Comments on: Perfect numbers, interlude (Challenge #11) http://www.mathlesstraveled.com/?p=90 Explorations in mathematical beauty Wed, 08 Sep 2010 14:28:08 -0400 http://wordpress.org/?v=2.9.1 hourly 1 By: Carnival of Mathematics #21: Bar-hopping at last « Secret Blogging Seminar http://www.mathlesstraveled.com/?p=90&cpage=1#comment-9823 Carnival of Mathematics #21: Bar-hopping at last « Secret Blogging Seminar Thu, 06 Dec 2007 05:39:45 +0000 http://www.mathlesstraveled.com/?p=90#comment-9823 [...] Less Travelled does a 4-part post on Perfect numbers, part I (and part II and part III, with a challenge in between). It’s a pretty elementary subject, but one that people have been unable unlock [...] [...] Less Travelled does a 4-part post on Perfect numbers, part I (and part II and part III, with a challenge in between). It’s a pretty elementary subject, but one that people have been unable unlock [...]

]]> By: Brent http://www.mathlesstraveled.com/?p=90&cpage=1#comment-9642 Brent Tue, 27 Nov 2007 17:30:30 +0000 http://www.mathlesstraveled.com/?p=90#comment-9642 DB: right. As it turns out, n does have to be prime, but that's not a sufficient condition. Steve: It is a nice way to pass the time, indeed. =) I don't think 9 works, though. 2^8 * 511 = 130816, the sum of the proper divisors of which is 171696, not 130816. That's because 511 is not prime; 2^(3*3) - 1 = (2^3 - 1)(2^6 + 2^3 + 1) = 7 * 73. DB: right. As it turns out, n does have to be prime, but that’s not a sufficient condition.

Steve: It is a nice way to pass the time, indeed. =) I don’t think 9 works, though. 2^8 * 511 = 130816, the sum of the proper divisors of which is 171696, not 130816. That’s because 511 is not prime; 2^(3*3) – 1 = (2^3 – 1)(2^6 + 2^3 + 1) = 7 * 73.

]]> By: Steve Gilberg http://www.mathlesstraveled.com/?p=90&cpage=1#comment-9638 Steve Gilberg Tue, 27 Nov 2007 16:21:04 +0000 http://www.mathlesstraveled.com/?p=90#comment-9638 I figured out the same thing as DB while I was on Thanksgiving break and away from the computer. Nice way to pass the time. But it is not required that n be prime. I'm pretty sure that it works when n is 9. I figured out the same thing as DB while I was on Thanksgiving break and away from the computer. Nice way to pass the time.

But it is not required that n be prime. I’m pretty sure that it works when n is 9.

]]> By: DB http://www.mathlesstraveled.com/?p=90&cpage=1#comment-9564 DB Fri, 23 Nov 2007 21:37:16 +0000 http://www.mathlesstraveled.com/?p=90#comment-9564 So far, these perfect numbers are of the form 2^(n-1) * (2^n - 1), but only when (2^n - 1) is prime. Works for n=2, 3, 5, and 7, so maybe n itself should be prime? So far, these perfect numbers are of the form 2^(n-1) * (2^n – 1), but only when (2^n – 1) is prime. Works for n=2, 3, 5, and 7, so maybe n itself should be prime?

]]>