%I
%S 1,2,4,6,8,16,24,28,32,40,64,96,120,128,224,256,288,360,384,496,512,
%T 640,672,1024,1536,1792,1920,2016,2048,2176,3744,4096,4320,4680,5632,
%U 5760,6144,6528,8128,8192,10240,10880,14336,15872,16384,16896,18432,18688
%N Numbers n that divide the sum of sums of elements of all subsets of divisors of n (A229335).
%C Harborth proved that this sequence is infinite. He showed that the terms are numbers n such that nsigma(n)*2^(d(n)  1), where d(n) is the number of divisors of n (A000005) and sigma(n) is their sum (A000203), and that the even terms, numbers of the form r*2^m where r is odd and m > 0, are those with m = ord_2(r/gcd(r, sigma(r)))*i with i = 1, 2, ... (ord_2(k) is the multiplicative order of 2 mod k, A002326). Thus this sequence includes all the powers of 2, all the numbers of the form n = 2^m*(2^(m + 1)  1) which include the even perfect numbers.
%H Amiram Eldar, <a href="/A316225/b316225.txt">Table of n, a(n) for n = 1..500</a>
%H Heiko Harborth, <a href="http://eudml.org/doc/141178">Eine Bemerkung zu den vollkommenen Zahlen</a>, Elemente der Mathematik, Vol. 31 (1976), pp. 119121 (in German).
%t divSumSubQ[n_] := Divisible[DivisorSigma[1, n] * 2^(DivisorSigma[0, n]  1), n]; Select[Range[100000], divSumSubQ]
%o (PARI) isok(n) = (sigma(n)*2^(numdiv(n)1) % n) == 0; \\ _Michel Marcus_, Dec 21 2018
%Y Cf. A000005, A000203, A000396, A002326, A229335.
%K nonn
%O 1,2
%A _Amiram Eldar_, Dec 21 2018
