A343978 Number of ordered 6-tuples (a,b,c,d,e,f) with gcd(a,b,c,d,e,f)=1 (1<= {a,b,c,d,e,f} <= n).
1, 63, 727, 4031, 15559, 45863, 116855, 257983, 526615, 983583, 1755143, 2935231, 4776055, 7407727, 11256623, 16498719, 23859071, 33434063, 46467719, 62949975, 84644439, 111486599, 146142583, 187854119, 240880239, 303814503, 382049919, 473813703, 586746719
Offset: 1
References
- Joachim von zur Gathen and Jürgen Gerhard, Modern Computer Algebra, Cambridge University Press, Second Edition 2003, pp. 53-54.
Links
- Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000
- Joachim von zur Gathen and Jürgen Gerhard, Extract from "3.4. (Non-)Uniqueness of the gcd" chapter, Modern Computer Algebra, Cambridge University Press, Second Edition 2003, pp. 53-54.
Crossrefs
Programs
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PARI
a(n)={sum(k=1, n+1, moebius(k)*(n\k)^6)} \\ Andrew Howroyd, May 08 2021 -
Python
from labmath import mobius def A343978(n): return sum(mobius(k)*(n//k)**6 for k in range(1, n+1))
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Python
from functools import lru_cache @lru_cache(maxsize=None) def A343978(n): if n == 0: return 0 c, j, k1 = 1, 2, n//2 while k1 > 1: j2 = n//k1 + 1 c += (j2-j)*A343978(k1) j, k1 = j2, n//j2 return n*(n**5-1)-c+j # Chai Wah Wu, May 17 2021
Formula
a(n) = Sum_{k=1..n} mu(k)*floor(n/k)^6.
Lim_{n->infinity} a(n)/n^6 = 1/zeta(6) = A343359 = 945/Pi^6.
a(n) = n^6 - Sum_{k=2..n} a(floor(n/k)). - Seiichi Manyama, Sep 13 2024
Extensions
Edited by N. J. A. Sloane, Jun 13 2021