# [seqfan] Re: A195264

jean-paul allouche jean-paul.allouche at imj-prg.fr
Tue Jun 6 07:27:14 CEST 2017

```Dear Neil

best wishes
jean-paul

Le 06/06/17 à 07:25, Neil Sloane a écrit :
> 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.
> 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."
>>
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
> --
> Seqfan Mailing list - http://list.seqfan.eu/

```