On Oct 17, 2009, at 3:43 PM, David Harvey wrote: > How about recursing into the same routine for the multiplication > modulo 2^(n/2) - 1? Do you think that would make much difference > for larger n? I suppose you would need n divisible by 4, or would > need to add some bit shifts. Sorry, I meant B^(n/2) - 1. david