[seqfan] Re: (b^k-1)/(b-1) == 1 (mod k)

Tomasz Ordowski tomaszordowski at gmail.com
Mon Apr 30 07:08:10 CEST 2018


Dear Max!

Thank you for your reply.

See https://oeis.org/history/view?seq=A302588&v=35

and https://oeis.org/history/view?seq=A271221&v=25

Best regards,

Thomas

2018-04-29 19:32 GMT+02:00 Max Alekseyev <maxale at gmail.com>:

> Thomas,
>
> It can be seen that a(n) <= smallest Carmichael number with all prime
> factors >= n.
> In particular,
> a(8..13) <= 29341
> a(14..17) <= 162401
> a(18..41) <= 252601
>
> I won't be surprised if the equality holds in all cases.
>
> Regards,
> Max
>
>
> On Tue, Apr 24, 2018 at 3:42 AM, Tomasz Ordowski <tomaszordowski at gmail.com
> >
> wrote:
>
> > Dear SeqFan!
> >
> > Let a(n) be the smallest composite k such that
> >
> >          b^k == b (mod (b-1)k)
> >
> > for all integer bases 2 <= b <= n = 2, 3, 4, 5, ...
> >
> > Conjecture: for n > 2, a(n) is a Carmichael number. Yes?
> >
> > Can also consider the congruence b^(k-1) == 1 (mod (b-1)k).
> >
> > Maybe someone will be interested in these sequences.
> >
> > Please give me some initial terms.
> >
> > Best regards,
> >
> > Thomas
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>


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