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
Keywords
Links
- Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000
Programs
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PARI
a(n)=sum(k=1,n,moebius(k)*floor(n/k)^4)
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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
Comments