Quote:
Each A had 6 factors allowing for 64 Bs per A.

Shouldn't that be 32 Bs per A given 6 factors? (2^(61) = 32)
For that 40digit number I mentioned above, my As have 5 factors, generating 16 bs per A. (2^(51) = 16)
Contini's paper says:
Quote:
To generate a polynomial, we need to find `b` satisfying `b^2 = N mod a`. There are actually 2^s distinct `b (mod a)` that satisfy this, but only half of the b's will be used since `g_{a,b}(x)` gives the same residues as `g_{a,b}(x)`. Hence we can get 2^{s1} polynomials.
