[seqfan] Re: Subject: Need help computing a number

[Neil Sloane]

PS

(I had asked for the smallest m such that 104917*2^m - 1 is prime.) Thanks
to everyone who replied, especially Robert Gerbicz, who pointed to the web
page


Ray Ballinger and Wilfrid Keller, <a href="
http://www.prothsearch.com/rieselprob.html">The Riesel Problem: Definition
and Status</a>, Proth Search Page.


(already cited in A050412), and Ed Pegg, who found the web page


 https://rieselprime.de/ziki/Riesel_prime_2_104917


Both pages assert that 104917*2^340181 - 1 is prime. But it isn't clear

whether m = 340181 is the /smallest/ m that gives a prime. The notation in
the second link is very unclear.  Can anyone clarify this?

Best regards
Neil

Neil J. A. Sloane, Chairman, OEIS Foundation.
Also Visiting Scientist, Math. Dept., Rutgers University,
Email: njasloane at gmail.com



On Tue, Jul 30, 2024 at 2:05 PM Neil Sloane <njasloane at gmail.com> wrote:

> Dear Math Fun, Sequence Fans,
>
> Start with an integer k, 13 say, and repeatedly double it and add 1 until
> reaching a prime:
>
> 13 -> 27 -> 55 -> 111 -> 223.
>
> This took 4 steps, so we set R(13) = 4. This is called Riesel's problem,
> and if we never reach a prime we set R(k) = 0. The sequence R(k) is A050412.
> I think Riesel showed R(509203) = 0, and it seems it is believed that R(k)
> != 0 for k<509203.
>
> For another sequence (A374965) that Harvey Dale and I have been studying,
> we badly need the value of R(104916). Can someone help?  If 104916 takes
> m steps, the prime reached will be 104917*2^m - 1,
>
> so we don't actually need to see the prime (just m).
>
> I ran a naive Mathematica program (from A050412) on my iMAC, but I killed
> it after nearly 24 hours.
>
>  I have no idea how far it got.  The bottleneck is presumably the
> primality testing, but I don't know who has the fastest program for that.
> Best regards
> Neil
>
> Neil J. A. Sloane, Chairman, OEIS Foundation.
> Also Visiting Scientist, Math. Dept., Rutgers University,
> Email: njasloane at gmail.com
>
>