[seqfan] Re: A000108(n) ≡ 1 (mod 6)
arndt at jjj.de
Sat Dec 12 18:32:35 CET 2015
* 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
> "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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