# [seqfan] Re: Glued primes

Artur grafix at csl.pl
Sun Nov 23 19:51:01 CET 2008

```Dear Max,
Thank you for example!
http://www.research.att.com/~njas/sequences/A003537
that mean that all 7 primes numbers from A003537 are glued only for
2^(29*k)-1
and these will be occured for every 2^(29*k)-1  where k=1,2,3,4,5,6,7,...
but what with 2^58*k+1=5*107367629*536903681 ????
Best wishes
Artur

Max Alekseyev pisze:
> On Sun, Nov 23, 2008 at 8:49 AM, Artur <grafix at csl.pl> wrote:
>
>> Dear Max,
>>
>> I was check few samples
>> e.g. pair of primes {10429407431911334611, 918125051602568899753}
>> which are two the biggest prime divisors of 2^243+1 and 2^486-1
>> 486=2*243
>> occured every time in pairs (never separately) and both in Fermat as in
>> Mersenne numbers (infinite many times)
>>
>
> You don't need to check that manually. For any number n that is
> multiple of 243,
> the number 2^n - 1 is divisible by both these primes. Similarly, for
> an odd multiple m of 243, the number 2^m + 1 is divisible by these
> primes as well.
>
> btw, usually Fermat numbers are those of the form 2^(2^n) + 1 not just
> 2^m + 1 - see http://en.wikipedia.org/wiki/Fermat_number
> It's very confusing when you call just arbitrary number of the form
> 2^m + 1 a Fermat number.
>
>
>> (I know few more such pairs) and many many products of such pairs but I
>> can't factorized these on separate primes because I was mentioned yesterday
>> these cases are most difficult to factorization in anyone method.
>> I don't know yet that triples and more primes product occured but I'm
>> examined these now.
>>
>
> I suggest a triple [233,1103,2089] which are the prime factors of 2^29
> - 1, all three of the multiplicative order 29.
> They always together divide or not divide Mersenne numbers 2^n - 1 as
> well as numbers of the form 2^m + 1.
>
> Regards,
> Max
>
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