog on 'Climb to a Prime':
> In a subsequent email he confided:
> "I'm not a mathematician by any stretch - the search that happened to work
> was hoping n = x*p=f(x)*10^y+p, where p is the largest prime factor of n.
> That requires that f(x)/(x-1) terminate and it's decimal expansion be a
> prime (p). That motivates looking for x of the form x=m*10^y+1 and hoping
> some common factors cancel between f(x) and (x-1). Turned out to be enough:
> m=1407 fell out immediately."
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