[seqfan] Re: n = number of divisors of 2^n -1

[Max Alekseyev]

On Wed, Apr 21, 2010 at 4:01 PM, Charles Greathouse
<charles.greathouse at case.edu> wrote:
> Well, 32 is the last power of two in the sequence since
> (2^64-1)/(2^32-1) = 641 * 6700417 is composite.

I don't understand your argument.
n = 2^k is in this sequence, if in particular 2^n - 1 is the product
of k distinct prime factors.

> Otherwise, I checked up to 16384 without finding any members.

How did you do that without knowing factorization of numbers 2^m + 1
for large m (for example, the complete factorization of 2^2881 + 1 is
currently unknown, according to Cunningham tables) ?

Regards,
Max