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

[Artur]

Another argument is that many from factors for big n are PRP (probably 
primes) not primes.
Best wishes
Artur


Max Alekseyev pisze:
> 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
>
>
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