[seqfan] Re: Anti-Carmichael numbers

[Tomasz Ordowski]

P.S. Note:

My proof that A121707 are "anti-Carmichael numbers" is not complete.*

So I gave it as a conjecture (see the last comment):

http://oeis.org/history/view?seq=A121707&v=35

http://oeis.org/draft/A121707

Thomas Ordowski
____________________
(*) It must be proved that
odd numbers n>1 divides 1^(n-1)+2^(n-1)+...+(n-1)^(n-1)
if and only if p-1 does not divide n-1 for every prime p dividing n.


2018-07-20 11:13 GMT+02:00 Ami Eldar <amiram.eldar at gmail.com>:

> Anti-Lucas-Carmichael numbers: 4, 6, 9, 10, 12, 16, 18, 21, 22, 24, 25, 28,
> 30, 33, 34, 36, 40, 42, 45, 46, 48, 49, 52, 54, 55, 57, 58, 60, 64, 66, 69,
> 70, 72, 76, 77,
> 78, 81, 82, 84, 85, 88, 90, 91, 93, 94, 96, 100, ...
> They are analogous to A121707 as A006972 is analogous to A002997.
>
> The are also numbers that are both Anti-Carmichael and
> Anti-Lucas-Carmichael: 55, 77, 115, 161, 187, 203, 209, 221, 235, 247, 253,
> 295, 299, ..
> They are analogous to Williams numbers (numbers that are both Carmichael
> and Lucas-Carmichael).
>
> None of these sequences are in OEIS. The last one (Williams numbers) is not
> there since there is no known example...
>
> Ami
>
>
>
>
>
> On Fri, Jul 20, 2018 at 11:38 AM, Tomasz Ordowski <
> tomaszordowski at gmail.com>
> wrote:
>
> > 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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