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Factor Convergence Conjecture

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  • kad
    Let N = pq be any semiprime such that a^2
    Message 1 of 2 , Jul 15, 2012
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      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.
    • 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 2 of 2 , Jul 16, 2012
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        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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