A015631 Number of ordered triples of integers from [ 1..n ] with no global factor.
1, 3, 8, 15, 29, 42, 69, 95, 134, 172, 237, 287, 377, 452, 552, 652, 804, 915, 1104, 1252, 1450, 1635, 1910, 2106, 2416, 2674, 3007, 3301, 3735, 4027, 4522, 4914, 5404, 5844, 6432, 6870, 7572, 8121, 8805, 9389, 10249, 10831, 11776, 12506
Offset: 1
Keywords
Examples
a(4) = 15 because the 15 triples in question are in lexicographic order: [1,1,1], [1,1,2], [1,1,3], [1,1,4], [1,2,2], [1,2,3], [1,2,4], [1,3,3], [1,3,4], [1,4,4], [2,2,3], [2,3,3], [2,3,4], [3,3,4] and [3,4,4]. - _Wolfdieter Lang_, Apr 04 2013 The a(4) = 15 triangles with at least two sides <= 4 and sides relatively prime (see _Henry Bottomley_'s comment above) are: [1,1,1], [1,2,2], [2,2,3], [1,3,3], [2,3,3], [2,3,4], [3,3,4], [3,3,5], [1,4,4], [2,4,5], [3,4,4], [3,4,5], [3,4,6], [4,4,5], [4,4,7]. - _Alois P. Heinz_, Feb 14 2020
Links
- R. J. Mathar, Table of n, a(n) for n = 1..10000
Programs
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Magma
[n eq 1 select 1 else Self(n-1)+ &+[MoebiusMu(n div d) *d*(d+1)/2:d in Divisors(n)]:n in [1..50]]; // Marius A. Burtea, Feb 14 2020
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Maple
with(numtheory): b:= proc(n) option remember; add(mobius(n/d)*d*(d+1)/2, d=divisors(n)) end: a:= proc(n) option remember; b(n) + `if`(n=1, 0, a(n-1)) end: seq(a(n), n=1..60); # Alois P. Heinz, Feb 09 2011
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Mathematica
a[1] = 1; a[n_] := a[n] = Sum[MoebiusMu[n/d]*d*(d+1)/2, {d, Divisors[n]}] + a[n-1]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Jan 20 2014, after Maple *) Accumulate[Table[Sum[MoebiusMu[n/d]*d*(d + 1)/2, {d, Divisors[n]}], {n, 1, 50}]] (* Vaclav Kotesovec, Jan 31 2019 *)
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PARI
a(n) = sum(k=1, n, sumdiv(k, d, moebius(k/d)*binomial(d+1, 2))); \\ Seiichi Manyama, Jun 12 2021
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PARI
a(n) = binomial(n+2, 3)-sum(k=2, n, a(n\k)); \\ Seiichi Manyama, Jun 12 2021
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PARI
my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, moebius(k)*x^k/(1-x^k)^3)/(1-x)) \\ Seiichi Manyama, Jun 12 2021
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Python
from functools import lru_cache @lru_cache(maxsize=None) def A015631(n): if n == 0: return 0 c, j = 1, 2 k1 = n//j while k1 > 1: j2 = n//k1 + 1 c += (j2-j)*A015631(k1) j, k1 = j2, n//j2 return n*(n-1)*(n+4)//6-c+j # Chai Wah Wu, Mar 30 2021
Formula
Partial sums of the Moebius transform of the triangular numbers (A007438). - Steve Butler, Apr 18 2006
Row sums of triangle A134543. - Gary W. Adamson, Oct 31 2007
a(n) ~ n^3 / (6*Zeta(3)). - Vaclav Kotesovec, Jan 31 2019
G.f.: (1/(1 - x)) * Sum_{k>=1} mu(k) * x^k / (1 - x^k)^3. - Ilya Gutkovskiy, Feb 14 2020
a(n) = n*(n+1)*(n+2)/6 - Sum_{j=2..n} a(floor(n/j)) = A000292(n) - Sum_{j=2..n} a(floor(n/j)). - Chai Wah Wu, Mar 30 2021
Comments