[seqfan] Re: Anti-Carmichael numbers

[Tomasz Ordowski]

Dear Ami,

thank you for your interest in the topic.

Are your anti-Lucas-Carmichael numbers in the OEIS?

Can you give me the first few such numbers?

Maybe their density D = 1/zeta(2) = 6/pi^2.

Best regards,

Thomas

P.S. By the way: https://oeis.org/draft/A121707


2018-07-20 0:24 GMT+02:00 Ami Eldar <amiram.eldar at gmail.com>:

> What about the Anti-Lucas-Carmichael numbers: numbers n such that p+1 non |
> n+1 for every prime p | n.
> It seems that their density is D > 0.6.
>
> On Thu, Jul 19, 2018 at 2:48 PM, Tomasz Ordowski <tomaszordowski at gmail.com
> >
> wrote:
>
> > Dear SeqFan!
> >
> > See http://oeis.org/A121707.
> >
> > Here's a simple definition of these numbers:
> >
> > Numbers n such that p-1 does not divide n-1 for every prime p dividing n.
> >
> > PROOF.
> >
> > Carmichael numbers: composite n coprime to 1^(n-1)+2^(n-1)+...+(n-1)^(n-
> > 1).
> >
> > Anti-Carmichael numbers: n such that p-1 non | n-1 for every prime p | n.
> >
> >  Odd numbers n>1 such that n | 1^(n-1)+2^(n-1)+...+(n-1)^(n-1). (**)
> >
> > Equivalently: Numbers n>1 such that n^3 | 1^n+2^n+...+(n-1)^n. (*)
> >
> > Cf. http://oeis.org/A121707 (see conjecture in the second comment).
> >
> > Professor Schinzel (2015) proved the equivalence (*) <=> (**). QED.
> >
> > Problem: What is the natural density D of the set of these numbers?
> >
> > Can be proved that 0.1 < D < 0.3.  My conjecture: D = 2/pi^2.
> >
> > Sincerely,
> >
> > Tomasz Ordowski
> > ______________
> > Szanowny Panie,
> > Podzielność (*) n^3|1^n+...+(n-1)^n=S_n(n) wymaga żeby n było
> nieparzyste,
> > bowiem z 2^a|n (a>0) wynika S_n(n)= n/2 mod 2^(a+2). Zatem (*) jest
> > rownoważna  podzielności n^3|(1^n+(n-1)^n)+(2^n+(n-2)^
> > n)+...+((n-1)^n+1^n).
> > Ale przy n nieparzystym dla każdego i :i^n+(n-i)^n=n^2*i^(n-1), zatem
> > podzielność (*) jest rownoważna podzielności (**) n|S_(n-1)(n). Z drugiej
> > strony podzielność (**) i warunek 4 nie dzieli n wymagają nieparzystości
> n.
> > Lącze pozdrowienia.
> >                 Andrzej Schinzel
> >
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> >
>
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