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?

>

> What I already know!

>

> '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

>