[seqfan] Re: A000108(n) ≡ 1 (mod 6)

[L. Edson Jeffery]

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