A256391 a(n) = number of tuples (a,b,c,d) of natural numbers a,b,c,d <= n with gcd(a,b)=gcd(b,c)=gcd(c,d)=gcd(d,a)=1.
1, 7, 35, 79, 243, 319, 787, 1155, 1859, 2295, 4267, 4891, 8295, 9743, 11851, 14539, 22191, 24359, 35427, 39387, 45915, 51687, 71171, 76407, 94911, 105047, 123251, 134447, 174003, 180835, 229783, 253007, 281447, 305111, 343315, 360215, 442547, 476115, 523111, 552307
Offset: 1
Keywords
Examples
For n=2, a(2)=7 counting the tuples (1,1,1,1), (2,1,1,1), (1,2,1,1), (1,1,2,1), (1,1,1,2), (2,1,2,1), (1,2,1,2).
Links
- Juan Arias-de-Reyna, Table of n, a(n) for n = 1..100
- J. Arias de Reyna and R. Heyman, Counting tuples restricted by pairwise primality, arXiv:1403.2769 [math.NT], 2014.
- J. Arias de Reyna, R. Heyman, Counting Tuples Restricted by Pairwise Coprimality Conditions, J. Int. Seq. 18 (2015) 15.10.4
Crossrefs
Cf. A256390.
Programs
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Mathematica
A[M_] := A[M] = Module[{X, a1, a2, a3, a4, K, count, k}, X = Flatten[ Table[{a1, a2, a3, a4}, {a1, 1, M}, {a2, 1, M}, {a3, 1, M}, {a4, 1, M}], 3]; K = Length[X]; count = 0; For[k = 1, k <= K, k++, {a1, a2, a3, a4} = X[[k]]; If[(GCD[a1, a2] == 1) && (GCD[a2, a3] == 1) && (GCD[a3, a4] == 1) && (GCD[a4, a1] == 1), count = count + 1]]; count]; Table[A[n], {n, 1, 20}]
Formula
a(n) = sum_a sum_b sum_c sum_d mu(a) mu(b) mu(c) mu(d) [n/gcd(a,b)][n/gcd(b,c)][n/gcd(c,d)][n/gcd(d,a)], where mu is Moebius function, a,b,c,d run through natural numbers.
Comments