%I A159912
%S A159912 0,1,4,7,14,17,24,31,46,49,56,63,78,85,100,115,146,149,156,163,178,185,
%T A159912 200,215,246,253,268,283,314,329,360,391,454,457,464,471,486,493,508,
%U A159912 523,554,561,576,591,622,637,668,699,762,769,784,799,830,845,876,907
%N A159912 Partial sums of A159913(k) = 2^bitcount(2k+1)-1 = A038573(2k+1), bitcount=A000120.
%C A159912 More precisely, a(n)=sum(i<n, A159913(i)), since we want the sequence 
               to start with a(0)=0 and not with A159913(0)=1.
%F A159912 a(n) = sum( i=0...n-1, A159913(i)) = sum(i=0..n-1, 2^A000120(i))*2-n
%o A159912 (PARI) A159912(n)=sum(i=0,n-1,1<<norml2(binary(i)))*2-n
%Y A159912 Cf. A000120, A038573.
%Y A159912 Sequence in context: A062380 A072031 A007437 this_sequence A049766 A161426 
               A115759
%Y A159912 Adjacent sequences: A159909 A159910 A159911 this_sequence A159913 A159914 
               A159915
%K A159912 nonn
%O A159912 0,3
%A A159912 M. F. Hasler (MHasler(AT)univ-ag.fr), May 03 2009