[seqfan] Re: A000108(n) ≡ 1 (mod 6)

[Emmanuel Vantieghem]

Maybe an example will show what I intend to say.
I take n = 3
The fist three columns will look like this :
2, 0, 0
1, 1, 0
2, 2, 0
1, 2, 1
2, 1, 0
1, 0, 1
2, 0, 2
1, 1, 1
2, 2, 2
1, 2, 2
2, 1, 2
1, 0, 2
2, 0, 1
1, 1, 2
2, 2, 1
1, 2, 0
2, 1, 1
1, 0, 0
2, 0, 0
1, 1, 0
2, 2, 0
1, 2, 1
2, 1, 0
1, 0, 1
 .  .   .

If now I ask you to find *all possible* trio's using symbols  0, 1, 2  such
that
   at the first place you must use only  1  or  2,
   at the second and third place you may use  0, 1 and  2.
I'm sure you will find the first 18 rows but in a different order since you
must use nine 1's and nine 2's for the first places and six 0's, six 1's
and six 2's.  This settles the equidistributions of the digits 0, 1, 2  in
the columns.
The periodicity of the columns lie in the fact that all these
configurations appear again in the same order.


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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