Mersenne Digest Sunday, May 14 2000 Volume 01 : Number 735 ---------------------------------------------------------------------- Date: Fri, 12 May 2000 08:25:45 -0700 From: "John R Pierce" <[EMAIL PROTECTED]> Subject: Re: Mersenne: Digest #732? > ... you can temporarily find it, and all > other issues of the Mersenne digest at: > > http://www.base.com/mersenne > > Note, this is a temporary not a permanent back issue archive. > It will be deleted May 19th. do you guys want to setup a permanent archive? I have plenty of disk space on my linux server (768/768kbps SDSL) - -jrp _________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers ------------------------------ Date: Fri, 12 May 2000 17:52:15 +0100 From: Michael Oates <[EMAIL PROTECTED]> Subject: Mersenne: I can't download the latest Prime95.zip Hi, I can't download the latest Prime95.zip or the p95setup.exe from web page... http://mersenne.org/freesoft.htm The links don't work. Is there a problem ? Can I get them from somewhere else ? Thanks, Mike, - -- ATLAS CELESTE - Bevis Star Atlas - & "The CD-ROM" A very rare atlas found at the Godlee Observatory http://www.u-net.com/ph/mas/bevis/ Astronomy in the UK http://www.ph.u-net.com _________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers ------------------------------ Date: Sat, 13 May 2000 02:01:43 -0400 From: "Fred W. Helenius" <[EMAIL PROTECTED]> Subject: Re: Mersenne: To crazy to be true, but it is.... At 04:22 PM 5/12/00 +0200, Pa'l La'ng wrote: > 1) The general task (including yours) is to have the unity, i.e. the number "1" >as the sum of reciprocals of natural numbers. > 2) In case the number of the summands is finite, the denominators are divisors >of a perfect number, - the LCM of them.- ( See: Daniel Shanks: Solved and >unsolved >problems in Number Theory, Vol.I.Spartan books, Washington DC.1962; page 25.) >Easy to prove: >the existance of such a finite summation is equivalent to the existance of >perfect numbers. No, this only works one way: Given a perfect number, the sum of the reciprocals of its divisors (excluding 1) is 1. It doesn't work the other way around; 1/2 + 1/4 + 1/6 + 1/12 = 1, but the LCM, 12, isn't perfect. > 3) So the existance (or non-existance) of producing the unity as the sum >of a finite number of reciprocals of odd numbers is equivalent with the >existance (or non-existance) >of odd perfect numbers. If only it were that easy: 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/27 + 1/35 + 1/63 + 1/105 + 1/135 = 1, but you can't get an odd perfect number out of it. - -- Fred W. Helenius <[EMAIL PROTECTED]> _________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers ------------------------------ Date: Sat, 13 May 2000 04:31:22 EDT From: [EMAIL PROTECTED] Subject: Mersenne: Proof of To crazy to be true, but it is... 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 ------------------------------ Date: Sun, 14 May 2000 15:23:23 +0200 From: "Pa'l La'ng" <[EMAIL PROTECTED]> Subject: Re: Mersenne: To crazy to be true, but it is.... "Fred W. Helenius" wrote: > No, this only works one way: Given a perfect number, the sum of the > reciprocals > of its divisors (excluding 1) is 1. It doesn't work the other way around; 1) Really, I cocentrated only to the words" weird"and " off topic" and unforgiveably I was unexact. The word "general" must have been replaced by the word "special". I apologize. > 1/2 + 1/4 + 1/6 + 1/12 = 1, but the LCM, 12, isn't perfect. 2) This is not counter-example, because multiplied by 2 we receive 1 + 1/2 +1/3+1/6 = 2 and 1/2 +1/3 +1/6 = 1 (This is the base-equation, the LCM is perfect, as the number "6" , too. This procedure could be repeated by any other perfect number) > > > > 3) So the existance (or non-existance) of producing the unity as the sum > >of a finite number of reciprocals of odd numbers is equivalent with the > >existance (or non-existance) > >of odd perfect numbers. 3) To the word "special". The unity as a sum of odd reciprocals has the following special necessery and sufficient conditions to be equivalent of the existance or non-existance of odd perfect numbers ( according to Euler): 31) The number of the summands must be odd 32) The reciprocals must be either those of squares, or those of such other prime-powers being not the divisors of any square(the exponents of this prime should be continued from the number "1" to an odd one). > 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/27 + 1/35 + 1/63 + 1/105 + 1/135 = 1, > but you can't get an odd perfect number out of it. Among these denominators you find no squares, so I really can't get a perfect number from this sum. And to find one fulfilling Euler's conditions -- , it is not easy at all. Best regards Pa'l La'ng , Budapest, Hungary _________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers ------------------------------ Date: Sun, 14 May 2000 13:13:32 -0700 From: "Simon J Rubinstein-Salzedo" <[EMAIL PROTECTED]> Subject: Re: Mersenne: To crazy to be true, but it is.... I have been studying odd perfect numbers, and I have noticed that there cannot be an odd perfect number below 45047 (no, I didn't try them all). Then I proved that there must be at least nine factors by some very limited trial and error (only about 10 cases were necessary). It seems quite likely that if there is an odd perfect number, then it will have a number of factors that comes close to what I call a triangular choose, meaning, for example, 4C1+4C2+4C3+4C4 for 4. It would probably be slightly less. If it helps. SJRS - ----- Original Message ----- From: Pa'l La'ng <[EMAIL PROTECTED]> To: Fred W. Helenius <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]> Sent: Sunday, May 14, 2000 6:23 AM Subject: Re: Mersenne: To crazy to be true, but it is.... > "Fred W. Helenius" wrote: > > > No, this only works one way: Given a perfect number, the sum of the > > reciprocals > > of its divisors (excluding 1) is 1. It doesn't work the other way around; > > 1) Really, I cocentrated only to the words" weird"and " off topic" and > unforgiveably I > was unexact. The word "general" must have been replaced by the word "special". > I apologize. > > > 1/2 + 1/4 + 1/6 + 1/12 = 1, but the LCM, 12, isn't perfect. > > 2) This is not counter-example, because multiplied by 2 we receive > > 1 + 1/2 +1/3+1/6 = 2 and 1/2 +1/3 +1/6 = 1 (This is the > base-equation, the LCM is perfect, > as the number "6" , too. This procedure could be repeated by any other perfect > number) > > > > > > > 3) So the existance (or non-existance) of producing the unity as the sum > > >of a finite number of reciprocals of odd numbers is equivalent with the > > >existance (or non-existance) > > >of odd perfect numbers. > > 3) To the word "special". The unity as a sum of odd reciprocals has the > following > special necessery and sufficient conditions to be equivalent of the existance or > non-existance > of odd perfect numbers ( according to Euler): > 31) The number of the summands must be odd > 32) The reciprocals must be either those of squares, or those of > such other > prime-powers being not the divisors of any square(the exponents of this prime > should be > continued from the number "1" to an odd one). > > > 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/27 + 1/35 + 1/63 + 1/105 + 1/135 = 1, > > but you can't get an odd perfect number out of it. > > Among these denominators you find no squares, so I really can't get a perfect > number from this sum. > And to find one fulfilling Euler's conditions -- , it is not easy at all. > > Best regards > > Pa'l > La'ng , Budapest, Hungary > _________________________________________________________________ > Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm > Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers > _________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers ------------------------------ End of Mersenne Digest V1 #735 ******************************
