What???? Of course you can verify a prime number by its ending digit oddness particularly if the odd ending is a 1, 3, 7 or 9!!! But what the heck are you talking about with the A(0-1), A(0+1) and B(0)? Are you trying to say that if B(0) is of some certain type, then A(0) is prime? Are you implying that the greatest math minds in the world didn't need to come up with Lucas-Lehmer, all they need to do is look at the last digit in the number in question? Huh?
B(0) is always prime. In your example what is the difference between example 1) B(0) = 331 making A(0) prime and example 3) B(0) = 127 making A(0) not prime? If B(0) is needed in order to do your proof, how do you find it? Isn't factoring numbers the more difficult proposition? Do you know what the factors are for M41-1??? Wouldn't significantly more computations be needed to find the factors of M41-1 then to use Lucas-Lehmer to check primality of M41? Yes, I must be missing something subtle also!! Greg -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Dipl.-Ing. K.Kudiabor Sent: Wednesday, July 21, 2004 4:37 PM To: The Great Internet Mersenne Prime Search list Subject: Re: [Prime] Re: You no longer need Lucas-Lehmer Double... Subject: [Prime] Re: You no longer need Lucas-Lehmer Double... A repeat of my proposal (now in email-format) :Given M8=2^31-1= 2 147 483 647 =A(0): Prove and certify the primality, so that all of us can check and verity at sight. Here is the proposal. 1) A(0-1) = 2 147 483 646 = 2 x 3 x 3 x 7x 11 x 31 x 151 x 331; B(0)=331 A(0 ) = 2 147 483 647 A(0+1) =2 147 483 648 = 2^31 (these prime factors are easily verifiable) A(0 ) is prime, certified by A(0-1), A(0+1) and B(0) below. (One and the same algorithm must be used to factorize the sequential integers A(0-1) to A(0+!) 2) In case B(0 ) = 331 is too large for sight prime verification this iteration would be necessary: B(0-1) = 330 = 2 x 3 x 5 x 11 (these prime factors are easily verifiable) B(0 ) = 331 B(0+1) = 332 = 2 x 2 x 83 (these prime factors are easily verifiable) B(0 ) is prime, certified by B(0-1) and B(0+1) 3) The Mersenne number M =2^29-1 = 536 870 911 = A(0) is patently no prime. A(0-1) = 536 870 910 = 2 x 3 x 5 x 29 x 43 x 113 x 127 ( prime factors verifiable) A(0 ) = 536 870 911 = 233 x 1103 x 2089 A(0+1) = 536 870 912 = 2^29 ( prime factors verifiable) 4) The same Adjacent Prime factor Criterion (APC) can be used fo verify John Findley's 7-million digit Mersenne: M41=2^24,036,583 - 1 =A(0). You'll save that much time and resources for the usual independent verification. http://www.mersenne.org A(0-1)= M41 -1 A(0) =M41 A(0+1)= M41+1 5) And this rule is always true: The rightmost digit of any prime p=>11 is an element of the odd-number set nodd=[1,3,7,9] CAREFUL !! 21, 39, 57, 99 are composites, So the odd-number rule is only useful for a quick sight primality check of newly found mega-digit primes, for instance; M40=2^20,996,011-1 = ..481331395421550326484866710969127787170820477533409300972948475231983471 676 653078163294714065762855682047 http://www.mersenne.org/prime6.txt M41= 2^24,036,583 - 1 = ..49549332624134295037485542595520771846437818325642314252685868703980055603 1 269118412915067436921882733969407 http://mersenne.org/prime7.txt Now to David A. Bartizal's question: Did I miss something subtle ? Check No. 1) above . It is the main point. No. 5) is just a short-cut of your quotation All primes > 2 are odd numbers, but not all odd numbers are prime. I use the short-cut (note excluding odd number 5) to check GIMPS-Mersennes, Maybe somebody will give away a trick how to do that otherwise. Kodjo Kudiabor > > _______________________________________________ > Prime mailing list > [EMAIL PROTECTED] > http://hogranch.com/mailman/listinfo/prime _______________________________________________ Prime mailing list [EMAIL PROTECTED] http://hogranch.com/mailman/listinfo/prime _______________________________________________ Prime mailing list [EMAIL PROTECTED] http://hogranch.com/mailman/listinfo/prime
