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Re: k^2+(k+1)^2 and k^4+(k+1)^4

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  • Robert
    ... After a good deal of lead swapping at small k, even numbers predominate from k=44886 until k=4245827 Regards Robert Smith
    Message 1 of 3 , Apr 30, 2009
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      --- In primenumbers@yahoogroups.com, "Robert" <robert_smith44@...> wrote:
      >
      > The series k^2+(k+1)^2 and k^4+(k+1)^4, k from 1 to infinity produce may primes (prps), as factors of these series (and all series of form k^((2^n))+(k+1)^((2^n))appear to be similar to those factors in cyclotomic polynomials.
      >
      > n=1 and 2 appear very rich, as the formula produces quite small primes and prps, but counting primes (prps) for the series k from 1 to k(x) it appear that n=2 has more than n=1, for low x.
      >
      > What is quite fascinating is if we look at the mod 2 values of k that produce prps for n=1,2 and race k=0mod2 against k=1mod2. I looked up to k=1200000 and over 100000 of these are prp, for both n=1 and 2, and the even values of k were in the lead for a goodly proportion of the time, in fact for 99%+. At k=1200000, n=2 there were approx 600 more evens than odds.
      >
      >
      After a good deal of lead swapping at small k, even numbers predominate from k=44886 until k=4245827

      Regards

      Robert Smith
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