[seqfan] Re: A000108(n) ≡ 1 (mod 6)

Emmanuel Vantieghem emmanuelvantieghem at gmail.com
Sun Dec 13 10:45:38 CET 2015


Dear Joerg,

I think you are misunderstanding something.
The question was not to prove the statement "2 is primitive root mod 3^n"
but to show that this statement implies the conjectures of Ed.

Emmanuel.


2015-12-12 18:32 GMT+01:00 Joerg Arndt <arndt at jjj.de>:

> * Emmanuel Vantieghem <emmanuelvantieghem at gmail.com> [Dec 09. 2015 13:46]:
> > Dear Ed,
> >
> > I sent you a personal mail in which I indicated how one can prove that
> the
> > theorem
> >      "2 is primitive root mod 3^n for all n"
> > implies your conjectures.  You asked me to post the proof here.
> >
> > The proof is immediately found when you write the numbers  2^k mod 3^n in
> > base 3.  All the numbers between 1  and  3^n-1
> > not divisible by 3 will appear once and only once if k is taken in the
> > interval  [1, 2*3^(n-1) ].
>
> You again just state as a fact what you want to prove.
> Have you checked the lists for all n, up to,
> but excluding, infinity?   8^)
>
> Best regards,  jj
>
> > [...]
>
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