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.
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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