[seqfan] Re: A000108(n) ≡ 1 (mod 6)

[L. Edson Jeffery]

William Keith recently stated the stronger conjecture that C(2^m-1) == 0
(mod 3), for all m>8.

I managed to arrive an equivalent conjecture based on what appears to be a
certain relation between the sequence of C(n) taken modulo 3 and Cantor's
ternary set. The approach also produces conjectures which are obviously
weaker but for which proof might be possible without resorting to
factorization or base p arithmetic (p a prime). An example is the following

Conjecture: C(2^A056576(n)-1) == 0 (mod 3), for all n >= 2  (A056576(n) is
the largest k such that 2^k <= 3^n, n >= 0).

If true, then we could at least say that Keith's conjecture is true for
most m. I want to send some details later. However, from the beginning,
everything depends on the following conjecture.

Conjecture: There exists no m_1 such that C(n) == 0 (mod 3), for all n >
m_1, and there exists no m_2 such that C(n) =!= 0 (mod 3), for all n > m_2
(=!= means "not congruent").

Does anyone know of a proof of this conjecture?

Ed Jeffery