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A015631
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Number of ordered triples of integers from [ 1..n ] with no global factor.
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12
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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
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OFFSET
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1,2
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COMMENTS
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Number of integer-sided triangles with at least two sides <= n and sides relatively prime. - Henry Bottomley, Sep 29 2006
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LINKS
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FORMULA
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Partial sums of the Moebius transform of the triangular numbers (A007438). - Steve Butler, Apr 18 2006
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
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EXAMPLE
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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
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MAPLE
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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:
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MATHEMATICA
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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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PROG
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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
(Python)
from functools import lru_cache
@lru_cache(maxsize=None)
if n == 0:
return 0
c, j = 1, 2
k1 = n//j
while k1 > 1:
j2 = n//k1 + 1
j, k1 = j2, n//j2
(PARI) a(n) = sum(k=1, n, sumdiv(k, d, moebius(k/d)*binomial(d+1, 2))); \\ Seiichi Manyama, Jun 12 2021
(PARI) a(n) = binomial(n+2, 3)-sum(k=2, n, a(n\k)); \\ Seiichi Manyama, Jun 12 2021
(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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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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