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Re: [PrimeNumbers] Digest Number 3737

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  • Kermit Rose
    On 7/27/2013 4:54 AM, Werner ... For any finite collection of primes, p1,p2,p3,...p_m, a polynomial,p(x), with rational coefficients, can be generated such
    Message 1 of 1 , Jul 27, 2013
      On 7/27/2013 4:54 AM,

      Werner

      wrote:
      > 1a. Re: Polynomials
      > Posted by: "Alexander"werner.sand@... werner.sand
      > Date: Fri Jul 26, 2013 5:45 am ((PDT))
      >
      >
      >
      > I sent my reply wrong, I'm afraid, so once more:
      >
      > I collect (each case n=1,2,3..100, I cannot say per cent):
      >
      > n²-n+41 (Euler) generates 86 primes
      > n²-n+27941 generates 77 primes
      > ¼* (n^5-133n^4+6729n^3 -158379n^2+1720294n-6823316) generates only 64 primes
      >
      > I checked n²-n+a up to a=10^6, no better result. Can anybody compute further? Can a better result actually be expected?
      >
      > Werner



      For any finite collection of primes, p1,p2,p3,...p_m, a polynomial,p(x),
      with

      rational coefficients,
      can be generated such p(x) = p_x.


      a0 + a1 x + a2 x^2 + ... a_m x^m = p_x


      a0 + a1 (x+1) + a2 (x+1)^2 + ... + a_m (x+1)^m = p_(x+1)

      By taking first differences,

      a1 + a2 (2x + 1) + a3 (3x^2 + 3x + 1) + ... + a_m( (x+1)^m - x^m)
      = p_(x+1) - p_x

      a1 + a2 (2(x+1) + 1) + a3 (3(x+1)^2 + 3(x+1) + 1) + ... + a_m( (x+2)^m -
      (x+1)^m)
      = p_(x+2) - p_(x+1)

      Then take second differences, third differences, etc

      until you reach a linear equation to be solved for a_m.

      Then solve previous equations for a_(m-1), a(m-2), etc until you solve

      a0 + a1 x + a2 x^2 + ... a_m x^m = p_x

      for a0.



      Kermit







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