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%I #22 Sep 12 2017 23:59:26
%S 1,3,4,6,4,8,5,6,12,14,8,6,18,20,6,8,12,24,6,14,10,8,30,32,9,12,18,8,
%T 38,20,14,42,8,44,12,24,48,18,14,54,10,12,20,30,60,8,62,32,14,12,68,
%U 18,24,72,74,38,20,12,14,80,42,84,18,44,30,12,90,9,14,24
%N Let k = A239161(n); then a(n) = (sum of divisors of k) /(sum of divisors d of k with d <= sqrt(k)).
%C A000203(k)/A066839(k) is an integer, where k = A239161(n).
%H Michel Lagneau, <a href="/A238502/b238502.txt">Table of n, a(n) for n = 1..10000</a>
%e a(10) = 14 because k = A239161(10) = 13 and A000203(13) = 14, A066839(13) = 1 and A000203(13)/A066839(13) is an integer.
%t lst={}; f[n_]:=DivisorSigma[1,n]/Plus@@Select[Divisors@n,#<=Sqrt@n&];Do[If[IntegerQ[f[n]],AppendTo[lst, f[n]]],{n,1,200}];lst
%o (PARI) lista(nn) = { for(n = 1, nn, q = sigma(n) / sumdiv(n, d, d*(d<=sqrt(n))); if (type(q) == "t_INT", print1(q, ", ")););} \\ _Michel Marcus_, Mar 05 2014
%Y Cf. A000203, A066839, A239161.
%K nonn
%O 1,2
%A _Michel Lagneau_, Feb 27 2014
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