2017-04-15, 18:52 | #397 |
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts |
sorry waking up an old thread I realized that I could find more information to help me make the LL test simpler potentially on this forum itself ( not this subforum in particular but mersenne forums). We can already prove that if doing the LL test with the modification mod p instead for mod 2^p-1 our answer is dependent on the other factor to multiply to s_(n-2) mod p I just realized there's a post that talks about the form this value must take and so I can now say it depends on 2*k+1 or 2*k-1 mod p for the k involved. the hard part now is determining this k value for any p value and then doing the modular math. edit: doh realized the a values i talked about are always even but we could factor out the powers of two out it and use that.
Last fiddled with by science_man_88 on 2017-04-15 at 19:02 |
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