A342841 Number of ordered triples (x, y, z) with gcd(x, y, z) = 1 and 1 <= {x, y, z} <= 10^n.
1, 841, 832693, 832046137, 831916552903, 831908477106883, 831907430687799769, 831907383078281024371, 831907373418800027750413, 831907372722449100147414487, 831907372589073124899487831735, 831907372581823023465031521920149, 831907372580768386561159867257319711
Offset: 0
Examples
For visualization, the set(x, y, z) is inscribed in a cube matrix. "o" stands for a gcd = 1. "." stands for a gcd > 1. . For n=1, the size of the cube matrix is 10 X 10 X 10: . / : : : : : : : : : : / 100 Sum (z = 1) z = 7 |/1 2 3 4 5 6 7 8 9 10 | --+--------------------- 75 Sum (z = 2) 1 /| o o o o o o o o o o 10 | 2/ | o o o o o o o o o o 10 91 Sum (z = 3) / 10 | z = 8 |/1 2 3 4 5 6 7 8 9 10 10 75 Sum (z = 4) --+--------------------- 10 / 1 /| o o o o o o o o o o 10 10 96 Sum (z = 5) 2/ | o . o . o . o . o . 5 9 / / 10 10 67 Sum (z = 6) z = 9 |/1 2 3 4 5 6 7 8 9 10 5 10 / --+--------------------- 10 10 / 1 /| o o o o o o o o o o 10 5 --/ 2/ | o o o o o o o o o o 10 10 99 Sum (z = 7) / 7 5 / z = 10 |/1 2 3 4 5 6 7 8 9 10 10 10 / --+--------------------- 10 5 / 1 | o o o o o o o o o o 10 7 --/ 2 | o . o . o . o . o . 5 10 75 Sum (z = 8) 3 | o o o o o o o o o o 10 10 / 4 | o . o . o . o . o . 5 7 / 5 | o o o o . o o o o . 8 10 / 6 | o . o . o . o . o . 5 --/ 7 | o o o o o o o o o o 10 91 Sum (z = 9) 8 | o . o . o . o . o . 5 / 9 | o o o o o o o o o o 10 / 10 | o . o . . . o . o . 4 / --/ 72 Sum (z = 10) / | ------ 841 Cube Sum (z = 1..10)
References
- Joachim von zur Gathen and Jürgen Gerhard, Modern Computer Algebra, Cambridge University Press, Second Edition 2003, pp. 53-54.
Links
- Chai Wah Wu, Table of n, a(n) for n = 0..15
- Karl-Heinz Hofmann, An animation of the cube with n = 1.
Crossrefs
Programs
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Python
import math for n in range (0, 10): counter = 0 for x in range (1, pow(10, n)+1): for y in range(1, pow(10, n)+1): for z in range(1, pow(10, n)+1): if math.gcd(math.gcd(y, x),z) == 1: counter += 1 print(n, counter)
Formula
Lim_{n->infinity} a(n)/10^(3*n) = 1/zeta(3) = 1/Apéry's constant.
a(n) = A071778(10^n).
Extensions
a(5)-a(10) from Hugo Pfoertner, Mar 25 2021
a(11) from Hugo Pfoertner, Mar 26 2021
a(12) from Bruce Garner, Mar 29 2021