[seqfan] Re: A000108(n) ≡ 1 (mod 6)

[Emmanuel Vantieghem]

Dear Ed,

In the meantime I found a simple proof of the fact that  2  is a primitive
root modulo  3^n  for every n > 0.  Hence, the numbers  2^N  run through
all numbers from  1  to  2^(3^(n-1))  that are not divisible by 3.  This
implies that the columns are periodic and that all but the first one
contain 0, 1, 2  with the same frequency.
Thus, your conjectures (1)  and (ii)  are true.

Emmanuel.

2015-12-05 9:28 GMT+01:00 L. Edson Jeffery <lejeffery2 at gmail.com>:

> Robert,
>
> In one of your other posts on this topic (see
> http://list.seqfan.eu/pipermail/seqfan/2015-November/015590.html), you
> gave
> a probabilistic argument. I wanted to try another version here.
>
> I just proposed A265210 which is an irregular triangle in which row n gives
> the base 3 digits of 2^n in reverse order. The first few rows are
>
> {1},
> {2},
> {1, 1},
> {2, 2},
> {1, 2, 1},
> {2, 1, 0, 1},
> {1, 0, 1, 2},
> {2, 0, 2, 1, 1},
> ...
>
> Let the columns be indexed by k=1,2,... .
>
> Conjecture: (i) The sequence in column k is periodic with period p(k) =
> 2*3^(k-1); (ii) The numbers 0,1,2 appear with equal frequency in each
> column (except column 1 which obviously contains only the numbers 1,2 with
> equal frequency).
>
> Assume that the conjecture is true. I imagined, after studying the triangle
> to some extent, that the distribution of the 2's throughout the triangle is
> such that, for fixed k, removing all rows which contain a 2 in column k
> does not affect the relative frequency of the digits 0,1,2 in columns k+1,
> k+2, ..., so let's further assume that is the case. Clearly, upon removing
> those rows, all columns remain periodic (proof is easy). Of course, the
> period is reduced.
>
> If we now iterate the process of removing all rows which contain a 2 in
> column k, for each k in {1,...,infinity}, then we will have removed
>
> 1/2 + 1/3 + 1/9 + 1/27 + ...
>
> of the rows. Since
>
> 1/2 + Sum_{j=1..infinity} 1/3^j = 1,
>
> I think we will have removed either all but finitely many of the rows or
> all rows except for some set of measure zero. Of course, the former case
> means that the set of all m such that 2^m is a sum of powers of 3 must be
> finite which is what we'd like to know. I don't think the latter case helps
> at all, but I see no way around it.
>
> Could proof of that conjecture be any easier than the original problem?
>
> Ed Jeffery
>
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