A256390 a(n) = number of triples (a,b,c) of natural numbers a,b,c <= n with gcd(a,b)=gcd(b,c)=gcd(c,a)=1.

1, 4, 13, 22, 55, 64, 133, 172, 247, 280, 469, 508, 781, 868, 997, 1144, 1621, 1714, 2323, 2488, 2785, 3010, 3907, 4078, 4837, 5176, 5833, 6178, 7627, 7798, 9463, 10102, 10927, 11530, 12631, 13006, 15379, 16150, 17311, 17926, 20863, 21256

Offset: 1

Author

Juan Arias-de-Reyna, Mar 27 2015

Keywords

• nonn

Comments

The sequence has asymptotics rho*n^3+O(n^2 log^2n) with rho=prod_p(1-3/p^2+2/p^3)=0.2867474284344...(product on primes). See A065473.

Examples

			a(3)=13 because the 13 triples (1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,1,3), (1,3,1), (3,1,1), (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1).
		

Links

• Juan Arias-de-Reyna, Table of n, a(n) for n = 1..1399
• J. Arias de Reyna and R. Heyman, Counting tuples restricted by pairwise primality, arXiv:1403.2769 [math.NT], 2014.
• J. Arias de Reyna, R. Heyman, Counting Tuples Restricted by Pairwise Coprimality Conditions, J. Int. Seq. 18 (2015) 15.10.4

Crossrefs

Cf. A256391.

Programs

Mathematica

A[M_] := A[M] = Module[{X, a1, a2, a3, K, count, k},
    X = Flatten[
      Table[{a1, a2, a3}, {a1, 1, M}, {a2, 1, M}, {a3, 1, M}], 2];
    K = Length[X];
    count = 0;
    For[k = 1, k <= K, k++,
     {a1, a2, a3} = X[[k]];
     If[(GCD[a1, a2] == 1) && (GCD[a2, a3] == 1) && (GCD[a3, a1] ==
         1), count = count + 1]];
    count];
Table[A[n], {n, 1, 100}]

Formula

a(n) = sum_a sum_b sum_c mu(a) mu(b) mu(c) [n/gcd(a,b)][n/gcd(b,c)][n/gcd(c,a)], where mu(.) is Moebius function [x] integer part of x, and a,b,c run through natural numbers.