[seqfan] Re: A195264
Neil Sloane
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),
so
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.
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 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':
>
> http://gladhoboexpress.blogspot.ca/2014/10/climb-to-prime.html
>
> 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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