<<I'm afraid that if you are correct, so is Wagstaff. The symbol "~", at least in mathematics means that if f(x)~g(x) then f(x)/g(x)=1 as x->infinity.>> So constants don't matter, of course. <<Your conjecture seems like it would yeild a better aproximation than Wagstaff's >> Nod, that's what I was aiming for. I wasn't trying to prove Wagstaff wrong - just find a closer way to estimate M(x). At least the ~ means that adding a constant doesn't change f(x)/g(x) = 1 as x -> infinity... right? (See, if something turns out to be faulty in the way I made my conjecture, that paper I mentioned will be a lot weaker.) S. "Glad he kept a scribbed piece of notebook paper so he could remember how he made his conjecture in the first place" L. _________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
