Hello! See my book "Mathematical Constants", pages 110-112, for more on this topic. You can browse my book freely online. Visit http://pauillac.inria.fr/algo/bsolve/ and click on the word "Google" in the second pararaph. In the search utility, type "carefree". Then select page 110. You'll be able to move ahead two pages as well. Please let me know if you run into any trouble. (I'm unsure why we disagree numerically.) Also see Pieter Moree's unpublished note "Counting carefree couples" at http://web.inter.nl.net/hcc/J.Moree/ The pairwise coprimality result conjected by Moree has been proved to be true: Cai, Jin-Yi; Bach, Eric. On testing for zero polynomials by a set of points with bounded precision, Theoret. Comput. Sci. 296 (2003), no. 1, 15--25. MR1965515 (2004m:68279). Best wishes! Steve Finch

From: "Daniel Asimov" <dasimov@earthlink.net> Reply-To: dasimov@earthlink.net, math-fun <math-fun@mailman.xmission.com> To: math-fun@mailman.xmission.com Subject: [math-fun] Probability of two (or three) integers being relativelyprime Date: Fri, 10 Dec 2004 23:14:04 -0500

Similar to the way I "computed" the probability that a "random pos. int." is quadratfrei, the probability that two "random pos. ints." can also be computed.

Call the two pos. ints. K and L.

Now (K,L) = 1 <=> For each prime p, it is false that p divides both K and L.

But Prob(p divides a random integer N) = 1/p.

Assuming "independence" between the prime factors of K and those of L,

Prob(p divides each of K and L) = 1/p^2, and so

Prob(It is false that p divides each of K and L) = 1 - 1/p^2.

Thus, Prob(For each prime p, it is false that p divides each of K and L) =

= Product over all primes p of (1 - 1/p^2).

Once again this product is the reciprocal of the Euler product for Zeta(2), and so the desired probability is 6/pi^2. ********************************************************************

NOW, how about three random integers K, L, M?

By same reasoning, for each p there must be at most one of the three random integers that p divides. The probability of this is

Prob(p divides none of K,L,M) + Prob(p divides just one of K,L,M) =

(1-1/p)^3 + 3(1/p)*(1-1/p)^2 =

1 - 3/(p^2) + 2/(p^3).

Thus Prob(K,L,M) are relatively prime =

Product over all prime p of (1 - 3/(p^2) + 2/(p^3)).

I don't know how to sum this in closed form, but numerically it turns out that

Prob( (K,L,M) = 1 ) ~ 0.1284

(taking the product over the first 1000 primes).

It would be nice to know that this jibes with the fraction of random triples of pos. ints. that are rel. prime.

--Dan Asimov