Hi, Another thought struck me - this could have useful applications in L-L testing programs.
If M is the Mersenne number being tested & R(i) is the L-L residue after i iterations, then R(i+1) = R(i) * R(i) - 2 (modulo M) (by the statement of the L-L algorithm) But note that (M - R(i))^2 - 2 = M^2 - 2MR(i) + R(i)^2 - 2 so (M-R(i))^2 - 2 (modulo M) is clearly equal to R(i+1). How can this be of any use? Well, when we have a dubious iteration (say an excessive roundoff or sum error) we can check the output by redoing the last iteration but starting from (M-R(i)) instead of R(i) - the output should be the same. Furthermore the action of calculating M-R(i) is very easy - just invert all the bits. Also, if we have suspicions about the accuracy of code when there is a high density of 1 bits, we can try just one iteration but starting at M-4 instead of 4. The output residual should be 14 irrespective of M (providing M>7 - as will often be the case!). The point here is that, just as the value 4 is represented by a string of p bits only one of which is set, M-4 is represented by a string of p bits only one of which is unset. Regards Brian Beesley _________________________________________________________________________ Unsubscribe & list info -- http://www.ndatech.com/mersenne/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
