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• 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.
• 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
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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

> **
>
>
> 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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