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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