[seqfan] Re: Anti-Carmichael numbers

Sat Jul 21 17:55:04 CEST 2018

P.S. Maybe in the next quarter Professor Schinzel will be able to prove my
last conjecture:

http://oeis.org/history/view?seq=A121707&v=48

So let's wait with the publication of this old email.

Thomas Ordowski

2018-07-21 16:02 GMT+02:00 Tomasz Ordowski <tomaszordowski at gmail.com>:

> Dear Neil,
>
> Here is the original email from Dec 29 2015:
>
> Szanowny Panie,
> Podzielnosc (*) n^3|1^n+...+(n-1)^n=S_n(n) wymaga,zeby n było
> nieparzyste,bowiem z 2^a|n (a>0) wynika S_n(n)= n/2 mod 2^(a+2).Zatem (*)
> jest rownowazna  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 rownowazna
> podzielności (**)n|S_(n-1)(n).Z drugiej strony podzielność (**) i warunek 4
> nie dzieli n wymagają nieparzystości n.
> Lacze pozdrowienia.
>                 Andrzej Schinzel
>
> In answer to my question from Dec 28 2015:
>
> Szanowny Panie Profesorze!
>
> Na koniec roku pragnę przypomnieć pewną zaskakującą koincydencję
> podzielności,
> której nie udało mi się wyjaśnić, więc zadam przed czasem takie pytanie
> kwartalne.
>
> Niech n > 1.
>
> (1) n^3 dzieli 1^n+2^n+...+(n-1)^n dla n = 35, 55, 77, 95, ...
>
> (2) 4 nie dzieli n dzieli 1^(n-1)+2^(n-1)+...+(n-1)^(n-1)
>
> dla tych samych n (sprawdzone aż do n = 10^4).
>
> Jak udowodnić tę równoważność?
>
> Należy zauważyć, że
>
> n^2 dzieli 1^n+2^n+...+(n-1)^n dla każdego n nieparzystego i tylko
> takiego.
>
> Proszę o odpowiedź w ramach naszej umowy, by mieć sprawę z głowy (rym).
>
> Z poważaniem,
>
> Tomasz Ordowski
> ___________________
> http://oeis.org/A121707
>
> Sincerely,
>
> Thomas Ordowski
>
>
> 2018-07-21 15:41 GMT+02:00 Neil Sloane <njasloane at gmail.com>:
>
>> 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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>> > > > >
>> > > >
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