[seqfan] Re: A000108(n) ≡ 1 (mod 6)

israel at math.ubc.ca israel at math.ubc.ca
Thu Dec 3 07:14:40 CET 2015

Yes, very little is known about the number of 2's in the base 3
representation of 2^n. One thing you can say, though, is that 2^n == 1 mod
3 if n is even and 2 if n is odd, so the number of 2's in 2^n-1 the same as
the number of 2's in 2^n if n is even, and one less if n is odd. The number
of 2's in the base 3 representation of 2^n is A260683.

Cheers,
Robert

On Dec 2 2015, L. Edson Jeffery wrote:

>Robert,
>
>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 (
>https://oeis.org/draft/A265157).
>
>Ed Jeffery
>
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