[seqfan] Re: A000108(n) ≡ 1 (mod 6)
emmanuelvantieghem at gmail.com
Sun Dec 13 10:45:38 CET 2015
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.
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
> > 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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