COMMENT on A137605

[Max Alekseyev]

On Wed, May 21, 2008 at 3:30 AM, Max Alekseyev <maxale at gmail.com> wrote:

> In general, I can prove that either A137605(n) = A002326(n-1) - 1, or
> A137605(n) = A002326(n-1)/2 - 1.
> But at the moment I don't see an efficient way to determine which case
> is taking place.
>
> Of course, if A002326(n-1) is odd then there is no other choice but
> A137605(n) = A002326(n-1) - 1.
> But if A002326(n-1) is even then there are two choices possible...

Actually, if A002326(n-1) is 2 modulo 4 then A137605(n) = A002326(n-1)/2 - 1.

So, the only remaining problematic case is when A002326(n-1) is a multiple of 4.

Regards,
Max