[seqfan] Re: Problem
Tomasz Ordowski
tomaszordowski at gmail.com
Tue Jul 28 10:11:49 CEST 2020
Dear Jack & all,
I checked it, congratulations!
Seek and you will find!
Conjecture: if p is a prime,
then there exists n such that (p+1)2^n-1 is prime.
Theorem: if, for every prime p,
(p+1)2^n-1 is prime for some n > 0,
then it is prime for infinitely many n.
We already know the proof of this conditional theorem.
By the dual Riesel conjecture;
if p is an odd prime, and an integer m > 0,
then there exists n such that (p+2^m)2^n-1 is prime.
Note: we consider the case of m = 0 separately.
By the dual Sierpinski conjecture;
if p is an odd prime, and an integer m > 0,
then there exists n such that (p-2^m)2^n+1 is prime.
Note: if m = 0, then p <> 2^(2^4)+1 the last Fermat primes.
Generally (both of these dual conjectures in one):
if p is an odd prime, and an integer m > 0,
then there exists n such that
|(p+/-2^m)2^n-/+1| is prime.
Exactly!
Thomas Ordowski
_______________________
https://en.wikipedia.org/wiki/Sierpinski_number#Dual_Sierpinski_problem
https://en.wikipedia.org/wiki/Riesel_number#The_dual_Riesel_problem
wt., 28 lip 2020 o 07:58 Jack Brennen <jfb at brennen.net> napisał(a):
> You will find that for p = 94603 and n = 237504, that you get a prime.
>
> I didn't do the prime search myself, just a web search. This problem,
> in a different restated way, was discussed on mersenneforum.org back in
> 2007:
>
> https://www.mersenneforum.org/showthread.php?p=116340#post116340
>
> You will see there that 23651*2^237506-1 was found to be prime, which is
> another way of writing the number I alluded to on the first line.
>
>
>
> On 7/27/2020 2:26 AM, W. Edwin Clark wrote:
> > My Maple program is still running for p = 94603 and now it has passed n =
> > 160000 without
> > finding any primes of the form (p+1)2^n - 1. To check the primality of
> the
> > number I
> > use Maple's probabilistic primality test (isprime) ---so I don't collect
> > any information
> > about LPF((p+1)2^n - 1).
> >
> > On Sun, Jul 26, 2020 at 11:46 PM Tomasz Ordowski <
> tomaszordowski at gmail.com>
> > wrote:
> >
> >> Hello Edwin and Allan!
> >>
> >> Thank you for your active interest in the topic.
> >>
> >> Let LPF(n) be the Least Prime Factor of n. The provable theorems:
> >> (1) There are no primes p such that LPF((p+1)2^n-1) < p for all n > 0.
> >> (2) There are no primes p such that LPF((p-1)2^n+1) < p for all n > 0.
> >>
> >> Have a nice Sunday!
> >>
> >> Thomas
> >>
> >> niedz., 26 lip 2020 o 09:41 Tomasz Ordowski <tomaszordowski at gmail.com>
> >> napisał(a):
> >>
> >>> Hello Edwin and Allan!
> >>>
> >>> Thank you for your active interest in the topic.
> >>>
> >>> Let LPF(n) be the Least Prime Factor of n. The provable theorems:
> >>> (1) There are no primes p such that LPF((p+1)2^n-1) < p for all n > 0.
> >>> (2) There are no primes p such that LPF((p-1)2^n+1) < p for all n > 0.
> >>>
> >>> Have a nice Sunday!
> >>>
> >>> Thomas
> >>>
> >>> pt., 24 lip 2020 o 23:14 W. Edwin Clark <wclark at mail.usf.edu>
> >> napisał(a):
> >>>> For the prime p = 94603 and for n from 1 to 100000, (p+1) 2^n - 1 is
> >>>> composite, says Maple.
> >>>> This prime appears twice in the OEIS if you don't count A094603. See
> >>>> http://oeis.org/search?q=94603&language=english&go=Search Note this
> >>>> search doesn't
> >>>> include things like the sequence of primes.
> >>>>
> >>>> On Sun, Jul 19, 2020 at 2:28 AM Tomasz Ordowski <
> >> tomaszordowski at gmail.com>
> >>>> wrote:
> >>>>
> >>>>> Dear SeqFans!
> >>>>>
> >>>>> Let a(0) = p and a(n) = 2 a(n-1) + 1. Note that a(n) = (p+1) 2^n - 1.
> >>>>> Are there primes p such that a(n) is composite for every n > 0 ?
> >>>>>
> >>>>> Best regards,
> >>>>>
> >>>>> Thomas Ordowski
> >>>>> _______________________
> >>>>> https://en.wikipedia.org/wiki/Riesel_number
> >>>>>
> >>>>> --
> >>>>> Seqfan Mailing list - http://list.seqfan.eu/
> >>>>>
> >> --
> >> Seqfan Mailing list - http://list.seqfan.eu/
> >>
> > --
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> >
> >
>
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