[seqfan] Re: Anti-Carmichael numbers

[Neil Sloane]

Thomas, What was the date of Andrzey Shinzel's letter that you mention in
A121707?

Can you please scan it and add the pdf file to A121707?  I can ask him for
permission.

Of course it does not matter that the letter is in Polish.



Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com


On Fri, Jul 20, 2018 at 8:40 AM, Tomasz Ordowski <tomaszordowski at gmail.com>
wrote:

> 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
> > > > >
> > > > > --
> > > > > Seqfan Mailing list - http://list.seqfan.eu/
> > > > >
> > > >
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