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Fermat factors?

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  • leavemsg1
    Hi, Group, et al. the famous Fermat function: F(x) = 2^(2^x) +1 let x be from the set of whole numbers: Does F(x) have Q= 2*floor(sqrt((x+3)/2)) number of
    Message 1 of 1 , Nov 12, 2007
      Hi, Group, et al.

      the famous Fermat function: F(x) = 2^(2^x) +1

      let x be from the set of whole numbers:

      Does F(x) have Q= 2*floor(sqrt((x+3)/2)) number of divisors ?,
      i.e. 'including 1 and F(x)';

      x=0; F(0) has 2*floor(sqrt(3/2)) == 2,... 1 & 3,
      x=1; F(1) has 2*floor(sqrt(4/2)) == 2,... 1 & 5,
      x=2; F(2) has 2*floor(sqrt(5/2)) == 2,... 1 & 17,
      x=3; F(3) has 2*floor(sqrt(6/2)) == 2,... 1 & 257,
      x=4; F(4) has 2*floor(sqrt(7/2)) == 2,... 1 & 65537,
      These account for All of the possible Fermat primes.

      other examples...
      x=5; F(5) has 2*floor(sqrt(8/2)) == 4 total,
      x=6; F(6) has 2*floor(sqrt(9/2)) == 4 again,
      x=7; F(7) has 2*floor(sqrt(10/2)) == 4 total,
      x=8; F(8) has 2*floor(sqrt(11/2)) == 4 again,
      x=9; F(9) has 2*floor(sqrt(12/2)) == 4; tried but couldn't factor it
      These account for the next few Fermat composites.

      ...

      Is the first time that we see 6 divisors of F(x), when x=15 ???
      i.e. Q(15)= 2*floor(sqrt((15+3)/2)) = 2*floor(sqrt(9)) = 6 ???

      I believe it is... Does anyone have a comment or factorization to
      offer?

      I'm trying to find a pattern/explanation for why F(x) is so hard to
      factor!

      Bill Bouris, Aurora, IL USA
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