%I
%S 6,24,112,1984,32512,171197,667879,780625,56513539,134201344,
%T 488265625,5203009849,9130639447,34359476224,47390685029,96595448129
%N Composite numbers whose sum of unitary divisors is a multiple of the sum of their aliquot parts.
%C A064591 is a subsequence of this sequence.
%C The ratios are 2, 1, 1, 1, 1, 12, 16, 4, 40, 1, 4, 100, 112, 1, 156, 180, ...
%C Up to 3*10^11 all the terms are of the form p^e*q. In particular, if 2^k1 is prime, then 2^(k+1)(2^k1) is a term. Similarly, if 2*5^k1 is prime, then 5^k*(2*5^k1) is a term. By solving appropriate Diophantine equations it is also possible to obtain large terms of the form p^2*q, like 1300253^2*1099140634496715133.  _Giovanni Resta_, Jun 01 2016
%e Unitary divisors of 6 are 1, 2, 3, 6 and their sum is 12. Aliquot parts are 1, 2, 3 and their sum is 6. Then, 12 / 6 = 2.
%e Unitary divisors of 24 are 1, 3, 8, 24 and their sum is 36. Aliquot parts are 1, 2, 3, 4, 6, 8, 12 and their sum is 36. Then, 36 / 36 = 1.
%e Unitary divisors of 171197 are 1, 169, 1013, 171197 and their sum is 172380. Aliquot parts are 1, 13, 169, 1013, 13169 and their sum is 14365. Then, 172380 / 14365 = 12.
%p with(numtheory): P:=proc(q) local a,b,k,n;
%p for n from 2 to q do if not isprime(n) then a:=ifactors(n)[2]; b:=mul(a[k][1]^a[k][2]+1,k=1..nops(a));
%p if type(b/(sigma(n)n),integer) then print(n); fi; fi; od; end: P(10^9);
%t Select[Range[10^6], Function[n, CompositeQ@ n && Mod[Total@ Select[Divisors@ n, GCD[#, n/#] == 1 &], DivisorSigma[1, n]  n] == 0]] (* _Michael De Vlieger_, Jun 01 2016 *)
%Y Cf. A001065, A034448, A064591.
%K nonn,more
%O 1,1
%A _Paolo P. Lava_, May 31 2016
%E a(9)a(16) from _Giovanni Resta_, Jun 01 2016
