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Re: A new series

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  • eharsh82
    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?
      >
      > 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
      >
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