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Old 2019-05-16, 10:01   #4
May 2019

23×3 Posts

Each A had 6 factors allowing for 64 Bs per A.
Shouldn't that be 32 Bs per A given 6 factors? (2^(6-1) = 32)

For that 40-digit number I mentioned above, my As have 5 factors, generating 16 bs per A. (2^(5-1) = 16)

Contini's paper says:

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^{s-1} polynomials.

Last fiddled with by tabidots on 2019-05-16 at 10:03
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