[seqfan] Re: Problem

Tomasz Ordowski tomaszordowski at gmail.com
Wed Jul 22 11:11:43 CEST 2020

Hello Allan,

I am pleased with the interest in this topic.

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.

Proof.
If (p+1)2^k-1 = q prime for some k > 0,
then (q+1)2^m-1 is prime for some m > 0,
because the assumption states that for every prime p,
(p+1)2^n-1 is prime for some n > 0 (without declaring truth).
Note that the prime (q+1)2^m-1 = (p+1)2^n-1, where n = k+m.
Hence, by induction, it is prime for infinitely many n (the thesis), qed.
So this conditional theorem is true. Is the condition (assumption)
provable?

Have a nice day!

Thomas
_________
Conjecture: For a prime p,
if (p+1)2^n-1 is prime for some n > 0,
then it is prime for infinitely many n.

śr., 22 lip 2020 o 10:38 Allan Wechsler <acwacw at gmail.com> napisał(a):

> I concur. But establishing the premise will be very hard. If it could be
> proved that every prime had a prime Sophie Germain extension, then the
> infinitude of the Mersenne primes would be a trivial corollary.
>
> On Wed, Jul 22, 2020, 12:40 AM Tomasz Ordowski <tomaszordowski at gmail.com>
> wrote:
>
> > 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.
> >
> > T. Ordowski
> >
> >
> > pon., 20 lip 2020 o 21:06 <israel at math.ubc.ca> napisał(a):
> >
> > > See sequence A257495. The b-file goes up to 7075 with no 0's, i.e. none
> > of
> > > the first 7075 primes has the property.
> > >
> > > Cheers,
> > > Robert
> > >
> > > On Jul 20 2020, Robert Dougherty-Bliss wrote:
> > >
> > > >Dear Thomas,
> > > >
> > > >You may already be aware, but none of the first 100 primes (<= 541)
> > > >satisfy this property.
> > > >
> > > >Amazingly, the earliest counterexample for p = 73 is the following
> > > integer:
> > > >
> > > >
> > > >
> > >
> >
> 12525084203259602214176345117827991857573063437151079650189656689252041617399
> > >
> > > >
> > >
> >
> 16118618976873174436648194378202145606096817433350319763375794132326993383200
> > >
> > > >
> > >
> >
> 14217732225003163760036417965916387747831867749318699104524437655151695087826
> > >
> > > >
> > >
> >
> 47278357731824391729532319069188907350539418959168425940169356532195426353195
> > >
> > > >
> > >
> >
> 84257183520755212129194474630919879413057346247800071524008686049488780942766
> > >
> > > >
> > >
> >
> 38123436651683349651892026768245860789398297612527549211852109219078820059778
> > >
> > > >
> > >
> >
> 19346432242814374609091413789240598598335924463948419947004368457022517766034
> > >
> > > >
> > >
> >
> 95591799870311650343246943884972083691195975663585667560716289785503524182355
> > >
> > > >
> > >
> >
> 53897768571561351251352502155056787443177087759615376430034900988921205572639
> > >
> > > >
> > >
> >
> 317118528079725593399200244440233458975807425711011346463660588817113315016703
> > > >
> > > >Robert
> > > >
> > > >
> > > >Robert
> > > >
> > > >
> > > >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/
> > > >
> > > >
> > >
> > > --
> > > Seqfan Mailing list - http://list.seqfan.eu/
> > >
> >
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> >
>
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