# [seqfan] Re: A000108(n) ≡ 1 (mod 6)

Emmanuel Vantieghem emmanuelvantieghem at gmail.com
Tue Dec 8 18:29:47 CET 2015

```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"

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) ].
This explains the equidistribution of the 0, 1, 2.  The periodicity is
explained by the fact that the same numbers appear in the same order when
you take bigger values of k.

The use of the term "equivalent" was a mistake of me, for which I want to
appologize.

Emmanuel.

2015-12-07 8:03 GMT+01:00 L. Edson Jeffery <lejeffery2 at gmail.com>:

> Emmanuel,
>
> You wrote:
>
> "The periodicity statement is equivalent with the statement that 2 is a
> primitive root modulo 3^n for every n. That seems rather easy to prove."
>
> I have been staring at this for twelve hours trying to work through it with
> no success. Would you please post the proof here?
>
> Ed Jeffery
>
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>
```