[seqfan] Fwd: Re: more

[Tomasz Ordowski]

Yes...

Conjecture:
for a Mersenne prime M = 2^p-1 > 3,
((M+1)^M - 1) / ((2^M-1)M^2) is prime
if and only if 2^M-1 is prime.

T. Ordowski


sob., 18 wrz 2021 o 16:28 Ami Eldar <amiram.eldar at gmail.com> napisał(a):

>  ((M+1)^M - 1) / ((2^M-1)M^2) with M=2^13-1 is composite.
> The next candidate, ((M+1)^M - 1) / ((2^M-1)M^2) with M=2^17 is a huge
> number with 631291 digits.
>
> Best regards,
> Amiram
>
>
> On Sat, Sep 18, 2021 at 12:21 PM Tomasz Ordowski <tomaszordowski at gmail.com
> >
> wrote:
>
> > Hello!
> >
> > Note that m^2 | (m+1)^m - 1 for every m > 0.
> > I noticed that if m = 2^n-1, then 2^m-1 | (m+1)^m - 1.
> > So if m = 2^n-1 and (m, n) = 1, then (2^m-1)m^2 | (m+1)^m - 1.
> > Hence, if m = 2^n-1 is squarefree, then (2^m-1)m^2 | (m+1)^m - 1.
> > In particular, this divisibility holds for Mersenne numbers. Let's
> define:
> >    Numbers M = 2^p-1 such that ((M+1)^M - 1) / ((2^M-1)M^2) is prime.
> > Such M must be a Mersenne prime. I found 2^3-1 = 7 and 2^7-1 = 127.
> > Maybe someone will find more (the "next" prime M = 2^127-1 is nice /;-).
> >
> >
> > Regards!
> >
> > T. Ordowski
> > _____________
> > A060073 - OEIS <https://oeis.org/A060073> (a(n+1) = ((n+1)^n - 1)/n^2).
> > A127837 - OEIS <https://oeis.org/A127837> (cannot be Mersenne primes >
> 3).
> >
> > --
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> >
>
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