Well, I learned two things from this. If you want a lot of mail, give your idea a catchy title. Second I learned of another crazy idea, which will be sent as soon as the first couple of terms of checked. It's on the same basic principle as the one I showed you, except the next one will no doubt be more useful when it comes to twin primes. I spent a whole day organizing the zeta function (which I learned 2 minutes before organizing …) to get: Z(n) = 1 + (1/(2^n - 1)) + (1/(3^n -1)) + (1/(5^n - 1)) + ... with the same x' group. This maybe way to obvious, or it might not (I need to check my math history... I think It's more fun to prove what other people have. It's much more fun : - D) But this is where I thought I might mess up since it's subtracting infinity from infinity. I set n = 1 and got: 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 +... = 1 + 1 + (1/2) + (1/4) + (1/5) + (1/6) + (1/9) + ... Where the left includes all terms, and the right includes all terms in the set x'', which is (x' - 1) Canceling like terms I got: (1/3) + (1/7) + (1/8) + (1/15) + ... = 1 Now, x' is the set that does not contain integral powers. So, the left side is the side that does contain integral powers, minus 1. So, we can put 1 = 1/(2^2 - 1) + (1/2^3 - 1) + ... + (1/3^2 - 1) + (1/3^3 -1) + ... And there you have your proof. Connie brought up an interesting point, which included a proof, so I can say my math was correct. I must say, though, that I didn't like subtracting infinity from infinity. Mark _________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
