[seqfan] Re: A000108(n) ≡ 1 (mod 6)
L. Edson Jeffery
lejeffery2 at gmail.com
Thu Dec 3 04:48:39 CET 2015
Amazing. Great work.
After the conjecture I wrote:
"If Conjecture 1 is true, then it should be enough to prove, for all m>8,
that the base 3 representations of both 2^m - 1 and 2^m contain at least
one 2, from which William Keith's conjecture that C(2^m-1) == 0 (mod 3)
would then follow."
At the time, I did not realize the difficulty of the problem. The comment
in A260683 says that Paul Erdős conjectured that there is at least one 2 in
the base 3 representation of 2^n, for all n>8; so, evidently, no proof is
known as of yet. A proof for the case 2^n - 1 is likely no easier.
By the way, the number of twos in the base 3 representation of 2n - 1 was
not in OEIS, so I just proposed it as A265157 (
More information about the SeqFan