[seqfan] Re: Problem

[Robert Dougherty-Bliss]

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
>
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