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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.
2

%I #19 Jan 19 2019 04:14:59

%S 1,4,13,22,55,64,133,172,247,280,469,508,781,868,997,1144,1621,1714,

%T 2323,2488,2785,3010,3907,4078,4837,5176,5833,6178,7627,7798,9463,

%U 10102,10927,11530,12631,13006,15379,16150,17311,17926,20863,21256

%N 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.

%C 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.

%H Juan Arias-de-Reyna, <a href="/A256390/b256390.txt">Table of n, a(n) for n = 1..1399</a>

%H J. Arias de Reyna and R. Heyman, <a href="http://arxiv.org/abs/1403.2769">Counting tuples restricted by pairwise primality</a>, arXiv:1403.2769 [math.NT], 2014.

%H J. Arias de Reyna, R. Heyman, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Heyman/heyman6.html">Counting Tuples Restricted by Pairwise Coprimality Conditions</a>, J. Int. Seq. 18 (2015) 15.10.4

%F 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.

%e 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).

%t A[M_] := A[M] = Module[{X, a1, a2, a3, K, count, k},

%t X = Flatten[

%t Table[{a1, a2, a3}, {a1, 1, M}, {a2, 1, M}, {a3, 1, M}], 2];

%t K = Length[X];

%t count = 0;

%t For[k = 1, k <= K, k++,

%t {a1, a2, a3} = X[[k]];

%t If[(GCD[a1, a2] == 1) && (GCD[a2, a3] == 1) && (GCD[a3, a1] ==

%t 1), count = count + 1]];

%t count];

%t Table[A[n], {n, 1, 100}]

%Y Cf. A256391.

%K nonn

%O 1,2

%A _Juan Arias-de-Reyna_, Mar 27 2015