[seqfan] Re: A195264

[Neil Sloane]

Dyslexics of the world, untie!

J-P, Thank you!

Now y=5 works, the prime is p = 96179,
and (1407*10^5+1)*p = 13532385396179
is indeed fixed by f !!!


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 Tue, Jun 6, 2017 at 1:27 AM, jean-paul allouche <
jean-paul.allouche at imj-prg.fr> wrote:

> Dear Neil
>
> Is "1047" instead of "1407" a misprint in your email or in your search?
> 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.
>> 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."
>>>
>>>
>>> --
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>> --
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>
>
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