> -----Original Message----- > From: Jud McCranie [mailto:[EMAIL PROTECTED]] > Sent: Monday, June 14, 1999 9:01 PM > To: Gary Diehl > Cc: [EMAIL PROTECTED] > Subject: Re: Mersenne: Ideas : maybe no new ones... > > > At 08:06 PM 6/14/99 -0500, Gary Diehl wrote: > > >When you square S over and over and over in one LL test, > does that same > >value of S come up in a test for another exponent? > > > It could, but I don't think that helps since for each > exponent the calculations > are done to a different modulus. > I was thinking about this last night. If you could keep track of the divisions for the last Mersenne, you could keep that as a starting point. So for example, take M37, and keep track of how many times you did your modulo and what the divisor was. Then you could conceptually get back to what S(M37) was w/o the modulo by multiplying M37*(sum of divisors) or something along those lines. It would at least save you some time, sure there's a big modulo calculation in the beginning, and I'm sure the number is friggin' huge, but it can be represented as (2^p-1)*(some number). I originally was trying to think if there was an easy way to get there from any number's residual, but then it becomes a hassle. On another note, I was thinking last night about a totally different approach, and would appreciate it if some math modulo guru out there could explain how to go from something like: (2^7+2^6+2)%(2^5-1) = 2^3 or how (2^12+2^8+2^7+2^2+2+1)%(2^7-1) = 2^5+2^3+2 without actually figuring out the number represented. I know that 2^7+2^6+2 = 194 and could then take the modulus from there... but that defeats the purpose. :) Thanks Jeremy P.S. Sorry for my part in the poaching thread. ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
