A015616 Number of triples (i,j,k) with 1 <= i < j < k <= n and gcd(i,j,k) = 1.

0, 0, 1, 4, 10, 19, 34, 52, 79, 109, 154, 196, 262, 325, 409, 493, 613, 712, 865, 997, 1171, 1336, 1567, 1747, 2017, 2251, 2548, 2818, 3196, 3472, 3907, 4267, 4717, 5125, 5665, 6079, 6709, 7222, 7858, 8410, 9190, 9748, 10609, 11299, 12127

Offset: 1

Author

Olivier Gérard

Keywords

• nonn

Examples

			For n=6, the a(6) = 19 solutions are the binomial(6,3) = (6*5*4)/(1*2*3) = 20 possible triples minus the triple (2,4,6) with GCD=2.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000292, A100448, A027430, A015631.

Programs

Maple

f:=proc(n) local i,j,k,t1,t2,t3; t1:=0; for i from 1 to n-2 do for j from i+1 to n-1 do t2:=gcd(i,j); for k from j+1 to n do t3:=gcd(t2,k); if t3 = 1 then t1:=t1+1; fi; od: od: od: t1; end;
# program based on Moebius transform, partial sums of A000741:
with(numtheory):
b:= proc(n) option remember;
      add(mobius(n/d)*(d-2)*(d-1)/2, d=divisors(n))
    end:
a:= proc(n) option remember;
      b(n) +`if`(n=1, 0, a(n-1))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 08 2011

Mathematica

a[n_] := (cnt = 0; Do[cnt += Boole[GCD[i, j, k] == 1], {i, 1, n-2}, {j, i+1, n-1}, {k, j+1, n}]; cnt); Table[a[n], {n, 1, 45}] (* Jean-François Alcover, Mar 05 2013 *)

PARI

print1(c=0);for(k=1,99,for(j=1,k-1, gcd(j,k)==1 && (c+=j-1) && next; for(i=1,j-1, gcd([i,j,k])>1 || c++)); print1(", "c))

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A015616(n):
    if n <= 1:
        return 0
    c, j = n*(n-1)*(n-2)//6, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c -= (j2-j)*A015616(k1)
        j, k1 = j2, n//j2
    return c # Chai Wah Wu, Mar 30 2021

Formula

a(n) = (A071778(n) - 3*A018805(n) + 2)/6. - Vladeta Jovovic, Dec 01 2004

a(n) = Sum_{i=1..n} A000741(i). - Alois P. Heinz, Feb 08 2011

For n > 1, a(n) = n(n-1)(n-2)/6 - Sum_{j=2..n} a(floor(n/j)) = A000292(n-2) - Sum_{j=2..n} a(floor(n/j)). - Chai Wah Wu, Mar 30 2021