A100448 Number of triples (i,j,k) with 1 <= i <= j < k <= n and gcd{i,j,k} = 1.

0, 1, 4, 9, 19, 30, 51, 73, 106, 140, 195, 241, 319, 388, 480, 572, 708, 813, 984, 1124, 1310, 1485, 1738, 1926, 2216, 2462, 2777, 3059, 3465, 3749, 4214, 4590, 5060, 5484, 6048, 6474, 7140, 7671, 8331, 8899, 9719, 10289, 11192, 11902, 12754, 13535, 14616

Offset: 1

Author

N. J. A. Sloane, Nov 21 2004

Keywords

• nonn
• easy

Comments

Probably the partial sums of A102309. - Ralf Stephan, Jan 03 2005

Links

• R. J. Mathar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A015616, A015631, A027430, A071778, A018805.

Programs

Maple

f:=proc(n) local i,j,k,t1,t2,t3; t1:=0; for i from 1 to n do for j from i to n 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;

Mathematica

f[n_] := Length[ Union[ Flatten[ Table[ If[ GCD[i, j, k] == 1, {i, j, k}], {i, n}, {j, i, n}, {k, j + 1, n}], 2]]]; Table[ If[n > 3, f[n] - 1, f[n]], {n, 47}] (* Robert G. Wilson v, Dec 14 2004 *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A100448(n):
    if n == 0:
        return 0
    c, j = 2, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*(6*A100448(k1)+1)
        j, k1 = j2, n//j2
    return (n*(n**2-1)-c+j)//6 # Chai Wah Wu, Mar 29 2021

Formula

a(n) = (A071778(n)-1)/6. - Vladeta Jovovic, Nov 30 2004

a(n) = (1/6)*(-1 + Sum_{k=1..n} moebius(k)*floor(n/k)^3). - Ralf Stephan, Jan 03 2005

Extensions

More terms from Robert G. Wilson v, Dec 14 2004

Edited by N. J. A. Sloane, Sep 06 2008 at the suggestion of R. J. Mathar