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Re: [PrimeNumbers] Factor Convergence Conjecture

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  • Maximilian Hasler
    It is not clear what you mean by converges to in your statement. Definition : A sequence s converges to a point x if for any neighborhood V of x almost all
    Message 1 of 2 , Jul 16, 2012
      It is not clear what you mean by "converges to" in your statement.

      Definition : A sequence s converges to a point x if for any neighborhood V
      of x almost all terms of s lie within V.


      I think in spite of the unclear definition i can give a
      Counter-example to what you say:

      let p=2, a=3 <=> L=3, m = [N/3] = [2q/3], b=3m <=> n=3m^2,

      Then all of your inequalities are satisfied (as soon as q>3), but
      a-n+m= 3 - 3m^2 + m
      does not at all "converge" (whatever this may mean) to p=2.

      Maximilian



      On Mon, Jul 16, 2012 at 1:43 AM, kad <yourskadhir@...> wrote:

      > **
      >
      >
      > Let N = pq be any semiprime such that a^2 < N < b^2 & p < q.
      >
      > Let 3l be the multiple of 3 nearer to a^2
      > and 3m be the multiple of 3 nearer to N
      > and 3n be the multiple of 3 nearer to b^2
      >
      > If (m - l) > (n - m) then
      >
      > (a - m - 3 + l) converges to 'p' else
      >
      > (a - n + m) converges to 'p'.
      >
      > Above conjecture will optimise the trial division method and Fermat's
      > Factorisation method.
      >
      >


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