[seqfan] Re: Set A121707 and its subsets

[Tomasz Ordowski]

P.S. Regarding my question, I noticed that n^{n-1} == 1 (mod 6^2(n-1)^2) if
and only if n == -1 (mod 72).
So if n == -1 (mod 72) with D_{n-1} = 6, then n^{n-1} == 1 (mod
D_{n-1}^2(n-1)^2),
where D_{n-1} is the denominator of Bernoulli number B_{n-1}.
Are there other odd solutions to the last congruence?
Cf. A157921 - OEIS <https://oeis.org/A157921> and A121707 - OEIS
<https://oeis.org/A121707>

T. Orddowski


wt., 14 wrz 2021 o 12:27 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):

> Dear readers!
>
> First, note that n^{n-1} == 1 (mod (n-1)^2) for every n > 1.
> Second, notice that n^{n-1} == 1 (mod D_{n-1}) if and only if (n, D_{n-1})
> = 1, *
> where D_{n-1} is the denominator of Bernoulli number B_{n-1}.
> For odd n > 1, D_{n-1} = Product_{p prime, p-1 | n-1} p.
> Let's define the subsets of "anti-Carmichael" numbers:
> {A} Odd numbers n > 1 such that n^{n-1} == 1 (mod D_{n-1}^2).
> {B} Odd numbers n > 1 such that n^{n-1} == 1 (mod (n-1)^2 D_{n-1}).
>  Are there odd numbers n > 1 such that n^{n-1} == 1 (mod (n-1)^2
> D_{n-1}^2) ?
> Odd numbers n > 1 such that n^{n-1} == 1 (mod (n-1) D_{n-1}) are all
> A121707.
>
> Best regards,
>
> Thomas Ordowski
> ________________________
> (*) Odd numbers n > 1 such that (n, D_{n-1}) = 1
> are those "anti-Carmichael" numbers A121707:
> A121707 - OEIS <https://oeis.org/A121707> (see my comments there).
>