## Re: A new series

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• The \$a means an integer and % means modulus (I was asked via email) Also I think a=2^n*p in most cases produces mersenne and fermat primes only, hence can be
Message 1 of 2 , Dec 20, 2005
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The \$a means an integer and % means modulus (I was asked via email)

Also I think a=2^n*p in most cases produces mersenne and fermat
primes only, hence can be ignored.(alteast when 2^n<p, though not
sure)

Harsh

--- In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
>
> Hi,
>
> I need some help with a new series.
>
> (((2^\$a)%\$a)*(2^\$a))-1 and (((2^\$a)%\$a)*(2^\$a))+1
>
> Here are the primes I have found.
> (((2^1)%1)*(2^1))+1
> (((2^3)%3)*(2^3))+1
> (((2^6)%6)*(2^6))+1
> (((2^7)%7)*(2^7))+1
> (((2^14)%14)*(2^14))+1
> (((2^27)%27)*(2^27))+1
> (((2^42)%42)*(2^42))+1
> (((2^81)%81)*(2^81))+1
> (((2^1869)%1869)*(2^1869))+1
> (((2^56)%56)*(2^56))-1
> (((2^98)%98)*(2^98))-1
> (((2^2555)%2555)*(2^2555))-1
> (((2^12025)%12025)*(2^12025))-1
> (((2^15778)%15778)*(2^15778))+1
> (((2^25479)%25479)*(2^25479))+1
>
> (Continuing to a million)
>
> Questions
>
> 1) Are there infinite primes? What is the distribution? Is there a
> pattern to the above primes?
>
>
> 'a' can never be a non-mersenne prime, because we get 2^(a+1)-1/+1
> which can never be prime, except when a= mersenne prime then we
get
> a fermat number.
>
> Alot of times 2^\$a%\$a produces a multiple of 2, which will give a
> fermat/ mersenne prime. (only a=1, a=3, a=7 and a=14 is known)
>
> Since mersenne primes/fermat primes have been searched to a deep
> depth, we can exclude these numbers.
>
> Let a =k*P, where p is a prime and
> 2^(2*k)-1%p==0, the number is not prime for the -1 case
> 2^(2*k)+1%p==0, the number is not prime for the +1 case
>
> Any other exceptions?
>
> 2) When all does a prime p divide the term? What is the
periodicity?
> No answers for my side eg. When would 3 divide the terms of a
> series? Shouldn't it be p*p-1, it does not appear to be true.
>
> 3) Is there any ways to write a fast sieve, faster than the one
used
> for Cullen/woodall?
>
> Thanks,
> Harsh
>
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