A015616 - OEIS

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,4

LINKS

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

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

EXAMPLE

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.

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 *)

PROG

(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