## Re: [PrimeNumbers] Primaliy test for a series

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• Pardon for the clumsy statement I made. Of course the summation I give is easy to simplify. Actually my interest is summations that are slightly more
Message 1 of 7 , Feb 22, 2007
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Pardon for the clumsy statement I made. Of course the summation I give is easy to simplify. Actually my interest is summations that are slightly more complicated.

Whenever I try to search for primes of the form
6*n*(2^p - 1- n) + 2^p - 1, where n = 1, 2, 3, ... and 2^p - 1 is a Mersenne prime, I find that the factors are of the form

2^(3*k) - 1 = (2^3 - 1)( (2^3)^(k - 1) + (2^3)^k - 2) + ...
(2^3)^2 + (2^3) + 1),

or alternatively

2^(3*k) + 1 = (2^3 + 1)( (2^3)^(k - 1) - (2^3)^(k - 2) + ...
+ (2^3)^2 - (2^3) + 1), where k is odd.

I wonder whether some of these factor summations cannot yield fairly large primes.

Peter.
----- Original Message -----
From: plesala
Sent: Wednesday, February 21, 2007 1:54 PM
Subject: [PrimeNumbers] Primaliy test for a series

Hi to all,

How does one test primality of a series like

1 + 2 + 2^2 + 2^3 + ... + 2^(2*n - 1),

for example, using primeform?

Thank you.
Peter.

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