A082540 Number of ordered quadruples (a,b,c,d) with gcd(a,b,c,d)=1 (1 <= {a,b,c,d} <= n).

1, 15, 79, 239, 607, 1199, 2303, 3823, 6223, 9279, 13919, 19183, 27007, 35743, 47519, 60735, 78719, 97103, 122447, 148527, 181839, 216959, 262543, 306863, 365343, 423855, 495855, 569055, 661679, 748527, 862047, 972191, 1104831, 1237247

Offset: 1

Author

Benoit Cloitre, May 11 2003

Keywords

• nonn

Links

• Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000

Crossrefs

Column k=4 of A344527.

Cf. A018805, A071778, A082544, A343978.

Cf. A015634.

Programs

PARI

a(n)=sum(k=1,n,moebius(k)*floor(n/k)^4)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A082540(n):
    if n == 0:
        return 0
    c, j = 1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*A082540(k1)
        j, k1 = j2, n//j2
    return n*(n**3-1)-c+j # Chai Wah Wu, Mar 29 2021

Formula

a(n) = Sum_{k=1..n} mu(k)*floor(n/k)^4.

a(n) is asymptotic to c*n^4 with c=0.92393....

Lim_{n->infinity} a(n)/n^4 = 1/zeta(4) = A215267 = 90/Pi^4. - Karl-Heinz Hofmann, Apr 11 2021

Lim_{n->infinity} n^4/a(n) = zeta(4) = A013662 = Pi^4/90. - Karl-Heinz Hofmann, Apr 11 2021

a(n) = n^4 - Sum_{k=2..n} a(floor(n/k)). - Seiichi Manyama, Sep 13 2024