%I A093653
%S A093653 1,2,3,3,3,6,4,4,5,6,4,9,4,8,9,5,3,10,4,9,9,8,5,12,6,8,9,12,5,18,6,6,8,
%T A093653 6,9,15,4,8,10,12,4,18,5,12,15,10,6,15,7,12,9,12,5,18,11,16,10,10,6,27,
%U A093653 6,12,17,7,8,16,4,9,10,18,5,20,4,8,16,12,11,20,6,15,12,8,5,27,9,10,12
%N A093653 Total number of 1's in binary expansion of all divisors of n.
%F A093653 a(n)=sum{k=0..n, if(mod(n, k)=0, A000120(k), 0)}. - Paul Barry (pbarry(AT)wit.ie), 
               Jan 14 2005
%e A093653 a(8)=4 because the divisors of 8 are [1, 2, 4, 8] and in binary: 1, 10, 
               100, 1000, so four 1's.
%Y A093653 Cf. A093687.
%Y A093653 Sequence in context: A087688 A126854 A115206 this_sequence A049982 A070167 
               A141479
%Y A093653 Adjacent sequences: A093650 A093651 A093652 this_sequence A093654 A093655 
               A093656
%K A093653 easy,nonn
%O A093653 1,2
%A A093653 Jason Earls (zevi_35711(AT)yahoo.com), May 16 2004