The probability that both numbers are square-free is (6/pi^2)^2, or 36/pi^4. The probability that a square-free number is divisible by a prime p is (1/p - 1/p^2)/(1 - 1/p^2) = 1/(p+1). Thus the probability that two square-free numbers are relatively prime is product_{p prime} 1 - 1/(p+1)^2. According to A065472, this is approximately .77588351000389549962040428442790, giving an overall probability of .28674742843447873410789271278984 (A065473). This is sometimes called the "strongly carefree" constant. See http://www.gn-50uma.de/alula/essays/Moree/Moree.en.shtml#r20-scfree "Moreover, 0.28675 is also the natural density of pairwise coprime triples of positive integers (relative to all such triples)." Franklin T. Adams-Watters -----Original Message----- From: dasimov@earthlink.net To: math-fun <math-fun@mailman.xmission.com> Sent: Fri, 14 Apr 2006 19:48:12 -0400 (GMT-04:00) Subject: [math-fun] Little number theory puzzle Given two integers selected independently and at random, what is the probability they are relatively prime *and* that each one is squarefree? --Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com