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%I #22 Sep 26 2018 12:25:52
%S 1,2,2,4,2,5,2,8,5,5,2,11,2,5,6,16,2,14,2,11,6,5,2,23,7,5,14,11,2,17,
%T 2,32,6,5,8,32,2,5,6,23,2,17,2,11,18,5,2,47,9,20,6,11,2,41,8,23,6,5,2,
%U 39,2,5,18,64,8,17,2,11,6,23,2,68,2,5,26,11,10
%N Sum of d/Gpf(d) for all divisors d of n, with Gpf(d) the greatest prime factor of d.
%C a(n) <= n (see Proposition 5.2 in Girard's paper, link below).
%C a(p) = 2, when p is prime.
%H Altug Alkan, <a href="/A217895/b217895.txt">Table of n, a(n) for n = 1..10000</a>
%H Benjamin Girard, <a href="https://arxiv.org/abs/1010.5042">On a combinatorial problem of Erdos, Kleitman and Lemke</a>, arXiv:1010.5042 [math.CO], 2010-2012.
%H Benjamin Girard, <a href="https://doi.org/10.1016/j.aim.2012.06.025">On a combinatorial problem of Erdos, Kleitman and Lemke</a>, Advances in Mathematics 231, 3-4 (2012) 1843-1857.
%e The divisors of 6 are : 1, 2, 3, 6; so a(6)=1/gpf(1)+2/gpf(2)+3/gpf(3)+6/gpf(6) = 1/1 + 2/2 + 3/3 + 6/3 = 5.
%t a[n_] := Sum[d/FactorInteger[d][[-1, 1]], {d, Divisors[n]}];
%t Array[a, 80] (* _Jean-François Alcover_, Sep 26 2018 *)
%o (PARI) gpf(n) = {if (n==1, return (1), return (vecmax(factor(n)[, 1])));}
%o a(n)= { my(d = divisors(n)); sum(j=1, length(d), d[j]/gpf(d[j]));} \\ revised by _Michel Marcus_, Sep 26 2018
%Y Cf. A052126 (Mobius transform).
%K nonn
%O 1,2
%A _Michel Marcus_, Oct 14 2012
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